TITLE: Underdog leading at least once in an infinite series of games
QUESTION [10 upvotes]: We are observing a tournament where 2 players play a series of games. Exactly one player wins each game. So we can keep count and 5:3 might be the standing after 8 games.
The first player is the favorite and will win a game with probability $p > 1 - p$.
What is the probability that the underdog will lead the standing at least once at some point in an infinite series of games? Or maybe more important than the specific value: Is it $1$ or not?
Somehow I think that it should be $1$, independent of $p$, but I am not sure how much my intuition is worth here. Maybe I can take it to the infinite case with a hint about finite series of $k$ games.

REPLY [4 votes]: A variant of Bertrand's ballot theorem says that in an election where candidate A receives $a$ votes and candidate B receives $b$ votes with $a \gt b$, the probability that A will be equal to or ahead of B throughout the count is $\dfrac{a+1-b}{a+1}$, so the probability B is ever strictly ahead is $\dfrac{b}{a+1}$.
The law of large numbers implies that here $\dfrac{a}{a+b} \to p$ and thus $\dfrac{b}{a+1} \to \dfrac{1-p}{p}$.