Gambler's ruin

Consider a coin-flipping game with two players where each player has a 50% chance of winning with each flip of the coin. After each flip of the coin the loser transfers one penny to the winner. The game ends when one player has all the pennies.

If there are no other limitations on the number of flips, the probability that the game will eventually end this way is 1. (One way to see this is as follows. Any given finite string of heads and tails will eventually be flipped with certainty: the probability of not seeing this string, while high at first, decays exponentially. In particular, the players would eventually flip a string of heads as long as the total number of pennies in play, by which time the game must have already ended.)

If player one has n₁ pennies and player two n₂ pennies, the probabilities P₁ and P₂ that players one and two, respectively, will end penniless are:

${\displaystyle {\begin{aligned}P_{1}&={\frac {n_{2}}{n_{1}+n_{2}}}\\[5pt]P_{2}&={\frac {n_{1}}{n_{1}+n_{2}}}\end{aligned}}}$

Two examples of this are if one player has more pennies than the other; and if both players have the same number of pennies. In the first case say player one ${\displaystyle (P_{1})}$ has 8 pennies and player two (${\displaystyle P_{2}}$) were to have 5 pennies then the probability of each losing is:

${\displaystyle {\begin{aligned}P_{1}&={\frac {5}{8+5}}={\frac {5}{13}}=0.3846{\text{ or }}38.46\%\\[6pt]P_{2}&={\frac {8}{8+5}}={\frac {8}{13}}=0.6154{\text{ or }}61.54\%\end{aligned}}}$

It follows that even with equal odds of winning the player that starts with fewer pennies is more likely to fail.

In the second case where both players have the same number of pennies (in this case 6) the likelihood of each losing is:

${\displaystyle {\begin{aligned}P_{1}&={\frac {6}{6+6}}={\frac {6}{12}}={\frac {1}{2}}=0.5\\[5pt]P_{2}&={\frac {6}{6+6}}={\frac {6}{12}}={\frac {1}{2}}=0.5\end{aligned}}}$

In the event of an unfair coin, where player one wins each toss with probability p, and player two wins with probability q = 1 − p, then the probability of each ending penniless is:

${\displaystyle {\begin{aligned}P_{1}&={\frac {1-({\frac {p}{q}})^{n_{2}}}{1-({\frac {p}{q}})^{n_{1}+n_{2}}}}\\[5pt]P_{2}&={\frac {1-({\frac {q}{p}})^{n_{1}}}{1-({\frac {q}{p}})^{n_{1}+n_{2}}}}\end{aligned}}}$

This can be shown as follows: Consider the probability of player 1 experiencing gamblers ruin having started with ${\displaystyle n>1}$ amount of money, ${\displaystyle P(R_{n})}$. Then, using the Law of Total Probability, we have

${\displaystyle P(R_{n})=P(R_{n}\mid W)P(W)+P(R_{n}\mid {\bar {W}})P({\bar {W}}),}$

where W denotes the event that player 1 wins the first bet. Then clearly ${\displaystyle P(W)=p}$ and ${\displaystyle P({\bar {W}})=1-p=q}$. Also ${\displaystyle P(R_{n}\mid W)}$ is the probability that player 1 experiences gambler's ruin having started with ${\displaystyle n+1}$ amount of money: ${\displaystyle P(R_{n+1})}$; and ${\displaystyle P(R_{n}\mid {\bar {W}})}$ is the probability that player 1 experiences gambler's ruin having started with ${\displaystyle n-1}$ amount of money: ${\displaystyle P(R_{n-1})}$.

Denoting ${\displaystyle q_{n}=P(R_{n})}$, we get the linear homogeneous recurrence relation

${\displaystyle q_{n}=q_{n+1}p+q_{n-1}q,}$

which we can solve using the fact that ${\displaystyle q_{0}=1}$ (i.e. the probability of gambler's ruin given that player 1 starts with no money is 1), and ${\displaystyle q_{n_{1}+n_{2}}=0}$ (i.e. the probability of gambler's ruin given that player 1 starts with all the money is 0.) For a more detailed description of the method see e.g. Feller (1970), An introduction to probability theory and its applications, 3rd ed.