*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 ` `?> ` |

**Output :**

30

**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.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.