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