## Finding pairs of dices that add up to 7

In my Algorithm class at Purdue University, I got an interesting probability question, and I brute forced it in C++ to check my answer.

Suppose we roll 5 dices, what is the expected number of pairs of dices that add up to 7?

Example: if we roll the dices and get: 1, 6, 5, 2, 6; we got 3 good pairs.

Here is the C++ code that gives the statistics:

``````#include <iostream>
#include <array>

using namespace std;

template<unsigned int N>
int compute_nb_good_pairs(const array<int, N>& dices)
{
int nb_good_pairs = 0;

for (int i = 0; i < dices.size(); i++)
{
for (int j = i + 1; j < dices.size(); j++)
{
if (dices[i] + dices[j] == 7)
{
nb_good_pairs++;
}
}
}

return nb_good_pairs;
}

int main()
{
array<int, 11> histogram = {0};

int total_nb_good_pairs = 0;
int total_nb_arrays = 0;

// Enumerate all possible outcomes with 5 dices
for (int a = 1; a <= 6; a++) {
for (int b = 1; b <= 6; b++) {
for (int c = 1; c <= 6; c++) {
for (int d = 1; d <= 6; d++) {
for (int e = 1; e <= 6; e++) {
const auto nb_good_pairs = compute_nb_good_pairs<5>({{a, b, c, d, e}});

histogram[nb_good_pairs]++;
total_nb_good_pairs += nb_good_pairs;
total_nb_arrays++;
}
}
}
}
}

// Display some interesting stats
cout << "Histogram:" << endl;
for (int i = 0; i < histogram.size(); i++)
{
cout << i << " => " << histogram[i] << endl;
}

cout << "Total number of tests: " << total_nb_arrays << endl;
cout << "Total number of good pairs: " << total_nb_good_pairs << endl;
cout << "Expected number of good pairs: " << double(total_nb_good_pairs) / total_nb_arrays << endl;

return 0;
}
``````

Output:

``````Histogram:
0 => 1566
1 => 1680
2 => 2880
3 => 1200
4 => 390
5 => 0
6 => 60
7 => 0
8 => 0
9 => 0
10 => 0
Total number of tests: 7776
Total number of good pairs: 12960
Expected number of good pairs: 1.66667
``````

The answer is 1.6667, which is equal to 5/3. Why? Probably because in a set of 5 dices, it’s possible to make 10 different pairs, and the probability of a pair of dices to add up to 7 is 1/6. So, 10/6 gives us the answer 5/3. Interestingly, it’s not possible to have exactly 5 pairs, I don’t know why, but it could be cool to investigate…