Player A has $$1$ and player B has $$2$. They repeatedly toss a coin that shows heads with probability $\tfrac23$: on heads B pays A $$1$, on tails A pays B $$1$. What is the probability that B runs out of money before A does?

Gambler's ruin with a two-thirds coin

Question

Player A has $\text{\char36}1$ and player B has $\text{\char36}2$ . They repeatedly toss a coin that shows heads with probability $\tfrac23$ : on heads B pays A $\text{\char36}1$ , on tails A pays B $\text{\char36}1$ . What is the probability that B runs out of money before A does?

Short answer

$4/7$ . A needs two net wins. A tail right away ruins A; heads-tails ( $\tfrac29$ ) returns to the start; heads-heads ( $\tfrac49$ ) wins. So $p=\tfrac49+\tfrac29p$ , giving $p=\tfrac47$ .

Adapted from QuantQuestions.io; question and answer rewritten by SelfMadeQuant.com.

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Lessons behind this question

Conditional Probability, Total Probability, and Bayes' Rule
Worked Examples — Probability Trees, Monty Hall, and Bayesian Updating
Sample Spaces, Events, and the Probability Axioms

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