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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?

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  • $\begingroup$ I know that I have 20 ways the 6 games can be won. I actually enumerated them. But I just want to learn the intuition behind the choose function in this case. $\endgroup$ – Note Sep 7 '18 at 21:17
  • $\begingroup$ Lets call the two gamblers "A" and "B". There are 6 games, 3 won by each player. To find the total number of ways that could happen, you could write down 6 "A"s and then choose three of them to be change to "B". $\endgroup$ – user247327 Sep 7 '18 at 22:49
  • $\begingroup$ (Note in passing that poker is a rather curious choice of a game to reduce to a binary "win or lose" outcome, since skill in poker is not about how many times you win vs lose, but about how cheaply you can lose when you lose, vs how much you can win when you win). $\endgroup$ – hmakholm left over Monica Sep 7 '18 at 23:08
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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)

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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.

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