2021 AMC 10 B Fall

Complete problem set with solutions and individual problem pages

Problem 20 Hard

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?(2021 AMC Fall 10B, Question #20)

  • A.

    \frac{61}{216}

  • B.

    \frac{367}{1296}

  • C.

    \frac{41}{144}

  • D.

    \frac{185}{648}

  • E.

    \frac{11}{36}

Answer:C

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( No one rolls a 6)-P( No one rolls a 5 or 6)=P( The highest roll is a 5), so \begin{gathered} P(\text { No one rolls a } 6)=\left(\frac{5}{6}\right)^{4} \\ P(\text { No one rolls a } 5 \text { or } 6)=\left(\frac{2}{3}\right)^{4} \end{gathered} P( The highest roll is a 5)=\left(\frac{5}{6}\right)^{4}-\left(\frac{4}{6}\right)^{4}=\frac{5^{4}-4^{4}}{6^{4}}=\frac{369}{1296}=(\mathbf{C}) \frac{41}{144}.

Solution 2:

The conditional probability formula states that P(A \mid B)=\frac{P(A \cap B)}{P(B)}, where A \mid B means \text{A} given \text{B} and A \cap B means \text{A} and \text{B}. Therefore the probability that Hugo rolls a five given he won is \frac{P(A \cap B)}{P(B)}, where \text{A} is the probability that he rolls a five and \text{B} is the probability that he wins. In written form, \text{P}(\text { Hugo rolled a } 5 \text { given he won })=\frac{\text{P}(\text { Hugo rolls a } 5 \text { and wins })}{\text{P}(\text { Hugo wins })} . The probability that Hugo wins is \frac{1}{4} by symmetry since there are four people playing and there is no bias for any one player. The probability that he gets a 5 and wins is more difficult; we will have to consider cases on how many players tie with Hugo\cdots

Case 1: No Players Tie In this case, all other players must have numbers from 1 through four. There is a \left(\frac{4}{6}\right)^{3}=\frac{8}{27} chance of this happening.

Case 2: One Player Ties In this case, there are \left(\begin{array}{l}3 \\ 1\end{array}\right)=3 ways to choose which other player ties with Hugo, and the probability that this happens is \frac{1}{6} \cdot\left(\frac{4}{6}\right)^{2}. The probability that Hugo wins on his next round is then \frac{1}{2} because there are now two players rolling die. Therefore the total probability in this case is 3 \cdot \frac{1}{2} \cdot \frac{1}{6} \cdot\left(\frac{4}{6}\right)^{2}=\frac{1}{9}.

Case 3: Two Players Tie In this case, there are \left(\begin{array}{l}3 \\ 2\end{array}\right)=3 ways to choose which other players tie with Hugo, and the probability that this happens is \left(\frac{1}{6}\right)^{2} \cdot \frac{4}{6}. The probability that Hugo wins on his next round is then \frac{1}{3} because there are now three players rolling the die. Therefore the total probability in this case is 3 \cdot \frac{1}{3} \cdot\left(\frac{1}{6}\right)^{2} \cdot \frac{4}{6}=\frac{1}{54}.

Case 4: All Three Players Tie In this case, the probability that all three players tie with Hugo is \left(\frac{1}{6}\right)^{3}. The probability that Hugo wins on the next round is \frac{1}{4}, so the total probability is \frac{1}{4} \cdot\left(\frac{1}{6}\right)^{3}=\frac{1}{864}. Finally, Hugo has a \frac{1}{6} probability of rolling a five himself, so the total probability is \frac{1}{6}\left(\frac{8}{27}+\frac{1}{9}+\frac{1}{54}+\frac{1}{864}\right)=\frac{1}{6}\left(\frac{369}{864}\right)=\frac{1}{6}\left(\frac{41}{96}\right) Finally, the total probability is this probability divided by \frac{1}{4} which is this probability times four; the final answer is 4 \cdot \frac{1}{6}\left(\frac{41}{96}\right)=\frac{2}{3} \cdot \frac{41}{96}=\frac{41}{48 \cdot 3}=\frac{41}{144}=C

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