If a sports team is down by 1-2 in a best of 7 series, what is their chance of winning (if both teams have a 1/2 chance of winning every game)? I know that the team has to win 3 of the 4 remaining games. So I thought that from the 4 remaining games, there are a total of $2^4 = 16$ total different outcomes. 
Could someone point me towards the right direction to determine the probability of the team winning 3 of the 4 remaining games? Would it be 4C3/16 since the three wins can be chosen amongst the 4 games left?
 A: The team that's behind needs to win three of the remaining four games.
One possibility is to explicitly write out all the possible scenarios: WWW, WWLW, WLWW, LWWW.  These have probability 1/8, 1/16, 1/16, and 1/16, so the probability of winning is the sum of these, 5/16.
Alternatively, imagine that all four remaining games get played even if they don't need to be.  Then the probability that the team that's behind wins at least three of its games is just the probability that a binomial(4, 1/2) random variable is at least 3 - that's 
$$ {{4 \choose 3} + {4 \choose 4} \over 2^4} = {4 + 1 \over 16} = {5 \over 16}$$
A: Starting with a reply to the OP's request for a tree:

Calculate the probabilities of getting to a "winning" end $42$ and $43$, being sure to add the probabilities along all paths:
$\underbrace{(1/2)^3}_{{\rm path~to~} 42} + \underbrace{3 (1/2)^4}_{{\rm paths~to~} 43} = 5/16$
A: Let $p(a,b)$ be the probability that team A is the first to win 4 games, given that team A has already won $a$ games and team B has already won $b$ games. By conditioning on the outcome of the next game, we see that $p$ satisfies the following recurrence relations:
\begin{align}
p(4,b) &= 1 \\
p(a,4) &= 0 \\
p(a,b) &= \frac{1}{2} p(a+1,b) + \frac{1}{2} p(a,b+1) &&\text{if $a<4$ and $b<4$}
\end{align}
You want to compute $p(1,2)$.  Now
\begin{align}
p(1,2)&=\frac{1}{2} p(2,2) + \frac{1}{2} p(1,3)\\
&=\frac{1}{2} \left(\frac{1}{2} p(3,2) + \frac{1}{2} p(2,3)\right) 
+ \frac{1}{2} \left(\frac{1}{2} p(2,3) + \frac{1}{2} p(1,4)\right)\\
&=\frac{1}{4} p(3,2) + \frac{1}{2} p(2,3) + \frac{1}{4} \cdot 0\\
&=\frac{1}{4} \left(\frac{1}{2} p(4,2) + \frac{1}{2} p(3,3)\right) + \frac{1}{2} \left(\frac{1}{2} p(3,3) + \frac{1}{2} p(2,4)\right) \\
&=\frac{1}{8} \cdot 1 + \frac{3}{8} p(3,3) + \frac{1}{4} \cdot 0 \\
&=\frac{1}{8} + \frac{3}{8} \left(\frac{1}{2} p(4,3) + \frac{1}{2} p(3,4)\right) \\
&=\frac{1}{8} + \frac{3}{8} \left(\frac{1}{2} \cdot 1 + \frac{1}{2} \cdot 0\right) \\
&=\frac{1}{8} + \frac{3}{16} \\
&=\frac{5}{16}
\end{align}
