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?
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3 Answers
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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.
answered Dec 1, 2017 at 6:13
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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.
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1$\begingroup$ Forgetting about the negative binomial the actual result would depend on the probability that team A wins a single game. Should it be 0.5? Are the results for one game independent of the others? $\endgroup$ Commented Nov 30, 2017 at 21:51
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3$\begingroup$ I think the part about 100 games / impossibility of winning $5$ or more games is misleading. The binomial random variable for the $6$ first games can get equally impossible values, such as $5$ wins for team A. Meanwhile, the negative-binomial computation gives the correct probability for "what is the probability that team A wins the series in Game 7" even though "that's not the way an actual seven game series works" - so the only issue is that it does not take into account that the other team is allowed to win Game 7, too. $\endgroup$ Commented Dec 1, 2017 at 6:23
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2$\begingroup$ As a thought experiment, imagine the teams play (an infinite sequence of) exhibition games after the series is over and define the case event as Game 7 being part of the actual series. Nothing changes if the exhibition games are then cancelled $\endgroup$ Commented Dec 1, 2017 at 6:24
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1$\begingroup$ I still don't understand. Do you agree that the N.B. is fully appropriate here if you want to know the probability that it takes 7 games for team A to win 4 games? And thus fully appropriate for the actual question as long as you add the probability that it takes 7 games for team B to win 4 games? To me, this answer sounds to claim that the probability that the $r$th failure occurs after exactly $k$ successes (for fixed $r,k$) cannot be computed using the negative-binomial pmf if the experiment is stopped after $N>k$ successes. $\endgroup$ Commented Dec 3, 2017 at 9:23
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1$\begingroup$ @jbowman it's unclear to me whether we disagree about the math or about something else. Is the following statement true or false? For $k\in \{0,1,2,3\}$, the probability that team A wins the series $4$--$k$ is equal to $P(X=k)$ where $X$ is negative binomial with $p=0.5$, $r=4$ (the version of NB that counts the number of successes before $r$th failure). $\endgroup$ Commented Dec 4, 2017 at 7:03
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You can also achieve the results using Permutations.
Let us say the two teams for this 7 game series are A and B.
In how ways A or B can be placed as winners in each of the 7 games? 2^7
For illustration purpose, we can depict the winning team for each game with a letter like the following,
A A B B A B A
A A B B A B B
Now, it is clear that for game to go till 7th game, one of the teams have to lose 3 times and must win on the 7th game.
What are the possible permutations?
Let's take for A,
B A B B A A A (This is one of the series that went to the 7th game and A won it)
B A B A B A B (This is one of the series that went to the 7th game and B won it)
That means how many possible combinations can happen for A and B? 2*C(6,3)
Probability of 7th game happening = # games that can go upto 7 / Total possible number of games
= 2*C(6,3)/2^7 = 0.31
