2018 AMC 12A Problem #24

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 1 2 \frac{1}{2} and 2 3 . \frac{2}{3}. Armed with this information, what number should Carol choose to maximize her chance of winning?

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13 24 \dfrac{13}{24} 7 12 \dfrac{7}{12} 2 3 \dfrac{2}{3} 1 2 \dfrac{1}{2} 5 8 \dfrac{5}{8}

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1 solution

This problem is a disguised version of a simple expected value problem. By the law of large numbers, which states that the arithmetic mean of a result approaches the expected value as more trials are performed, we can assume that maximizing our chances of winning means finding the expected value of winning. Thus, if we take the arithmetic mean of each range of numbers and then take the arithmetic mean of that, we should reach our answer.

Alice's expected value: 0 + 1 2 = 1 2 \frac{0 + 1}{2} = \frac{1}{2}

Bob's expected value: 1 2 + 2 3 2 = 7 12 \frac{\frac{1}{2} + \frac{2}{3}}{2} = \frac{7}{12}

Optimized expected value: 1 2 + 7 12 2 = 13 24 \frac{\frac{1}{2} + \frac{7}{12}}{2} = \boxed{\frac{13}{24}}

If you're interested in further research:

Law of Large Numbers

Expected Value

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