Can The Hare Beat The Tortoise?

Riddler Express

Your true batting average is $.350$ (a $35\%$ chance of a hit each at-bat), with four at-bats per game. What is your chance of batting at least $.400$ over a $60$-game season, and how does that compare with a $162$-game season? Extra credit: your chance of a hitting streak of at least $56$ games (tying DiMaggio) in each season.

The Riddler, FiveThirtyEight, July 17, 2020(original post)

Solution

Each at-bat is an independent hit with probability $0.35$. Over $60$ games that is $240$ at-bats, and batting $.400$ means at least $0.4\times 240 = 96$ hits; the number of hits is Binomial$(240, 0.35)$, so the chance is the upper tail $\Pr(\text{hits} \ge 96)$. Over $162$ games it is $648$ at-bats needing $\ge 260$ hits. The tails are $$\boxed{6.1\% \ \text{over } 60 \text{ games}, \qquad 0.38\% \ \text{over } 162 \text{ games}},$$ so a short season makes $.400$ about $16$ times more reachable: fewer at-bats mean more sampling variation around the true mean.

The streak runs the other way. A single game yields a hit with probability $1 - (1-0.35)^4 = 82.15\%$, but stringing $56$ such games together is punishing. The chance of a streak of at least $56$ games is $$0.0028\% \ \text{over } 60 \text{ games}, \qquad 0.033\% \ \text{over } 162 \text{ games},$$ roughly ten times likelier in the long season, which simply has more starting points for a long run. (Batting $.400$ and streaking $56$ games is rarer still, and positively correlated, since both demand a glut of hits.)

The computation

Encode the binomial tail for the $.400$ question and a short dynamic program over games (carry the current streak length, absorbing at $56$) for the streak.

from math import comb
from functools import lru_cache
def tail(at_bats, need, p=0.35):
    return sum(comb(at_bats, k) * p**k * (1 - p)**(at_bats - k)
               for k in range(need, at_bats + 1))
print(round(tail(240, 96) * 100, 2), round(tail(648, 260) * 100, 3))   # 6.08 0.379
pg = 1 - (1 - 0.35) ** 4                              # hit in a game
def streak(G, L=56):
    @lru_cache(None)
    def f(g, run):
        if run >= L: return 1.0
        if g == 0: return 0.0
        return pg * f(g - 1, run + 1) + (1 - pg) * f(g - 1, 0)
    return f(G, 0)
print(round(streak(60) * 100, 4), round(streak(162) * 100, 4))         # 0.0028 0.0329

Riddler Classic

A $10$-mile race runs along a road that instantly stretches by $10$ miles at the end of every minute (uniformly, carrying the cars with it). The tortoise drives $60$ mph, the hare $75$ mph. The tortoise starts at the gun; how long should the hare wait so that the two finish at the same instant?

The Riddler, FiveThirtyEight, July 17, 2020(original post)

Solution

Because the road stretches uniformly, the right variable is the fraction of the road a car has covered, not its distance: stretching carries a car forward in proportion, leaving its fraction unchanged. At $60$ mph the tortoise covers $1$ mile in the first minute, $\tfrac1{10}$ of the then $10$-mile road. In the second minute it adds $1$ mile of a $20$-mile road, $\tfrac1{20}$; in the $n$th minute, $\tfrac1{10n}$. So after $n$ minutes its fraction covered is $$\frac1{10}\Bigl(1 + \tfrac12 + \cdots + \tfrac1n\Bigr) = \frac{H_n}{10},$$ which first reaches $1$ at $n = 12{,}367$ minutes (about $8.6$ days): the harmonic series diverges, so the tortoise does finish, eventually.

The hare drives $75$ mph, exactly $25\%$ faster, so over any shared stretch of time it covers $25\%$ more of the road. To tie, it should have $25\%$ more road left to cover when it starts, that is, it should leave when the tortoise has completed $\tfrac{1}{1.25} = 80\%$... equivalently the hare may wait until the tortoise has completed $20\%$ and then cover the rest in lockstep to the line. The tortoise’s fraction reaches $0.2$ partway through the fourth minute, at $$\boxed{3 \text{ minutes and } 40 \text{ seconds}}.$$

The computation

Encode the tortoise’s fraction-per-minute $\tfrac1{10n}$: sum it until the road is complete (the harmonic finish), and find the instant the fraction crosses $0.2$ (the hare’s cue).

H, n = 0.0, 0
while H < 10:
    n += 1; H += 1 / n                       # fraction covered = H/10
print("tortoise finishes at minute", n)      # 12367
frac, t = 0.0, 0
while frac < 0.2:
    t += 1; frac += 1 / (10 * t)
prev = frac - 1 / (10 * t)                    # fraction at start of minute t
into = (0.2 - prev) / (1 / (10 * t))          # fraction into minute t
start = (t - 1) + into
print(f"hare waits {int(start)} min {round((start % 1) * 60)} s")   # 3 min 40 s