Question

Baseball's division series is a best of five game series. That is, the first team to win 3 games is the winner. Team A and team B are playing in the series, and A is the clear favorite. In fact, you believe that A will defeat B in any given game with probability .65, and assume each game played is independent.
1. What the probability that A wins the series in 3 games?
2. What is the probability A wins the series in 4 games? Hint: which game did B win? Note also that things like AAAB are impossible (once A wins three games, the series is over!).
3.What is the probability A wins the series in 5 games?
4. What is the probability A wins the series (period)?
5. What is the probability A wins the series if a "best-of-three" game series is played?

Answer 1

Answer:

1. 27.46% probability that A wins the series in 3 games.

2. There is a 28.84% probability A wins the series in 4 games.

3. There is a 20.18% probability A wins the series in 5 games.

4. There is a 76.48% probability A wins the series.

5. There is a 71.825% probability A wins the series if a "best-of-three" game series is played.

Step-by-step explanation:

For each game, there are these following probabilities:

A 65% probability team A wins.

A 35% probability team B wins.

The combination formula is important to solve this problem:

This is because for example, team A winning games 1,2 and 4 and team B winning game 3 is the same as team A winning games 1,3,4 and team B winning game 2. That is, the order is not important.

$C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

1. What the probability that A wins the series in 3 games?

This is team A winning all 3 games. For each game, there is a 65% probability that team A wins. So

$P = (0.65)^{3} = 0.2746$

There is a 27.46% probability that A wins the series in 3 games.

2. What is the probability A wins the series in 4 games?

This is the team A winning three games and the team B 1. The team B win cannot happen in the fourth game, so the number of possibilities is a combination of 3 by 2. So

$C_{3,2} = \frac{3!}{2!1!} = 3$

The probability that team A wins the series in 4 games is:

$P = 3*(0.65)^{3}*(0.35) = 0.2884$

There is a 28.84% probability A wins the series in 4 games.

3.What is the probability A wins the series in 5 games?

This is the team A winning three games and the team B 2. The team B second win cannot happen in the fifth game, so the number of possibilities is a combination of 4 by 2. So

$C_{4,2} = \frac{3!}{2!1!} = 6$

The probability that team A wins the series in 5 games is:

$P = 6*(0.65)^{3}*(0.35)^{2} = 0.2018$

There is a 20.18% probability A wins the series in 5 games.

4. What is the probability A wins the series (period)?

They can win the series in 3,4 or 5 games. So this is the sum of the answers for 1,2,3.

So

$P = 0.2746 + 0.2884 + 0.2018 = 0.7648$

There is a 76.48% probability A wins the series.

5. What is the probability A wins the series if a "best-of-three" game series is played?

These three following outcomes are accepted:

A - A

A - B - A

B - A - A

So

$P = (0.65)^{2} + 2*(0.65)^{2}*(0.35) = 0.71825$

There is a 71.825% probability A wins the series if a "best-of-three" game series is played.