Birthday Paradox

How many people must be there in a room to make the probability 100% that at-least two people in the room have same birthday?
Answer: 367 (since there are 366 possible birthdays, including February 29). 
The above question was simple. Try the below question yourself.
How many people must be there in a room to make the probability 50% that at-least two people in the room have same birthday?
Answer: 23 
The number is surprisingly very low. In fact, we need only 70 people to make the probability 99.9 %.
Let us discuss the generalized formula.
What is the probability that two persons among n have same birthday?
Let the probability that two people in a room with n have same birthday be P(same). P(Same) can be easily evaluated in terms of P(different) where P(different) is the probability that all of them have different birthday.
P(same) = 1 – P(different)
P(different) can be written as 1 x (364/365) x (363/365) x (362/365) x …. x (1 – (n-1)/365)
How did we get the above expression?
Persons from first to last can get birthdays in following order for all birthdays to be distinct: 
The first person can have any birthday among 365 
The second person should have a birthday which is not same as first person 
The third person should have a birthday which is not same as first two persons. 
……………. 
…………… 
The n’th person should have a birthday which is not same as any of the earlier considered (n-1) persons.
Approximation of above expression
The above expression can be approximated using Taylor’s Series. 

provides a first-order approximation for ex for x << 1:

To apply this approximation to the first expression derived for p(different), set x = -a / 365. Thus,

The above expression derived for p(different) can be written as 
1 x (1 – 1/365) x (1 – 2/365) x (1 – 3/365) x …. x (1 – (n-1)/365)
By putting the value of 1 – a/365 as e^-a/365, we get following.



Therefore,
p(same) = 1- p(different) 

An even coarser approximation is given by
p(same) 

By taking Log on both sides, we get the reverse formula.

Using the above approximate formula, we can approximate number of people for a given probability. For example the following C++ function find() returns the smallest n for which the probability is greater than the given p.
Implementation of approximate formula.
The following is program to approximate number of people for a given probability. 

C++

// C++ program to approximate number of people in Birthday Paradox
// problem
#include <cmath>
#include <iostream>
using namespace std;

// Returns approximate number of people for a given probability
int find(double p)
{
    return ceil(sqrt(2*365*log(1/(1-p))));
}

int main()
{
   cout << find(0.70);
}

Java

// Java program to approximate number
// of people in Birthday Paradox problem
class GFG {

    // Returns approximate number of people
    // for a given probability
    static double find(double p) {

        return Math.ceil(Math.sqrt(2 *
            365 * Math.log(1 / (1 - p))));
    }

    // Driver code
    public static void main(String[] args)
    {

        System.out.println(find(0.70));
    }
}

// This code is contributed by Anant Agarwal.

Python3

# Python3 code to approximate number
# of people in Birthday Paradox problem
import math

# Returns approximate number of
# people for a given probability
def find( p ):
    return math.ceil(math.sqrt(2 * 365 *
                     math.log(1/(1-p))));

# Driver Code
print(find(0.70))

# This code is contributed by "Sharad_Bhardwaj".

C#

// C# program to approximate number
// of people in Birthday Paradox problem.
using System;

class GFG {

    // Returns approximate number of people
    // for a given probability
    static double find(double p) {

        return Math.Ceiling(Math.Sqrt(2 *
            365 * Math.Log(1 / (1 - p))));
    }

    // Driver code
    public static void Main()
    {
    Console.Write(find(0.70));
    }
}

// This code is contributed by nitin mittal.

PHP

<?php
// PHP program to approximate
// number of people in Birthday
// Paradox problem

// Returns approximate number
// of people for a given probability
function find( $p)
{
    return ceil(sqrt(2 * 365 *
                     log(1 / (1 - $p))));
}

// Driver Code
echo find(0.70);

// This code is contributed by aj_36
?>

Javascript

<script>

// JavaScript program to approximate number
// of people in Birthday Paradox problem

// Returns approximate number of
// people for a given probability
function find( p){
    return Math.ceil(Math.sqrt(2*365*Math.log(1/(1-p))));
}
document.write(find(0.70));

</script>

Output :

30

Time Complexity: O(log n)

Auxiliary Space: O(1)

Source:
http://en.wikipedia.org/wiki/Birthday_problem
Applications:
1) Birthday Paradox is generally discussed with hashing to show importance of collision handling even for a small set of keys. 
2) Birthday Attack
Below is an alternate implementation in C language : 

C

#include<stdio.h>
int main(){

    // Assuming non-leap year
    float num = 365;

    float denom = 365;
    float pr;
    int n = 0;
    printf("Probability to find : ");
    scanf("%f", &pr);

    float p = 1;
    while (p > pr){
        p *= (num/denom);
        num--;
        n++;
    }

    printf("\nTotal no. of people out of which there "
          " is %0.1f probability that two of them "
          "have same birthdays is %d ", p, n);

    return 0;
}

C++

// CPP program for the above approach
#include <bits/stdc++.h>
using namespace std;


int main(){

    // Assuming non-leap year
    float num = 365;

    float denom = 365;
    float pr;
    int n = 0;
    cout << "Probability to find : " << endl;
    cin >> pr;

    float p = 1;
    while (p > pr){
        p *= (num/denom);
        num--;
        n++;
    }

    cout <<  " Total no. of people out of which there is " << p
    << "probability that two of them have same birthdays is  "  << n << endl;

    return 0;
}

// This code is contributed by sanjoy_62.

Java

class GFG{
  public static void main(String[] args){

    // Assuming non-leap year
    float num = 365;

    float denom = 365;
    double pr=0.7;
    int n = 0;


    float p = 1;
    while (p > pr){
      p *= (num/denom);
      num--;
      n++;
    }

    System.out.printf("\nTotal no. of people out of which there is ");
    System.out.printf( "%.1f probability that two of them "
                      + "have same birthdays is %d ", p, n);
  }
}

// This code is contributed by Rajput-Ji

Python3

if __name__ == '__main__':

    # Assuming non-leap year
    num = 365;

    denom = 365;
    pr = 0.7;
    n = 0;

    p = 1;
    while (p > pr):
        p *= (num / denom);
        num -= 1;
        n += 1;


    print("Total no. of people out of which there is ", end="");
    print ("{0:.1f}".format(p), end="")
    print(" probability that two of them " + "have same birthdays is ", n);

# This code is contributed by Rajput-Ji

C#

using System;
public class GFG {
  public static void Main(String[] args) {

    // Assuming non-leap year
    float num = 365;

    float denom = 365;
    double pr = 0.7;
    int n = 0;

    float p = 1;
    while (p > pr) {
      p *= (num / denom);
      num--;
      n++;
    }

    Console.Write("\nTotal no. of people out of which there is ");
    Console.Write("{0:F1} probability that two of them have same birthdays is {1} ", p, n);
  }
}

// This code is contributed by Rajput-Ji

Javascript

<script>

        // Assuming non-leap year
        var num = 365;

        var denom = 365;
        var pr = 0.7;
        var n = 0;

        var p = 1;
        while (p > pr) {
            p *= (num / denom);
            num--;
            n++;
        }

        document.write("\nTotal no. of people out of which there is ");
        document.write(p.toFixed(1)+" probability that two of them " + "have same birthdays is "+ n);

// This code is contributed by Rajput-Ji
</script>

Time Complexity: O(log n)

Auxiliary Space: O(1)

This article is contributed by Shubham. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.