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2021 Fall AMC 10B Problems/Problem 20

Revision as of 23:10, 11 October 2024 by Mathboy282 (talk | contribs) (→‎Solution 2 (Conditional Probability))

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 (see remark, solution is incorrect)

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 $P(\text{No one rolls a 6})-P(\text{No one rolls a 5 or 6})=P(\text{The highest roll is a 5})$,

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

Remark (Bayes' Theorem)

Although Solution $1$ gives the right answer, it is incorrect, since it only considered the first round, and it didn't consider that there might be one or more tiebreak rounds.

Solutions $2$ and $3$ use Bayes' theorem, and they are correct.

$P \left( A | B \right) = \frac{P \left( B | A \right) P \left( A \right) }{P \left( B \right)}$

In this problem

$P \left( \text{Hugo first rolled 5 } | \text{ Hugo won} \right) = \frac{P \left( \text{Hugo won } | \text{ Hugo first rolled 5} \right) \cdot P \left( \text{Hugo first rolled 5} \right) }{P \left( \text{Hugo won} \right)}$

~isabelchen

Solution 2 (Conditional Probability)

The conditional probability formula states that $P(A|B) = \frac{P(A\cap B)}{P(B)}$, where $A|B$ means A given B and $A\cap B$ means A and B. Therefore the probability that Hugo rolls a five given he won is $\tfrac{P(A \cap B)}{P(B)}$, where A is the probability that he rolls a five and B is the probability that he wins. In written form, \[\text{P(Hugo rolled a 5 given he won)}=\frac{\text{P(Hugo rolls a 5 and wins)}}{\text{P(Hugo wins)}}.\]

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. We consider the cases in which Hugo wins first, find the probabilities of those, and then account for the probability that Hugo rolled a 5, at the end.


$\textbf{Case 1:}$ No Players Tie

In this case, all other players must have numbers from 1 through four, and Hugo had a probability of 1 to win.

There is a $\left(\frac{4}{6}\right)^{3} \cdot 1 = \frac{8}{27}$ chance of this happening.


$\textbf{Case 2:}$ One Player Ties

In this case, there are $3 \choose 1$ $= 3$ ways to choose which other player ties with Hugo, and the probability that this happens is $\tfrac{1}{6} \cdot \left(\tfrac{4}{6}\right)^2$. The probability that Hugo wins on his next round is then $\tfrac{1}{2}$ because there are now two players rolling die.

Therefore the total probability in this case is $3 \cdot \frac{1}{2} \cdot \frac{1}{6} \cdot \left(\frac{4}{6}\right)^2= \frac{1}{9}$.

$\textbf{Case 3:}$ Two Players Tie

In this case, there are $3 \choose 2$ $= 3$ ways to choose which other players tie with Hugo, and the probability that this happens is $\left(\tfrac{1}{6}\right)^2 \cdot \tfrac{4}{6}$. The probability that Hugo wins on his next round is then $\tfrac{1}{3}$ because there are now three players rolling the die.

Therefore the total probability in this case is $3 \cdot \frac{1}{3} \cdot \left(\frac{1}{6}\right)^2 \cdot \frac{4}{6} = \frac{1}{54}$.

$\textbf{Case 4:}$ All Three Players Tie

In this case, the probability that all three players tie with Hugo is $\left(\tfrac{1}{6}\right)^3$. The probability that Hugo wins on the next round is $\tfrac{1}{4}$, so the total probability is $\frac{1}{4} \cdot \left(\frac{1}{6}\right)^3 = \frac{1}{864}$.

Hugo has a $\frac{1}{6}$ probability of rolling a five himself, so the total probability that Hugo rolls a 5 and wins 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

\[4 \cdot \frac{1}{6}\left(\frac{41}{96}\right) = \frac{2}{3} \cdot \frac{41}{96} = \frac{41}{48 \cdot 3} = \frac{41}{144} = \boxed{C}.\]


~KingRavi

~Edits by BakedPotato66, mathboy282

Solution 3

We use $H$ to refer to Hugo. We use $H_1$ to denote the outcome of Hugo's $t$th toss. We denote by $A$, $B$, $C$ the other three players. We denote by $N$ the number of players among $A$, $B$, $C$ whose first tosses are 5. We use $W$ to denote the winner.

We have \begin{align*} P \left( H_1 = 5 | W = H \right) & = \frac{P \left( H_1 = 5 , W = H \right)}{P \left( W = H \right)} \\ & = \frac{P \left( H_1 = 5 \right) P \left( W = H | H_1 = 5 \right) }{P \left( W = H \right)} \\ & = \frac{\frac{1}{6} P \left( W = H | H_1 = 5 \right)}{\frac{1}{4}} \\ & = \frac{2}{3} P \left( W = H | H_1 = 5 \right) . \end{align*}

Now, we compute $P \left( W = H | H_1 = 5 \right)$.

We have \begin{align*} & P \left( W = H | H_1 = 5 \right) \\ & = P \left( W = H | H_1 = 5 , \max \left\{ A_1, B_1, C_1 \right\} \leq 4 \right) P \left( \max \left\{ A_1, B_1, C_1 \right\} \leq 4 | H_1 = 5 \right) \\ & \quad + P \left( W = H | H_1 = 5 , \max \left\{ A_1, B_1, C_1 \right\} = 6 \right) P \left( \max \left\{ A_1, B_1, C_1 \right\} = 6 | H_1 = 5 \right) \\ & \quad + \sum_{N = 1}^3 P \left( W = H | H_1 = 5 , \max \left\{ A_1, B_1, C_1 \right\} = 5 , N \right) P \left( \max \left\{ A_1, B_1, C_1 \right\} = 5 , N | H_1 = 5 \right) \\ & = P \left( W = H | H_1 = 5 , \max \left\{ A_1, B_1, C_1 \right\} \leq 4 \right) P \left( \max \left\{ A_1, B_1, C_1 \right\} \leq 4 \right) \\ & \quad + P \left( W = H | H_1 = 5 , \max \left\{ A_1, B_1, C_1 \right\} = 6 \right) P \left( \max \left\{ A_1, B_1, C_1 \right\} = 6 \right) \\ & \quad + \sum_{N = 1}^3 P \left( W = H | H_1 = 5 , \max \left\{ A_1, B_1, C_1 \right\} = 5 , N \right) P \left( \max \left\{ A_1, B_1, C_1 \right\} = 5 , N \right) \\ & = 1 \cdot P \left( \max \left\{ A_1, B_1, C_1 \right\} \leq 4 \right) + 0 \cdot P \left( \max \left\{ A_1, B_1, C_1 \right\} = 6 \right) \\ & \quad + \sum_{N = 1}^3 P \left( W = H | H_1 = 5 , \max \left\{ A_1, B_1, C_1 \right\} = 5 , N \right) P \left( \max \left\{ A_1, B_1, C_1 \right\} = 5 , N \right) \\ & = P \left( \max \left\{ A_1, B_1, C_1 \right\} \leq 4 \right) \\ & \quad + \sum_{N = 1}^3 P \left( W = H | H_1 = 5 , \max \left\{ A_1, B_1, C_1 \right\} = 5 , N \right) P \left( \max \left\{ A_1, B_1, C_1 \right\} = 5 , N \right) \\ & = \left( \frac{4}{6} \right)^3 + \sum_{N = 1}^3 \frac{1}{N + 1} \cdot \binom{3}{N} \left( \frac{1}{6} \right)^N \left( \frac{4}{6} \right)^{3 - N} \\ & = \frac{41}{96} . \end{align*} The first equality follows from the law of total probability. The second equality follows from the property that Hugo's outcome is independent from other players' outcomes.

Therefore, \begin{align*} P \left( H_1 = 5 | W = H \right) & = \frac{2}{3} P \left( W = H | H_1 = 5 \right) \\ & = \frac{2}{3} \frac{41}{96} \\ & = \frac{41}{144} , \end{align*}

so the answer is $\boxed{\textbf{(C) }\frac{41}{144}}$.

~Steven Chen (www.professorchenedu.com)

Video Solution 1

https://youtu.be/kfn0Bq1-Y5I

Video Solution 2 (by Interstigation)

https://youtu.be/rFYTtzlN1Bw

~Interstigation

Video Solution 3 by WhyMath

https://youtu.be/V7yTQ8CU-JI

~savannahsolver

See Also

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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