What is the mathematical equation/formula to calculate the probability of this statistics problem: if you take 42 random people, what is the chance that 2 of those people will have the same birthday?

For the sake of the argument, let's assume that there are 365 days in a year. It is obvious that: P(at least 2 people have the same birthday) = 1 - P(all have different birthdays) Let's calculate the probability of the event "all have different birthdays": Again, there are 365 days in a year. The first person of the 42 can have any day of the 365 days as his birthday. The second person, however, cannot have a birthday that's the same as that of the first person, so his birthday can be any 364 days of the 365 days. And the third person's birthday can be any 363 days of the 365 days... You get the idea! Now because there are 42 people, and according to the multiplication rule of independent events, we have: P(all have different birthdays) = 365/365 * 364/365 * 363/365 * ...... * 325/365 * 324/365 (Note that there are 42 terms, corresponding with the 42 people) .. = (365*364*..*324) / 365^42 .. = (365!/323!) / 365^42 Now just type that into your calculator, you get: P(all have different birthdays) = 0.0859695... which means that the probability of the event "at least two people have the same birthday" is: P(at least 2 people have the same birthday) = 1 - P(all have different birthdays) = 1 - 0.0859695 = 91.4% Yep, there you have it. A whopping 91.4%!

Well, if we think of the probability of a pair of birthdays in a set of x members as P(x), its easier to calculate ~P(x). P(x)=1-~P(x) If the group size is: 1, the probability is 0, 2, ~P(2)=364/365 3, ~P(3)=363/365*364/365 4, ~P(4)=362/365*363/365*364/365 . . . x, ~P(x)=364!/((365-x)!*365^x) So P(x)=1-364!/((365-x)!*365^x)

Its quite straight forward. Assume there are n random dudes under consideration. The probability we are interested is P(n) = the probability of at least 2 dudes among the n having the same birthday. But its complement P'(n) is much more easier to calculate. P'(n) = No two dudes among the n dudes has same birthday.         = [math]\frac{365}{365}*\frac{365-1}{365}*\frac{365-2}{365}*..*\frac{365-(n-1)}{365}[/math]         = [math]\frac{365!}{(365-n)!*365^n}[/math] Of course. P(n)     = 1-P'(n) In your case,p(42) =  1 - 0.086                            =  0.914