Fall 2021 AMC 10 B

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 $$P(A_j)=\left(\genfrac{}{}{0ex}{}{k-1}{j}\right){\left(\frac{1}{n}\right)}^j{\left(\frac{m-1}{n}\right)}^{k-1-j}.$$ 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, $$\begin{array}{rl}P(\text{Hugo won}|\text{Hugo’s first roll was an }m)& =\sum _{j=0}^{k-1}P(A_j)\cdot \frac{1}{j+1}\\ & =\sum _{j=0}^{k-1}\frac{1}{j+1}\left(\genfrac{}{}{0ex}{}{k-1}{j}\right){\left(\frac{1}{n}\right)}^j{\left(\frac{m-1}{n}\right)}^{k-1-j}\end{array}$$ $$\begin{array}{rl}& =\sum _{j=0}^{k-1}\frac{1}{k}\left(\genfrac{}{}{0ex}{}{k}{j+1}\right){\left(\frac{1}{n}\right)}^j{\left(\frac{m-1}{n}\right)}^{k-1-j}\\ & =\frac{n}{k}\sum _{j=1}^k\left(\genfrac{}{}{0ex}{}{k}{j}\right){\left(\frac{1}{n}\right)}^j{\left(\frac{m-1}{n}\right)}^{k-j}\\ & =\frac{n}{k}\left({\left(\frac{1}{n}+\frac{m-1}{n}\right)}^k-{\left(\frac{m-1}{n}\right)}^k\right)\\ & =\frac{m^k-(m-1{)}^k}{k{n}^{k-1}},\end{array}$$

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 $$\begin{array}{rl}P(\text{Hugo’s first roll was }\le m\mid \text{Hugo won})& =P(\text{all first rolls were }\le m\mid \text{Hugo won})\\ & =P(\text{all first rolls were }\le m)={\left(\frac{m}{n}\right)}^k,\end{array}$$ 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.