{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0001","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bf1425b150e731e859da634207406a690dd90ec310ac53f950aa6e883ddb97cb","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256001,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,4],[3,4]],\"steps\":1,\"transition\":[[[1,5],[4,5]],[[1,4],[3,4]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/4, 3/4] to be nonnegative with exact total 1; steps=1. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/4, 3/4], 1, [[1/5, 4/5], [1/4, 3/4]], [19/80, 61/80])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 1, [[1/5, 4/5], [1/4, 3/4]], [19/80, 61/80])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0002","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"323b6d6ef836a11782d7e30aa1ec004c7bd3cdf94021af975a367c72c5b3fb06","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256002,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,4],[3,4]],\"steps\":2,\"transition\":[[[2,5],[3,5]],[[1,4],[3,4]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/4, 3/4] to be nonnegative with exact total 1; steps=2. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/4, 3/4], 2, [[31/100, 69/100], [23/80, 57/80]], [469/1600, 1131/1600])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 2, [[31/100, 69/100], [23/80, 57/80]], [469/1600, 1131/1600])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0003","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c37e1787102dc40492c53ff1b6f298db9cdcc0273e88b922d54c72d2e883004a","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256003,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,4],[3,4]],\"steps\":3,\"transition\":[[[3,5],[2,5]],[[1,4],[3,4]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/4, 3/4] to be nonnegative with exact total 1; steps=3. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/4, 3/4], 3, [[411/1000, 589/1000], [589/1600, 1011/1600]], [12123/32000, 19877/32000])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 3, [[411/1000, 589/1000], [589/1600, 1011/1600]], [12123/32000, 19877/32000])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0004","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"35bce9febde44688d1ddb52d506e5fa34222594273d70f800af4da3640927edb","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256004,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,4],[3,4]],\"steps\":1,\"transition\":[[[4,5],[1,5]],[[1,4],[3,4]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/4, 3/4] to be nonnegative with exact total 1; steps=1. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/4, 3/4], 1, [[4/5, 1/5], [1/4, 3/4]], [31/80, 49/80])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 1, [[4/5, 1/5], [1/4, 3/4]], [31/80, 49/80])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0005","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"604cad0234a81d2dadcdeae4cf5fc8c80b985dc95d48dd553c42c5bc8b45c8f8","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256005,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,4],[3,4]],\"steps\":2,\"transition\":[[[1,5],[4,5]],[[1,2],[1,2]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/4, 3/4] to be nonnegative with exact total 1; steps=2. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/4, 3/4], 2, [[11/25, 14/25], [7/20, 13/20]], [149/400, 251/400])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 2, [[11/25, 14/25], [7/20, 13/20]], [149/400, 251/400])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0006","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0788e8096d2eeabccb556dfbfc6064e0159d5604837735c8ae082550ebe9c861","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256006,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,4],[3,4]],\"steps\":3,\"transition\":[[[2,5],[3,5]],[[1,2],[1,2]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/4, 3/4] to be nonnegative with exact total 1; steps=3. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/4, 3/4], 3, [[227/500, 273/500], [91/200, 109/200]], [1819/4000, 2181/4000])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 3, [[227/500, 273/500], [91/200, 109/200]], [1819/4000, 2181/4000])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0007","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b034e3733085045d098eb93f750e129130f61d3062c1084f773eaebc4dab7887","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256007,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,4],[3,4]],\"steps\":1,\"transition\":[[[3,5],[2,5]],[[1,2],[1,2]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/4, 3/4] to be nonnegative with exact total 1; steps=1. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/4, 3/4], 1, [[3/5, 2/5], [1/2, 1/2]], [21/40, 19/40])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 1, [[3/5, 2/5], [1/2, 1/2]], [21/40, 19/40])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0008","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4bed8ae85d0e0c6780527edcbbe2224dd3d96cfbe6a8e7c6c4fe6a1324829e46","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256008,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,4],[3,4]],\"steps\":2,\"transition\":[[[4,5],[1,5]],[[1,2],[1,2]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/4, 3/4] to be nonnegative with exact total 1; steps=2. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/4, 3/4], 2, [[37/50, 13/50], [13/20, 7/20]], [269/400, 131/400])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 2, [[37/50, 13/50], [13/20, 7/20]], [269/400, 131/400])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0009","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"950554f8a39a8a2e4860acdf13304ef6d665ecdfbc71a4c4ad14d0c70382a545","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256009,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,4],[3,4]],\"steps\":3,\"transition\":[[[1,5],[4,5]],[[3,4],[1,4]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/4, 3/4] to be nonnegative with exact total 1; steps=3. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/4, 3/4], 3, [[199/500, 301/500], [903/1600, 697/1600]], [16729/32000, 15271/32000])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 3, [[199/500, 301/500], [903/1600, 697/1600]], [16729/32000, 15271/32000])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0010","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3a4932258a60285db652289e5f47024c96e87d2812d416c88e54c99bd2327298","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256010,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,4],[3,4]],\"steps\":1,\"transition\":[[[2,5],[3,5]],[[3,4],[1,4]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/4, 3/4] to be nonnegative with exact total 1; steps=1. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/4, 3/4], 1, [[2/5, 3/5], [3/4, 1/4]], [53/80, 27/80])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 1, [[2/5, 3/5], [3/4, 1/4]], [53/80, 27/80])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0011","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fa3c8ebb2df836f4b9c53d6d59fb7defd5e453b01be61bab531b834e8267237e","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256011,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,4],[3,4]],\"steps\":2,\"transition\":[[[3,5],[2,5]],[[3,4],[1,4]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/4, 3/4] to be nonnegative with exact total 1; steps=2. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/4, 3/4], 2, [[33/50, 17/50], [51/80, 29/80]], [1029/1600, 571/1600])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 2, [[33/50, 17/50], [51/80, 29/80]], [1029/1600, 571/1600])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0012","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1012e3ba44eecf9de07b3c5192f48f9af6f32c43e7edea15c6af6bab0cda02ca","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256012,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,4],[3,4]],\"steps\":3,\"transition\":[[[4,5],[1,5]],[[3,4],[1,4]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/4, 3/4] to be nonnegative with exact total 1; steps=3. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/4, 3/4], 3, [[1579/2000, 421/2000], [1263/1600, 337/1600]], [25261/32000, 6739/32000])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 3, [[1579/2000, 421/2000], [1263/1600, 337/1600]], [25261/32000, 6739/32000])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0013","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e678dea5447079491f49f2b1dd0568f382bdeae2c51760a0b06ab57a2f84e0e8","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256013,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,2],[1,2]],\"steps\":1,\"transition\":[[[1,5],[4,5]],[[1,4],[3,4]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/2, 1/2] to be nonnegative with exact total 1; steps=1. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/2, 1/2], 1, [[1/5, 4/5], [1/4, 3/4]], [9/40, 31/40])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 1, [[1/5, 4/5], [1/4, 3/4]], [9/40, 31/40])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0014","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c69ca57a1acb284375e116ca80e9cdc6ea637641ac8d421c0fb2e0ca6f25586","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256014,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,2],[1,2]],\"steps\":2,\"transition\":[[[2,5],[3,5]],[[1,4],[3,4]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/2, 1/2] to be nonnegative with exact total 1; steps=2. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/2, 1/2], 2, [[31/100, 69/100], [23/80, 57/80]], [239/800, 561/800])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 2, [[31/100, 69/100], [23/80, 57/80]], [239/800, 561/800])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0015","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"af2c9eff4b3deea7e7fb741efe3dbb509eb5b45e4edfdb2193f0c266362a227b","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256015,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,2],[1,2]],\"steps\":3,\"transition\":[[[3,5],[2,5]],[[1,4],[3,4]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/2, 1/2] to be nonnegative with exact total 1; steps=3. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/2, 1/2], 3, [[411/1000, 589/1000], [589/1600, 1011/1600]], [6233/16000, 9767/16000])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 3, [[411/1000, 589/1000], [589/1600, 1011/1600]], [6233/16000, 9767/16000])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0016","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a3046beb332c5c73c9614665c7de484755a4916430bf8634dcee88172e0b3be8","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256016,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,2],[1,2]],\"steps\":1,\"transition\":[[[4,5],[1,5]],[[1,4],[3,4]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/2, 1/2] to be nonnegative with exact total 1; steps=1. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/2, 1/2], 1, [[4/5, 1/5], [1/4, 3/4]], [21/40, 19/40])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 1, [[4/5, 1/5], [1/4, 3/4]], [21/40, 19/40])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0017","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2a446d3045120619bb8290db8b9939cd48d0fd99e4b7b79ef178f2787edc3998","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256017,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,2],[1,2]],\"steps\":2,\"transition\":[[[1,5],[4,5]],[[1,2],[1,2]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/2, 1/2] to be nonnegative with exact total 1; steps=2. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/2, 1/2], 2, [[11/25, 14/25], [7/20, 13/20]], [79/200, 121/200])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 2, [[11/25, 14/25], [7/20, 13/20]], [79/200, 121/200])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0018","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c23c0438b367c24824cdf240e437185be4bdedec9be58f6048ee57a01b68ac61","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256018,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,2],[1,2]],\"steps\":3,\"transition\":[[[2,5],[3,5]],[[1,2],[1,2]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/2, 1/2] to be nonnegative with exact total 1; steps=3. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/2, 1/2], 3, [[227/500, 273/500], [91/200, 109/200]], [909/2000, 1091/2000])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 3, [[227/500, 273/500], [91/200, 109/200]], [909/2000, 1091/2000])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0019","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3cfd75ca21025bff8d0fc339d7b4bda7bd8947b514d5a957063b758fd4cfba35","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256019,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,2],[1,2]],\"steps\":1,\"transition\":[[[3,5],[2,5]],[[1,2],[1,2]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/2, 1/2] to be nonnegative with exact total 1; steps=1. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/2, 1/2], 1, [[3/5, 2/5], [1/2, 1/2]], [11/20, 9/20])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 1, [[3/5, 2/5], [1/2, 1/2]], [11/20, 9/20])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0020","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eaa1912bc73aac49a0092788b27f6e9897d9237af7158a98fb1059a5ba538a1e","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256020,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,2],[1,2]],\"steps\":2,\"transition\":[[[4,5],[1,5]],[[1,2],[1,2]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/2, 1/2] to be nonnegative with exact total 1; steps=2. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/2, 1/2], 2, [[37/50, 13/50], [13/20, 7/20]], [139/200, 61/200])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 2, [[37/50, 13/50], [13/20, 7/20]], [139/200, 61/200])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0021","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"01a150648e6e3a841a01c03e1cf533c5e936c16d29b8898cb7b1c0020c33b459","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256021,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,2],[1,2]],\"steps\":3,\"transition\":[[[1,5],[4,5]],[[3,4],[1,4]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/2, 1/2] to be nonnegative with exact total 1; steps=3. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/2, 1/2], 3, [[199/500, 301/500], [903/1600, 697/1600]], [7699/16000, 8301/16000])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 3, [[199/500, 301/500], [903/1600, 697/1600]], [7699/16000, 8301/16000])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0022","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e3236166b5dd2e5f7c6edbaa86532af9f3b6b1aca7c36c1fcb98d15ce246df82","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256022,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,2],[1,2]],\"steps\":1,\"transition\":[[[2,5],[3,5]],[[3,4],[1,4]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/2, 1/2] to be nonnegative with exact total 1; steps=1. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/2, 1/2], 1, [[2/5, 3/5], [3/4, 1/4]], [23/40, 17/40])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 1, [[2/5, 3/5], [3/4, 1/4]], [23/40, 17/40])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0023","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4062dc9c855be11d57c8c8726cc705110571eb36978e0f1a88ef1acde75e6561","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256023,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,2],[1,2]],\"steps\":2,\"transition\":[[[3,5],[2,5]],[[3,4],[1,4]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/2, 1/2] to be nonnegative with exact total 1; steps=2. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/2, 1/2], 2, [[33/50, 17/50], [51/80, 29/80]], [519/800, 281/800])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 2, [[33/50, 17/50], [51/80, 29/80]], [519/800, 281/800])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0024","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5e5f0bf5604bf3c7397a5cf1f027dfdd780653c63d7682f44fec3a3befc6de96","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256024,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[1,2],[1,2]],\"steps\":3,\"transition\":[[[4,5],[1,5]],[[3,4],[1,4]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[1/2, 1/2] to be nonnegative with exact total 1; steps=3. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([1/2, 1/2], 3, [[1579/2000, 421/2000], [1263/1600, 337/1600]], [12631/16000, 3369/16000])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 3, [[1579/2000, 421/2000], [1263/1600, 337/1600]], [12631/16000, 3369/16000])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0025","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"440db6bdb3c1e9d21035cb67b5279a0ef20955a9014969e7be9d937aaff98efa","family_id":"qf-mod25-markovdeep-n-step-law","seed":202610256025,"semantic_template":"qf-mod25-markovdeep-n-step-law/{\"initial\":[[3,4],[1,4]],\"steps\":1,\"transition\":[[[1,5],[4,5]],[[1,4],[3,4]]]}"},"fields":{"title":"An Exact n-Step Row Law, Including Zero Steps","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. Require initial law=[3/4, 1/4] to be nonnegative with exact total 1; steps=1. Give (initial law,steps,P^steps,initial*P^steps).","answer":{"kind":"expression","value":"([3/4, 1/4], 1, [[1/5, 4/5], [1/4, 3/4]], [17/80, 63/80])","explanation":"Compute the finite matrix power exactly. At zero steps the identity leaves the initial law unchanged. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([3/4, 1/4], 1, [[1/5, 4/5], [1/4, 3/4]], [17/80, 63/80])."},"skills":["An Exact n-Step Row Law, Including Zero Steps"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0026","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d756681bc5caa6a14f7292872fef065bb4fbe443feafb93fb39ab6084bc212fb","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256026,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":0,\"steps\":1,\"to\":1,\"transition\":[[[1,5],[4,5]],[[1,4],[3,4]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=0, to=1, steps=1. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(0, 1, 1, [[1/5, 4/5], [1/4, 3/4]], 4/5)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (0, 1, 1, [[1/5, 4/5], [1/4, 3/4]], 4/5)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0027","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"31b2fca07e6cfd41c1e9f1f9ff3a25f6bb83851697cbdd282ca2a53f5abf2190","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256027,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":1,\"steps\":1,\"to\":0,\"transition\":[[[2,5],[3,5]],[[1,4],[3,4]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=1, to=0, steps=1. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(1, 0, 1, [[2/5, 3/5], [1/4, 3/4]], 1/4)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (1, 0, 1, [[2/5, 3/5], [1/4, 3/4]], 1/4)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0028","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"02015928a91c6e4013e00dffcd3cda7156b7b02d70e090b6bbd193b7aa89f526","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256028,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":0,\"steps\":1,\"to\":1,\"transition\":[[[3,5],[2,5]],[[1,4],[3,4]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=0, to=1, steps=1. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(0, 1, 1, [[3/5, 2/5], [1/4, 3/4]], 2/5)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (0, 1, 1, [[3/5, 2/5], [1/4, 3/4]], 2/5)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0029","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b5496aa5f184233ab0cb07a4db5efe7ec4051d7118cbabf4cc34e51c45746bf8","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256029,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":1,\"steps\":1,\"to\":0,\"transition\":[[[4,5],[1,5]],[[1,4],[3,4]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=1, to=0, steps=1. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(1, 0, 1, [[4/5, 1/5], [1/4, 3/4]], 1/4)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (1, 0, 1, [[4/5, 1/5], [1/4, 3/4]], 1/4)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0030","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dc920547a5f2c7cf0bc919d724cf64fe0ee2ec42c8747ecce718c569d79c1038","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256030,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":0,\"steps\":1,\"to\":1,\"transition\":[[[1,5],[4,5]],[[1,2],[1,2]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=0, to=1, steps=1. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(0, 1, 1, [[1/5, 4/5], [1/2, 1/2]], 4/5)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (0, 1, 1, [[1/5, 4/5], [1/2, 1/2]], 4/5)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0031","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c784fdab3b86a82b445413959f2f79308872ba31b80d1e9d8f0b8e454be76eda","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256031,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":1,\"steps\":1,\"to\":0,\"transition\":[[[2,5],[3,5]],[[1,2],[1,2]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=1, to=0, steps=1. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(1, 0, 1, [[2/5, 3/5], [1/2, 1/2]], 1/2)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (1, 0, 1, [[2/5, 3/5], [1/2, 1/2]], 1/2)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0032","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2c743f6d77f672161ca96356fb90de96025613489113a3183694da0906c79e37","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256032,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":0,\"steps\":1,\"to\":1,\"transition\":[[[3,5],[2,5]],[[1,2],[1,2]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=0, to=1, steps=1. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(0, 1, 1, [[3/5, 2/5], [1/2, 1/2]], 2/5)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (0, 1, 1, [[3/5, 2/5], [1/2, 1/2]], 2/5)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0033","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8c5aecee5474c83735e4690080c731c25652966c3b5587dc41ce5ef89b42aac1","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256033,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":1,\"steps\":1,\"to\":0,\"transition\":[[[4,5],[1,5]],[[1,2],[1,2]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=1, to=0, steps=1. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(1, 0, 1, [[4/5, 1/5], [1/2, 1/2]], 1/2)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (1, 0, 1, [[4/5, 1/5], [1/2, 1/2]], 1/2)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0034","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8d78cd8bd932ee0ab9f9b1a780bf799aba75688564d8bdfa804e8cd3b58af8ce","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256034,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":0,\"steps\":1,\"to\":1,\"transition\":[[[1,5],[4,5]],[[3,4],[1,4]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=0, to=1, steps=1. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(0, 1, 1, [[1/5, 4/5], [3/4, 1/4]], 4/5)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (0, 1, 1, [[1/5, 4/5], [3/4, 1/4]], 4/5)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0035","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"93a50f935698e82a2a92bd42ac5da7cb1cea876f9d11056cdd16ee0cf8b98338","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256035,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":1,\"steps\":1,\"to\":0,\"transition\":[[[2,5],[3,5]],[[3,4],[1,4]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=1, to=0, steps=1. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(1, 0, 1, [[2/5, 3/5], [3/4, 1/4]], 3/4)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (1, 0, 1, [[2/5, 3/5], [3/4, 1/4]], 3/4)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0036","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f86974bafdabd134d5858ff0a4677baa4d0c0b2658cff8cc309bf064b73dee86","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256036,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":0,\"steps\":1,\"to\":1,\"transition\":[[[3,5],[2,5]],[[3,4],[1,4]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=0, to=1, steps=1. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(0, 1, 1, [[3/5, 2/5], [3/4, 1/4]], 2/5)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (0, 1, 1, [[3/5, 2/5], [3/4, 1/4]], 2/5)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0037","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4ebef8624917a2ec205260ce63ea4d94cb013e4cf8a68df417fcce2b0566b593","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256037,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":1,\"steps\":1,\"to\":0,\"transition\":[[[4,5],[1,5]],[[3,4],[1,4]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=1, to=0, steps=1. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(1, 0, 1, [[4/5, 1/5], [3/4, 1/4]], 3/4)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (1, 0, 1, [[4/5, 1/5], [3/4, 1/4]], 3/4)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0038","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"512711a2ea0fa48ea689511f53ebe50153681f1249c9440b91c6e091a1033cfc","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256038,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":0,\"steps\":2,\"to\":1,\"transition\":[[[1,5],[4,5]],[[1,4],[3,4]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=0, to=1, steps=2. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(0, 1, 2, [[6/25, 19/25], [19/80, 61/80]], 19/25)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (0, 1, 2, [[6/25, 19/25], [19/80, 61/80]], 19/25)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0039","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cd96bdcca0af1eac46c68c1405124959cb062cefb76e5f5605e369cec7c5903e","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256039,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":1,\"steps\":2,\"to\":0,\"transition\":[[[2,5],[3,5]],[[1,4],[3,4]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=1, to=0, steps=2. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(1, 0, 2, [[31/100, 69/100], [23/80, 57/80]], 23/80)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (1, 0, 2, [[31/100, 69/100], [23/80, 57/80]], 23/80)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0040","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c3c8705df47ef5c41329df5a123301a90907b82415ec24c33f9e6d9926d9ccb2","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256040,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":0,\"steps\":2,\"to\":1,\"transition\":[[[3,5],[2,5]],[[1,4],[3,4]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=0, to=1, steps=2. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(0, 1, 2, [[23/50, 27/50], [27/80, 53/80]], 27/50)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (0, 1, 2, [[23/50, 27/50], [27/80, 53/80]], 27/50)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0041","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dd53be11adc58c9bf557b62ee1d58d1f745eb688e7322d99aa9673bdf75a74ed","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256041,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":1,\"steps\":2,\"to\":0,\"transition\":[[[4,5],[1,5]],[[1,4],[3,4]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=1, to=0, steps=2. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(1, 0, 2, [[69/100, 31/100], [31/80, 49/80]], 31/80)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (1, 0, 2, [[69/100, 31/100], [31/80, 49/80]], 31/80)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0042","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"09bedc70a9cb7a689a0168c9f9902567c16ca4b6eb734ef4a776f43f7890e277","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256042,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":0,\"steps\":2,\"to\":1,\"transition\":[[[1,5],[4,5]],[[1,2],[1,2]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=0, to=1, steps=2. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(0, 1, 2, [[11/25, 14/25], [7/20, 13/20]], 14/25)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (0, 1, 2, [[11/25, 14/25], [7/20, 13/20]], 14/25)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0043","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1e9ab5ebd68265296df9703cdfa911471ae2226afd45ed3c9f5621b6fa3bde8c","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256043,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":1,\"steps\":2,\"to\":0,\"transition\":[[[2,5],[3,5]],[[1,2],[1,2]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=1, to=0, steps=2. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(1, 0, 2, [[23/50, 27/50], [9/20, 11/20]], 9/20)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (1, 0, 2, [[23/50, 27/50], [9/20, 11/20]], 9/20)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0044","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a4f1293bd7088e113e1520a5a2c09fc31f32545b503bd0025e6c3a096f35bd0a","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256044,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":0,\"steps\":2,\"to\":1,\"transition\":[[[3,5],[2,5]],[[1,2],[1,2]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=0, to=1, steps=2. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(0, 1, 2, [[14/25, 11/25], [11/20, 9/20]], 11/25)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (0, 1, 2, [[14/25, 11/25], [11/20, 9/20]], 11/25)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0045","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"68085e1e8884efd747455a02cfc8fdacd2125fd035b567d4409a111f7985e42d","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256045,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":1,\"steps\":2,\"to\":0,\"transition\":[[[4,5],[1,5]],[[1,2],[1,2]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=1, to=0, steps=2. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(1, 0, 2, [[37/50, 13/50], [13/20, 7/20]], 13/20)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (1, 0, 2, [[37/50, 13/50], [13/20, 7/20]], 13/20)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0046","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8546dbe5e37099d82a2d4ff909b96d122e59e54062e9680a62011b1b5edf39a2","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256046,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":0,\"steps\":2,\"to\":1,\"transition\":[[[1,5],[4,5]],[[3,4],[1,4]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=0, to=1, steps=2. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(0, 1, 2, [[16/25, 9/25], [27/80, 53/80]], 9/25)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (0, 1, 2, [[16/25, 9/25], [27/80, 53/80]], 9/25)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0047","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"874105c81ae5781337cc4cbfc5c3b2e2318ec0a919a13a3fbc0db65accdd6555","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256047,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":1,\"steps\":2,\"to\":0,\"transition\":[[[2,5],[3,5]],[[3,4],[1,4]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=1, to=0, steps=2. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(1, 0, 2, [[61/100, 39/100], [39/80, 41/80]], 39/80)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (1, 0, 2, [[61/100, 39/100], [39/80, 41/80]], 39/80)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0048","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7b3359c9f74122e6da12e1c985416ea6ae9b19430e496e4518f482f6ec4c4897","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256048,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":0,\"steps\":2,\"to\":1,\"transition\":[[[3,5],[2,5]],[[3,4],[1,4]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=0, to=1, steps=2. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(0, 1, 2, [[33/50, 17/50], [51/80, 29/80]], 17/50)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (0, 1, 2, [[33/50, 17/50], [51/80, 29/80]], 17/50)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0049","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7aed9a860a574f7ffd96f1300c6e225a12ee4d6eae96adf647402aef8ccd3d50","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256049,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":1,\"steps\":2,\"to\":0,\"transition\":[[[4,5],[1,5]],[[3,4],[1,4]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=1, to=0, steps=2. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(1, 0, 2, [[79/100, 21/100], [63/80, 17/80]], 63/80)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (1, 0, 2, [[79/100, 21/100], [63/80, 17/80]], 63/80)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0050","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1032f789db0d9326e8345c44f7541dcef37a644d28ff09778b7459d427e76e20","family_id":"qf-mod25-markovdeep-n-step-entry","seed":202610256050,"semantic_template":"qf-mod25-markovdeep-n-step-entry/{\"from\":0,\"steps\":3,\"to\":1,\"transition\":[[[1,5],[4,5]],[[1,4],[3,4]]]}"},"fields":{"title":"A Conditional n-Step Entry without Inventing an Initial Law","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. from=0, to=1, steps=3. Give (from,to,steps,P^steps,its from-to entry).","answer":{"kind":"expression","value":"(0, 1, 3, [[119/500, 381/500], [381/1600, 1219/1600]], 381/500)","explanation":"Keep row/column direction. At zero steps the entry is 1 exactly when the two state labels agree. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (0, 1, 3, [[119/500, 381/500], [381/1600, 1219/1600]], 381/500)."},"skills":["A Conditional n-Step Entry without Inventing an Initial Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0051","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2869d69adc872d97f9b4b064a932e72b9b48f56d254cfd2a1a9ab6e784c5b30d","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256051,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":0,\"initial\":[[1,4],[3,4]],\"steps\":1,\"to\":1,\"transition\":[[[1,5],[4,5]],[[1,4],[3,4]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/4, 3/4], from=0, to=1, steps=1. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/4, 3/4], 0, 1, 1, 1/4, 4/5, 1/5)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 0, 1, 1, 1/4, 4/5, 1/5)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0052","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"15a5b7f7dde5bf79c8fb3dec34201445bedf01a577baa0e6301c0bf2894fe93b","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256052,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":1,\"initial\":[[1,4],[3,4]],\"steps\":2,\"to\":0,\"transition\":[[[2,5],[3,5]],[[1,4],[3,4]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/4, 3/4], from=1, to=0, steps=2. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/4, 3/4], 1, 0, 2, 3/4, 23/80, 69/320)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 1, 0, 2, 3/4, 23/80, 69/320)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0053","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0ac3a835bfd44a749256ca5c44bf2e5eebd383c0bee355ec8d5d4963bbc528ae","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256053,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":0,\"initial\":[[1,4],[3,4]],\"steps\":3,\"to\":1,\"transition\":[[[3,5],[2,5]],[[1,4],[3,4]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/4, 3/4], from=0, to=1, steps=3. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/4, 3/4], 0, 1, 3, 1/4, 589/1000, 589/4000)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 0, 1, 3, 1/4, 589/1000, 589/4000)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0054","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f389e18762b50b689309e9ada30c014e88592497ad368a8c8456667179b068bb","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256054,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":1,\"initial\":[[1,4],[3,4]],\"steps\":1,\"to\":0,\"transition\":[[[4,5],[1,5]],[[1,4],[3,4]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/4, 3/4], from=1, to=0, steps=1. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/4, 3/4], 1, 0, 1, 3/4, 1/4, 3/16)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 1, 0, 1, 3/4, 1/4, 3/16)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0055","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4b5faff3208b911eeab18a328bcc32e6efc389e4511b79db98bf82c5deb9c152","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256055,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":0,\"initial\":[[1,4],[3,4]],\"steps\":2,\"to\":1,\"transition\":[[[1,5],[4,5]],[[1,2],[1,2]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/4, 3/4], from=0, to=1, steps=2. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/4, 3/4], 0, 1, 2, 1/4, 14/25, 7/50)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 0, 1, 2, 1/4, 14/25, 7/50)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0056","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"12f1a726d3af71f13944c19cea6bdc63cb29c97f7c934760c5e1d83aa7491e51","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256056,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":1,\"initial\":[[1,4],[3,4]],\"steps\":3,\"to\":0,\"transition\":[[[2,5],[3,5]],[[1,2],[1,2]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/4, 3/4], from=1, to=0, steps=3. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/4, 3/4], 1, 0, 3, 3/4, 91/200, 273/800)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 1, 0, 3, 3/4, 91/200, 273/800)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0057","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c3970ddf83ac195b4bda2e4e180a43240e1b6f01a5a4ab7d185f14a77452c04c","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256057,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":0,\"initial\":[[1,4],[3,4]],\"steps\":1,\"to\":1,\"transition\":[[[3,5],[2,5]],[[1,2],[1,2]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/4, 3/4], from=0, to=1, steps=1. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/4, 3/4], 0, 1, 1, 1/4, 2/5, 1/10)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 0, 1, 1, 1/4, 2/5, 1/10)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0058","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"15bc9e8d0c99660156014bb4d5ea46d625e2beeacdc32fd41fd20811f057a969","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256058,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":1,\"initial\":[[1,4],[3,4]],\"steps\":2,\"to\":0,\"transition\":[[[4,5],[1,5]],[[1,2],[1,2]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/4, 3/4], from=1, to=0, steps=2. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/4, 3/4], 1, 0, 2, 3/4, 13/20, 39/80)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 1, 0, 2, 3/4, 13/20, 39/80)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0059","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b1cc38b752a77902b9fcc6e795a4615246194672add6393fefb2e1847685fdad","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256059,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":0,\"initial\":[[1,4],[3,4]],\"steps\":3,\"to\":1,\"transition\":[[[1,5],[4,5]],[[3,4],[1,4]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/4, 3/4], from=0, to=1, steps=3. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/4, 3/4], 0, 1, 3, 1/4, 301/500, 301/2000)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 0, 1, 3, 1/4, 301/500, 301/2000)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0060","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c5c5867432c1e4c339f7bbc6be2382084a6bc60dd729292ed590cfc25a840e87","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256060,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":1,\"initial\":[[1,4],[3,4]],\"steps\":1,\"to\":0,\"transition\":[[[2,5],[3,5]],[[3,4],[1,4]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/4, 3/4], from=1, to=0, steps=1. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/4, 3/4], 1, 0, 1, 3/4, 3/4, 9/16)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 1, 0, 1, 3/4, 3/4, 9/16)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0061","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eef8532eb84ebaf8c1068ae6ad10cf1038472bb12be33f87325e8249782fd6ec","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256061,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":0,\"initial\":[[1,4],[3,4]],\"steps\":2,\"to\":1,\"transition\":[[[3,5],[2,5]],[[3,4],[1,4]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/4, 3/4], from=0, to=1, steps=2. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/4, 3/4], 0, 1, 2, 1/4, 17/50, 17/200)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 0, 1, 2, 1/4, 17/50, 17/200)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0062","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c2082d2f9072cfe3ed53616622ebd0ba9178f66735d6289fee4489281837bb75","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256062,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":1,\"initial\":[[1,4],[3,4]],\"steps\":3,\"to\":0,\"transition\":[[[4,5],[1,5]],[[3,4],[1,4]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/4, 3/4], from=1, to=0, steps=3. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/4, 3/4], 1, 0, 3, 3/4, 1263/1600, 3789/6400)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/4, 3/4], 1, 0, 3, 3/4, 1263/1600, 3789/6400)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0063","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"477046baaa485347f77df97b7693e531e2856eed486e8dea5306bd0bb3bb726c","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256063,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":0,\"initial\":[[1,2],[1,2]],\"steps\":1,\"to\":1,\"transition\":[[[1,5],[4,5]],[[1,4],[3,4]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/2, 1/2], from=0, to=1, steps=1. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/2, 1/2], 0, 1, 1, 1/2, 4/5, 2/5)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 0, 1, 1, 1/2, 4/5, 2/5)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0064","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"81cb01adb2ce58f184d2dbec432cb998d9a33f8b3b6d51af8b1dc1e1e7ce2461","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256064,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":1,\"initial\":[[1,2],[1,2]],\"steps\":2,\"to\":0,\"transition\":[[[2,5],[3,5]],[[1,4],[3,4]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/2, 1/2], from=1, to=0, steps=2. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/2, 1/2], 1, 0, 2, 1/2, 23/80, 23/160)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 1, 0, 2, 1/2, 23/80, 23/160)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0065","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d55baa5ab1040e9a628a12a70a68ce99b6e3f111d8f919c3e79dbd60f5b0970b","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256065,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":0,\"initial\":[[1,2],[1,2]],\"steps\":3,\"to\":1,\"transition\":[[[3,5],[2,5]],[[1,4],[3,4]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/2, 1/2], from=0, to=1, steps=3. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/2, 1/2], 0, 1, 3, 1/2, 589/1000, 589/2000)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 0, 1, 3, 1/2, 589/1000, 589/2000)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0066","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f0afe3c49ddc94d3f509cb8beb1188c3f8789663397ede2a357285ce38088fe2","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256066,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":1,\"initial\":[[1,2],[1,2]],\"steps\":1,\"to\":0,\"transition\":[[[4,5],[1,5]],[[1,4],[3,4]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/2, 1/2], from=1, to=0, steps=1. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/2, 1/2], 1, 0, 1, 1/2, 1/4, 1/8)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 1, 0, 1, 1/2, 1/4, 1/8)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0067","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9b934336be6839b8533463828ad31aa4f133d3dd4a865375dacb4680e0cece93","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256067,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":0,\"initial\":[[1,2],[1,2]],\"steps\":2,\"to\":1,\"transition\":[[[1,5],[4,5]],[[1,2],[1,2]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/2, 1/2], from=0, to=1, steps=2. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/2, 1/2], 0, 1, 2, 1/2, 14/25, 7/25)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 0, 1, 2, 1/2, 14/25, 7/25)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0068","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ba995155a5fb8989696f1b414d9753cd5cedc5e121935b4e105a0f207bff6f47","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256068,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":1,\"initial\":[[1,2],[1,2]],\"steps\":3,\"to\":0,\"transition\":[[[2,5],[3,5]],[[1,2],[1,2]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/2, 1/2], from=1, to=0, steps=3. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/2, 1/2], 1, 0, 3, 1/2, 91/200, 91/400)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 1, 0, 3, 1/2, 91/200, 91/400)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0069","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0d557a94dfed30398c8fa7ebdb223e6ef73f52fd2f0c9c0948f9b5d41dd2f60d","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256069,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":0,\"initial\":[[1,2],[1,2]],\"steps\":1,\"to\":1,\"transition\":[[[3,5],[2,5]],[[1,2],[1,2]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/2, 1/2], from=0, to=1, steps=1. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/2, 1/2], 0, 1, 1, 1/2, 2/5, 1/5)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 0, 1, 1, 1/2, 2/5, 1/5)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0070","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a9e554ab173e6c261019483d65f96d5ee81aae52681e46c13f17cfcfdb527e2c","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256070,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":1,\"initial\":[[1,2],[1,2]],\"steps\":2,\"to\":0,\"transition\":[[[4,5],[1,5]],[[1,2],[1,2]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/2, 1/2], from=1, to=0, steps=2. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/2, 1/2], 1, 0, 2, 1/2, 13/20, 13/40)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 1, 0, 2, 1/2, 13/20, 13/40)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0071","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"beba98ae59d229923f103fe1432f308ccdeaac428e503a7b21da05180c14ff5b","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256071,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":0,\"initial\":[[1,2],[1,2]],\"steps\":3,\"to\":1,\"transition\":[[[1,5],[4,5]],[[3,4],[1,4]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/2, 1/2], from=0, to=1, steps=3. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/2, 1/2], 0, 1, 3, 1/2, 301/500, 301/1000)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 0, 1, 3, 1/2, 301/500, 301/1000)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0072","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e00ac0b6d7f4560f5ac6f7d3156fb2ff004190815a990dd3fdb8566af3b55b56","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256072,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":1,\"initial\":[[1,2],[1,2]],\"steps\":1,\"to\":0,\"transition\":[[[2,5],[3,5]],[[3,4],[1,4]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/2, 1/2], from=1, to=0, steps=1. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/2, 1/2], 1, 0, 1, 1/2, 3/4, 3/8)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 1, 0, 1, 1/2, 3/4, 3/8)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0073","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"573418613750d4e69d8bb6ccf78075c596059068b380a18ba786231877cfa604","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256073,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":0,\"initial\":[[1,2],[1,2]],\"steps\":2,\"to\":1,\"transition\":[[[3,5],[2,5]],[[3,4],[1,4]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/2, 1/2], from=0, to=1, steps=2. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/2, 1/2], 0, 1, 2, 1/2, 17/50, 17/100)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 0, 1, 2, 1/2, 17/50, 17/100)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0074","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4194674b3dbdb3367510bfbda706732d08dac841e3ac3b4ea71170d0731e5887","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256074,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":1,\"initial\":[[1,2],[1,2]],\"steps\":3,\"to\":0,\"transition\":[[[4,5],[1,5]],[[3,4],[1,4]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[1/2, 1/2], from=1, to=0, steps=3. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([1/2, 1/2], 1, 0, 3, 1/2, 1263/1600, 1263/3200)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1/2, 1/2], 1, 0, 3, 1/2, 1263/1600, 1263/3200)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0075","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a9ba196a1c63c760847891f993818f97f5ce67860b66f2015c6a9d351e380b01","family_id":"qf-mod25-markovdeep-endpoint-joint-probability","seed":202610256075,"semantic_template":"qf-mod25-markovdeep-endpoint-joint-probability/{\"from\":0,\"initial\":[[3,4],[1,4]],\"steps\":1,\"to\":1,\"transition\":[[[1,5],[4,5]],[[1,4],[3,4]]]}"},"fields":{"title":"Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. initial law=[3/4, 1/4], from=0, to=1, steps=1. Give (initial,from,to,steps,initial[from],P^steps[from][to],product).","answer":{"kind":"expression","value":"([3/4, 1/4], 0, 1, 1, 3/4, 4/5, 3/5)","explanation":"This is the joint probability of the initial and later endpoint, not the conditional transition entry alone or the later-time marginal. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([3/4, 1/4], 0, 1, 1, 3/4, 4/5, 3/5)."},"skills":["Initial Mass Times an n-Step Entry Gives an Endpoint Joint Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0076","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c30c273a31e3c53bd60f04d6b2531ebf60c5d91f813e8ab4ce26df5e15120837","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256076,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[0,1,0],\"transition\":[[[1,5],[4,5]],[[1,4],[3,4]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[0, 1, 0] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([0, 1, 0], [4/5, 1/4], 1/5, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 1, 0], [4/5, 1/4], 1/5, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0077","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ca722b50c1a7b6e46481f9e3316b7280c3c80e63046082379164c01712659a7e","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256077,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[1,0,1],\"transition\":[[[2,5],[3,5]],[[1,4],[3,4]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[1, 0, 1] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([1, 0, 1], [1/4, 3/5], 3/20, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, 0, 1], [1/4, 3/5], 3/20, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0078","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6c229fb108547d26bc13f1734bc2902be91cd68d64eaad31341bd2965c42075c","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256078,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[0,1,0],\"transition\":[[[3,5],[2,5]],[[1,4],[3,4]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[0, 1, 0] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([0, 1, 0], [2/5, 1/4], 1/10, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 1, 0], [2/5, 1/4], 1/10, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0079","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d5cce871126be2144a5c734001e0b3f4428fe4d43a187405a157ec2c952c5370","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256079,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[1,0,1],\"transition\":[[[4,5],[1,5]],[[1,4],[3,4]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[1, 0, 1] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([1, 0, 1], [1/4, 1/5], 1/20, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, 0, 1], [1/4, 1/5], 1/20, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0080","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f6e8c34afbd16548c990eccb29b69f03981054727cb4a4fcf8fce15ef906120e","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256080,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[0,1,0],\"transition\":[[[1,5],[4,5]],[[1,2],[1,2]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[0, 1, 0] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([0, 1, 0], [4/5, 1/2], 2/5, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 1, 0], [4/5, 1/2], 2/5, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0081","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"753b91be8ae566c74586dd45fae32130c1e37d4eea696efc94d9127670ee611c","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256081,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[1,0,1],\"transition\":[[[2,5],[3,5]],[[1,2],[1,2]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[1, 0, 1] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([1, 0, 1], [1/2, 3/5], 3/10, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, 0, 1], [1/2, 3/5], 3/10, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0082","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e6552b4569c31a555783cf6f06b950a2d3fa07509a081b90d408ce616d11a663","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256082,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[0,1,0],\"transition\":[[[3,5],[2,5]],[[1,2],[1,2]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[0, 1, 0] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([0, 1, 0], [2/5, 1/2], 1/5, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 1, 0], [2/5, 1/2], 1/5, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0083","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ceb23711b42c1dbf974cef8bc2e3ec760379740df3b054941c59075569d52a8b","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256083,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[1,0,1],\"transition\":[[[4,5],[1,5]],[[1,2],[1,2]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[1, 0, 1] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([1, 0, 1], [1/2, 1/5], 1/10, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, 0, 1], [1/2, 1/5], 1/10, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0084","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"494ec58596a43e55f4fdf918244e6fa2ea4e39f148eeb5c632eb5294c7599d0c","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256084,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[0,1,0],\"transition\":[[[1,5],[4,5]],[[3,4],[1,4]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[0, 1, 0] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([0, 1, 0], [4/5, 3/4], 3/5, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 1, 0], [4/5, 3/4], 3/5, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0085","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"13feb38a97c2fe5f958a986dd62d6f765482d75cf535b1a181f44d500c26648e","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256085,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[1,0,1],\"transition\":[[[2,5],[3,5]],[[3,4],[1,4]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[1, 0, 1] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([1, 0, 1], [3/4, 3/5], 9/20, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, 0, 1], [3/4, 3/5], 9/20, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0086","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a1a9674ba9f2eda207b97996c1a3dab6aaf362001f6f6ded622f4fe6a98cf558","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256086,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[0,1,0],\"transition\":[[[3,5],[2,5]],[[3,4],[1,4]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[0, 1, 0] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([0, 1, 0], [2/5, 3/4], 3/10, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 1, 0], [2/5, 3/4], 3/10, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0087","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1fd3c8cacee4fb18d50f433d3616ffe1ce3f5a89fe6cd023de39caeaac05c35c","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256087,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[1,0,1],\"transition\":[[[4,5],[1,5]],[[3,4],[1,4]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[1, 0, 1] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([1, 0, 1], [3/4, 1/5], 3/20, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, 0, 1], [3/4, 1/5], 3/20, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0088","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cdd2f1e8fb26430aa745771d32a88cccd3a3fe7362e398b1b9ed0652f2590e4a","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256088,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[0,1,0,1],\"transition\":[[[1,5],[4,5]],[[1,4],[3,4]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[0, 1, 0, 1] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([0, 1, 0, 1], [4/5, 1/4, 4/5], 4/25, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 1, 0, 1], [4/5, 1/4, 4/5], 4/25, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0089","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8d529af06bfc8af8a3f2e3278a006671173e3e35ff5f0e6704140d3acb4833e6","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256089,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[1,0,1,0],\"transition\":[[[2,5],[3,5]],[[1,4],[3,4]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[1, 0, 1, 0] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([1, 0, 1, 0], [1/4, 3/5, 1/4], 3/80, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, 0, 1, 0], [1/4, 3/5, 1/4], 3/80, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0090","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4d6eaf647f9ba3db98e54a98235bf66d798e5714dc72fcb4057511faff0940cb","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256090,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[0,1,0,1],\"transition\":[[[3,5],[2,5]],[[1,4],[3,4]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[0, 1, 0, 1] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([0, 1, 0, 1], [2/5, 1/4, 2/5], 1/25, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 1, 0, 1], [2/5, 1/4, 2/5], 1/25, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0091","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ceddbddadaf9aef384f07280cb39d3f0f094d36b79eaa379f1e7711a8d83882f","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256091,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[1,0,1,0],\"transition\":[[[4,5],[1,5]],[[1,4],[3,4]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[1, 0, 1, 0] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([1, 0, 1, 0], [1/4, 1/5, 1/4], 1/80, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, 0, 1, 0], [1/4, 1/5, 1/4], 1/80, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0092","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"08a751561fef365b8ec0c561607302c813c941fbd17e8da81f4363301ff44970","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256092,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[0,1,0,1],\"transition\":[[[1,5],[4,5]],[[1,2],[1,2]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[0, 1, 0, 1] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([0, 1, 0, 1], [4/5, 1/2, 4/5], 8/25, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 1, 0, 1], [4/5, 1/2, 4/5], 8/25, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0093","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1c228dda3145cff0722b7f708a6c8a4745818d59bac549951402a060f1de8252","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256093,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[1,0,1,0],\"transition\":[[[2,5],[3,5]],[[1,2],[1,2]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[1, 0, 1, 0] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([1, 0, 1, 0], [1/2, 3/5, 1/2], 3/20, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, 0, 1, 0], [1/2, 3/5, 1/2], 3/20, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0094","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"68e3cafa60a727b13a4d2d32fd7536970b5e55b6869a63b69f79e4a8a0bd3cd2","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256094,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[0,1,0,1],\"transition\":[[[3,5],[2,5]],[[1,2],[1,2]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[0, 1, 0, 1] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([0, 1, 0, 1], [2/5, 1/2, 2/5], 2/25, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 1, 0, 1], [2/5, 1/2, 2/5], 2/25, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0095","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"738b24331e735f0b4faf2d880bf3f9428079eaef38fab2b33e1ff9917dac1d98","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256095,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[1,0,1,0],\"transition\":[[[4,5],[1,5]],[[1,2],[1,2]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [1/2, 1/2]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[1, 0, 1, 0] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([1, 0, 1, 0], [1/2, 1/5, 1/2], 1/20, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, 0, 1, 0], [1/2, 1/5, 1/2], 1/20, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0096","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8401645fdb70589512c1daad639c02fa13b4d265d6cc6577afb484961da9178c","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256096,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[0,1,0,1],\"transition\":[[[1,5],[4,5]],[[3,4],[1,4]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[0, 1, 0, 1] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([0, 1, 0, 1], [4/5, 3/4, 4/5], 12/25, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 1, 0, 1], [4/5, 3/4, 4/5], 12/25, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0097","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"477cfe6326066aeff160a4a3702680dec6149e951363d58c4b1f18a4dccc69e0","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256097,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[1,0,1,0],\"transition\":[[[2,5],[3,5]],[[3,4],[1,4]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[2/5, 3/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[1, 0, 1, 0] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([1, 0, 1, 0], [3/4, 3/5, 3/4], 27/80, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, 0, 1, 0], [3/4, 3/5, 3/4], 27/80, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0098","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f2ae95178bb9f9379246e8af267e746d037333633dfb969e44f2a3e8e9d868ee","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256098,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[0,1,0,1],\"transition\":[[[3,5],[2,5]],[[3,4],[1,4]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[3/5, 2/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[0, 1, 0, 1] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([0, 1, 0, 1], [2/5, 3/4, 2/5], 3/25, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 1, 0, 1], [2/5, 3/4, 2/5], 3/25, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0099","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"529bd6d4a7ef15e0a0450a3f63b0824314ce78ab7b10c50a1da290ee799fcc3d","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256099,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[1,0,1,0],\"transition\":[[[4,5],[1,5]],[[3,4],[1,4]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[4/5, 1/5], [3/4, 1/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[1, 0, 1, 0] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([1, 0, 1, 0], [3/4, 1/5, 3/4], 9/80, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, 0, 1, 0], [3/4, 1/5, 3/4], 9/80, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261025-r06-finite-markov-chain-0100","batch_id":"batch-20261013-advance-20261025-r06-finite-markov-chain-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6369839ed4612faaa02d26afe70eba3d5fc88f3ce3a3b3add03348c836364e87","family_id":"qf-mod25-markovdeep-conditional-path-probability","seed":202610256100,"semantic_template":"qf-mod25-markovdeep-conditional-path-probability/{\"path\":[0,1,0,1,0],\"transition\":[[[1,5],[4,5]],[[1,4],[3,4]]]}"},"fields":{"title":"A Kernel-Started Path Excludes Any Initial-Law Mass","stem":"Use the fixed exact finite time-homogeneous row-stochastic transition table=[[1/5, 4/5], [1/4, 3/4]]. Rows=current state; columns=next state. Matrix powers keep state order and P^0 is the identity. Use exact rational arithmetic with reduced positive denominators; omit /1 for integers. Finite-step laws and conditional paths imply no stationarity, irreducibility, ergodicity or limit theorem. nonempty path=[0, 1, 0, 1, 0] is conditional on the specified initial state. Give (path,successive transition factors,their product,conditional_on_initial_state). Do not include an initial-law mass; none is supplied. A one-state path has no transitions and empty-product value 1.","answer":{"kind":"expression","value":"([0, 1, 0, 1, 0], [4/5, 1/4, 4/5, 1/4], 1/25, conditional_on_initial_state)","explanation":"Multiply only the transitions in their path order. This differs from an unconditional finite-path probability, which additionally requires an initial distribution. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 1, 0, 1, 0], [4/5, 1/4, 4/5, 1/4], 1/25, conditional_on_initial_state)."},"skills":["A Kernel-Started Path Excludes Any Initial-Law Mass"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}