Combination issue I understand almost all parts of this problem. The thing I don't get is where the 5 comes from in the combination set-up (same for 6 in the next part). Why isn't it 7 choose 3 in Pr(x=7) , vice versa ?



 A: The teams compete until one team wins four games.
$\mathsf P(X=4)$ is the probability that one team won their forth game on game four.   This is the event of winning all in the first three games then the next game.  $$\mathsf P(X=4) = 2(0.5)^4$$
$\mathsf P(X=5)$ is the probability that one team won their forth game on game five.   This is the event of winning three games in the first four, then the next game.  $$\mathsf P(X=5) = 2\binom{4}{3}(0.5)^5\\ \qquad\qquad = 2\binom{4}{1}(0.5)^5$$
$\mathsf P(X=6)$ is the probability that one team won their forth game on game six.   This is the event of winning three games in the first five, then the next game.  $$\mathsf P(X=6) = 2\binom{5}{3}(0.5)^6\\ \qquad\qquad = 2\binom{5}{2}(0.5)^6$$
$\mathsf P(X=7)$ is the probability that one team won their forth game on game 
seven.   This is the event of winning three games in the first six, then the next game.  $$\mathsf P(X=7) = 2\binom{6}{3}(0.5)^7$$
$\mathsf P(X=8)$ is the probability that one team won their forth game on game eight.   This is the event cannot happen because someone must win four games earlier than that.  $$\mathsf P(X=8) = 0$$
A: Think of it this way: The losing team can't win the last game. So if you want the series to go six games you have to have the losing team lose their two games within the first five games, not within six games. Same for if the series goes for five games or seven games.
A: Assume the teams are $A,B$, with respectively probabilities of winning of $p, q (=1-p)$. 
Assume tournament ends after $n$ games. 
If $A$ wins the tournament, then $A$ has to win $k-1$ games in the first $n-1$ games, and also win the last game. Probability of this happening is 
$$\binom {n-1}{k-1}p^{k-1}q^{n-k}p=\binom {n-1}{k-1}p^kq^{n-k}=\binom{n-1}3(0.5)^n$$
as $k=4, p=q=0.5$.
Similarly, if B wins the tournament, probability is
$$\binom {n-1}{k-1}q^kp^{n-k}=\binom{n-1}3(0.5)^n$$
Combining both gives probability of tournament ending in $n$ games (with either $A$ or $B$ winning), i.e. 
$$p(n)=2\binom{n-1}3(0.5)^n$$
Now consider all possible values of $n$ (i.e. $4$ to $7$). 
Expected number of games is given by 
$$\sum_{n=4}^7 n\cdot \underbrace{2\binom{n-1}3(0.5)^n}_{p(n)}=2\left[4\binom 33(0.5)^4+5\binom 43(0.5)^5+6\binom 53(0.5)^6+7\binom 63(0.5)^7\right]\qquad\blacksquare$$
