The Red Sox play the Yankees in a best-of-seven series that ends as soon as one team wins four games. Suppose that the probability that the Red Sox win Game $n$ is $\frac{n-1}{6}$. What is the probability that the Red Sox will win the series?
The clever solution here is to assume that there will be $7$ games played, then the probability that the Yankees win game $8-n$ is $$1 - \frac{8 - n - 1}{6} = \frac{6 - 8 + n + 1}{6} = \frac{n - 1}{6}.$$ So there is symmetry, specifically $$\mathbb{P}(\text{Red Sox win game } n) = \mathbb{P}(\text{Yankees win game } 8-n).$$ But how does this let us conclude that $\frac{1}{2}$ must therefore be the probability that Red Sox win the best-of-seven in any number of games?
As far as I can tell, the real point is that "ends as soon as one team wins four games" is a distraction, because it doesn't matter. Once one team has won four games, they'll win the entire series whether you end the series or not! So the problem doesn't change if you just ignore that sentence and assume they play the entire series, and now you can apply the symmetry argument.