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.

Part 1:

As the two teams are equally matched, the probability that Mudville will win the next four games is (1/2)^4, or 1/16, or 6.25%.

Part 2:

We first must use Bayes' rule to find the probability that Mudville is in fact the better team (60% probability of winning a given game), despite its poor record in the series thus far:

P(better) = (1/2)*((2/5)^3) / ((1/2)*((2/5)^3) + (1/2)*((3/5)^3)) = 8/35

And the probability they are the less likely to win a given game is therefore (35-8)/35 = 27/35.

The probabilities of going on to win the next four games are (3/5)^4 and (2/5)^4 respectively for the two cases.

Therefore the overall probability that Mudville will win the next four games is

(8/35)*((3/5)^4) + (27/35)*((2/5)^4) = 216/4375 = 0.0493714285714285..., with the repeating cycle of digits being the 714285, or, not quite 5%.

Posted by Charlie on 2010-07-16 13:35:03