!b4 Note:
By same birthday we mean same day and month but not necessarily
the same year. We assume further that the birthdays of the k people
are unrelated and each equally likely to be in any of the 365 days
of the year.
!b2 Solution
Define the proposition:
!c3 E = "at least 2 equal birthdays"
Let S denote the statement of the problem above. Then,
the probability of E given S is,
!c3 Pr(E|S) = 1 - Pr(not E |S)
But
!c3 (not E) = "all different birthdays"
So the problem reduces to the computation of the number
of ways in which k people can have different birthdays.
There are three cases. First, if k > 365 then necessarily
at least 2 people will have the same birthday. Second,
if k < 2 then there is no problem. Third, when 1^{k}
The probability we are looking for is then,
> answer := 1 - a!/((a-k)!*a^k);
365!
answer := 1 - ----------------
k
(365 - k)! 365
# This is a function of k that can be easily computed with
> p := k -> evalf((1 - 365!/((365-k)!*365^k)),4):
> p10 := p(10); p20:=p(20); p23 := p(23); p40:=p(40); p50:=p(50);
p10 := .1169
p20 := .4114
p23 := .5073
p40 := .8912
p50 := .9704
>