Thread: Best-of-Seven Probability Question

Hi again,

I am struggling once again (no surprise) with calculating probabilities...

Two baseball games play a best-of-seven series in which the series ends as soon as one team wins four games. The first two games are played on A's field, the next 3 on B's and the last 2 on A's. Given that the probability A wins on their home field is 0.7, and on B's field is 0.5, what is the probability the series does not go to 6 games?

I have that the probability they in in 4 is 0.1225, and 5 games is 0.175... Are these in any way related to the answer?

I have thought about calculating the probability that the game goes to and taking the complement of that, but that would not include the series that went to 7 games. Kind of unsure where to piece this info together. Any insight would be much appreciated.

Thanks in advance.

Hello, KelvinScale!

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 played on A's field, the next 3 on B's and the last 2 on A's.
Given that the probability A wins on their home field is 0.7, and on B's field is 0.5,
what is the probability the series does not go to 6 games?

There is no formula for this problem.
I had to make a brute-force LIST of the possible outcomes.


Suppose wins the series.

could win the first four games: .

could win 3 of the first 4 games, then win the 5th game.
. .

Hence: .
. . . . . . . . . .


Suppose wins the series.

could win the first four games: .

could win 3 of the first 4 games, then win the 5th game.
. .

Hence: .
. . . . . . . . . .


Therefore: .

Thanks a lot, it is clear now, I forgot to consider the case where B won in 4 and 5 games.