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Difference between revisions of "2018 AMC 12A Problems/Problem 24"

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== Problem ==
 
== 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 <math>\tfrac{1}{2}</math> and <math>\tfrac{2}{3}.</math> Armed with this information, what number should Carol choose to maximize her chance of winning?
+
Alice, Bob, and Carol play a game in which each of them chooses a real number between <math>0</math> and <math>1.</math> 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 <math>0</math> and <math>1,</math> and Bob announces that he will choose his number uniformly at random from all the numbers between <math>\tfrac{1}{2}</math> and <math>\tfrac{2}{3}.</math> Armed with this information, what number should Carol choose to maximize her chance of winning?
−
 
 
  
 
<math>
 
<math>
Line 12: Line 11:
 
</math>
 
</math>
  
−
== Solution 1==
+
== Solution 1 (Expected Values) ==
 +
The expected value of Alice's number is <math>\frac12\left(0+1\right)=\frac12,</math> and the expected value of Bob's number is <math>\frac12\left(\frac12+\frac23\right)=\frac{7}{12}.</math> To maximize her chance of winning, Carol should choose the midpoint between these two expected values. So, the answer is <math>\frac12\left(\frac12+\frac{7}{12}\right)=\boxed{\textbf{(B) }\frac{13}{24}}.</math>
 +
 
 +
Alternatively, once we recognize that the answer lies in the interval <math>\left(\frac12,\frac{7}{12}\right),</math> we should choose <math>\textbf{(B)}</math> since no other answer choices lie in this interval.
 +
 
 +
~Random_Guy ~MRENTHUSIASM
 +
 
 +
==Solution 2 (Piecewise Function)==
 +
Let <math>a,b,</math> and <math>c</math> be the numbers that Alice, Bob, and Carol choose, respectively.
 +
 
 +
Based on the value of <math>c,</math> we construct the following table:
 +
<cmath>\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<\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\leq 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}</cmath>
 +
Let <math>P(c)</math> be Carol's probability of winning when she chooses <math>c.</math> We write <math>P(c)</math> as a piecewise function:
 +
<cmath>P(c) = \begin{cases}
 +
c & \mathrm{if} \ 0<c<\frac12 \\
 +
-12c^2+13c-3 & \mathrm{if} \ \frac12\leq c\leq\frac23 \\
 +
1-c & \mathrm{if} \ \frac23<c<1
 +
\end{cases}.</cmath>
 +
Note that <math>P(c)</math> is continuous in the interval <math>(0,1),</math> increasing in the interval <math>\left(0,\frac12\right),</math> increasing and then decreasing in the interval <math>\left(\frac12,\frac23\right),</math> and decreasing in the interval <math>\left(\frac23,1\right).</math> The graph of <math>y=P(c)</math> is shown below.
 +
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(200);
 +
 
 +
real f(real x) { return x; }
 +
real g(real x) { return -12x^2+13x-3; }
 +
real h(real x) { return 1-x; }
 +
 
 +
draw((1/2,0)--(1/2,1.25),dashed);
 +
draw((2/3,0)--(2/3,1.25),dashed);
 +
draw(graph(f,0,1/2),red);
 +
draw(graph(g,1/2,2/3),red);
 +
draw(graph(h,2/3,1),red);
 +
 
 +
real xMin = -0.25;
 +
real xMax = 1.25;
 +
real yMin = -0.25;
 +
real yMax = 1.25;
 +
 
 +
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
 +
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
 +
label("$c$",(xMax,0),(2,0));
 +
label("$y$",(0,yMax),(0,2));
 +
 
 +
pair A[];
 +
A[0] = (0,0);
 +
A[1] = (1/2,1/2);
 +
A[2] = (2/3,1/3);
 +
A[3] = (1,0);
 +
 
 +
dot(A[1],red+linewidth(3.5));
 +
dot(A[2],red+linewidth(3.5));
 +
 
 +
label("$0$",A[0],(-1.5,-1.5));
 +
label("$\frac12$",(1/2,0),(0,-1.5));
 +
label("$\frac23$",(2/3,0),(0,-1.5));
 +
label("$1$",A[3],(0,-1.5));
 +
label("$1$",(0,1),(-1.5,0));
 +
 
 +
draw((1/2,-0.02)--(1/2,0.02),linewidth(1));
 +
draw((2/3,-0.02)--(2/3,0.02),linewidth(1));
 +
draw((1,-0.02)--(1,0.02),linewidth(1));
 +
draw((-0.02,1)--(0.02,1),linewidth(1));
 +
</asy>
 +
Therefore, the maximum point of <math>P(c)</math> occurs in the interval <math>\left[\frac12,\frac23\right],</math> namely at <math>c=-\frac{13}{2\cdot(-12)}=\boxed{\textbf{(B) }\frac{13}{24}}.</math>
 +
 
 +
~MRENTHUSIASM
  
−
Plug in all the answer choices to get <math>\boxed{\textbf{(B)}}</math>.
+
== Solution 3 (Answer Choices)==
 +
Let <math>a,b,</math> and <math>c</math> be the numbers that Alice, Bob, and Carol choose, respectively.
  
−
==Solution 2==
+
From the answer choices, we construct the following table:
−
Let the value we want be <math>x</math>. The probability that Alice's number is less than Carol's number and Bob's number is greater than Carol's number is <math>x(\frac{2}{3}-x)</math>. Similarly, the probability that Bob's number is less than Carol's number and Alice's number is greater than Carol's number is <math>(x-\frac{1}{2})(1-x)</math>. Adding these together, the probability that Carol wins given a certain number <math>x</math> is <math>-2x^2+\frac{13}{6}x-\frac{1}{2}</math>. Using calculus or the fact that the extremum of a parabola occurs at <math>\frac{-b}{2a}</math>, the maximum value occurs at <math>x=\frac{13}{24}</math>, which is <math>\boxed{\textbf{(B)}}</math>.
+
<cmath>\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}</cmath>
 +
Therefore, Carol should choose <math>\boxed{\textbf{(B) }\frac{13}{24}}</math> to maximize her chance of winning.
  
−
== Solution 3 ==
+
~MRENTHUSIASM
−
The expected value of Alice's number is <math>\frac{1}{2}</math> and the expected value of Bob's number is <math>\frac{7}{12}</math>. To maximize her chance of winning, Carol would choose number exactly in between the two expected values, giving:<math>\frac{6+7}{12*2}=\frac{13}{24}</math>. This is <math>\boxed{\textbf{(B)}}</math>. (Random_Guy)
 
  
−
EDIT: I believe this method is incorrect. Assume Bob can only choose <math>\frac{7}{12}</math> but Alice chooses from the same range as before. The answer, using the above method, remains <math>\frac{13}{24}</math> which is clearly wrong in this case. Correct me if I misunderstood the solution.
+
== Solution 4 (Calculus) ==
−
(turnip123)
+
There are two cases of winning:
  
−
EDIT2: It would make sense for the answer to remain the same, given that Bob's expected value stays the same. Why should the answer change in your scenario? (KenV)
+
Case 1: Alice choose a number that is smaller than Carol's, and Bob choose a number that is bigger.  
  
−
EDIT3: I realized my mistake. The method works in all situations where the expected value falls within both of their range. In my case, Carol's number would be less than Bob's number if Carol chooses any number from the range [0, 7/12). She would then want to maximize the chances of picking a number greater than Alice, which is achieved by picking the largest number possible from the range [0, 7/12), which is not 13/24.
+
Case 2: Alice choose a number that is bigger, and Bob choose a number that is smaller.
  
−
== Solution 4 ==
+
Let Carol's number be <math>\frac{1}{2}+x</math>, then the probability of Case 1 can be expressed by <math>\frac{1/2 + x}{1}\cdot\frac{1/3 - x }{1/3}</math>, and the probability of Case 2 can be expressed by <math>\frac{1/3 + 1/6 - x}{1}\cdot\frac{x}{1/3}</math>.
  
−
Let’s call Alice’s number a, Bob’s number b, and Carol’s number c. Then, in order to maximize her chance of choosing a number that is in between a and b, she should choose c = (a+b)/2.
+
Then the probability of Carol winning can be expressed by the sum of the probability of Case 1 and the probability of Case 2: <math>3\left(\frac{1}{2} + x\right)\left(\frac{1}{3} - x\right)+ 3x\left(\frac{1}{2} - x\right)</math>, which simplify to <math>\frac{1}{2} + x - 6x^2</math>. Taking the first derivative, we get <math>1 - 12x</math>, which yield a maximum point at <math>x = \frac{1}{12}</math>. Hence, Carol should select <math>\frac{1}{2} + \frac{1}{12} = \boxed{\textbf{(B) }\frac{13}{24}}</math>.
  
−
We need to find the average value of (a+b)/2 over the region [0, 1] x[1/2, 2/3] in the a-b plane.
+
== Video Solution by Richard Rusczyk ==
  
−
We can set up a double integral with bounds 0 to 1 for the outer integral and 1/2 to 2/3 for the inner integral with an integrand of (a+b)/2. We need to divide our answer by 1/6, the area of the region of interest. This should yield 13/24, B.
+
https://artofproblemsolving.com/videos/amc/2018amc12a/474
  
−
== Solution 5 ==
+
~ dolphin7
  
−
Have Carol pick a point <math>x</math> on the real line. The probability that Alice's number is below is thus <math>x</math>, and the probability that Alice's number is above is <math>1 - x</math>. Now, Carol must pick a point between <math>\frac{1}{2}</math> and <math>\frac{2}{3}</math>, exclusive. If she does not, then it is impossible for Carol's point to be between both Alice and Bob's point. Now what is the probability that Bob's point is below Carol's? To make it simple, scale Bob's number to the normal number line by subtracting <math>\frac{1}{2}</math> and multiplying by <math>6</math>. So thus the chance Bob's number is below is <math>6(x - \frac{1}{2})</math> and above is <math>1 - 6(x-\frac{1}{2})</math>.
+
== Video Solution (Meta-Solving Technique) ==
 +
https://youtu.be/GmUWIXXf_uk?t=926
  
−
Now we want to maximixe the probability of (Alice number below <math>x</math>) AND (Bob number above <math>x</math>) + (Alice number above <math>x</math>) AND (Bob number below <math>x</math>).  This gives the formula <math>x(1 - 6(x-\frac{1}{2})) + (1 -x)6(x-\frac{1}{2})  = -3 + 13 x - 12 x^2</math>.
+
~ pi_is_3.14
−
Then proceed using the process of completing the square or averaging the roots to get that the maximum of this is <math>\frac{13}{24}</math>.
 
  
 
==See Also==
 
==See Also==

Revision as of 11:39, 20 August 2023

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 $\frac12\left(0+1\right)=\frac12,$ and the expected value of Bob's number is $\frac12\left(\frac12+\frac23\right)=\frac{7}{12}.$ 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 lies in the interval $\left(\frac12,\frac{7}{12}\right),$ we should choose $\textbf{(B)}$ since no other answer choices lie in this interval.

~Random_Guy ~MRENTHUSIASM

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: \[\begin{array}{c|c|c} & & \\ [-2ex] \textbf{Case} & \textbf{Conditions for }\boldsymbol{a}\textbf{ and }\boldsymbol{b} & \textbf{Carol's Probability of Winning} \\ [0.5ex] \hline & & \\ [-1.5ex] 0<c<\frac12 & 0<a<c \text{ and } \frac12<b<\frac23 & \hspace{1.25mm}\frac{c}{1}\cdot\frac{1/6}{1/6}=c \\ [1.5ex] \frac12\leq c\leq\frac23 & \left(0<a<c \text{ and } c<b<\frac23\right) \text{ or } \left(c<a<1 \text{ and } \frac12<b<c\right) & \hspace{1.25mm}\frac{c}{1}\cdot\frac{2/3-c}{1/6}+\frac{1-c}{1}\cdot\frac{c-1/2}{1/6}=-12c^2+13c-3 \\ [1.5ex] \frac23<c<1 & c<a<1 \text{ and } \frac12<b<\frac23 & \hspace{4.375mm}\frac{1-c}{1}\cdot\frac{1/6}{1/6}=1-c \\ [1.5ex] \end{array}\] Let $P(c)$ be Carol's probability of winning when she chooses $c.$ We write $P(c)$ as a piecewise function: \[P(c) = \begin{cases} c & \mathrm{if} \ 0<c<\frac12 \\ -12c^2+13c-3 & \mathrm{if} \ \frac12\leq c\leq\frac23 \\ 1-c & \mathrm{if} \ \frac23<c<1 \end{cases}.\] Note that $P(c)$ 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 $\left(\frac23,1\right).$ The graph of $y=P(c)$ is shown below. [asy] /* Made by MRENTHUSIASM */ size(200); real f(real x) { return x; } real g(real x) { return -12x^2+13x-3; } real h(real x) { return 1-x; } draw((1/2,0)--(1/2,1.25),dashed); draw((2/3,0)--(2/3,1.25),dashed); draw(graph(f,0,1/2),red); draw(graph(g,1/2,2/3),red); draw(graph(h,2/3,1),red); real xMin = -0.25; real xMax = 1.25; real yMin = -0.25; real yMax = 1.25; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$c$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); pair A[]; A[0] = (0,0); A[1] = (1/2,1/2); A[2] = (2/3,1/3); A[3] = (1,0); dot(A[1],red+linewidth(3.5)); dot(A[2],red+linewidth(3.5)); label("$0$",A[0],(-1.5,-1.5)); label("$\frac12$",(1/2,0),(0,-1.5)); label("$\frac23$",(2/3,0),(0,-1.5)); label("$1$",A[3],(0,-1.5)); label("$1$",(0,1),(-1.5,0)); draw((1/2,-0.02)--(1/2,0.02),linewidth(1)); draw((2/3,-0.02)--(2/3,0.02),linewidth(1)); draw((1,-0.02)--(1,0.02),linewidth(1)); draw((-0.02,1)--(0.02,1),linewidth(1)); [/asy] Therefore, the maximum point of $P(c)$ occurs 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

Solution 4 (Calculus)

There are two cases of winning:

Case 1: Alice choose a number that is smaller than Carol's, and Bob choose a number that is bigger.

Case 2: Alice choose a number that is bigger, and Bob choose a number that is smaller.

Let Carol's number be $\frac{1}{2}+x$, then the probability of Case 1 can be expressed by $\frac{1/2 + x}{1}\cdot\frac{1/3 - x }{1/3}$, and the probability of Case 2 can be expressed by $\frac{1/3 + 1/6 - x}{1}\cdot\frac{x}{1/3}$.

Then the probability of Carol winning can be expressed by the sum of the probability of Case 1 and the probability of Case 2: $3\left(\frac{1}{2} + x\right)\left(\frac{1}{3} - x\right)+ 3x\left(\frac{1}{2} - x\right)$, which simplify to $\frac{1}{2} + x - 6x^2$. Taking the first derivative, we get $1 - 12x$, which yield a maximum point at $x = \frac{1}{12}$. Hence, Carol should select $\frac{1}{2} + \frac{1}{12} = \boxed{\textbf{(B) }\frac{13}{24}}$.

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)
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All AMC 12 Problems and Solutions

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