Problem 142 Solution

Recall from problem 141 that the probability that A wins the first time is 7/13. Thus the probability B will win the first time is 6/13. In fact if the last roll was not a 7 the probability that A and B will win are still 7/13 and 6/13 respectively. If the last roll was a 7 the probability B will win is:

1/6 + (29/36)*(6/13) = 252/468 = 7/13.

So the probability that A will win if the last roll was a 7 is 6/13, otherwise it is 7/13. Likewise the probability that B will win if the last roll was a 7 is 7/13, and 6/13 otherwise.

By jumping ahead x rolls the probability that the (x-1)^th roll was a 7 is 1/6 and the probability of a non 7 is 5/6. Thus the probability of of A winning the next bet resolved is:

pr((x-1)^th roll = 7)*(6/13) + pr((x-1)^th roll != 7)*(7/13) =

(1/6)*(6/13) + (5/6)*(7/13) = 41/78 =~ 0.525641 .

Thanks to Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 6.

Michael Shackleford, ASA, August 19 1999

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