, 2021

First observe that if $q$ players tie on the initial roll, the probability that any one of these $q$ players will ultimately win is $\frac{1}{q}$. Let $N$ be the value of Hugo's first roll.

Consider four cases based on the number of highest scoring rolls in the first round. The probability that Hugo will roll a number larger than the other three players is
$$\frac{(N - 1)^3}{6^3} = \frac{N^3 - 3N^2 + 3N - 1}{216}.$$
The probability that Hugo will tie one other player, beat the other two players, and ultimately win is
$$3 \cdot \frac{1}{6} \cdot \frac{(N - 1)^2}{6^2} \cdot \frac{1}{2} = \frac{N^2 - 2N + 1}{144}.$$
Similarly, the probability that Hugo will tie two other players, beat the other player, and ultimately win is
$$3 \cdot \frac{1}{6^2} \cdot \frac{N-1}{6} \cdot \frac{1}{3} = \frac{N-1}{216}.$$
Finally, the probability that Hugo will tie all three players and ultimately win is
$$\frac{1}{6^3} \cdot \frac{1}{4} = \frac{1}{864}.$$
The sum of these four probabilities is
$$\frac{4N^3 - 6N^2 + 4N - 1}{864}.$$
Evaluating this expression for $N$ from 1 to 6 yields $\frac{1}{864}$, $\frac{15}{864}$, $\frac{65}{864}$, $\frac{175}{864}$, $\frac{369}{864}$, and $\frac{671}{864}$, respectively. Hence the probability that Hugo rolled a 5 on his initial roll given that he won is
$$\frac{369}{1 + 15 + 65 + 175 + 369 + 671} = \frac{369}{1296} = \frac{41}{144}.$$

OR

This can also be solved using Bayes' Theorem. The probability that Hugo rolled a 5 on his initial roll given that he won, written $P(5 | W)$, is
$$\frac{P(W | 5) \cdot P(5)}{P(W)} = \frac{\frac{41}{96} \cdot \frac{1}{6}}{\frac{1}{4}} = \frac{41}{144},$$
where $\frac{41}{96}$ is the number $\frac{369}{864}$ computed in the solution above for $N = 5$.

This solution is an application of the following general formula. Suppose there were $k$ players and an $n$-sided die was rolled. Then, for any $1 \le m \le n$, using the notation in the second solution,
$$P(\text{Hugo's first roll was } m \mid \text{Hugo won}) = \frac{m^k - (m-1)^k}{n^k}.$$
In the context of the original problem, this makes the answer $\frac{5^4-4^4}{6^4} = \frac{41}{144}$.

Two proofs are presented. The first proof is straightforward but somewhat computational. The second proof is harder to motivate but more elegant.

Proof 1: Algebra For each $0 \le j \le k-1$, let $A_j$ be the event that $j$ of the remaining $k-1$ players rolled an $m$ while the other $k-1-j$ players rolled less than $m$. Note that
$$P(A_j) = \binom{k-1}{j} \left(\frac{1}{n}\right)^j \left(\frac{m-1}{n}\right)^{k-1-j}.$$
After this first round, the remaining $j+1$ players went into the tiebreaker, and the probability of winning there is $\frac{1}{j+1}$ by symmetry. As a result,
$$\begin{align*} P(\text{Hugo won} | \text{Hugo's first roll was an } m) &= \sum_{j=0}^{k-1} P(A_j) \cdot \frac{1}{j+1} \\ &= \sum_{j=0}^{k-1} \frac{1}{j+1} \binom{k-1}{j} \left(\frac{1}{n}\right)^j \left(\frac{m-1}{n}\right)^{k-1-j} \end{align*}$$
$$\begin{align*} &= \sum_{j=0}^{k-1} \frac{1}{k} \binom{k}{j+1} \left(\frac{1}{n}\right)^j \left(\frac{m-1}{n}\right)^{k-1-j} \\ &= \frac{n}{k} \sum_{j=1}^{k} \binom{k}{j} \left(\frac{1}{n}\right)^j \left(\frac{m-1}{n}\right)^{k-j} \\ &= \frac{n}{k} \left( \left(\frac{1}{n} + \frac{m-1}{n}\right)^k - \left(\frac{m-1}{n}\right)^k \right) \\ &= \frac{m^k - (m-1)^k}{k n^{k-1}}, \end{align*}$$

Proof 2: Combinatorics Observe that, given that Hugo won, Hugo's roll was at most $m$ if and only if everyone's rolls were at most $m$. Indeed, Hugo's roll must be the highest roll to even have a chance at winning.
Therefore
$$\begin{align*} P(\text{Hugo's first roll was } \le m \mid \text{Hugo won}) &= P(\text{all first rolls were } \le m \mid \text{Hugo won}) \\ &= P(\text{all first rolls were } \le m) = \left(\frac{m}{n}\right)^k, \end{align*}$$
where the fact is used that the events “all first rolls were $\le m$” and “Hugo won” are independent. Therefore the
probability that Hugo’s first roll was an $m$ given that Hugo won is
$$P(\text{Hugo's first roll was } \le m | \text{Hugo won}) - P(\text{Hugo's first roll was } \le m-1 | \text{Hugo won}),$$
which equals
$$\frac{m^k - (m-1)^k}{n^k},$$
as desired.