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2018 AMC 12A Problems/Problem 24

Revision as of 11:46, 22 August 2021 by MRENTHUSIASM (talk | contribs) (Removed repetitive solutions, and rearranged solutions by similar thoughts.)

Problem

Alice, Bob, and Carol play a game in which each of them chooses a real number between $0$ and $1.$ The winner of the game is the one whose number is between the numbers chosen by the other two players. Alice announces that she will choose her number uniformly at random from all the numbers between $0$ and $1,$ and Bob announces that he will choose his number uniformly at random from all the numbers between $\tfrac{1}{2}$ and $\tfrac{2}{3}.$ Armed with this information, what number should Carol choose to maximize her chance of winning?

$\textbf{(A) }\frac{1}{2}\qquad \textbf{(B) }\frac{13}{24} \qquad \textbf{(C) }\frac{7}{12} \qquad \textbf{(D) }\frac{5}{8} \qquad \textbf{(E) }\frac{2}{3}\qquad$

Solution 1 (Expected Values)

The expected value of Alice's number is $\frac12\left(1-0\right)=\frac12,$ and the expected value of Bob's number is $\frac12\left(\frac23-\frac12\right)\frac{7}{12}.$ To maximize her chance of winning, Carol should choose the midpoint between these two expected values. So, the answer is $\frac12\left(\frac12+\frac{7}{12}\right)=\boxed{\textbf{(B) }\frac{13}{24}}.$

Alternatively, once we recognize that the answer is in the interval $\left(\frac12,\frac{7}{12}\right),$ we should choose $\textbf{(B)}$ since no other answer choices are in this interval.

~Random_Guy (Solution)

~MRENTHUSIASM (Revision)

Solution 2 (Piecewise Function)

Let $a,b,$ and $c$ be the numbers that Alice, Bob, and Carol choose, respectively.

Based on the value of $c,$ we construct the following table: \[\begin{array}{c|c|c} & & \\ [-2ex] \textbf{Case} & \textbf{Conditions for }\boldsymbol{a}\textbf{ and }\boldsymbol{b} & \textbf{Carol's Probability of Winning} \\ [0.5ex] \hline & & \\ [-1.5ex] 0<c\leq\frac12 & 0<a<c \text{ and } \frac12<b<\frac23 & \hspace{1.25mm}\frac{c}{1}\cdot\frac{1/6}{1/6}=c \\ [1.5ex] \frac12<c\leq\frac23 & \left(0<a<c \text{ and } c<b<\frac23\right) \text{ or } \left(c<a<1 \text{ and } \frac12<b<c\right) & \hspace{1.25mm}\frac{c}{1}\cdot\frac{2/3-c}{1/6}+\frac{1-c}{1}\cdot\frac{c-1/2}{1/6}=-12c^2+13c-3 \\ [1.5ex] \frac23<c<1 & c<a<1 \text{ and } \frac12<b<\frac23 & \hspace{4.375mm}\frac{1-c}{1}\cdot\frac{1/6}{1/6}=1-c \\ [1.5ex] \end{array}\] Let $P(c)$ be Carol's probability of winning when she chooses $c.$ We write $P(c)$ as a piecewise function: \[P(c) = \begin{cases} c & \mathrm{if} \ 0<c\leq\frac12 \\ -12c^2+13c-3 & \mathrm{if} \ \frac12<c\leq\frac23 \\ 1-c & \mathrm{if} \ \frac23<c<1 \end{cases}.\] Note that the graph of $P(c)$ is continuous in the interval $(0,1),$ increasing in the interval $\left(0,\frac12\right),$ increasing and then decreasing in the interval $\left(\frac12,\frac23\right),$ and decreasing in the interval $\left(\frac23,1\right).$ Therefore, the maximum point of $P(c)$ is in the interval $\left(\frac12,\frac23\right),$ namely at $c=-\frac{13}{2\cdot(-12)}=\boxed{\textbf{(B) }\frac{13}{24}}.$

~MRENTHUSIASM

Solution 3 (Answer Choices)

Let $a,b,$ and $c$ be the numbers that Alice, Bob, and Carol choose, respectively.

From the answer choices, we construct the following table: \[\begin{array}{c|c|c} & & \\ [-2ex] \boldsymbol{c} & \textbf{Conditions for }\boldsymbol{a}\textbf{ and }\boldsymbol{b} & \textbf{Carol's Probability of Winning} \\ [0.5ex] \hline & & \\ [-1.5ex] \frac12 & 0<a<\frac12 \text{ and } \frac12<b<\frac23 & \hspace{23.375mm}\frac{1/2}{1}\cdot\frac{1/6}{1/6}=\frac12 \\ [1.5ex] \frac{13}{24} & \left(0<a<\frac{13}{24} \text{ and } \frac{13}{24}<b<\frac23\right) \text{ or } \left(\frac{13}{24}<a<1 \text{ and } \frac12<b<\frac{13}{24}\right) & \frac{13/24}{1}\cdot\frac{1/8}{1/6}+\frac{11/24}{1}\cdot\frac{1/24}{1/6}=\frac{25}{48} \\ [1.5ex] \frac{7}{12} & \left(0<a<\frac{7}{12} \text{ and } \frac{7}{12}<b<\frac23\right) \text{ or } \left(\frac{7}{12}<a<1 \text{ and } \frac12<b<\frac{7}{12}\right) & \frac{7/12}{1}\cdot\frac{1/12}{1/6}+\frac{5/12}{1}\cdot\frac{1/12}{1/6}=\frac12 \\ [1.5ex] \frac58 & \left(0<a<\frac58 \text{ and } \frac58<b<\frac23\right) \text{ or } \left(\frac58<a<1 \text{ and } \frac12<b<\frac58\right) & \hspace{5.625mm}\frac{5/8}{1}\cdot\frac{1/24}{1/6}+\frac{3/8}{1}\cdot\frac{1/8}{1/6}=\frac{7}{16} \\ [1.5ex] \frac23 & \frac23<a<1 \text{ and } \frac12<b<\frac23 & \hspace{23.25mm}\frac{1/3}{1}\cdot\frac{1/6}{1/6}=\frac13 \\ [1.5ex] \end{array}\] Therefore, Carol should choose $\boxed{\textbf{(B) }\frac{13}{24}}$ to maximize her chance of winning.

~MRENTHUSIASM

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2018amc12a/474

~ dolphin7

Video Solution (Meta-Solving Technique)

https://youtu.be/GmUWIXXf_uk?t=926

~ pi_is_3.14

See Also

2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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 12 Problems and Solutions

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