probability that yet another person selected at random shares a birthday with the
first two is 2/365 so the probability this person does not share a birthday with
either of the first two is one minus this (or 363/365). The probability of none of
these three sharing a birthday is the multiplication of these two probabilities, or
(364/365) × (363/365) which is 0.9918.
Continuing this line of thought for 4, 5, 6,... people unravels the birthday
problem paradox. When we get as far as 23 people with our pocket calculator we
get the answer 0.4927 as the probability that none of them shares a birthday.
The negation of ‘none of them sharing a birthday’ is ‘at least two people share a
birthday’ and the probability of this is 1 – 0.4927 = 0.5073, just greater than the
crucial ½.
If n = 22, the probability of two people sharing a birthday is 0.4757, which is
less than ½. The apparent paradoxical nature of the birthday problem is bound
up with language. The birthday result makes a statement about two people
sharing a birthday, but it does not tell us which two people they are. We do not
know where the matches will fall. If Mr Trevor Thomson whose birthday is on 8
March is in the room, a different question might be asked.
How many birthdays coincide with Mr Thomson’s?
For this question, the calculation is different. The probability of Mr Thomson
not sharing his birthday with another person is 364/365 so that the probability
that he does not share his birthday with any of the other n – 1 people in the
room is (364/365)n – 1. Therefore the probability that Mr Thomson does share
his birthday with someone will be one minus this value.
If we compute this for n = 23 this probability is only 0.061151 so there is
only a 6% chance that someone else will have their birthday on 8 March, the
same date as Mr Thomson’s birthday. If we increase the value of n, this
probability will increase. But we have to go as far n = 254 (which includes Mr
Thomson in the count) for the probability to be greater than ½. For n = 254, its
value is 0.5005. This is the cutoff point because n = 253 will give the value
0.4991 which is less than ½. There will have to be a gathering of 254 people in
the room for a chance greater than ½ that Mr Thomson shares his birthday with
someone else. This is perhaps more in tune with our intuition than with the
startling solution of the classic birthday problem.