p****3
发帖数: 448 | 1 足够多的, 互无联系的人, 随机排成一队, 挨个报出自己生日.
直到有重复的生日出现.
问平均报出了多少生日? | m***n
发帖数: 62 | 2 You can do repeated expectation:
let the expectation you want to be x;
fix the first person;
conditioning on the 2nd person has the same bday as the 1st person, you got
2;
conditioning on the 2nd person has a different bday as the 1st person, you
can condition on the 3rd person; if the 3rd person has the same bday as the
previous two people, you got 3;
otherwise keep going...
so x = 1/365*2+364/365*2/365*3+363/365*3/365*4+...
then x = 1/365* sum_{n = 1}^{364} n*(n+1)*(365-n+1)/365 and it is a power
summation | z****p
发帖数: 18 | 3
got
the
The solution seems to have a problem:
Look at the 4th person. The condition for the 4th person to have an
opportunity is that "the 2nd person failed and the 3rd person failed as well
". The probability for this to happen is
(364/365)*(363/365) instead of 363/365
【在 m***n 的大作中提到】
: You can do repeated expectation:
: let the expectation you want to be x;
: fix the first person;
: conditioning on the 2nd person has the same bday as the 1st person, you got
: 2;
: conditioning on the 2nd person has a different bday as the 1st person, you
: can condition on the 3rd person; if the 3rd person has the same bday as the
: previous two people, you got 3;
: otherwise keep going...
: so x = 1/365*2+364/365*2/365*3+363/365*3/365*4+...
| d*****n
发帖数: 44 | 4 Every person is Bernoulli distribution with various success probabilities.
Denote the i-th person has success probability Pi. And I assume we do not
count the last repeated b-day.
1st: P1 = 1
2nd: P2 = p(1st and 2nd have different b-days) = 1*(364/365)
3rd: P3 = p(1st, 2nd, and 3rd have different b-days) = p(3rd person
different b-day|1st and 2nd people different b-days)*p(1st and 2nd people
different b-days) = (363/365) *[1*(364/365)]
4th: P4 = p(4th person different b-day|1st, 2nd, and 3rd people different b-
days)*p(1st, 2nd, and 3rd people different b-days) = (362/365) *[1*(364/365)
*(363/365)]
...
Result: 1+1*(364/365)+1*(364/365)*(363/365)+1*(364/365)*(363/365)*(362/365)
+... | s*****n
发帖数: 134 | 5 根据抽屉原理，不考虑闰年的话，最多有365个人没有相同的生日，到366的时候至少有
一对相同的生日。所以你列出的期望 应该是从第二个人到第366人 N×P（N）的和。
具体的计算好像可以用泰勒级数近似，但是这个Wiki里给出的结果只是 N个人里没有两
个人有相同生日的 CDF， 感觉上和计算从第几个人开始出现相同生日的期望不一定一
样。
http://en.wikipedia.org/wiki/Birthday_problem
b-
365)
【在 d*****n 的大作中提到】
: Every person is Bernoulli distribution with various success probabilities.
: Denote the i-th person has success probability Pi. And I assume we do not
: count the last repeated b-day.
: 1st: P1 = 1
: 2nd: P2 = p(1st and 2nd have different b-days) = 1*(364/365)
: 3rd: P3 = p(1st, 2nd, and 3rd have different b-days) = p(3rd person
: different b-day|1st and 2nd people different b-days)*p(1st and 2nd people
: different b-days) = (363/365) *[1*(364/365)]
: 4th: P4 = p(4th person different b-day|1st, 2nd, and 3rd people different b-
: days)*p(1st, 2nd, and 3rd people different b-days) = (362/365) *[1*(364/365)
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