PUMaC 2016 · 团队赛 · 第 10 题
PUMaC 2016 — Team Round — Problem 10
题目详情
- (8) Chad and Chad2 run competing rare candy stores at Princeton. Chad has a large supply of boxes of candy, each box containing three candies and costing him $3 to purchase from his supplier. He charges $1.50 per candy per student. However, any rare candy in an opened box must be discarded at the end of the day at no profit. Chad knows that at each of 8am, 10am, noon, 2pm, 4pm, and 6pm, there will be one person who wants to buy one candy, and that they choose between Chad and Chad2 at random. (He knows that those are the only times when he might have a customer.) Chad may refuse sales to any student who asks for candy. m If Chad acts optimally, his expected daily profit can be written in simplest form as . Find n m + n . (Chad’s profit is 3 per box he opens.) 2
解析
- If there’s 2 students left, Chad shouldn’t open another box, because the chance that he can 1 2 1 make 0 is , the chance he makes -$1.5 is , so 4 4 4 there’s no expected gain. If there are any more students left, then he should continue to offer sales. We can thus separate consideration of the first 3 students (for which Chad always opens a box) and the last 3 students (for which Chad never opens a box). Of the first 3, if 0 go to Chad, his profit is 0. 1 If 1, his profit is 1 . 5 if at least two of the last three go to him (with probability ), 0 if 2 1 exactly one go to him, and − 1 . 5 if none do (with probability ). The total expected profit is 8 1 1 9 · 1 . 5 − · 1 . 5 = . 2 8 16 If 2, his profit is 1 . 5 if at least one of the last three go to him, and 0 otherwise. The to tal 7 21 expected profit is · 1 . 5 = . 8 16 If 3, his profit is 1 . 5. 3 9 3 21 1 3 57 So his expected profit is · + · + · = . Thus, the answer is 57 + 64 = 121 . 8 16 8 16 8 2 64