Can You Solve The Puzzle Of The Baseball Division Champs?

In a sport where each team plays $162$ games a season, take a division of five teams of exactly equal ability: each has a $50\%$ chance of winning any given game. What is the expected number of wins for the team that finishes first?

The Riddler, FiveThirtyEight (Nick Keenan)(original post)

Solution

A single team of average ability wins half its games, $81$. The division champion is not an average team though, it is the best of five, and being the best of a group pulls the number up. The expected first-place total is about $$\boxed{88.4}.$$

The honest caveat first: a real schedule couples the teams, since one club’s win is another’s loss, so the five totals are not quite independent. But each team plays only $76$ of its $162$ games inside the division and $86$ outside it, and those out-of-division games swamp the coupling. The clean model that the column settles on treats every game as its own coin flip, making each team’s season an independent $\mathrm{Binomial}(162,\tfrac12)$ and the champion their maximum.

For the maximum of five independent counts, lean on the tail rather than the bell. The champion wins at least $w$ games unless all five fall short of $w$, so with $F$ the single-team cumulative distribution, $$\mathbb{E}[\text{champ}]=\sum_{w\ge 1}\Pr(\text{champ}\ge w) =\sum_{w\ge 0}\bigl(1-F(w)^5\bigr).$$ Summing this against the exact binomial $F$ gives $88.39$, so the first-place team averages a little over $88$ wins, some seven games above the $81$ an average team manages. (A real big-league schedule nudges this only slightly, to about $88.8$.)

The computation

Compute the single-team distribution exactly, raise its cumulative function to the fifth power for the champion, and sum the tail. A Monte Carlo season cross-checks the closed form.

from math import comb
import numpy as np

n, teams = 162, 5
cdf = [sum(comb(n, k) for k in range(w + 1)) / 2**n for w in range(n + 1)]
exact = sum(1 - cdf[w] ** teams for w in range(n + 1))   # E[max] via the tail sum
print(round(exact, 4))                                   # 88.3943

rng = np.random.default_rng(0)
sim = rng.binomial(n, 0.5, size=(2_000_000, teams)).max(axis=1).mean()
print(round(sim, 3))                                     # 88.391