I\'m having some trouble with this one. I\'ve been trying to figure it out all w

Question

I'm having some trouble with this one. I've been trying to figure it out all week.

Using a hash set to do a Monte Carlo analysis of the birthday paradox. The paradox states
that the odds of 2 people in a group sharing a birthday is surprisingly high.
A Monte Carlo analysis is using random numbers to simulate actual probable outcomes.
Test your code for various numbers of people.
Here is pseudo code of the algorithm:
Set number of collisions to zero
Loop (10 to100 by 10)
Loop:
Generate a birthday. (use 1-365)
See if it is already in the set. If it is count as a collision and end the loop.
Add to the set.
Print out the number of collisions as a probability P.
For example, if you ran 50 people and got 25 collisions P = 0.5
The probability of 100 is around 1.0. P can never exceed 1.0. If you exceed 1.0 you are
counting collisions in a group more than once.

Example Output:
--------------------Configuration: <Default>--------------------
After 100000 tests there were 11446 occurrences of shared birthdays in a set of 10 people.
Probability: 0.11446
After 100000 tests there were 41031 occurrences of shared birthdays in a set of 20 people.
Probability: 0.41031
After 100000 tests there were 70455 occurrences of shared birthdays in a set of 30 people.
Probability: 0.70455
After 100000 tests there were 89061 occurrences of shared birthdays in a set of 40 people.
Probability: 0.89061
After 100000 tests there were 97043 occurrences of shared birthdays in a set of 50 people.
Probability: 0.97043
After 100000 tests there were 99454 occurrences of shared birthdays in a set of 60 people.
Probability: 0.99454
After 100000 tests there were 99926 occurrences of shared birthdays in a set of 70 people.
Probability: 0.99926
After 100000 tests there were 99986 occurrences of shared birthdays in a set of 80 people.
Probability: 0.99986
After 100000 tests there were 99999 occurrences of shared birthdays in a set of 90 people.
Probability: 0.99999
After 100000 tests there were 100000 occurrences of shared birthdays in a set of 100 people.
Probability: 1.0
Process completed.

Explanation / Answer

Here you go :)

import java.util.HashSet;
import java.util.Random;
import java.util.Set;

public class MonteCarlo
{
   public static void main(String[] args)
   {
       int collisions=0;
       HashSet<Integer> bag=new HashSet<Integer>();
       Random r=new Random();
       for(int i=10;i<=100;i=i+10)
       {
           for(int j=0;j<100000;j++)
           {
               for(int k=0;k<i;k++)
               {
                   int day=r.nextInt(365)+1;
                   if(bag.contains(day)){collisions++;break;}
                   else bag.add(day);
               }
               bag.clear();
           }
           System.out.println("After 100000 tests there were "+collisions+" occurrences of shared birthdays in a set of "+i+" people.");
           System.out.println("Probability: "+(float)collisions/100000);
           collisions=0;  
       }
       System.out.println("Process completed.");
   }
}

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