Problem 50 Solution

In general the probability of either team winning exactly 3 out of n (n>3) games is (n-1)*(n-2)*(n-3)/(3*2n).

For a series to last exactly n games at least one team must have won exactly 3 out of the last (n-1). Thus:

The probability of the series lasting 4 games is 1/8.

The probability of the series lasting 5 games is 1/4.

The probability of the series lasting 6 games is 5/16.

The probability of the series lasting 7 games is 5/16.

The expected number of games played in team A's home field is: 2*(1/8) + 2*(1/4) + 3*(5/16) + 4*(5/16) = 188/64.

The expected number of games played in team B's home field is: 2*(1/8) + 3*(1/4) + 3*(5/16) + 3*(5/16) = 184/64.

Michael Shackleford, A.S.A.