There was some angst in Yankeeland after the team failed to close out the Angels in game 5. However the Yankees would appear to still be in pretty good shape returning home up 3 games to 2 in the best of 7 American League Championship Series (ALCS).

This suggests the following puzzle. Suppose the Yankees will win each game with probability p (independent of the results of the other games). Then in a best of 7 playoff how large does p have to be for the Yankees to be worse off leading 3 to 2 than they were at the beginning before any games were played.

Friday Cat Blogging – 23 March 2018

3 hours ago

I get p=0.772. This is done by setting B_cum(3;7,p) = 2p-p^2, and solving for p. The first expression is the probability of yankees winning from the outset (and B_cum refers to the cumulative binomial function), and the second expression is the probability of winning given being up 3 to 2. I used a numerical solver - i'm not sure whether it can be solved analytically given the degrees of polynomials involved.

ReplyDeleteLet q = 1-p. The probability that the Yankees will win n >= 1 more games before their opponents win m >= 1 more games is

ReplyDeleteW(p,n,m) = p**n*Sum(from k=0 to k=m-1)of

[(n-1+k)!/((n-1)!*k!)]*q**k ,

where the kth term expresses the probability that the opponents win exactly k more games. One can verify that W satisfies the following equalities:

W(p,n,m) = p*W(p,n-1,m) + q*W(p,n,m-1) ,

which expresses the effect of playing one more game;

W(p,n,m) + W(q,m,n) = 1 ,

which expresses the fact that exactly one team wins;

W(0,n,m) = 0 ; and W(1,n,m) = 1 .

We are interested in W(p,4,4) (before the series begins) and W(p,1,2) (when the Yankees lead 3 to 2). With computer assistance the following is quickly established:

W(p,1,2) > W(p,4,4) when 0 < p < 0.7721 ,

W(p,1,2) = W(p,4,4) = 0.9481 when p = 0.7721 , and

W(p,1,2) < W(p,4,4) when 0.7721 < p < 1 .

So the desired value is p = 0.7721. QED

I like the puzzle and the solutions, thanks.

ReplyDeleteIt would be interesting (and somewhat useful IMO) to have a table generalizing this [3,2]-puzzle to an [a,b]-puzzle. Such a table, plus an assumed or calculated a priori value of p, could give one a crude sense of whether a team has improved its lot in life mid-Series.

(The crudeness is due not only to the roughness of the estimate of p, but also to the fact that such an approach ignores significant factors such as pitching, home field advantage, momentum/morale, etc.)