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++ 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 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 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# 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 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 ?> |
<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 implementation:
#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;
} |
// 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. |
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 |
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 |
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 |
<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> |
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)