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Birthday Paradox in C++
The birthday paradox is a very famous problem in the section of probability. The problem statement of this problem is stated as,
There are several people at a birthday party, some are having the same birthday collision. We need to find the approximate no. of people at a birthday party on the basis of having the same birthday.
In the probability we know that the chance of getting ahead is 1/2 same as if we have some coins, the chance of getting 10 heads is 1/100 or 0.001.
Let us understand the concept,
The chance of two people having the different birthday is,
364/365 which is 1-1/365 in a Non-leap year.
Thus we can say that the first person having the probability of a specific birthday is ‘1’ and for others, it would be different which is,
P(different)= 1×(1-1/365)× (1-2/365)× (1-3/365) × (1-4/365).......
Thus P(same)= 1- P(different)
For Example,
No. of people having the same birthday for which probability is 0.70.
N= √2×365×log(1-1/p).
N= √2×365×log(1-1/0.70)= 30
Thus total approximate no. of people having the same birthday is 30.
Example
#include<bits/stdc++.h>
using namespace std;
int findPeople(double p){
return ceil(sqrt(2*365*log(1/(1-p))));
}
int main(){
printf("%d",findPeople(0.70));
}Output
30
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