Birthday Paradox

Difficulty Level : Medium
Last Updated : 13 Apr, 2021

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.
$e^{x}=1+x+\frac{x^{2}}{2!}+...$
provides a first-order approximation for ex for x << 1:
$e^{x}\approx 1+x$
To apply this approximation to the first expression derived for p(different), set x = -a / 365. Thus,
$\Large{e^{\frac{-a}{365}}\approx 1-\frac {a}{365}}$
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.
$\approx 1\times e^{\frac{-1}{365}}\times e^{\frac{-2}{365}}...\times e^{\frac{-(n-1)}{365}}$
$\approx 1\times e^{\frac{-(1+2+...+(n-1))}{365}}$
$\approx 1\times e^{\frac {-(n(n-1))/2}{365}}$
Therefore,
p(same) = 1- p(different)
$\approx 1-e^{-n(n-1)/(2 \times 365)}$
An even coarser approximation is given by
p(same)
$\approx 1-e^{-n^{2}/(2 \times 365)}$
By taking Log on both sides, we get the reverse formula.
$n \approx \sqrt{2 \times 365 ln\left ( \frac{1}{1-p(same)} \right )}$
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 :

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;
}

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.

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