2018 AMC 12A Problems/Problem 24

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

Plug in all the answer choices to get $\boxed{\textbf{(B)}}$ .

Solution 2

Let the value we want be $x$ . The probability that Alice's number is less than Carol's number and Bob's number is greater than Carol's number is $x(\frac{2}{3}-x)$ . Similarly, the probability that Bob's number is less than Carol's number and Alice's number is greater than Carol's number is $(x-\frac{1}{2})(1-x)$ . Adding these together, the probability that Carol wins given a certain number $x$ is $-2x^2+\frac{13}{6}x-\frac{1}{2}$ . Using calculus or the fact that the extremum of a parabola occurs at $\frac{-b}{2a}$ , the maximum value occurs at $x=\frac{13}{24}$ , which is $\boxed{\textbf{(B)}.}$

Solution 3

The expected value of Alice's number is $\frac{1}{2}$ and the expected value of Bob's number is $\frac{7}{12}$ . To maximize her chance of winning, Carol would choose number exactly in between the two expected values, giving: $\frac{6+7}{12*2}=\frac{13}{24}$ . This is $\boxed{\textbf{(B)}}$ . (Random_Guy)

EDIT: I believe this method is incorrect. Assume Bob can only choose $\frac{7}{12}$ but Alice chooses from the same range as before. The answer, using the above method, remains $\frac{13}{24}$ which is clearly wrong in this case. Correct me if I misunderstood the solution. (turnip123)

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)

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.

Solution 4

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.

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.

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.

Solution 5

Have Carol pick a point x on the real line. The probability that Alice's number is below is thus X, and the probability that Alice's number is above is (1-X). Now, Carol must pick a point between 1/2 and 2/3, exclusive. If she does not, then it is impossible for Carol's point to be between both Alice and Bobs point. Now what is the probability that Bob's point is below Carols? Scale Bob's to the normal number line to make it simple. (1/2) to (2/3) has 2 conversion to make it to the (0-1) line. Subtract by 1/2, and then multiply by 6. So thus the chance B is below is 6(x-1/2) and above is 1-6(x-1/2).

Now we want to maximixe the probability of (Alice number below X) AND (Bob number above X) + (Alice number above X) AND (Bob number below X). This gives the formula x*(1-6*(x-1/2))+(1-x)*6*(x-1/2) = -3 + 13 x - 12 x^2. Then proceed using the process of completing the square to get that the maximum of this is 13/24.

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

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