Perfectly Balanced Dice Rolls
You roll a fair six-sided die 36 times. What is the probability that each face (1 through 6) appears exactly 6 times?
Hints
- Think about how many distinct ways you can arrange 36 rolls into six groups of equal size. What generalization of the binomial handles more than two outcomes?
- The multinomial distribution gives $P = \frac{36!}{n_1! \cdots n_6!} \cdot (1/6)^{36}$ where each $n_i = 6$. Write out the formula and simplify using the symmetry.
- Plug in directly: $P = \frac{36!}{(6!)^{6} \cdot 6^{36}}$. Use Stirling's approximation or a calculator to confirm the answer is on the order of $10^{-4}$, roughly 1 in 3,863.
Worked Solution
How to Think About It: You roll a fair die 36 times, so each face shows up $6$ times on average. The question asks how likely it is that every face hits its expected count exactly. Your gut should say "pretty unlikely" -- even though $6$ is the mean count per face, the variance is large enough that perfect balance across all six faces simultaneously is rare. This is a textbook multinomial probability calculation.
Quick Estimate: Each face count $X_i \sim \text{Binomial}(36, 1/6)$ with mean $6$ and standard deviation $\sqrt{36 \cdot \frac{1}{6} \cdot \frac{5}{6}} = \sqrt{5} \approx 2.24$. The probability a single face lands exactly on its mean is roughly $1/(\sqrt{2\pi} \cdot 2.24) \approx 0.18$. But we need all six faces to hit their means simultaneously, and the counts are dependent (they sum to 36). A multinomial normal approximation gives a probability on the order of $10^{-4}$, which lines up with the exact answer.
Approach: Apply the multinomial distribution directly. Count the ways to arrange 36 rolls so each face appears exactly 6 times, then multiply by the probability of any single such sequence.
Formal Solution:
The 36 rolls follow a $\text{Multinomial}(36;\, 1/6,\, 1/6,\, 1/6,\, 1/6,\, 1/6,\, 1/6)$ distribution. The probability that face $i$ appears exactly $n_i$ times is:
$$P(X_1 = n_1, \ldots, X_6 = n_6) = \frac{36!}{n_1! \cdot n_2! \cdots n_6!} \left(\frac{1}{6}\right)^{36}$$
With $n_1 = n_2 = \cdots = n_6 = 6$:
$$P = \frac{36!}{(6!)^{6}} \cdot \left(\frac{1}{6}\right)^{36}$$
The multinomial coefficient evaluates to:
$$\binom{36}{6,6,6,6,6,6} = \frac{36!}{(6!)^{6}} \approx 2.67 \times 10^{24}$$
and $6^{36} \approx 1.031 \times 10^{28}$, so:
$$P = \frac{2.67 \times 10^{24}}{1.031 \times 10^{28}} \approx 0.000259$$
Answer:
$$P = \frac{36!}{(6!)^{6} \cdot 6^{36}} \approx 0.000259 \approx \frac{1}{3{,}863}$$
About a 0.026% chance -- roughly 1 in 3,863 attempts will produce a perfectly balanced outcome.
Intuition
This problem tests whether you understand the multinomial distribution, which is the natural generalization of the binomial to experiments with more than two outcomes. Each roll assigns a ball to one of 6 bins, and we want every bin to have exactly 6 balls. The multinomial coefficient counts the number of orderings, and the factor $(1/6)^{36}$ reflects that each ordering is equally likely.
The key takeaway is how rare perfect balance actually is. Even though 6 is the expected count for each face, the probability of hitting the mean exactly for all faces simultaneously is tiny -- about 1 in 3,863. This comes up constantly in quant work: just because an outcome is the "most likely" single outcome does not mean it is likely in absolute terms. The probability mass is spread across an enormous number of configurations, and the mode captures only a small fraction. This is the same intuition behind why the peak of a high-dimensional Gaussian is negligible in probability -- the mode is the most likely point, but almost all the mass lives in a thin shell away from it.