What is the probability that a best of seven series goes to the seventh game with negative binomial Why cant we use negative binomial to calculate the probability of getting a 7th game in best of 7 game series? 
P(X=7, r=4) = {7-1 C 4 -1} * (.5)^(7-4) * (.5)^4 = .15 , so there is 15 % chance of 7 trails to get 4 success or 15% chance its a 7 game series. 
Another way to look at this would be : Binomial , you need exactly 3 wins in 6 games to go to 7 game series, so that puts Binomial (n=6,x=3)= .31, so there is 31% chance that there is a 7 game series. 
Why don't they come out equal? 
 A: Summary: the negative-binomial based approach in the question ignores that either team can win Game 7. After correcting for this the results agree. 
Assumption
Not explicitly stated in the question, but it seems we are assuming the games are iid. with probability 0.5 for either team to win (a sequence of fair coin flips). 
The probability of Game 7 happening, using the negative binomial distribution
The series goes to Game $7$ if and only if either team obtains its $4$th win in the $7$th game. The events "A wins $4$th time in Game $7$`"  and "B wins $4$th time in Game $7$" are mutually exclusive, so 
\begin{equation}
\mathbb{P}(\textrm{Game $7$ is played}) = \\ \mathbb{P}(\textrm{A obtain its $4$th win in the $7$th game}) +  \mathbb{P}(\textrm{B obtains its $4$th win in the $7$th game}).
\end{equation}
Each term on the right-hand-side is the probability of the case team obtaining $4$th wins after $3$ losses. Or, equivalently, the probability of the losing team winning (exactly) $3$ games before the $4$th loss. As reasoned in the question, this is the probability of $3$ in a negative-binomial distribution with parameters $p=0.5,~r=4$ (where $r$ is the number of failures). Thus, the total probability 
\begin{equation}
= 2 {4+3-1 \choose 3 }\,0.5^4\,0.5^3  \approx 0.31,
\end{equation}
 exactly the same answer as derived in the question using the binomial distribution.
Additional remark
With a general series played to $r$ wins going to $2r-1$th game, the approaches can be seen to give the same result: the binomial coefficient is the same and the factor $2$ in the negative-binomial approach cancels one $0.5$ factor. This cancellation is very closely related to the fact that the binomial distribution approach "naturally" ignores the result of the final game while in the negative-binomial distribution approach both cases need to be taken into account.
A: The negative binomial would be appropriate if you wanted to know how many games it would take before team A won 4 games.  However, that might be, say, 104 games, in which case team B would have won 100 games.  Obviously that's not the way an actual seven game series works!
Your calculation - $P(X=7 | r=4)$ - using the negative binomial distribution calculates the probability that team B would have won exactly 3 games before team A wins 4, given that team B might have won any number of games before team A wins 4.  It ignores the possibility that team B is the one that wins 4 games first and the impossibility of either team winning 5 or more games. 
