A and B have equal chances of winning a single game. A wants n games and B wants n+1 games to win the match. find the odds in favour of A

Question

I am stuck with this question can someone please give me a hint, how to go about this one. Is the total number of matches 2n and 2n +1?, I'm confused by this one

Answer 1

Just some hints to get you started. Half the time, player B wins the first game, and both players need $n$ games to win. In this case, they each have probability of winning $\frac12$ so this case contributes $\frac14$ to B's probability of winning.

One quarter of the time, B loses the first game, but wins the second. This brings us to the case where B need $n$ games and A needs $n-1$ so if we were doing it by induction we could substitute the probability from the $n-1$ case.

The other quarter of the time, A wins the first two games, so that B needs $n+1$ games and A needs $n-2$. It's clear now that simple induction on $n$ is not going to work, because this is not one of the prior cases. I think you will need to find a more general formula that works when A need $n$ games and B needs $m$ games. I suggest that you make a number of small examples and see if you can guess then general formula. Then you can prove it by recursion.

Answer 2

One of the hard parts of this problem is that the total number of games can be as small as $n$ (if A wins all $n$ of them) or as large as $2n$ (if the wins alternate B,A,B,A,..., for example).

We can avoid this by extending the match to $2n$ games, even if the winner has already been determined. That way:

• One player is guaranteed to satisfy their victory condition: whenever neither A nor B has won yet, the total number of games is at most $(n-1) + n = 2n-1$.
• However, both players cannot have satisfied their victory conditions: for that, we need at least $n + (n+1) = 2n+1$ games.

So whoever has the required number of wins after $2n$ games was the one to reach their target first, and win the match.

After the number of games is fixed, you can solve the problem with binomial probabilities.