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?
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 â€¦. x (1 â€“ (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 #include using namespace std; // Returns approximate number of people for a given probabilityint 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 problemclass 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 problemimport math # Returns approximate number of # people for a given probabilitydef find( p ):    return math.ceil(math.sqrt(2 * 365 *                     math.log(1/(1-p)))); # Driver Codeprint(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

 

## Javascript

 

Output
30

Time Complexity: O(log n)
Auxiliary Space: O(1)

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 implementation:

## C

 #includeint 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 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

 

Output
Probability to find :
Total no. of people out of which there  is 0.0 probability that two of them have same birthdays is 239 

Time Complexity: O(log n)
Auxiliary Space: O(1)

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