Difference between revisions of "2021 Fall AMC 10B Problems/Problem 20"

Revision as of 08:00, 21 September 2022

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

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

Note: The solution above doesn't fully justify the claim. Here is more detail:

$P(\text{Hugo first rolled a 5})/P(\text{Hugo won})$

$= P(\text{Hugo first rolled a 5, and Hugo won})/P(\text{Hugo won})$ (redundancy)

$= P(\text{The highest first roll was a 5, and Hugo won})/P(\text{Hugo won})$ (winner first role matches highest first roll)

$= P(\text{The highest roll was a 5}) * P(\text{Hugo won})/P(\text{Hugo won})$ (separating independent events)

$= P(\text{The highest roll was a 5})$ .

(Hugo's chance of winning, is independent on the values of the first roll (but *not* who got which value). It is the same $1/4$ regardless of which a $N$ -way tie there is for highest first roll: Hugo has an $N/4$ chance of being in that tie, and a $1/N$ chance of winning the tiebreak.)

~oinava

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...

$\textbf{Case 1:}$ 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.

$\textbf{Case 2:}$ 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 $\frac{1}{4} \cdot \left(\frac{1}{6}\right)^3 = \frac{1}{864}$ .

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

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

~KingRavi

~Edits by BakedPotato66

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

@@ Line 17: / Line 17: @@
 ~kingofpineapplz
-Note: This is a tricky analysis. The solution above seems to ignore the different probabilities of Hugo winning after each possible first first roll (and possibly is incomplete reasoning). But that turns out to not matter!
+Note: The solution above doesn't fully justify the claim. Here is more detail:
-As noted above,
-P(*Hugo rolled a 5*, and Hugo won)/P(Hugo won)
+<cmath>P(\text{Hugo first rolled a 5})/P(\text{Hugo won})</cmath>
-= P(*The highest roll was a 5* and Hugo won)/P(Hugo won).
+<cmath>= P(\text{Hugo first rolled a 5, and Hugo won})/P(\text{Hugo won})</cmath> (redundancy)
-Furthermore, The value of the highest roll is independent of who wins, by symmetry across permutations of the players. (Observe in more detail: Hugo’s chance of winning is the same 1/4, regardless of what the *values* of the first rolls were, before considering who rolled which value. It is the same 1/4 whether he was in a 4 way tie for first roll and had a 1/4 chance of winning tiebreake; or there was a N-way tie for another highest roll value, and Hugo had an N/4 chance of being in that tie, and a 1/N chance of winning the tiebreaker.)
+<cmath>= P(\text{The highest first roll was a 5, and Hugo won})/P(\text{Hugo won})</cmath> (winner first role matches highest first roll)
-So, the overall conditional probability
+<cmath>= P(\text{The highest roll was a 5}) * P(\text{Hugo won})/P(\text{Hugo won})</cmath> (separating independent events)
-= P(The highest roll was a 5) * P(Hugo won)/P(Hugo won)
+<cmath>= P(\text{The highest roll was a 5})</cmath>.
-and thus doesn’t depend on Hugo’s winning chance, and the only thing that matters is the probability of the acceptable first rolls.
+(Hugo's chance of winning, is independent on the values of the first roll (but *not* who got which value). It is the same <math>1/4</math> regardless of which a <math>N</math>-way tie there is for highest first roll: Hugo has an <math>N/4</math> chance of being in that tie, and a <math>1/N</math> chance of winning the tiebreak.)
 ~oinava

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

Page

Toolbox

Search