Plotting the first four options,
Sports personality Stephen A. Smith said something earlier this week that caught my attention—for mathematical reasons!
On the ESPN show, “First Take,” the discussion turned to the NBA’s New York Knicks, who would be facing the favored Boston Celtics in the best-of-seven Eastern Conference Semifinals. (Note that this segment aired prior to the beginning of the series.) The question was whether the Knicks were more likely to be “swept” (i.e., lose the series in four games) or for the series to go to seven games. Here’s what Stephen had to say:
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 < p < b.
Determine the values of a and b.
The following table summarizes all outcomes in terms of $p$.
| Winner | Number of Games | Probability | Expression |
|---|---|---|---|
| Celtics | $4$ | ${4 \choose 0}p^4(1-p)^4$ | $= p^4$ |
| Celtics | $5$ | ${4 \choose 1}p^4(1-p)^1{1 \choose 1}$ | $= 4p^4(1-p)$ |
| Celtics | $6$ | ${5 \choose 2}p^4(1-p)^2{1 \choose 1}$ | $= 10p^4(1-p)^2$ |
| Celtics | $7$ | ${6 \choose 3}p^4(1-p)^3{1 \choose 1}$ | $= 20p^4(1-p)^3$ |
| Knicks | $7$ | ${6 \choose 3}p^3(1-p)^4{1 \choose 1}$ | $= 20p^3(1-p)^4$ |
| Knicks | $6$ | ${5 \choose 2}p^2(1-p)^4{1 \choose 1}$ | $= 10p^2(1-p)^4$ |
| Knicks | $5$ | ${4 \choose 1}p^1(1-p)^4{1 \choose 1}$ | $= 4p(1-p)^4$ |
| Knicks | $4$ | ${4 \choose 0}p^0(1-p)^4$ | $= (1-p)^4$ |
Plotting the first four options,
Zooming in, $R$ and $S$ correspond to $a$ and $b$, respectively.
Solving for $R_x$,
\begin{align*} 10p^4(1−p)^2 &= 4p^4(1−p) \\ \frac{(1-p)^2}{(1-p)} &= \frac{4p^4}{10p^4} \\ 1-p &= \frac{2}{5} \\ \end{align*}Solving for $S_x$,
\begin{align*} 4p^4(1−p) &= p^4 \\ 1-p &= \frac{1}{4} \\ \end{align*}Thus,
$$\boxed{a = \frac{3}{5} \text{ and } b = \frac{3}{4}}$$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 p4 represent the probability that the Celtics sweep the Knicks in four games. And let p7 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 p4 is greater than p7? In other words, how often will it be the case that probably losing in five games means a sweep is more likely than a seven-game series?
The two 7 game outcomes have a combined probability of $20p^3(1-p)^3$.
Plotting the relevant lines,
Solving for $W_x$,
\begin{align*} 20p^3(1-p)^3 &= p^4 \\ 20(1-p)^3 &= p \\ \end{align*}Using an equation solver, $p = 0.676582$
Solving for the probability the Knicks win in 4 games given $\frac{3}{5} \le p \le \frac{3}{5}$,
\begin{align*} p_4 &= \frac{s_x-p}{s_x-r_x} \\ &= \frac{0.75-0.67582}{0.75-0.6} \\ &= \frac{0.07418}{0.15} \\ &\approx 0.49453\\ \end{align*}The likelihood the Knicks will lose in four games is $\boxed{49.45\%}$.