<s>[INST] 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?
 [/INST] 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}
</s>