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2021 AMC Fall 10B Problem 20

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2021 AMC Fall 10B Problem 20AMC 10 Step by Step
Name:                             Date: September 13, 2026
Conditional Probability & States · difficulty 4 of 5 · about 8 min · position P16-20

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?

  1. A)

    61216\frac{61}{216}

  2. B)

    3671296\frac{367}{1296}

  3. C)

    41144\frac{41}{144}

  4. D)

    185648\frac{185}{648}

  5. E)

    1136\frac{11}{36}

Answer

C

Key insight

P(Hugo wins) is 1/4 by symmetry, and given a first roll of 5 he wins with probability sum over k other 5s of C(3,k)(1/6)^k(4/6)^(3-k)/(k+1).

Solution

We want
P(first roll 5∣won)=P(first roll 5)⋅P(won∣first roll 5)P(won).P(\text{first roll } 5 \mid \text{won}) = \frac{P(\text{first roll } 5)\cdot P(\text{won} \mid \text{first roll } 5)}{P(\text{won})}.

By symmetry among the four players, P(won)=14P(\text{won}) = \frac14, and P(first roll 5)=16P(\text{first roll } 5) = \frac16.

Now suppose Hugo rolled a 55. Each other player independently rolls a 66 (probability 16\frac16, Hugo loses), a 55 (probability 16\frac16, tie), or at most 44 (probability 46\frac46). If nobody rolls a 66 and exactly kk others roll 55, then k+1k + 1 players re-roll from scratch and Hugo wins with probability 1k+1\frac{1}{k+1} by symmetry. Therefore
P(won∣5)=(46)3+3(16)(46)2⋅12+3(16)2(46)⋅13+(16)3⋅14.P(\text{won} \mid 5) = \left(\tfrac46\right)^3 + 3\left(\tfrac16\right)\left(\tfrac46\right)^2 \cdot \tfrac12 + 3\left(\tfrac16\right)^2\left(\tfrac46\right)\cdot\tfrac13 + \left(\tfrac16\right)^3\cdot\tfrac14.
Over the denominator 864=4⋅216864 = 4 \cdot 216: the terms are 256864+96864+16864+1864=369864=4196\frac{256}{864} + \frac{96}{864} + \frac{16}{864} + \frac{1}{864} = \frac{369}{864} = \frac{41}{96}.

Finally
P(5∣won)=16⋅419614=4⋅41576=41144.P(5 \mid \text{won}) = \frac{\frac16 \cdot \frac{41}{96}}{\frac14} = \frac{4 \cdot 41}{576} = \frac{41}{144}.

The answer is (C) 41144\boxed{\textbf{(C)}\ \frac{41}{144}}.

Why this works

Bayes' rule turns "given that he won" into a forward computation, and symmetry supplies two of the three ingredients for free. The re-roll process looks infinite, but a tie among jj players is just a fresh, fair game among them, so its outcome is 1j\frac1j for each without any geometric series.

The trap

Ignoring the re-roll rule and requiring the other three players to roll strictly less than 5, which gives (4/6)^3 and loses the tie cases.

Common mistakes

  • Ignoring the re-roll rule and requiring the other three players to roll strictly less than 5, which gives (4/6)^3 and loses the tie cases.
  • Forgetting the binomial factors 33 for which players tie, or dividing by P(won)=14P(\text{won}) = \frac14 incorrectly (multiplying by 14\frac14 instead).

Techniques

Split into exhaustive cases and handle each · Exploit symmetry to reduce work or pair up objects