What is the probability that on campus of 730 students there exists at least one triplet of students celebrating their birthday on the same date?
Rem: Assume a year of 365 days
The only way to avoid such a triplet is for each possible birthday to be represented exactly twice. Any birthday which was represented only once would leave 364 birthdays for the remaining 2*364+1=729 students. The extra (+1) student would have to match one of the 364 pairs.
So the sought probability is the complement of the probability (very small) that each birthday is represented by exactly 2 students.
The probability that Jan. 1 is represented by exactly 2 students is:
(1/365)^2 * (364/365)^728 * C(730,2)
Given the above, the probability that Jan. 2 is represented by exactly 2 students is:
(1/364)^2 * (363/364)^726 * C(728,2)
etc. through
Dec. 30:
(1/2)^2 * (1/2)^2 * C(2,2)
After that it's certain Dec. 31 falls into line.
5 P=1
10 for I=0 to 363
20 Fctr=1/(365I)^2*((364I)/(365I))^?(7282*I)*combi(7302*I,2)
30 P=P*Fctr
40 next
50 print P
evaluates this but finds the result is essentially zero, making the sought answer the complement of this, 1. To be more exact, it comes to
.999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
5303464934762603251693408597388055673175174032016877752433351478562873590196228567525980691466868940650
44294693483441139112975532973, approximately.
(The probability that a given birthdaysuch as the Jan. 1 mentioned above occurs exactly twice is about 27%. The conditional probabilities rise, as more birthdays are found to be in pairs, rising to 3/8 at the endeven split between Dec 30 and 31, above.)

Posted by Charlie
on 20141009 13:48:49 