- #1

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## Homework Statement

Write a program that approximates the probability that at least two people in a room of 50 people have the same birthday. Run 5,000 trials, count up the number at least two people have the same birthday, and divide by the number of trials to get a probability.

## Homework Equations

## The Attempt at a Solution

Code:

```
#include <iostream>
#include <ctime>
#include <cstdlib>
using namespace std;
int bDayParadox (int people[]);
///////////////////////////////////////////////////
int main ()
{
int probability;
int sum = 0;
const int trials = 5000;
int people[49];
for (int j = 0; j < trials; j++)
{
sum = sum + bDayParadox(people);
}
probability = (sum / trials);
cout << bDayParadox(people) << endl;
cout << sum << endl;
cout << trials << endl;
cout << probability << endl;
}
/////////////////////////////////////////////////
int bDayParadox (int people[])
{
int coincidence = 0;
srand(time(NULL)); //generates seed for random numbers
for (int a = 0; a < 50; a++)
{
people[a] = rand() % 365 + 1;
}
for (int b = 0; b < 50; b++)
{
for (int c = b+1; c < 50; c++)
{
if (people[c] == people[b])
{
coincidence = 1;
}
}
}
return coincidence;
}
```

The function I made, bDayParadox, puts random numbers (from 1 - 365) into an array of 50. The function then checks to see if at least two of the values are similar, and if so, counts it as 1 (coincidence variable). Up top, I'm running the function 5000 times and adding the values together to get sum. Sum divided by trials should give me the probability.

My output looks like this:

1 (bDayParadox run once)

5000 (sum)

5000 (trials, the constant)

353 (probability)

The value that is spit out for probability is unique each time.

My main problem with this is that probability should not be what it says. Also, my sum shouldn't always be 5000.