Home Field Advantage

five games, three games… one invariant

So let’s say that there’s some sports final, where 2 teams play, and the winner is the best 3 out of 5 (first to get to a total of 3 wins). Let’s call these two teams A and B respectively. Now let’s say that these two teams are equally skilled, and the only advantage comes from being able to play on their home field. The team with strictly fewer wins (the team currently ‘losing’ the tournament) will always get to play at home. In the event of a tie, team A gets to play at home because they were the underdog going in or [some other silly reason to make this question work]. Let’s say that the home field advantage makes a team have a 60% probability of winning. What is the probability that team A wins?

Solution & Generalization

Normally I would say that it’s easier to do questions by breaking them down into tiny cases, but I think that my explanation here is literally identical to my explanation for the general case so let’s just jump right into the general case.

The general case is: this tournament consists of N games, where N is some odd number >= 1, and the winner is the best (N-1)/2 +1 out of N games and the home team has some probability, p, of winning.

For the sake of simplicity, let’s denote (N-1)/2 +1 as K. Let’s define our current state S(a,b) where a and b are the number of games won by teams A and B respectively. By the problem definition, we end when a=K or when b=K.

Suppose that we are in state S(K-1,K-1), then the probability of A winning is simply p. Now let’s say we are at state S(K-2,K-2), there are 2 ways to get to S(K-1,K-1): a+1 then b+1 or b+1 then a+1. The first happens with probability p^2 and the latter happens with probability (1-p)p. and we know that team A wins in this state with probability p. Now what about the other cases? We can go to S(K,K-2) or S(K-2,K). These happen with probability p(1-p) and (1-p)(1-p). I’ve (likely) thought about this problem for longer than the reader at this point, but can you see how the paths where we end up back to a tie differ by p? In other words, the “upper paths” happen with probability p and the lower paths happen with probability (1-p). If we end in an upper path, team A wins, and if we end in a lower state then team B wins. Anytime we are in a “tie” state we go to an upper path with probability p and go to a lower path with probability (1-p).

We can therefore conclude that regardless of the number of rounds that this is the best of (the value of N), or the value of p, team A will always win with probability p. Kinda cool right? Okay maybe not that cool.