Statistical-logical problem with rand ()

Do not use rand() !

If you are looking for precision in the randomness distributions, leave rand() . Using rand() is not only inappropriate because it is a C language utility (being your question about C ++) but it is also because the use you are giving is distorting the distribution.

Distribution of rand() .

The rand() function returns an integer, pseudorandom number between 0 and RAND_MAX (both included). The distribution of the resulting values to call rand() is uniform , that is: all the numbers between 0 and RAND_MAX have the same probability of being obtained.

Break the distribution of rand() .

Assuming that the interval of rand() is [0, 32767], if the result you operate it with modulus ( % ) on a number that is not divisor of the maximum value of the interval (in your case 100 ) you will be falsifying the distribution: since the remainder of dividing 32767 by 100 is 67, the numbers from 0 to 67 are more likely to appear than numbers 68 to 99.

What should you do?

As I said, you should stop using rand() , which besides being not a C ++ utility, you were using it badly! The appropriate way to approach your problem is to use <random> the library of pseudo-random C ++ numbers that allows you to choose the probability distribution (uniform, Bernoulli, Poisson, normal, discrete, constant, linear ...) , the underlying type of the generated value and even the algorithm to be used (minstd, mt19937, ranlux, knuth ...).

In your case, since your probabilities are divided in two:

For the first probability, you could use a Bernoulli distribution , which allows you to set the probability of an event (take out ball) happen:

std::random_device device;
std::mt19937 generador(device());
std::bernoulli_distribution distribucion(0.05);

if (distribucion(generador));
{
    // Entramos en este if un 5% de las veces.
}

For the second probability, you could use a discrete distribution , which allows you to distribute the weight of each probability (the probability of appearance of each ball):

std::random_device device;
std::mt19937 generador(device());
std::discrete_distribution<> distribucion({15, 15, 30, 40});

switch(distribucion(generador))
{
    case 0: std::cout << "Pelota dorada\n"; break;
    case 1: std::cout << "Pelota roja\n"; break;
    case 2: std::cout << "Pelota azul\n"; break;
    case 3: std::cout << "Pelota verde\n"; break;
}

You can see an example of this working in Wandbox 三 へ (へ ਊ) へ ハ ッ ハ ッ .

The odds, as you comment, are:

• 15% gold
• 15% red
• 30% blue
• 40% green

You can verify that the sum gives 100.

So dividing the random number by 100 is relatively correct.

Now then. How do we distribute the odds? Very easy:

enum Color
{
  Indefinido,
  Dorado,
  Rojo,
  Azul,
  Verde
};

for(d=0;d<5;d++){
  Color color = Indefinido;
  rnd=rand() % 100;
  if( rnd < 15 )
    color = Dorado;
  else if( rnd < 30 )
    color = Rojo;
  else if( rnd < 60 )
    color = Azul
  else
    color = Verde;

  // Hacer algo con el color
  // ...
}

How does it work? It's simple to understand:

• If the number is less than 15 (between 0 and 14, that is, 15 values over 100) it is Golden
• If the number is greater than 14 and less than 30 (that's why the else if ), then it's Red (15-29 are another 15 values out of 100).
• The number is less than 60 then it is blue (Range 30-59 -> 30 values)
• If none of the above is met, that is, if the number is equal to or greater than 60 then the ball is green (Range 60-99 -> 40 values).