糖果收藏家
Candy Collectors
题目详情
五个孩子一起玩“不给糖就捣蛋”游戏，并决定随机 晚上结束时分享他们的糖果。事实证明，他们 总共获得25颗糖果，5种不同类型各5份 （他们住在一个小镇）。他们通过选择来分发糖果 从所有洗牌中均匀随机排列 25 个，以及 然后将前 5 个给第一个孩子，第二个 5 个给第一个孩子 第二，依此类推。
每个孩子拥有一种糖果的概率是多少 他们的“不给糖就捣蛋”游戏比其他所有游戏都多？给予 你的（准确的！）答案用最低的分数表示。
**11 月更新：**此难题的正确解决方案已出现更多 比其他人慢，所以我们要再坚持一个月 并将于 12 月初在网站上推出新的谜题。
Five children went trick-or-treating together and decided to randomly split their candy haul at the end of the night. As it turned out, they got a total of 25 pieces of candy, 5 copies each of 5 different types (they live in a small town). They distribute the candies by choosing an ordering of the 25 uniformly at random from all shufflings, and then giving the first 5 to the first child, the second 5 to the second, and so on.
What is the probability that each child has one type of candy that they have strictly more of than every other trick-or-treater? Give your (exact!) answer in a lowest terms fraction.
November update: correct solutions to this puzzle have come in more slowly than others, so we are going to keep it up for an extra month and will have a new puzzle on the site in early December.
解析
每个孩子严格意义上拥有更多糖果的概率 比其他的“不给糖就捣蛋”的人恰恰是 318281087/8016470462，略低于 4%。最 计算这个的简单方法是分配特定的糖果 给每个孩子并计算其概率（编写一些代码 当然这里有帮助），然后将结果乘以 120 得到 解释糖果的不相交的其他分配 不给糖就捣蛋。
恭喜以下求解者成功提交了精确的 概率正确！
Original Explanation
The probability that every child would have strictly more of a candy than each other trick-or-treater was precisely 318281087/8016470462, which comes to just under 4%. The most straightforward way to compute this was to assign a particular candy to each child and compute the probability of that (writing some code certainly helped here) and then multiplying the result by 120 to account for the disjoint other assignments of candy to trick-or-treater.
Congrats to the following solvers who managed to submit the precise probability correctly!