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.

(In reply to re: Solution by Charlie)

I was under the understanding that 60/40 split applied to each game individually - not to a specific team.  If indeed it applies to a team, and if a probability is estimated from the first three games, then there is a 22.86% chance that Mudville has the 60% edge and a 77.14% chance that they have the 40% edge.

The team with the 60% edge has a 12.96% chance of winning the final four in a row while the team with the 40% edge has a 2.56% chance.

Multiply the probabilities and add the results, and you get 4.94%.

Posted by hoodat on 2010-07-19 15:42:02