Simple gambling problem

Originally posted by eldragonfly
can you show me the calculations.

With B going first: the prob that B wins first roll is 1/6 and prob that A will win first roll is (5/6)*(5/36)=25/216. So the total prob that one of them will win in the first round is 61/216.
If no one wins first round then the odds are the same again for the second round and so on. Since there must be a winner eventually, the question could be thought of as; what is the prob that A wins given that there is a winner in the first round.
Therefore p(A)=5/6*5/36/(61/216)=25/61.

Odds of Player A winning with one of his specific rolls are 5/36.
Odds of Player B winning with one of his specific rolls are 1/6.

Assume B starts, and look at only the next 2 rolls.

B wins if his roll wins. 1/6 * 36/36 (second roll doesn't matter) = 36/216.
A wins if B's fails and A's succeeds. 5/6 * 5/36 = 25/216.
The game is a push if both rolls fail. 5/6 * 31/36 = 155/216.

If the game is a push, we end up with an identical situation as before, with identical odds, and so we can safely ignore a push and just take the definite end results, because we don't care how often this has rolled around or how long this takes.

Of the 61 (36+25) times the game is over in 2 rolls, A only wins 25 times, so his chances are 25/61.


Now, with A starting (or up next) the numbers work this way with identical reasoning.

A wins - 5/36 * 6/6 = 30/216
B wins - 31/36 * 1/6 = 31/216
Push - 31/36 * 5/6 = 155/216

A wins 30 of 61 times.

The other way to do it is sum the probabilities of A winning at each round.
This gives 25/216+(25/216*155/216)+((25/216*(155/216)^2)...etc.
That is 25/216* limit of infinite series (155/216)^n as n goes from zero to infinity. Again leading to 25/216*216/61.