Can You Sweep the Series?

In a best-of-seven series the favoured team wins each game independently with probability $p$. For which range $a<p<b$ is “winning in exactly five games” the single most likely of the eight possible series results?

The Fiddler, Zach Wissner-Gross, May 9, 2025(original post)

Solution

Writing $q=1-p$, the favoured team wins in $4,5,6,7$ games with probabilities $$p^4,\quad 4p^4q,\quad 10p^4q^2,\quad 20p^4q^3,$$ and the losing results are smaller in this regime. Win-in-five beats win-in-four when $4q>1$ and beats win-in-six when $4>10q$, i.e. $$q<\tfrac14\ \text{and}\ q>\tfrac25\ \Longrightarrow\ \tfrac35<p<\tfrac34.$$ So the range is $\boxed{\tfrac35<p<\tfrac34}$ (that is, $0.6<p<0.75$).

The computation

Tabulate all eight series outcomes (win or lose in $4$ to $7$ games) and, scanning $p$, mark where “win in five” is the single largest. The marked band runs from $0.6$ to $0.75$.

import numpy as np
def best(p):
    q = 1 - p
    d = {'w4': p**4, 'w5': 4*p**4*q, 'w6': 10*p**4*q**2, 'w7': 20*p**4*q**3,
         'l4': q**4, 'l5': 4*q**4*p, 'l6': 10*q**4*p**2, 'l7': 20*q**4*p**3}
    return max(d, key=d.get)
ps = [p for p in np.linspace(.001, .999, 99999) if best(p) == 'w5']
print(round(min(ps), 3), round(max(ps), 3))     # 0.6  0.75

Extra Credit

Choose $p$ uniformly from $(\tfrac35,\tfrac34)$. Let $p_4$ be the chance of a four-game sweep (either team) and $p_7$ the chance the series reaches seven games. How often is $p_4>p_7$?

Solution

With $p_4=p^4+q^4$ and $p_7=20p^3q^3$, the inequality $p_4>p_7$ flips once inside the interval, at $p\approx0.671$, so $$P(p_4>p_7)=\frac{0.75-0.671}{0.75-0.60}\approx\boxed{52\%}.$$ (The source’s value is paywalled; this is my own.)

The computation

Sample $p$ across the interval and test the sweep-vs-seven inequality directly, taking the fraction where the four-game sweep is likelier.

p = np.linspace(0.6, 0.75, 2_000_000); q = 1 - p
print(round(((p**4 + q**4) > 20*p**3*q**3).mean(), 3))   # ~0.524