數數數．．．．．．．．．．．．．．．．

For those who do not wish to follow the above link, the question is asking for the probability of having at lease two students in a class of 23 students having the same birthday, ignoring the case of those born on the 29th of February. An answer is required also for a class of 40 students.

It turns out that it is easier to calculate the case where NO TWO PERSONS have the same birthday.

We will proceed with the case of one student. The first student can have his birthday on any of the 365 days without a clash, so the probability is
365/365=1

For two students, the second can only have choose among the remaining 364 days without conflict, so the probability for two students is
365/365 * 364/365

For three students, the probability is
365/365 * 364/365 * 363/365

So for n students (where n<=365), the mathematical expression becomes
365/365 * 364/365 * 363/365 ... *(365+n-1)/365
=365!/(365-n)!/365^n

To find the probability that at least two out of n students have the same birthday, we just need to subtract the above expression from 1, namely:
f(n)=1-365!/(365-n)!/365^n

For 23 students, the expression turns out to be:
f(23)=38093904702297390785243708291056390518886454060947061/75091883268515350125426207425223147563269805908203125
=0.507297
For 40 students,
f(40)=19393328682731619082071437723120714139262612692853031533269435559474912532712558219274716441893/21760139695522172283205169322668538168792903146260712847284222971593416002579033374786376953125
=0.891232

Note:
f(1)=0, since duplicate birthdays cannot occur alone.
f(365)=1-1.45496*10^-157, i.e. it is highly likely that at least two students have the same birthday out of 365.
f(366)=1 It is sure to have at least a pair of students out of 366 with the same birthday, ignoring leap years.

Here's a puzzle:
How can two persons be born on the same day, same year and to the same mother, but they are not twins?

2007-10-12 03:09:46 補充：
The following link discusses how to include the case of people born on the 29th of February.http://home.att.net/~numericana/answer/counting.htm#siblingsNear the end of the article, it also gives the answer to the above puzzle.