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/(365-I)^2*((364-I)/(365-I))^?(728-2*I)*combi(730-2*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

44294693483441139112975532973, approximately.

(The probability that a given birthday--such 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 end--even split between Dec 30 and 31, above.)