# How to compute the number of games that must be played for the better team to win 70% of the time?

I've been stuck on this sample statistics problem for a couple days now:

Assume that two baseball teams have win records of $58\%$ and $54\%$. Demonstrate how to compute or estimate the number of games that must be played in order for the better team to win $70 \%$ of the time, $80\%$ of the time, and $90\%$ of the time.

I think the question is asking how many games must each team play against each other in order for the better team (team with the $0.58$ winning percentage) to win $X\%$ of the time. If I use a binomial distribution, the Central Limit Theorem and the $Z$ score formula I can work backwards and get a formula for the number of games. However, this approach only uses the winning percentage of the better team (i.e. $0.58$) which makes me think that I'm missing something important. Perhaps I'm misinterpreting the question?

Any thoughts or suggestions for how to approach this problem would be greatly appreciated!!!

I think the question has been garbled, either by the original source or in the quote. Please check the original source again.

I think the question meant is, "Assuming $X$ wins $58\%$ of the time and $Y$ wins $54\%$ of the time, how many games do they have to play against other teams before there is a $70\%$ chance that $X$ has a better record than $Y$? This does not say the better record has to be any particular percentage.

In baseball, it is perfectly reasonable that teams $X$ and $Y$ are in different leagues and don't play each other. So, let's assume that they never play each other, which allows us to assume that their records are independent, too.

It sounds like you are doing problems where you apply a $Z$-score. You can use that method on the difference between the (binomial) number of wins by $X$ and the (binomial) number of wins by $Y$. The difference does not have a binomial distribution, but you should still be able to compute the mean and standard deviation. When this difference is at least $1$, $X$ has the better record, although a better whole number adjustment may be to say the difference should be greater than $0.5$.

• +1 DouglasZare for a well thought out answer. But I would like to point out that with interleague play now in the MLB teams in opposing leagues can meet during the season. Commented Jul 10, 2012 at 10:51

The problem is with the question not the solution. If the better has only a 58% chance of winning then in the long run they will win only 58% and there is a good possibility that they will never reach 70%. There is of course a probability of 0.58 that they will win the first game and that will be 100%. But if they lose the first game and win the next 3 they will reach 75% and that has probability (.42) (.58)$^3$. You could ask what is the expected number of games and calculate it in this way. However the probability of long runs might be high enough so that the expectation is infinite.

• I agree. The question is either bogus or has been misinterpreted in some way from some other source into the form presented. It doesn't make much sense as it stands. Commented Jul 10, 2012 at 2:45

I agree with Douglas Zare; I don't think there's an answer to this problem as stated, although I will approach my discussion more laboriously.

Suppose team A and team B are rated on an Elo system (originally used for chess but now used in many sports contexts) and have a 58% and 54% win percentage against teams with an average rating of X. For discussion, let X=1500, although the number doesn't matter. Then the rating of A is 1556 and the rating of B is 1527. But if A and B play each other, A is expected to win versus B 54% of the time, not 70% of the time. In fact, the more often they play the less likely it is that A beats B in at least 70% of the games.

For more discussion and the key equations, see http://en.wikipedia.org/wiki/Elo_rating_system