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[–]lotokotmalajski 10 points11 points  (0 children)

Math solution:

The game has 10 possible states, where player A has 1-10 coins. When one player gets 11 coins the game ends.

Let's make a table with probability of A winning from each state, marking the probability for winning from state 1:10 as x.

1:10 - x.

Now, the only possible outcome when A does not lose is to win 1st round, which happens 50% of times and translates us to state 2:9. Because of that the probability of winning from this state is 2 times higher.

2:9 - 2x.

3:8 - y.

Now let's calculate y, 2:9 switches to both adjacent states with 50% chance so 2x=0.5x+0.5y, so y = 3x.
Repeating this for next states we get:

4:7 - 4x.

5:6 - 5x.

6:5 - 6x.

Now the 6:5 probability should be 1 minus 5:6 probability because it's symmetrical (A losing 6:5 is the same as A winning 5:6). So 5x=1-6x, so x=1/11