Can You Survive Squid Game?

Riddler Express

A spherical pumpkin has the same numerical value for its volume (in cubic inches) and its surface area (in square inches). What is its radius? Extra credit: for an $n$-dimensional ball, when do the numerical volume and surface “area” agree, and what is the radius then?

Solution

Set the two numbers equal. For the sphere, $$\frac{4}{3}\pi r^3 = 4\pi r^2 \;\Longrightarrow\; r = \boxed{3}.$$ The general case falls out of one fact: the surface of an $n$-ball is the derivative of its volume with respect to the radius, and the volume scales as $r^n$, so $$A_n(r) = \frac{d}{dr}V_n(r) = \frac{n}{r}\,V_n(r).$$ Setting $A_n(r) = V_n(r)$ gives $\tfrac{n}{r} = 1$, that is $$r = \boxed{n}.$$ The sphere ($n = 3$) is just the case $r = 3$, and a circle ($n = 2$) matches at $r = 2$.

The computation

Form the $n$-ball volume from the standard formula, differentiate to get the surface, and solve volume $=$ surface symbolically.

import sympy as sp
r, n = sp.symbols('r n', positive=True)
V = sp.pi**(n/2) / sp.gamma(n/2 + 1) * r**n          # n-ball volume
A = sp.diff(V, r)                                      # surface = dV/dr
print(sp.solve(sp.Eq(V, A), r))                       # [n]
print(sp.solve(sp.Eq(V.subs(n, 3), A.subs(n, 3)), r)) # [3]

Riddler Classic

Sixteen competitors must cross a bridge of $18$ pairs of glass squares. In each pair one square is tempered (safe) and one is normal (breaks); nobody knows which. A competitor stepping onto an unrevealed pair guesses, and a wrong guess eliminates them but reveals that pair to everyone behind. On average, how many of the $16$ survive?

Solution

Each pair is revealed exactly once, by whoever first steps on it, and that first step is a fair coin: tails (probability $\tfrac12$) eliminates the guesser, heads lets them carry on. Once a pair is known, everyone behind walks it safely. So if competitors never ran out, the number eliminated would be the number of “tails” among $18$ independent flips, a $\text{Binomial}(18, \tfrac12)$ with mean $9$. The only complication is that you cannot lose more than $16$ people, so survivors are $$\mathbb{E}[\text{survivors}] = \mathbb{E}\big[\max(16 - B,\ 0)\big], \qquad B \sim \text{Binomial}(18, \tfrac12).$$ The cap almost never binds (needing $16$ tails out of $18$), so the answer is a hair above $16 - 9 = 7$: $$\frac{458757}{65536} \approx \boxed{7.0001}.$$

The computation

Solve the survivor recurrence exactly: with $n$ competitors and $m$ pairs left, the next pair is safe or fatal with equal chance, $S(n,m) = \tfrac12 S(n, m-1) + \tfrac12 S(n-1, m-1)$.

from fractions import Fraction as F
from functools import lru_cache
@lru_cache(None)
def S(n, m):
    if n == 0: return F(0)
    if m == 0: return F(n)
    return (S(n, m - 1) + S(n - 1, m - 1)) / 2
print(S(16, 18), float(S(16, 18)))            # 458757/65536, 7.0001