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Smallest Possible Chance to Win? - Mochy

Posted By: Havard Raddum
Date: Tuesday, 18 May 2010, at 6:38 a.m.

In Response To: Smallest Possible Chance to Win? - Mochy (Stick)

Since nobody has attempted to answer the question, I'll give my two cents on the first position.

Black has at most 15 rolls left in the game, and this requires only rolling 21 fifteen times in a row. The probability of this is 1/18^15.

White has 60 crossovers left before he has taken all checkers off. It is then clear that white has to make four crossovers in every roll, 15 times in succession, to win. This requires double three or higher on every roll, but not all of these combinations work, and it's tricky to enumerate which sequences that work and which that don't. But assume that all sequences of double 3 or higher work. The chance for white to bear off in 15 rolls would then be 1/9^15.

So the winning chance for white in the first position is less than 1/(18*9)^15 = 1/1389073474362284556756399987130368. That's a 34-digit number in the denominator.

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