Find the least positive integer 'N' so that, in a room containing 'N' people, the odds that two of the people have the same birthday (not the year, just the day) are more than 50%.
[Note: Forget about Leap Year. Assume that a year has 365 days and that people are born randomly throughout the year].
(In reply to
Puzzle Answer by K Sengupta)
The probability that a person does not have the same birthday as another person is 364/365
In a group of N people there are comb(N, 2) comparisons (or, combinations). Each one of the N(N-1)/2 comparisons has the probability of not being a match.
Therefore, for N people, the probability of not a match is:
(364/365)^(N(N-1)/2)
So, the probability of a match is:
1 - (364/365)^(N(N-1)/2)
By the given problem:
1 - (364/365)^(N(N-1)/2) =1/2
=> (364/365)^(N(N-1)2) =1/2
=> N(N-1)/2 = (log 2)/log(365/364) ~= 252.6519888
=> N ~= 22.9845
=> N ~= 23
Consequently, the least value of the positive integer N is 23.
As a CHECK:
1 - (364/365 )^253
~= 1 - 0.49952
= 0.50048
which is indeed very close to 1/2.
Explanation to Puzzle Answer
