Best of seven series in ping pong. Finding probability. You are playing ping pong with a friend and your chance to win any point is $P$. This is a world series.
Find the probability that you score 4 points before your friend has a score of 4. Evaluate this expressions for $P=\frac{1}{2}$ and $P=\frac{2}{3}$.
I looked at this link: Ping Pong Winning Probability (World Series) but the answers nor the comments made any sense to me.
If someone can help me understand how to solve this, I would greatly appreciate it. 
 A: If you don't like the slightly sneaky approach taken in the linked answer, we can break the problem into cases. It is clear that there will be either $4$ games in total, or $5$, or $6$, or $7$. We find the associated probabilities, and add them. We will assume independence.
(i) The series lasts $4$ games, and you win: So you win $4$ games in a row. The probability of this is $p^4$.
(ii) The series lasts $5$ games, and you win:  So you win exactly $3$ of the first $4$ games, and win the fifth game. The probability you win $3$ of the first $4$ games is $\binom{4}{3}p^3(1-p)$. If that happens, you have probability $p$ of winning the fifth game. So the probability the series lasts $5$ games and you win is $\binom{4}{3}p^3(1-p)p$.
(iii) The series lasts $6$ games, and you win:  So you win exactly $3$ of the first $5$ games, and win the sixth game. The probability you win $3$ of the first $5$ games is $\binom{5}{3}p^3(1-p)^2$. So the probability the series lasts $6$ games and you win is $\binom{5}{3}p^3(1-p)^2p$.
(iv) The series lasts $7$ games, and you win:  Using the same technique, we find that the probability the series lasts $7$ games and you win is $\binom{6}{3}p^3(1-p)^3p$.
Add up and simplify a little. We get $p^4\left(1+(1-p)+(1-p)^2+(1-p)^3\right)$. By summing the short geometric series, we get the somewhat simpler expression $p^3\left(1-(1-p)^4\right)$.
