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Suppose that two teams are playing a series of games, each of which is independently won by team A with probability p and by team B with probability (1-p). The winner of the series is the first team to win i games. If i = 4, find the probability that a total of 7 games are played. Also show that the probability is maximized when p = 1/2.

Given the above problem, I applied the negative binomial distribution.

$$P(X=7)=\binom{7-1}{4-1}p^{4}(1-p)^{7-4}$$ $$P(X=7)=20p^{4}(1-p)^{3}$$

To find the number of optimal $p$, I thought of taking the derivative of $P$ (when $n$ is fixed at 7) with respect to $p$ to find any local maxima.

$$0=\frac{dP(X=7)}{dp}=20(1-p)^{2}(4p^{3}-7p^{4})$$ $$p=\frac{4}{7}\ne\frac{1}{2}$$

Is my reasoning flawed? I’ve come across other solutions where they use the expected value of $X$ to find that $p=\frac{1}{2}$ is indeed optimal.

Daniel
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  • I don't understand your probability. The only way the series can require $7$ games is if the first $6$ split evenly. Work from there. – lulu Nov 13 '18 at 13:19
  • I think the first equation is only halve of the probability. In that occasion team A wins, but it's also possible team B wins in 7 games. So you have to add those probabilities. –  Nov 13 '18 at 14:06

1 Answers1

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My initial probability calculation is flawed. @lulu helped me see that it should be like so

$$P(X=7)=\binom{6}{3}p^{3}(1-p)^{3}$$

After that, taking the derivative with respect to $p$ shows that there are roots at $p=0,1,\frac{1}{2}$.

Daniel
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