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