2018 AMC 12A Problems/Problem 24
Contents
Problem
Alice, Bob, and Carol play a game in which each of them chooses a real number between and 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 and and Bob announces that he will choose his number uniformly at random from all the numbers between and Armed with this information, what number should Carol choose to maximize her chance of winning?
Solution 1 (Expected Values)
The expected value of Alice's number is and the expected value of Bob's number is To maximize her chance of winning, Carol should choose the midpoint between these two expected values. So, the answer is
Alternatively, once we recognize that the answer lies in the interval we should choose since no other answer choices lie in this interval.
~Random_Guy ~MRENTHUSIASM
Solution 2 (Piecewise Function)
Let and be the numbers that Alice, Bob, and Carol choose, respectively.
Based on the value of we construct the following table: Let be Carol's probability of winning when she chooses We write as a piecewise function: Note that is continuous in the interval increasing in the interval increasing and then decreasing in the interval and decreasing in the interval The graph of is shown below. Therefore, the maximum point of occurs in the interval namely at
~MRENTHUSIASM
Solution 3 (Answer Choices)
Let and be the numbers that Alice, Bob, and Carol choose, respectively.
From the answer choices, we construct the following table: Therefore, Carol should choose 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 , then the probability of Case 1 can be expressed by , and the probability of Case 2 can be expressed by .
Then the probability of Carol winning can be expressed by the sum of the probability of Case 1 and the probability of Case 2: , which simplify to . Taking the first derivative, we get , which yield a maximum point at . Hence, Carol should select .
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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