Answer: The Pepys–Newton Problem

Key Idea / Intuition

Each scenario has the same expected number of sixes (1, 2, and 3 respectively, since the expected number of sixes in $n$ dice is $n/6$). So at first glance, all three might seem equally likely. But expectation and probability are not the same thing! The key is that in each case, you need to meet or exceed the mean — and the probability of doing so depends on the skewness and spread of the binomial distribution. With fewer dice, the distribution is more spread and asymmetric, so the probability of hitting at least the mean is higher. As $n$ grows, the distribution becomes more symmetric and the probability of reaching the mean from below approaches $1/2$. So (A) > (B) > (C).

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Formal Proof / Solution

Each scenario follows a Binomial distribution with success probability $p = 1/6$.

Option (A): At least one 6 in 6 dice

$P(A) = 1 - P(\text{no 6s in 6 dice}) = 1 - \left(\frac{5}{6}\right)^6$

$= 1 - \frac{5^6}{6^6} = 1 - \frac{15625}{46656} = \frac{31031}{46656} \approx \mathbf{0.6651}$

Option (B): At least two 6s in 12 dice

$P(B) = 1 - P(0 \text{ sixes}) - P(1 \text{ six})$

$P(0) = \left(\frac{5}{6}\right)^{12} \approx 0.1122$

$P(1) = \binom{12}{1}\left(\frac{1}{6}\right)^1\left(\frac{5}{6}\right)^{11} = 12 \cdot \frac{1}{6} \cdot \left(\frac{5}{6}\right)^{11} \approx 12 \cdot \frac{1}{6} \cdot 0.1346 \approx 0.2692$

$P(B) = 1 - 0.1122 - 0.2692 \approx \mathbf{0.6187}$

Option (C): At least three 6s in 18 dice

$P(C) = 1 - P(0) - P(1) - P(2)$

$P(0) = \left(\frac{5}{6}\right)^{18} \approx 0.0376$

$P(1) = \binom{18}{1}\frac{1}{6}\left(\frac{5}{6}\right)^{17} \approx 18 \cdot \frac{1}{6} \cdot 0.0451 \approx 0.1353$

$P(2) = \binom{18}{2}\frac{1}{36}\left(\frac{5}{6}\right)^{16} \approx 153 \cdot \frac{1}{36} \cdot 0.0541 \approx 0.2299$

$P(C) = 1 - 0.0376 - 0.1353 - 0.2299 \approx \mathbf{0.5973}$

Summary

| Option | Dice | Mean sixes | Probability | |--------|------|-----------|-------------| | A | 6 | 1 | ≈ 0.6651 | | B | 12 | 2 | ≈ 0.6187 | | C | 18 | 3 | ≈ 0.5973 |

Option (A) wins, despite all having the same expected number of sixes.

Why does this happen?

The binomial distribution $\text{Bin}(n, 1/6)$ is right-skewed for small $n$: the median lies below the mean. As $n \to \infty$, by the CLT the distribution becomes symmetric and $P(\text{at least mean}) \to 1/2$. For small $n$, the probability of meeting the mean is significantly above $1/2$, and this effect is strongest for $n = 6$. Newton correctly identified (A) as the best bet.