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2021 Fall AMC 10B Problem 20

Problem 20 of 25HarderCounting & Probability

In a particular game, each of 44 players rolls a standard 66-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 5,5, given that he won the game?

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Solution

By symmetry, P(Hugo wins)=14.P(\text{Hugo wins})=\frac14. Therefore, the requested conditional probability is four times the probability that Hugo first rolls 55 and eventually wins. If Hugo rolls 5,5, then no other player can roll 6.6. Case on how many of the other three players also roll 5.5. If tt other players tie Hugo, then Hugo wins the eventual tiebreaker with probability 1t+1.\frac1{t+1}. Thus P(Hugo rolls 5 and wins)=164t=03(3t)43tt+1=64+24+4+1464=36941296. \begin{gathered} P(\text{Hugo rolls }5\text{ and wins})\\ {}=\frac{1}{6^4}\sum_{t=0}^3 \binom3t\frac{4^{3-t}}{t+1}\\ {}=\frac{64+24+4+\frac14}{6^4}\\ {}=\frac{369}{4\cdot1296}. \end{gathered} Multiplying by 44 gives 3691296=41144.\frac{369}{1296}=\frac{41}{144}. Thus, the answer is C .

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Concepts: conditional probability · dice (probability) · casework · symmetry

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.