Problem 1: In a best of 7 World Championship Series, the Mudville Nine have lost the first 3 games. What are the chances that they will rally against the Louisville Sluggers to win the next 4 games and the series, assuming the two teams are known to have an equal chance of winning any given game?

Problem 2: What are the chances that Mudville prevails if one team has a 60% chance of winning any given game? Before the series started it was equally likely that either team was the one with that advantage.

Assume that all games are played at a neutral venue, with no home field advantage. Also, there is no psychological advantage to being either on the brink of elimination, or a game away from winning the World Championship.

Either way, the odds are 1 in 16 (0.625%).  The first is solved by calculating 0.5^4 = 0.0625.  The first three games have no bearing on the results of 4 thru 7.

In problem 2, there is an equal probability that either team will have a 60% likelihood of winning a game.  So there are 16 different combinations ranging from 0.4-0.4-0.4-0.4 to 0.6-0.6-0.6-0.6.  Solving all combinations, adding the results, and dividing by 16 yields 0.0625.

0.4 0.4 0.4 0.4  0.0256
0.4 0.4 0.4 0.6  0.0384
0.4 0.4 0.6 0.4  0.0384
0.4 0.4 0.6 0.6  0.0576
0.4 0.6 0.4 0.4  0.0384
0.4 0.6 0.4 0.6  0.0576
0.4 0.6 0.6 0.4  0.0576
0.4 0.6 0.6 0.6  0.0864
0.6 0.4 0.4 0.4  0.0384
0.6 0.4 0.4 0.6  0.0576
0.6 0.4 0.6 0.4  0.0576
0.6 0.4 0.6 0.6  0.0864
0.6 0.6 0.4 0.4  0.0576
0.6 0.6 0.4 0.6  0.0864
0.6 0.6 0.6 0.4  0.0864
0.6 0.6 0.6 0.6  0.1296
     
     0.0625

Edited on July 19, 2010, 3:45 am

Edited on July 19, 2010, 3:46 am

Posted by hoodat on 2010-07-19 03:43:52