Two players, 9 versus 21, winner takes all

Question

You bring 9 tokens and your rival brings 21 tokens, for 30 tokens in play. Each round is a fair game: one token moves from the loser to the winner, each of you equally likely to win. Play continues until one person owns all 30 tokens.

What is the probability you end up with all 30 tokens?

Accepted Answer

Your token count does a fair random walk between 0 (you are wiped out) and 30 (you have everything). Winning the whole pot means your count reaches 30 before 0. Let hi be that probability starting from i tokens, with h0 = 0 and h{30} = 1. A fair round gives hi = 	frac12 h{i+1} + 	frac12 h{i-1}, so hi is linear, hi = i/N. Starting at i = 9 with N = 30, h9 = \frac{9}{30} = \frac{3}{10}. Equivalently, expected final holdings equal your start of 9 tokens, and 30p = 9 gives p = 3/10. Takeaway. In a fair winner-take-all match your chance of sweeping the table is simply your share of the tokens: 9…

Two players, 9 versus 21, winner takes all

Your answer

More Probability questions