TITLE: The Birthday Problem # !b2 Problem What is the chance that at least two people in a group of k people will have the same birthday?

!b4 Note: By same birthday we mean same day and month but not necessarily the same year. We assume further that the birthdays of the k people are unrelated and each equally likely to be in any of the 365 days of the year. !b2 Solution Define the proposition: !c3 E = "at least 2 equal birthdays" Let S denote the statement of the problem above. Then, the probability of E given S is, !c3 Pr(E|S) = 1 - Pr(not E |S) But !c3 (not E) = "all different birthdays" So the problem reduces to the computation of the number of ways in which k people can have different birthdays. There are three cases. First, if k > 365 then necessarily at least 2 people will have the same birthday. Second, if k < 2 then there is no problem. Third, when 1 Hence, the total number of ways is, !c3 365*364*...*(365-k+1). The total number of ways in which k people can have their birthdays (with out constraints) is, !c3 365k The probability we are looking for is then, > answer := 1 - a!/((a-k)!*a^k); 365! answer := 1 - ---------------- k (365 - k)! 365 # This is a function of k that can be easily computed with > p := k -> evalf((1 - 365!/((365-k)!*365^k)),4): > p10 := p(10); p20:=p(20); p23 := p(23); p40:=p(40); p50:=p(50); p10 := .1169 p20 := .4114 p23 := .5073 p40 := .8912 p50 := .9704 >