Question on Binomial Probability 
The Grunters and the Screamers are playing for the Grand Championship, which is a best of 7 series. The first team to win 4 games wins the Championship. Each team has a $\frac{1}{2}$ probability of winning any individual game. If the Grand Championship series lasts exactly 6 games, what is the probability that the Grunters win?

I attempted the question as follows. If the series lasts 6 games and the Grunters need 4 games to win, that means the Screamers won 2 games. Further, the Grunters must have won the last game, otherwise the series would be less than 6 games. Thus, out of the five games which are not the final game, two of those must have been won by the screamers, which can be done in $\binom{5}{2}$ ways. The probability of the screamers winning those two games is $\left(\frac{1}{2}\right)^2$ and the probability of the Grunters winning the other 4 games is $\left(\frac{1}{2}\right)^4$. This gives the final probability of $\binom{5}{2}\cdot\left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^4$ which is $\frac{5}{32}$. 
However, this given answer is $\frac{1}{2}$. Where did I go wrong? 

The key observation is that either team has an equal probability of winning any single game, so the overall probability that any given team wins the series is also $1/2$:  simply imagine switching the roles of the two teams.
The calculation you made is a joint probability.  Suppose $G$ is the event that the Grunters win.  Let $X$ be the random variable that counts the total number of games played in the series to determine the winner.  You then calculated $$\Pr[G \cap (X = 6)],$$ the probability that the Grunters win and do so in exactly 6 games. But the question is asking for $$\Pr[G \mid (X = 6)],$$ which is the probability that the Grunters win given a winner was determined in 6 games.  Indeed, if you were to calculate $\Pr[X = 6]$, you would find that it equals $5/16$, from which it follows that $\Pr[G \mid (X = 6)] = (5/32)/(5/16) = 1/2$.
