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by Jtsummers
4594 days ago
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Since this seems to only care about the probability of any transition would you need the 2 invisible squares, or wouldn't the "Finish" square suffice? That is, in any square that can advance to 12 (10, 11) 10 has a 1/3 probability of advancing to 11, and 2/3 of hitting 12/13. While 11 has a probability of reaching 12 (or beyond) of 1. Also, thanks for that link. Markov chains are something I know about, but never really set to the task of learning. It's also a much quicker solution than my initial concept, and neatly handles the chute/ladder squares (that is, on square 5 you'd have a 1/3 chance of ending on 11, 1/3 on 2 and 1/3 on 8; 6 and 7 are never actually reached - which also means they can be eliminated from the set of states). |
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