Guess we'll still never know ...

Now that you’ve determined the values of $a$ and $b$, let’s analyze the rest of Stephen’s statement. Is it true that losing in five games is “closer to a sweep than a seven-game series”? Let $p_4$ represent the probability that the Celtics sweep the Knicks in four games. And let $p_7$ represent the probability that the series goes to seven games (with either team winning).

Suppose $p$ is randomly and uniformly selected from the interval $(a, b),$ meaning we take it as a given that the most likely outcome is that the Knicks will lose the series in five games. How likely is it that $p_4$ is greater than $p_7$? In other words, how often will it be the case that losing in five games means a sweep is more likely than a seven-game series?

As we saw earlier, for any value of $p \in \left(\frac{3}{5}, \frac{3}{4}\right),$ we have $p_4 = p^4.$ Similarly, we have $$p_7 = P(\text{C},7\mid p) + P(\text{K}, 7\mid p) = 20p^3 (1-p)^3.$$ We can solve for the point at which we have $p_4(p) = p_7(p)$ and get the cubic equation $20(1-p)^3 - p = 0,$ or substituting $t = 1-p$ we get $$t^3 + \frac{1}{20} t - \frac{1}{20} = 0.$$ Using Cardano's formula, given that the discriminant when $p = \frac{1}{20}$ and $q = -\frac{1}{20}$ is $$\Delta = \frac{q^2}{4} + \frac{p^3}{27} = \frac{1}{1600} + \frac{1}{432000} = \frac{17}{2700} \gt 0,$$ we have the solution $$t^* = \sqrt[3]{\frac{1}{40} + \frac{1}{30} \sqrt{\frac{17}{30}}} + \sqrt[3]{\frac{1}{40} - \frac{1}{30}\sqrt{\frac{17}{30}}},$$ so the equilibrium point between $p_4$ and $p_7$ occurs at $$p^* = 1 - t^* = 1 - \sqrt[3]{\frac{1}{40} + \frac{1}{30} \sqrt{\frac{17}{30}}} - \sqrt[3]{\frac{1}{40} - \frac{1}{30}\sqrt{\frac{17}{30}}} \approx 0.676582453403\dots.$$ So if $p$ were uniformly samples from $U \left( \frac{3}{5}, \frac{3}{4} \right)$ then the probability that a sweep is more probable than a seven-game series is $$\frac{p^* - \frac{3}{5}}{\frac{3}{4} - \frac{3}{5}} = \frac{20p^* - 12}{3} \approx 0.510549689351\dots.$$