Question

uppose two equally matched baseball teams (Team A and Team B) are playing in a World Series in which the winner is the 1st team to win 4 games. After 2 games, the series must be halted due to unforeseen circumstances. At that time, Team A leads Team B (2 games to 0 games). The Baseball Commissioner decides to divide the $36.4 million that would normally go to the winner between the two teams. He calculates P(Team A would have won the series) and P(Team B would have won the series) assuming that the teams are exactly evenly matched. He divides the money up proportionately using these probabilities. How much money should Team A receive?

Answer 1

Answer:

Team A should receive $29.57 million. The rest should go to team B

Step-by-step explanation:

Lets see, in order for team A to win the series, it needs to win 2 more games from a total of 5 remaining (this is a best of seven. We are assuming that all seven games are being played even if the series is finished; that shouldnt change the probability that team A wins the series).

Since both teams are even, the probability for team A to win a game is 0.5. We assume that 2 different matches are independent within each other, so the total amount of games that teams A would have won if the series would have continued is a binomial random variable, lets call it X, with parameters n=5, p=0.5. We want to know the probability that X is equal or greater than 2, that is, X=2, X=3, X=4 or X=5.

$P(X=2) = {5 \choose 2} * 0.5^2 * 0.5^3 = 0.3125$

$P(X=3) = {5 \choose 3} * 0.5^3 * 0.5^2 = 0.3125$

$P(X=4) = {5 \choose 4} * 0.5^4 * 0.5^1 = 0.15625$

$P(X=5) = {5 \choose 5} * 0.5^5 * 0.5^0 = 0.03125$

Therefore

$P(X \geq 2) = 0.3125+0.3125+0.15625+0.03125 = 0.8125$

This means that Team A should receive 81.25% of the money, that is 0.8125*36.4 = $29.57 million. The rest would go to team B.