Why do we use the Choose function here?

Question

Assume two gamblers playing a series of poker games, which is independently won with prob. $p$ for player $1$ and $1-p$ for player $2$. The ultimate winner of the series is the first to win k card games.

What is $Pr[\text{a total of 7 games are played}]$, for $k=4$?

I can see that to have 7 games, we need to be tied at 3 wins for each player after 6 games. So the $$Pr[\text{a total of 7 games are played}] = Pr[\text{each player wins 3 games after 6}] = \binom{6}{3}p^3(1-p)^3 $$ I just don't understand quietly the use of the choose function to solve this?

What makes this dst different from a neg. Bin. Dst?

Answer 1

Once you know that the first player has won three games out of six, there are $\binom 63$ choices/possibilities for which three games they won.

(The second player won the others)

Answer 2

It is not a negative binomial distribution because there are two different stopping conditions: either that the first player has won 4 games or that the second player has won 4 games. That is a more complex situation than the negative binomial distribution is made to model.