find the probabilty

Question

Two baseball teams are in the World Series, first team to win four games ends the series. Team A has won 2 games and team B has won one. If the probability of team A winning any one game is .6 (60%). Compute the probability team A will win the series.

Answer 1

Team A can either win the series in 5 games (they win the next 2 games), 6 games (2-1 in the next 3 games), or 7 games (2-2 in the next 4 games). So we can use the binom distribution formula to solve for the probability of each.

 

However, we need to keep one thing in mind, which is that there is actually a trick to the combinations portion. For example, the binom dist formula for team A winning in 7 games would say there are 4C2 combos of ways they can win twice in 4 games.  However, there are actually only 3 ways they can do this (because if they won their 4th game in game 5 or 6 then the series is over and thus the final win has to be game 7). So just note that slight deviation off the binom dist formula.

 

P(win game 4 and game 5)

= 1 * .6^2 * .4^0

= .36

 

P(win either game 4/5, win game 6)  

= 2 * .6^2 * .4^1

= 2 * .36 * .4

=.288

 

P(win 1 game in games 4-6, lose the other 2, then win game 7)

= 3 * .6^2 * .4^2

= 3 * .36 * .16

=.1728

 

Add them together and you get 82.08%.