Calculating Birthday Probability for large numbers

Question

The probability that two people have the same birthday in a room full of n people is 1-p. Where:

p = 365! / 365^n(365 - n)!

Obviously the numbers will be too big to solve this equation, what is a creative way to go about this?

I already solved this in a different way using simulation, but I figured the formula might be more elegant.

Answer 1

Holly Macaroni! What a show!

Anyway, right way to compute such things with large intermediates is to log() them

p = exp(log(p))

log(p) = log(365!) - n*log(365) - log((365 - n)!)

For factorial, use Gamma function, G(n+1)=n!, and there is very handy function in C library which computes log(G(x)): lgamma(x)

No more loops, no long doubles, no bignum libraries, no overflows...

Code

#include <math.h>
#include <stdio.h>

double b(int n) {
    double l = lgamma(365.0 + 1.0) -
               (double)n * log(365.0) -
               lgamma(365.0 - (double)n + 1.0);

    return exp(l);
}

int main() {
    double p = b(20);
    printf("%e %e\n", p, 1.0 - p);

    return 0;
}

Answer 2

You can take advantage of 365!/(365-n)! = 365 * 364 * ... * (365-(n-1))

So to calculate this term ( let it be A=365!/(365-n)! ) you can simply the above numbers like this:

unsinged double A=1; // to make sure there is no overflow
for(int i=0;i<n;i++) A*=365-i;

To take it one step further : p=A/365^n = (364*363*...*(365-(n-1)))/365^(n-1)= 364/365 * 363/365 * ... (365-(n-1))/365.

so p can be calcuated like this:

unsigned double p=1;
for(int i=0;i<n;i++) p*= (365-i)/365.0;

in linear time

I think this should work :P

Answer 3

You don't want to calculate the full factorial. Instead, calculate each term and multiply to the result.

The probability you don't share a birthday with:

• 1 person: 364/365
• 2 people: 364/365 * 363/365
• 3 people: 364/365 * 363/365 * 362/365
• ...

Given this, you calcuate p as follows.

int n = 30;
int i;
double p = 1;
for (i = 1; i < n; i++) {
    p *= (365 - i) / 365.0;
    printf("i=%d, p=%f\n", i, 1-p);
}