Difference between revisions of "2018 AMC 12A Problems/Problem 24"
| Line 27: | Line 27: | ||
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) | 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: 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 | + | 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. |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2018|ab=A|num-b=23|num-a=25}} | {{AMC12 box|year=2018|ab=A|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 14:40, 10 February 2018
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 
and 
Armed with this information, what number should Carol choose to maximize her chance of winning?
Solution 1
Plug in all the answer choices to get 
Solution 2
Let the value we want be 
. The probability that Alice's number is less than Carol's number and Bob's number is greater than Carol's number is 
. Similarly, the probability that Bob's number is less than Carol's number and Alice's number is greater than Carol's number is 
. Adding these together, the probability that Carol wins given a certain number is 
. Using calculus or the fact that the extremum of a parabola occurs at 
, the maximum value occurs at 
, which is
Solution 3
The expected value of Alice's number is 
and the expected value of Bob's number is 
. To maximize her chance of winning, Carol would choose number exactly in between the two expected values, giving:
. This is 
. (Random_Guy)
EDIT: I believe this method is incorrect. Assume Bob can only choose but Alice chooses from the same range as before. The answer, using the above method, remains 
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.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
