Guess we'll never know ...

In a recent First Take episode, co-host Stephen A. Smith said:

I got [the Knicks] losing this in five games, which means they’re closer to a sweep than a seven-game series. That’s how I’m looking at it right now.

Let’s look at the first part of Stephen’s statement, that he believed the Knicks would lose to the Celtics in five games.

Let $p$ represent the probability the Celtics win any given game in the series. You should assume that $p$ is constant (which means there’s no home-court advantage) and that games are independent.

For certain values of $p$, the likeliest outcome is indeed that the Celtics will win the series in exactly five games. While this probability is always less than $50$ percent, this outcome is more likely than the Celtics winning or losing in some other specific number of games. In particular, this range can be specified as $a \lt p \lt b.$

Determine the values of $a$ and $b$.

Let $P(t, n \mid p)$ be the probability of team $t \in \{ \text{C}, \text{K} \}$ winning the series in $n$ games, given that the probability of the Celtics winning any individual game is $p.$ Then since team $t$ would have to win the final game of the series, the probability is given by $$P(t, n \mid p) = \begin{cases}\binom{n-1}{3} p^4 (1-p)^{n-4}, &\text{if $t = \text{C}$;}\\ \binom{n-1}{3} p^{n-4}(1-p)^4, &\text{if $t = \text{K}.$}\end{cases}$$

We are trying to find the maximal interval $(a,b) \subseteq [0,1],$ such that $$P(\text{C}, 5 \mid p ) = \max_{t \in \{\text{C}, \text{K}\}, n \in \{4,5,6,7\}} P(t, n \mid p),$$ for all $p \in (a,b).$ Let's just brute force this thing: $$\begin{align*} P(\text{C}, 5 \mid p) = 4p^4(1-p) \gt P(\text{C}, 4 \mid p) = p^4 &\Rightarrow p \in \left(0,\frac{3}{4}\right) \\ P(\text{C}, 5 \mid p) = 4p^4(1-p) \gt P(\text{C}, 6 \mid p) = 10p^4(1-p)^2 &\Rightarrow p \in \left(\frac{3}{5},1\right) \\ P(\text{C}, 5 \mid p) = 4p^4(1-p) \gt P(\text{C}, 7 \mid p) = 20p^4(1-p)^3 &\Rightarrow p \in \left(1 - \frac{1}{\sqrt{5}},1\right), \end{align*}$$ so since as long as $p \gt \frac{1}{2}$ we have $P(\text{C}, n \mid p) \gt P(\text{K}, n \mid p)$ for each $n \in \{4,5,6,7\},$ we see that the region where a Celtics win in five games is the most probable occurs when $p \in \left(\frac{3}{5}, \frac{3}{4}\right).$