# Equivalence between “gambler's ruin” and seemingly different game

My question concerns two experiments with different rules, but with the same probabilities. I was wondering, is there is an intuitive explanation for this equality, or is it is a coincidence?

Suppose that when Alice and Bob play chess, Alice wins with probability $p$ independently of previous games.

Game 1: Alice and Bob start with $n$ dollars each. They play chess over and over. Each time, the loser pays the winner a dollar, until someone runs out of money.

Let $q=1-p$. Using the classic gambler's ruin formula, $$P(\text{Alice wins Game 1}) = \frac{1-(\frac{q}p)^n}{1-(\frac{q}p)^{2n}} = \frac{1}{1+(\frac{q}p)^n} = \frac{p^n}{p^n+q^n}$$

Game 2: Alice and Bob play $n$ games of chess. If one of them wins all $n$ games, they immediately win the series. Otherwise, they repeat, playing blocks of $n$ games until someone wins them all.

Obviously, $$P(\text{Alice wins Game 2}) = \frac{p^n}{p^n+q^n}$$

• Interesting question (+1) – Peter Sep 28 '17 at 22:54
• A slight generalization of what you wrote: if we condition on the length of gambler's ruin to be $n + 2k$, then for any $k$ the probability Alice wins is $\frac{p^n}{p^n + q^n}$; the second game is the case when $k = 0$. I don't see why this is true combinatorially, but it's not too bad to write out the conditional probability statement to prove that. – Marcus M Sep 29 '17 at 0:21
• @MarcusM That's a very illuminating comment, thank you! – Mike Earnest Sep 29 '17 at 1:54

The probability that Alice wins each game in a block is $p^n$ and the probability that neither win each game in a block is $1-(p^n + q^n)$. Let $\tau$ be the probability Alice eventually wins each game in a block. Since each block of games is independent, we have $$\tau = p^n + (1-(p^n+q^n))\tau,$$ which yields $$\tau = \frac{p^n}{p^n+q^n}.$$
I think there is nothing too deep going on here. Consider the set of paths where the game's length is $$2k$$. Let $$A$$ be the set of paths where Alice wins, and $$B$$ be the set of paths where Bob wins. There is an obvious bijection between $$A$$ and $$B$$ (vertical reflection). Furthermore, all paths in $$A$$ have the same unconditional probability, as do all paths in $$B$$, and the probability of the latter is $$q^n/p^n$$ times that of the former. It immediately follows that, conditional on the game lasting $$2k$$ rounds, the probability that $$A$$ wins is $$1/(1+(q/p)^n)$$. Since the conditional probability is independent of $$k$$, it follows the unconditional probability is equal to this same value.