Two 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 games are independent. Find the probability, that the series does not go to $6$ games.
My Answer is
$$2\cdot 0.7\cdot 0.7\cdot 0.5\cdot 0.5\cdot 0.5\\+2\cdot 0.7\cdot 0.3\cdot 0.5\cdot 0.5\cdot 0.5\\+2\cdot 0.3\cdot 0.3\cdot 0.5\cdot 0.5\cdot 0.5\\+2\cdot 0.7\cdot 0.3\cdot 0.5\cdot 0.5\cdot 0.5\\=0.25$$
But the correct answer is $0.397$.
How to get the correct answer?
Thank you very much.
Best Answer
We want the probability that the series lasts $4$ or $5$ games.
Team A sweeps in $4$ with probability $(0.7)^2(0.5)^2$, and Team B sweeps with probability $(0.3)^2(0.5)^2$. The probability of a sweep is $0.145$.
The $5$ game analysis is lengthier. Let's first deal with Team A winning in $5$. Team A has to win the fifth game (probability $0.5$, and $3$ of the first $4$. So it has to either (i) win its $2$ home games, and $1$ of the $2$ away games or (ii) it has to win $1$ home game, and $2$ away games.
Event (i) has probability $(0.7)^2(2)(0.5)^2$. Event (ii) has probability $(2)(0.7)(0.3)(0.5)^2$, for a total of $(0.5)((0.7)^2(2)(0.5)^2+(2)(0.7)(0.3)(0.5)^2)$. This is $0.175$.
A similar calculation for B winning in $5$ gives $(0.5)((0.3)^2(2)(0.5)^2+(2)(0.7)(0.3)(0.5)^2)$. This is $0.075$.
Add up. We get $0.395$.