Question

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?

$\dfrac{61}{216}$

$\dfrac{367}{1296}$

$\dfrac{41}{144}$

$\dfrac{185}{648}$

$\dfrac{11}{36}$

Solution:

By symmetry, $P(\text{Hugo wins})=\frac14$ . So the desired probability is $4\cdot P(\text{Hugo first rolls }5\text{ and wins})$ .

If Hugo rolls $5$ , then no other player can roll $6$ . Case on how many of the other three players also roll $5$ . If $t$ other players tie Hugo, then Hugo wins the eventual tiebreaker with probability $\frac1{t+1}$ .

Thus $P(\text{Hugo rolls }5\text{ and wins})=\frac{64+24+4+\frac14}{6^4}=\frac{369}{4\cdot1296}.$

Multiplying by $4$ gives $\frac{369}{1296}=\frac{41}{144}$ .

Thus, the answer is C .

2021 AMC 10B Fall Problem 20

Solution:

Problem 20 in Other Years