Two players, 6 versus 9 dollars, expected number of rounds

Question

You hold 6 dollars and your rival holds 9 dollars, for 15 dollars total. Each round is a fair game: one dollar passes from loser to winner, each of you equally likely to win. Play continues until one player is broke.

On average, how many rounds until someone is wiped out?

Accepted Answer

Your money does a fair random walk between 0 and N = 15. The game ends when your holdings hit either wall. Let Di be the expected number of rounds from i dollars, with D0 = D{15} = 0. A fair round gives Di = 1 + 	frac12 D{i+1} + 	frac12 D{i-1}, solved by Di = i(N - i). Starting at i = 6, D6 = 6 	imes (15 - 6) = 6 	imes 9 = 54. Conveniently, this is just your stake times your rival's stake, 6 	imes 9. Takeaway. A fair match between stakes a and b lasts a 	imes b rounds on average: 6 	imes 9 = 54.

Two players, 6 versus 9 dollars, expected number of rounds

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