My code

… was written in C++ and can be compiled with G++, Clang++, Visual C++. You can download it, too. Or just jump to my GitHub repository.

       
#include <iostream>
       
#include <iomanip>
       
#include <vector>
       
 
       
// roll dices, store at count[x] how often the sum of all dices was exactly x

       
void roll(unsigned int dices, unsigned int sides, std::vector<unsigned int>& count, unsigned int sum = 0)
       
{
       
  // rolled all dices, increment the sum's counter
       
  if (dices == 0)
       
  {
       
    count[sum]++;
       
    return;
       
  }
       
 
       
  // all combinations of a dice ...
       
  for (unsigned int i = 1; i <= sides; i++)
       
    roll(dices - 1, sides, count, sum + i);
       
}
       
 
       
int main()
       
{
       
  unsigned int dicesPeter, sidesPeter;
       
  std::cin >> dicesPeter >> sidesPeter; // 6 6
       
  unsigned int dicesColin, sidesColin;
       
  std::cin >> dicesColin >> sidesColin; // 9 4
       
 
       
  // "high score", by default 9*4 = 6*6 = 36
       
  unsigned int maxTotal = std::max(dicesPeter * sidesPeter, dicesColin * sidesColin);
       
 
       
  // roll all 6x6 dices (Colin)
       
  std::vector<unsigned int> colin(maxTotal + 1, 0);
       
  roll(dicesPeter, sidesPeter, colin);
       
  unsigned int sumColin = 0; // will be  46656 for 6,6
       
  for (auto x : colin)
       
    sumColin += x;
       
 
       
  // roll all 9x4 dices (Peter)
       
  std::vector<unsigned int> peter(maxTotal + 1, 0);
       
  roll(dicesColin, sidesColin, peter);
       
  unsigned int sumPeter = 0; // will be 262144 for 9,4
       
  for (auto x : peter)
       
    sumPeter += x;
       
 
       
  // for each of Peter's potential dice sums, find how often it wins against Colin
       
  double winPeter = 0;
       
  for (unsigned int sum = 1; sum <= maxTotal; sum++)
       
  {
       
    // win if total is higher
       
    unsigned int numWins = 0;
       
    // => add all combinations of Colin that are smaller
       
    for (unsigned int j = 1; j < sum; j++)
       
      numWins += colin[j];
       
 
       
    // compute percentage that current sum (Peter) wins against Colin
       
    double beats = numWins / double(sumColin);
       
    // add all of them ...
       
    winPeter += beats * peter[sum] / sumPeter;
       
  }
       
 
       
  // done !
       
  std::cout << std::fixed << std::setprecision(7) << winPeter << std::endl;
       
  return 0;
       
}

This solution contains 9 empty lines, 12 comments and 3 preprocessor commands.