*How many people must be there in a room to make the probability 100% that 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 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.

**C++ 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); }

## 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".

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

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.