When a betting game allows draws

Question

You start with 5 chips and play until you reach 10 chips or go broke. Each round you win a chip with probability 0.4, lose a chip with probability 0.4, or tie (nothing changes) with probability 0.2. What is your probability of reaching 10 chips? And how do the ties affect the expected number of rounds?

Accepted Answer

The win probability. Ties leave your chip count unchanged, so they cannot influence where the walk eventually ends up, only when. Ignore the tie rounds and look only at rounds where something happens. Conditioned on a non-tie round, you win with probability \frac{0.4}{0.4 + 0.4} = \frac12, and lose with probability 	frac12. So the sequence of actual chip changes is a fair symmetric random walk. By symmetric gambler's ruin, the probability of reaching N = 10 from i = 5 is \frac{i}{N} = \frac{5}{10} = \frac12. The ties are completely irrelevant to your 50\% chance of success. The duration. Tie…

When a betting game allows draws

Your answer

More Probability questions