What is the probability of a world series ending after 4 games? After 5 games? 6 games? 7 games?

Assume that each team has equal probability of winning each game, regardless of who has won the previous games.

(*For the baseball-challenged, the World Series ends after one team has won 4 games. Ties are not possible.*)

(In reply to

Answer by K Sengupta)

Let us denote the winning team as A and the losing team as B.

By the problem, since a World Series concludes after one team has won 4 games with a score of 4-n, where n=0,1,2,3 so that max(n) = 3, it follows that the respective minimum number and the maximum number of games are 4 and 7.

Now, for a (4+n) game series to be feasible, the final game will have to be won by A while games 1,2,...., (3+n) must be split 3-n with n=1,2,3

Since the probability of a win or loss of a game is 0.5 for each of A and B, and we have precisely n succedsses in (n+3) trials in terms of a Binomial probability Distribution, we must have the probability of having a (4+n) game in a World Series equal to:

comb(n+3, n)*(1/2)^(n+3)......(*)

*Four Game Case*

For a four game world series, substituting n=0, in (*), we have the required probability equal to: comb(3, 0)/(2^3) = 1/8.

**Five Game Case**

For a five game world series, substituting n=1, in (*), we have the required probability equal to: comb(4, 1)/(2^4) = 1/4.

*Six Game Case*

For a six game world series, substituting n=2, in (*), we have the required probability equal to: comb(5, 2)/(2^5) = 5/16.

*Seven Game Case*

For a seven game world series, substituting n=3, in (*), we have the required probability equal to: comb(6, 3)/(2^6) = 5/16.

As a check, we observe that these probabilities are indeed

exhaustive as : 1/8 + 1/4 + 5/16 + 5/16 = 1