Bernoulli random variables in c++

Question

What is the easiest way to get a Bernoulli distributed random variable in C++? I I have a non-const p. p starts from 0, getting p+=0.01 , until p=1. ("for" loop) I have this idea:

declare int aray in size of 100, intilized with zeros. evrey time the p gets +0.01, I change one zero to "1"... and get a random number%100.

is it good? thanks

By using one of the already existing [Bernoulli distributions](http://en.cppreference.com/w/cpp/numeric/random#Bernoulli_distributions)? — Some programmer dude, May 08 '15 at 16:22
Did you try [`std::bernoulli_distribution`](http://en.cppreference.com/w/cpp/numeric/random/bernoulli_distribution)? — Mike Seymour, May 08 '15 at 16:22

score 4 · Accepted Answer · answered May 08 '15 at 17:43

You should use modern C++ (C++11 or later if possible). You can do this:

#include <random>
#include <iostream>

int main()
{
    std::random_device rd{}; // use to seed the rng 
    std::mt19937 rng{rd()}; // rng

    double p = 0.2; // probability
    std::bernoulli_distribution d(p);

    // generate 5 runs
    for(std::size_t i = 0; i < 5; ++i)
        std::cout << d(rng) << " ";
}

score 3 · Answer 2 · answered May 08 '15 at 16:23

3

you can get a random number between 0 and 1, and calculate the bernoulli random number from it

double p = 0.5;
double r = ((double) rand() / (RAND_MAX));
unsigned int br = 0;
if (r >= p) 
   br = 1;

answered May 08 '15 at 16:23

dlavila

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Bernoulli random variables in c++

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