C++ Birthday Probability - c++

Question

I am trying to teach myself C++ in preparation for graduate school this coming fall but I am having some trouble with this birthday paradox problem. My code seems to run ok but I am not getting the correct output. If anyone has any suggestions please let me know.
#include <cstdlib>
#include <iostream>
#include <ctime>
using namespace std;
int main()
{
srand(time(NULL));
const int trials = 100000;
int birthdays[50];
int numMatches;
for(int i = 2; i <= 50; i++)
{
numMatches = 0;
for(int j = 1; j <= trials; j++)
{
for(int k = 1; k <= i; k++)
{
birthdays[k] = (rand() % 365) + 1;
}
int m = 1;
bool matched = false;
while(m < i && !matched){
int n = m + 1;
while(n <= i && !matched){
if(birthdays[m] == birthdays[n]){
numMatches++;
matched = true;
}
n++;
}
m++;
}
}
cout << "Probability of " << i << " people in a room sharing a birthday is \t"
<< ( float(numMatches) / float(trials) ) << endl;
}
}

Answer 1

Your code is not computing the probability of two people in a room of 50 sharing a birthday. There's several bugs, mostly with indexing, but here's the biggest issue:
for(int j = 1; j <= trials; j++) {
// assigns a random birthday to the first i people (should be 0 indexed)
for(k = 1; k <= i; k++)
birthdays[k] = (rand() % 365) + 1;
// Does *exactly* the same thing as the previous loop, overwriting what
// the initial loop did. Useless code
for(m = 1; m <= i; m++)
birthdays[m] = (rand() % 365) + 1;
// At this point, m = k = i + 1. Here you check if
// the i + 1st array value has the same b-day. It will, because they're
// the same thing. Note you never set the i + 1st value so the loops
// above did nothing
if(birthdays[k] == birthdays[m])
++numMatches;
}
So what you've got here is something like:
Perform 48 iterations of the following (from your first loop which goes from 2 to 50: no idea where those values came from)
For each of those 48 iterations, perform 10k iterations of:
assign a bunch of random stuff to an array overwriting stuff
Ignore the values you wrote in the array, do a comparison that's always true and increment numMatches by 1

Answer 2

Consider what's going on here:
for(int j = 1; j <= trials; j++) {
for(k = 1; k <= i; k++)
birthdays[k] = (rand() % 365) + 1;
for(m = 1; m <= i; m++)
birthdays[m] = (rand() % 365) + 1;
if(birthdays[k] == birthdays[m])
++numMatches;
}
You go through i birthdays and assign a random number, then you go through the same i birthdays and assign them a new random number. Then you try to find a match for just one value of k and m (which both happen to equal i+1, which isn't one of the values set!).
My suggestion is to break the problem down into smaller units that will make it easier to figure out how to code - here are the functions I would try to write.
/* randomizeBirthdays()
* Put n random birthdays into the pre-allocated array birthdays.
* birthdays must of course be of length <= n.
*/
void randomizeBirthdays(int * birthdays, int n);
/* hasMatchingBirthdays()
* Check if birthdays array has two people with the same birthday
* in the first n entries.
* Return value is boolean.
*/
bool hasMatchingBirthdays(int * const birthdays, int n);
/* probabilityOfMatch()
* Calculate the probability that at least 2 out of n people will
* have the same birthday, using nTrials number of trials.
* Return value is double.
*/
double probabilityOfMatch(int n, int nTrials);
If you break it down like this it becomes easier to write and easier to troubleshoot.

Answer 3

As I said in comments already:
I think your aim is to test if 2 people in room of 2-50 people share
birthday, not if 2-50 people share birthday as you say in output. And
that's 2 people out of 23 have 50.7%, not 24.
I completely reworked your code:
#include <iostream>
#include <cstdlib>
#include <ctime>
using namespace std;
#define DAYS_IN_YEAR 365
#define TRIALS 10000
void clearArray (bool * array)
{
for (int i = 0; i < DAYS_IN_YEAR; i++)
array[i] = false;
}
int main()
{
srand(time(NULL));
bool birthdays[DAYS_IN_YEAR]; //we are trying to hit same day in year twice
int r, numMatches;
for(int i = 2; i < 50; i++)
{
numMatches = 0;
for(int j = 0; j < TRIALS; j++)
{
clearArray(birthdays);
for(int k = 0; k < i; k++)
{
r = rand() % DAYS_IN_YEAR; // == 0-364
if (birthdays[r])
{
numMatches++;
break; // 2 people already have same birthdays here
}
birthdays[r] = true;
}
}
cout << "Probability of 2 people having same birthday in room of " << i << " people is "
<< (float)numMatches / TRIALS << endl;
}
}
Output:
Probability of 2 people having same birthday in room of 23 people is 0.516

Answer 4

I think the code must be something like this.
#include <cstdlib>
#include <iostream>
#include <ctime>
using namespace std;
int main() {
srand(time(NULL));
int birthdays[10000][50];
int numMatches;
int trials=10000,check;
for(int n=0;n<trials;n++)
{
for(int j=0;j<50;j++)
{
birthdays[n][j]=rand()%365+1;
}
}
for(int i=2;i<=50;i++)
{
numMatches=0;
for(int n=0;n<trials;n++)
{
check=1;
for(int j=0;j<i;j++)
{
for(int k=j+1;k<=i;k++)
{
if(birthdays[n][j]==birthdays[n][k]&&check)
{
numMatches++;
check=0;
}
}
}
}
cout << "Probability of " << i << " people in a room sharing a birthday is \t" <<
(static_cast<float>(numMatches) / (trials)) << endl;
}
}