Problem
In the World Series of baseball, two teams (call them A and B) play a
sequence of games against each other, and the first team to win four games
wins the series. Let p be the probability that A wins an individual game, and
assume that the games are independent. What is the probability that team A
wins the series?
There will be at maximum seven games played to decide a clear winner.
This is a practice problem from a course on Probability. Its solution provided in the course gives two approaches.
First approach:
The Game will stop as any one of team wins.
$$ P(A) = P(\text{A winning in 4 games}) + P(\text{A winning in 5 games})
+ P(\text{A winning in 6 games}) + P(\text{A winning in 7 games}) $$
For A to win, the last game must be won by team A.
$$ \Rightarrow P(A) = p^4 + { 4 \choose 3}p^4q + { 5 \choose 3}p^4q^2 + { 6 \choose 3}p^4q^3 $$
Second approach
Imagine telling the players to continue playing the games even
after the match has been decided, then the outcome of the match won’t
be affected by this, and this also means that the probability that A wins the
match won’t be affected by assuming that the teams always play 7 games.
$$ P(A) = P(\text{A winning 4 times in 7 games}) + P(\text{A winning 5 times in 7 games})
+ P(\text{A winning 6 times in 7 games}) + P(\text{A winning 7 times in 7 games}) $$
I am not able to follow the second approach. Why is second approach correct?
Update:
I am looking for an intuitive explanation for the second approach because in the second approach it appears that probabilities for winning 5, 6 and 7 matches are used which were not required for team A to win.
Best Answer
Team $A$ wins the game if they win four matches times before $B$ wins four matches. Thus the first approach measures the probability of team A doing so when team $B$ wins zero, one, two, or three matches before $A$'s fourth victory.
$$P(A) = \left(\binom 30 p^3q^0+\binom 41 p^3q^1+\binom 52 p^3q^2+\binom 63 p^3q^3\right)p$$
Now imagine the teams kept playing for a full seven matches even after one of them wins the game. Team $A$ wins the game if they win at least four matches among those seven, since if they do so then team $B$ can win at most three matches before $A$ wins their fourth.
$$\begin{align}P(A) & = \binom 77 p^7q^0+\binom 76 p^6q^1+\binom 75 p^5q^2+\binom 74p^4q^3\end{align}$$
We can see these are equal by taking the terms of the first equation, and including the imagined games after victory. Then if you use the binomial theorem to expand...
$$\begin{align}P(A) &= \binom 30 p^4q^0(p+q)^3+\binom 41 p^4q^1(p+q)^2+\binom 52 p^4q^2(p+q)+\binom 63 p^4q^3 \\[1ex] &= \binom 30 p^4q^0(p+q)^3+\binom 41 p^4q^1(p+q)^2+\binom 52 p^5q^2+\left(\binom 52+\binom 63\right) p^4q^3\\[1ex] &\vdots \\[1ex] &= \binom 77 p^7q^0+\binom 76 p^6q^1+\binom 75 p^5q^2+\left(\binom 30+\binom 41+\binom 52+\binom 63\right) p^4q^3 \\[1ex]& = \binom 77 p^7q^0+\binom 76 p^6q^1+\binom 75 p^5q^2+\binom 74p^4q^3\end{align}$$