How Many Turkey Trotters Can You Pass?

I have five kinds of fair Platonic dice: tetrahedra (whose faces are numbered $1-4$), cubes (numbered $1-6$), octahedra (numbered $1-8$), dodecahedra (numbered $1-12$) and icosahedra (numbered $1-20$). When I roll two of the cubes, there is a single most likely sum: seven. But when I roll one cube and two tetrahedra, there is no single most likely sum — eight and nine are both equally likely.

Which whole numbers are never the single most likely sum, no matter which combinations of dice I pick?

Solution

The whole numbers that can never be the single most likely sum are $$\boxed{1,\ 2,\ 3,\ 4,\ 6,\ \text{and}\ 8};$$ every other whole number is the unique most likely sum of some combination of these dice.

Why nothing below 5. A single die has a flat distribution, so it has no single most likely sum; in particular $1,2,3,4$ cannot come from one die. With $n \geq 2$ dice the most likely sum sits at the centre of the distribution, which is at least the centre for $n$ tetrahedra, $2.5n \geq 5$. So no combination at all has its peak below $5$.

Why 6 and 8 fail. The same centring argument bounds how many dice could possibly peak at a small target: a sum peaking at $6$ or $8$ can use at most $3$ dice. Checking every combination of at most three dice (code below): the unique peaks available are $5$ (two tetrahedra), $7$ (two cubes), $9$ (two octahedra, or $4+4+6$ and friends), $13$, $21$, and various larger values, but never $6$ or $8$. Two different dice (say $4$ and $6$ faces) always produce a plateau of tied sums rather than a single peak, and the three-dice combinations near these targets all tie two central values (three tetrahedra tie $7$ and $8$).

Why everything else works. A pair of identical dice is symmetric about a whole number: two tetrahedra peak at $5$, two cubes at $7$, two octahedra at $9$, two dodecahedra at $13$, two icosahedra at $21$, each a unique peak. Dice distributions are log-concave, and sums of independent symmetric log-concave lattice variables stay symmetric and single-peaked about the sum of the centres. So stacking pairs adds the peaks: any sum $5a + 7b + 9c + 13d + 21e$ (with $a,\dots,e \geq 0$, not all zero) is a unique most likely sum. Since $5$ and $7$ are coprime, every whole number from $24$ upward is $5a+7b$ (the Frobenius number of $\{5,7\}$ is $23$), and a direct search over combinations of at most eight dice exhibits unique peaks for every value from $5$ to $39$ except $6$ and $8$. Together: every whole number except $1,2,3,4,6,8$ is achievable.

The computation

Convolve the face distributions for every combination of up to eight dice, record the sums that have a single mode, and report which small whole numbers never appear.

from itertools import combinations_with_replacement

DICE = [4, 6, 8, 12, 20]

def conv(d1, d2):
    out = {}
    for a, pa in d1.items():
        for b, pb in d2.items():
            out[a+b] = out.get(a+b, 0) + pa*pb
    return out

unique_peaks = set()
for n in range(1, 9):
    for combo in combinations_with_replacement(DICE, n):
        d = {0: 1}
        for faces in combo:
            d = conv(d, {i: 1 for i in range(1, faces+1)})
        top = max(d.values())
        modes = [s for s, c in d.items() if c == top]
        if len(modes) == 1:
            unique_peaks.add(modes[0])

print(sorted(k for k in range(1, 40) if k not in unique_peaks))
# [1, 2, 3, 4, 6, 8]