What is the Probability of Two Sharing Birthday?

Assume a year has 365 days, what is the probability among 50 students, that at least two having the same birthday?

Let’s consider this: what is the probability that all 50 students are having different birthdays?

.

P is the permutation function, so P(365, 50) is the total permutation of choosing 50 out of 365. We are choosing 50 days out of a year which is the students having different birthdays. The total number of possibilities is 365⁵⁰.

So the answer will be 1 – 0.03 = 97%.

Let’s consider this: what is the probability that all only two (exactly two) share the birthday?

Let’s solve this step by step:

1. Pick two out of 50 students, which is C(50, 2) i.e. C is the combination function.
2. Pick one out of 365 days, which is used as the same birthday, that is 365 solutions.
3. Among the 48 students, their birthdays should be among the rest 364 days, which is P(364, 48)

Therefore, the probability that all only two (exactly two) share the birthday is:

C++ solution to compute the probability

Let’s create a simple console application in C/C++.

1
2
3
4
5
6
7
8
9
10
11
12
13

// birthday.c
// https://helloacm.com/what-is-the-probability-of-two-sharing-birthday/
#include <stdio.h>
 
int main() {
  double r = 1.0;
  for (int i = 0; i < 48; i ++) {
     r *= (364 - i) / 365.0;
  }
  r = r * (25.0 * 49) / 365.0;
  printf("%.5lf\n", r);
  return 0;
}

// birthday.c
// https://helloacm.com/what-is-the-probability-of-two-sharing-birthday/
#include <stdio.h>

int main() {
  double r = 1.0;
  for (int i = 0; i < 48; i ++) {
     r *= (364 - i) / 365.0;
  }
  r = r * (25.0 * 49) / 365.0;
  printf("%.5lf\n", r);
  return 0;
}

Compile above using gcc birthday.c -o birthday. Run ./birthday:

0.11485

That means the probability of exactly two out of 50 having the same birthday is 11.485%.

Useful Link

• Birthday Problem (Wiki)

–EOF (The Ultimate Computing & Technology Blog) —