Question

Two baseball teams play a best-of-seven series, in which the series ends as soon as one team wins four games. The first two games are to be played on A’s field, the next three games on B’s field, and the last two on A’s field. The probability that A wins a game is 0:7 at home and 0:5 away. Assume that the results of the games are independent. Find the probability that:

(a) A wins the series in 4 games; in 5 games;

(b) the series does not go to 6 games.

Answer 1

I will refer to team A as A and to team B as B.

The probability that A wins a game at home is 0.7.

The probability that A wins a game away is 0.5.

The first 2 games are played at A's home and the next three at B's home and the last two back on A's.

a. Find the probability that A will win the series in 4 games

For this to happen, A must win the first 4 games.

Since the events are independent of each other, we can get the probability by multiplying the probability of A winning the games.

Thus, the probability is 0.7*0.7*0.5*0.5= 0.1225 or 12.25%

a. Find the probability that A will win the series in 5 games

This can happen in many ways. Let me state down the combinations below:

1. AAABA

2. AABAA

3. ABAAA

4. BAAAA

( Note, that A cannot win the first 4 games since the series will end there itself)

Let us calculate the probabilities of each of these and add them to get the final answer.

1. 0.7*0.7*0.5*0.5*0.5= 0.06125

2. 0.7*0.7*0.5*0.5*0.5= 0.06125

3. 0.7*0.3*0.5*0.5*0.5= 0.02625

4. 0.3*0.7*0.5*0.5*0.5= 0.02625

Thus, the required probability is 1+2+3+4 = 0.175 or 17.5%

b. The series does not go to 6 games

This again can happen in multiple ways. Let us state all of them and then add them all.

1. A wins first 4

2. A wins the 5th game and series ends

3. B wins first 4

4. B wins the 5th game and series ends

We have already calculated 1 and 2 in the previous question.

Let us calulate 3 for 4.

For 3 to happen, the porbability is a straightforward multiplication.

Thus, the probability for 3 is 0.3*0.3*0.5*0.5= 0.0225

For 4 to happen, the same combinations can be applied switching A and B.

i. BBBAB

ii. BBABB

iii. BABBB

iv. ABBBB

i. 0.3*0.3*0.5*0.5*0.5= 0.01125

ii. 0.3*0.3*0.5*0.5*0.5= 0.01125

iii. 0.3*0.7*0.5*0.5*0.5*0.5= 0.013125

iv. 0.7*0.3*0.5*0.5*0.5*0.5= 0.013125

i+ii+iii+iv= 0.04875

Now, finally, 1+2+3+4 = 0.1225+0.175+0.0225+0.04875= 0.36875 or 36.875%