Problem 141 Solution

Let p be the probability that A wins. Consider the following four events:

If the first roll is a 12 then A wins.
If the first roll is a anything other than a 7 and 12 then neither side has improved their odds and the probability of A winning is still p.
If the first roll is a 7 and the second roll is a 12 then A wins.
If the first roll is a 7 and the second roll is anything other than a 7 and 12 then neither side has improved their odds and the probability of A winning is still p.

We can now express p as follows:

p = 1/36 + (29/36)*p + (1/6)*(1/36) + (1/6)*(29/36)* p

p = 7/216 + (203/216)*p

13p = 7

p = 7/13

So the probability that A will win is 7/13 =~ 0.538462 .

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

Michael Shackleford, ASA, August 19 1999