What is the probability that 2 out of 15 students will have the same birthday?

Answer 1

P( 2 equal birthday in group of 15 ) ≈ 22.3%

The probability that 2 persons and only 2 persons in a random group of n persons can be calculated using the following expression*:

P(2 equal bd in group of n persons) = nC2(1-1/1461)(4/1461)Π2n-1[1-4(i-1)/1461] + nC2(1/1461)2Π2n-1[1-(4i-3)/1461]

with n=15, 30C2 = 105,

P(2 eq bd in group of 15) = 105(1-1/1461)(4/1461)[1-4/1461]

[1-8/1461][1-12/1461]∙∙∙[1-52/1461]

P(...) = 0.223263549... ≈ 22.3%

*You find this expression discussed in question "What is the probability that in a room of 8 people 2 have the same birthday ?". A simpler form of this expression that neglects February 29 of the leap day that gives a good

approximation is:

P(2 eq bd in group of n persons) = nC2(1/365)Π1n-1[1-(i-1)/365]

for n=15, this expression gives P(...) = 0.214918798... ≈ 21.5%