# Probability of Game Series

A world series is a best of $7$ series between team $A$ and team $B.$ It takes $4$ wins to win the series. How many ways can a team win the World Series?

I said:

Suppose that a World Series is completed leading to an extended series where extra exhibition games are played (between the same two teams $A$ and $B$) after the series has already been won. These exhibition games will always be won by the team that lost the official World Series and are to continue until that losing team also has a total of $4$ wins. In this way, if the results of the World Series read $ABAABA$, then the results of the extended series read $ABAABABB$.

The sequence of results for the extended series is just an arrangement of 4 copies of the letter A and $4$ copies of the letter B. There are $_8C_4$ = $8*7*6*5/(4*3*2*1) = 70$ such arrangements. (We may choose $4$ positions for the letter $A$ from $8$ possible positions.)

Each sequence of results for the extended series determines exactly one possible sequence of results for the World Series (by omitting the exhibition games).

I am not sure if this is correct. Is there a simpler way of doing this? Can someone show me?

Another approach would be to ask how many ways are there for team A to win the series. To count this, arrange $7$ slots and pick $4$ slots for $A$ to win. This can be done in $35$ ways. There are an equal number of ways for team B to win the series. This agrees with your answer of $70$.
The number of distinct outcomes for the series of $7$ matches is $2^{7}$. If the probability of a team winning the series is $p$, then the number of ways in which that team can do so is $p*2^{7}$. Assuming equal skill, both teams win with probability $\frac{1}{2}$ giving you $128$ ways for a given team to win the series.