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A happy trio (Posted on 2014-10-09) Difficulty: 3 of 5
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

No Solution Yet Submitted by Ady TZIDON    
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Solution A better approximation | Comment 2 of 3 |
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

.999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
5303464934762603251693408597388055673175174032016877752433351478562873590196228567525980691466868940650
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.)


  Posted by Charlie on 2014-10-09 13:48:49
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