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.
(a) What is the probability that team A wins the series?
(b) Give a clear intuitive explanation of whether the answer to (a) depends on whether the teams always play 7 games (and whoever wins the majority wins the series), or the teams stop playing more games as soon as one team has won 4 games (as is actually the case in practice: once the match is decided, the two teams do not keep playing more games).
I fully understand part (a); below is my solution.
(a) $$P(\text{A wins}) = P(\text{A winning in 4 games}) + P(\text{A winning in 5 games}) +P(\text{A wins in 6 games}) $$ $$ + P(\text{A winning in 7 games}) = p^4 + \binom{4}{3}p^3qp + \binom{5}{3}p^3q^2p + \binom{6}{3}p^3q^3p$$
Another solution: Let $X \sim Bin(7,p)$ and $q=1-p$.
\begin{align*}
P(\text{A wins}) &= P(X=4) + P(X=5) + P(X=6)+P(X=7) \\
&= \binom{7}{4}p^4q^3 + \binom{7}{5}p^5q^2+ \binom{7}{6}p^6q + p^7
\end{align*}
However, I am quite confused on part (b). I know the two solutions above give the same answer, but I'm not sure why this works. Here is the solution provided by the professor, but I don't totally follow.
"Intuitively, the answer to (a) does not depend on whether the teams play all seven games no matter what. Imagine telling the players to continue playing the games even after the match has been decided, just for fun: 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!"
I know there's a question on this already, but it didn't provide an intuitive explanation.
Best Answer
Four ways of intuitively understanding the apparent paradox
Once $A$ has won the series, it is impossible for $A$ to lose the series in the remaining games, so their results don't matter
When you stop at $4$ wins for $A$, you implicitly include all possibilities of win/loss for $A$ subsequently, and in the overall picture, $A$ will always have lost $\leq 3$ matches
When $A$ wins the series at $4-1$, say, considering all "after-sequences" mathematically amounts to multiplying the winning probability at $4-1$ by $1$
"First to four" is just another way of saying "best of seven"