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.

Re: Best-of-Seven Probability Question

Hello, KelvinScale!

Quote:

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 $\displaystyle A$ wins the series.

$\displaystyle A$ could win the first four games: .$\displaystyle AA|AA \:=\:(0.7)^2(0.5)^2 \:=\:0.12250$

$\displaystyle A$ could win 3 of the first 4 games, then win the 5th game.

. . $\displaystyle \begin{array}{cccccc}AA|ABA &=& (0.7)^2)0.5)^3 &=& 0.06125 \\ AA|BAA &=& (0.7)^2(0.5)^3 &=& 0.06125 \\ AB|AAA &=& (0.7)(0.3)(0.5)^3 &=& 0.02625 \\ BA|AAA &=& (0.3)(0.7)(0.5)^3 &=& 0.02625 \end{array}$

Hence: .$\displaystyle P(A\text{ wins in less than 6 games})$

. . . . . . . . . . $\displaystyle =\:0.12250 + 0.06125 + 0.06125 + 0.02625 + 0.02625 \:=\:0.2975$

Suppose $\displaystyle B$ wins the series.

$\displaystyle B$ could win the first four games: .$\displaystyle BB|BB \:=\:(0.3)^2(0.5)^2 \:=\:0.02250$

$\displaystyle B$ could win 3 of the first 4 games, then win the 5th game.

. . $\displaystyle \begin{array}{cccccc}BB|BAB &=& (0.3)^2(0.5)^3 &=& 0.01125 \\ BB|ABB &=& (0.3)^2(0.5)^3 &=& 0.01125 \\ BA|BBB &=& (0.3)(0.7)(0.5)^3 &=& 0.02625 \\ AB|BBB &=& (0.7)(0.3)(0.5)^3 &=& 0.02625 \end{array}$

Hence: .$\displaystyle P(B\text{ wins in less than 6 games})$

. . . . . . . . . . $\displaystyle =\:0.02250 + 0.01125 + 0.01125 + 0.02625 + 0.02625 \:=\:0.0975$

Therefore: .$\displaystyle P(\text{Series ends in 5 games or less}) \;=\; 0.2975 + 0.0975 \;=\;\boxed{0.395}$

Re: Best-of-Seven Probability Question

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