Can You Eat An Apple Like A Toddler?

Riddler Express

A set of dominoes has $28$ tiles, each tile showing two ends with zero to six dots, every unordered pair of values (including doubles) appearing exactly once. Question 1: what is the probability of drawing a “double,” a tile with the same number on both ends? Question 2: you draw a random tile and uncover one end, seeing six dots. What is the probability the tile is the double six?

The Riddler, FiveThirtyEight, May 8, 2020(original post)

Solution

Of the $28$ tiles, exactly seven are doubles (zero-zero up to six-six), so $$\Pr(\text{double}) = \frac{7}{28} = \frac14.$$

The second question is the classic trap. Seven tiles carry a six, only one of which is the double, which tempts an answer of $\tfrac17$. But you revealed a random end, and the double six has two six-ends, so it is twice as likely as a single-six tile to have shown you a six. Count ends, not tiles: there are eight ends bearing a six (one on each of the six “six-and-something” tiles, two on the double six), and of those eight ends, the two belonging to the double six have a six on the far side. So $$\Pr(\text{double six} \mid \text{a six is showing}) = \frac{2}{8} = \frac14.$$ Both answers are $$\boxed{\tfrac14}.$$ That they agree is no accident: seeing a six on the revealed end tells you nothing extra, since whatever number you uncovered, the chance the tile is a double is unchanged. The reveal carries no information.

The computation

Build the $28$ tiles. For Question 1 count doubles. For Question 2 walk over the two ends of every tile, restrict to ends showing a six, and ask how often the other end is also a six.

from fractions import Fraction as F
tiles = [(i, j) for i in range(7) for j in range(i, 7)]
assert len(tiles) == 28
print("Q1:", F(sum(i == j for i, j in tiles), 28))        # 1/4

shown = same = 0
for i, j in tiles:
    for near, far in ((i, j), (j, i)):     # reveal each end in turn
        if near == 6:
            shown += 1
            same += (far == 6)
print("Q2:", F(same, shown))                              # 1/4

Riddler Classic

A two-year-old eats an apple as follows. Each minute he rotates it to a uniformly random orientation and looks down; if there is still skin where he means to bite, he bites, removing a circular patch; if that spot is already eaten, he waits a minute and rotates again. The apple is a sphere of radius $4$ cm, and each bite is a circle whose radius, measured along the curved surface, is $1$ cm. On average, how many minutes to eat all the skin?

The Riddler, FiveThirtyEight, May 8, 2020(original post)

Status

This is the problem of covering a sphere with randomly placed caps, counting every rotation including the wasted ones, and it has no elementary closed form. A clean lower bound does follow from area: the sphere’s surface is $4\pi R^2 = 64\pi$ square centimetres, while each bite is a spherical cap of angular radius $\theta = 1/4$ radian and area $2\pi R^2(1-\cos\theta) \approx 0.995\pi$, just under $\pi$. So at least $64\pi / \pi = 64$ bites are needed, and the number of minutes is larger still because late rotations usually land on already-eaten skin.

The official answer is a simulation result of roughly $565$ minutes, but with an important caveat that the column itself stressed: a finite cloud of sample points is always fully consumed slightly before the continuous surface is, so every such simulation underestimates. The published runs bear this out, rising with resolution: about $550$ minutes at $10{,}000$ points, $565$ at $250{,}000$, $568$ at $500{,}000$, with the true expectation a little higher. Because a faithful problem-encoding computation here is precisely the delicate, resolution-sensitive covering simulation whose value we could not pin down to the published figure (our continuous-area model runs markedly higher than the point-cloud estimates), we record the result rather than box a number we cannot independently reproduce. The lower bound of $64$ bites is exact; the headline of “about $565$ minutes” is the official’s acknowledged-underestimate simulation value.