2021 Fall AMC 10B Problems/Problem 20

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 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

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2021 Fall AMC 10B Problems/Problem 20

Problem 20

Solution 1

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