生日狂欢
Birthday Bash
题目详情
随机选择的一组人的最小规模是多少，使得 他们有超过 50% 的机会为至少一个人庆祝生日 一年中的每一天，小组中的人？假设所有年份有 365 天，并且生日在一年中均匀分布。
一如既往，您可以使用您想要尝试和回答的任何技术/资源 这个谜题，我们不希望您提交除答案以外的任何内容。 也就是说，如果你想挑战自己，看看你能走多远 你的直觉！如果加上铅笔和纸怎么样？如果你添加一个怎么样？ 计算器？通过这些可能很难得出准确的答案 方法，但如果是这样，你能将答案约束得有多紧？
What is the minimum size of a randomly selected group of people such that there is a greater than 50% chance they can celebrate the birthday of at least one person in the group on every day of the year? Assume that all years have 365 days, and that birthdays are evenly spread throughout the year.
As always, you can use whatever techniques/resources you want to try and answer this puzzle, and we don’t expect you to submit anything other than your answer. That said, if you want to challenge yourself, see how far you can get just using your intuition! How about if you add pencil & paper? How about if you add a calculator? It could be difficult to arrive at an exact answer via these methods, but if so, how tightly can you bound the answer?
解析
本月难题的答案是需要2287人才能 使每个生日被覆盖的概率大于50%。
这个数字不一定容易计算，但有一些近似值 我们可以让我们知道答案应该是这个顺序。
如果我们询问所需的平均人数（而不是中位数）， 这相当于“优惠券收集者问题”。平均 所需人数为 365 * ( + … + 1) = 大约 2365。鉴于平均情况包含应该很长的内容 尾部位于分布的右侧，平均值应大于 中位数，因此 2365 将是上限。
我们还可以得到一个下限。对于给定的 D 天，至少 1 的概率 在这一天过生日的 N 个人中有 [1 – (364/365)^N]。 考虑到至少有 1 个人的生日在 D，这使得它稍微少一些 可能其他日子都被覆盖了。但如果我们忽略这种影响并假设 独立性，我们可以问 N 需要多大才能得到乘积 所有 365 天的个体概率超过 0.5。表达式 [1 – 当 N 达到 2285 时，(364/365)^N]^365 变得大于 0.5，因此 2285 将是 下限。
恭喜所有解决本月难题的人！
Original Explanation
The answer to this month’s puzzle is that 2287 people are needed in order to make the probability that each birthday is covered greater than 50%.
This number isn’t necessarily easy to compute, but there are some approximations we can make which show us the answer should be on that order.
If we had asked for the average number of people needed (instead of the median), that would’ve been equivalent to the “coupon collector problem”. The average number of people needed would’ve been 365 * (1/365 + 1/364 + 1/363 + … + 1) = roughly 2365. Given that the average case incorporates what should be a long tail on the right side of the distribution, the average should be greater than the median, and thus 2365 would be an upper bound.
We can also get a lower bound. For a given day D, the probability of at least 1 out of N people having their birthday on that day would be [1 – (364/365)^N]. Given that at least 1 person has their birthday on D, it makes it slightly less likely that the other days are covered. But if we ignore this effect and assume independence, we could ask how large N would need to be in order for the product of all 365 days’ individual probabilities to exceed 0.5. The expression [1 – (364/365)^N]^365 becomes greater than 0.5 when N reaches 2285, so 2285 would be a lower bound.
Congratulations to everyone who solved this month’s puzzle!