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2021 Fall AMC 10B Problems/Problem 20

Revision as of 22:59, 25 November 2021 by Stevenyiweichen (talk | contribs)

Problem 20

In a particular game, each of $4$ players rolls a standard $6{ }$-sided die. The winner is the player who rolls the highest number. If there is a tie for the highest roll, those involved in the tie will roll again and this process will continue until one player wins. Hugo is one of the players in this game. What is the probability that Hugo's first roll was a $5,$ given that he won the game?

$(\textbf{A})\: \frac{61}{216}\qquad(\textbf{B}) \: \frac{367}{1296}\qquad(\textbf{C}) \: \frac{41}{144}\qquad(\textbf{D}) \: \frac{185}{648}\qquad(\textbf{E}) \: \frac{11}{36}$

Solution 1

Since we know that Hugo wins, we know that he rolled the highest number in the first round. The probability that his first roll is a $5$ is just the probability that the highest roll in the first round is $5$.

Let $P(x)$ indicate the probability that event $x$ occurs. We find that $P(\text{No one rolls a 6})-P(\text{No one rolls a 5 or 6})=P(\text{The highest roll is a 5})$,

so \[P(\text{No one rolls a 6})=\left(\frac56\right)^4,\] \[P(\text{No one rolls a 5 or 6})=\left(\frac23\right)^4,\] \[P(\text{The highest roll is a 5})=\left(\frac56\right)^4-\left(\frac46\right)^4=\frac{5^4-4^4}{6^4}=\frac{369}{1296}=\boxed{(\textbf{C}) \: \frac{41}{144}}.\]

~kingofpineapplz

Solution 3

We use $H$ to refer to Hugo. We use $H_1$ to denote the outcome of Hugo's $t$th toss. We denote by $A$, $B$, $C$ the other three players. We denote by $N$ the number of players among $A$, $B$, $C$ whose first tosses are 5. We use $W$ to denote the winner.

We have \begin{align*} P \left( H_1 = 5 | W = H \right) & = \frac{P \left( H_1 = 5 , W = H \right)}{P \left( W = H \right)} \\ & = \frac{P \left( H_1 = 5 \right) P \left( W = H | H_1 = 5 \right) }{P \left( W = H \right)} \\ & = \frac{\frac{1}{6} P \left( W = H | H_1 = 5 \right)}{\frac{1}{4}} \\ & = \frac{2}{3} P \left( W = H | H_1 = 5 \right) . \end{align*}

Now, we compute $P \left( W = H | H_1 = 5 \right)$.

We have \begin{align*} & P \left( W = H | H_1 = 5 \right) \\ & = P \left( W = H | H_1 = 5 , \max \left\{ A_1, B_1, C_1 \right\} \leq 4 \right) P \left( \max \left\{ A_1, B_1, C_1 \right\} \leq 4 | H_1 = 5 \right) \\ & \quad + P \left( W = H | H_1 = 5 , \max \left\{ A_1, B_1, C_1 \right\} = 6 \right) P \left( \max \left\{ A_1, B_1, C_1 \right\} = 6 | H_1 = 5 \right) \\ & \quad + \sum_{N = 1}^3 P \left( W = H | H_1 = 5 , \max \left\{ A_1, B_1, C_1 \right\} = 5 , N \right) P \left( \max \left\{ A_1, B_1, C_1 \right\} = 5 , N | H_1 = 5 \right) \\ & = P \left( W = H | H_1 = 5 , \max \left\{ A_1, B_1, C_1 \right\} \leq 4 \right) P \left( \max \left\{ A_1, B_1, C_1 \right\} \leq 4 \right) \\ & \quad + P \left( W = H | H_1 = 5 , \max \left\{ A_1, B_1, C_1 \right\} = 6 \right) P \left( \max \left\{ A_1, B_1, C_1 \right\} = 6 \right) \\ & \quad + \sum_{N = 1}^3 P \left( W = H | H_1 = 5 , \max \left\{ A_1, B_1, C_1 \right\} = 5 , N \right) P \left( \max \left\{ A_1, B_1, C_1 \right\} = 5 , N \right) \\ & = 1 \cdot P \left( \max \left\{ A_1, B_1, C_1 \right\} \leq 4 \right) + 0 \cdot P \left( \max \left\{ A_1, B_1, C_1 \right\} = 6 \right) \\ & \quad + \sum_{N = 1}^3 P \left( W = H | H_1 = 5 , \max \left\{ A_1, B_1, C_1 \right\} = 5 , N \right) P \left( \max \left\{ A_1, B_1, C_1 \right\} = 5 , N \right) \\ & = P \left( \max \left\{ A_1, B_1, C_1 \right\} \leq 4 \right) \\ & \quad + \sum_{N = 1}^3 P \left( W = H | H_1 = 5 , \max \left\{ A_1, B_1, C_1 \right\} = 5 , N \right) P \left( \max \left\{ A_1, B_1, C_1 \right\} = 5 , N \right) \\ & = \left( \frac{4}{6} \right)^3 + \sum_{N = 1}^3 \frac{1}{N + 1} \cdot \binom{3}{N} \left( \frac{1}{6} \right)^N \left( \frac{4}{6} \right)^{3 - N} \\ & = \frac{41}{96} . \end{align*} The first equality follows from the law of total probability. The second equality follows from the property that Hugo's outcome is independent from other players' outcomes.

Therefore, \begin{align*} P \left( H_1 = 5 | W = H \right) & = \frac{2}{3} P \left( W = H | H_1 = 5 \right) \\ & = \frac{2}{3} \frac{41}{96} \\ & = \frac{41}{144} . \end{align*}

Therefore, the answer is $\boxed{\textbf{(C) }\frac{41}{144}}$.

~Steven Chen (www.professorchenedu.com)


See Also

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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