Random numbers with different probabilities [duplicate]

Answer 1

You can easily implement this using the rand function:

bool TrueFalse = (rand() % 100) < 75;

The rand() % 100 will give you a random number between 0 and 100, and the probability of it being under 75 is, well, 75%. You can substitute the 75 for any probability you want.

Answer 2

Check at the C++11 pseudo random number library.

http://en.cppreference.com/w/cpp/numeric/random

http://en.wikipedia.org/wiki/C%2B%2B11#Extensible_random_number_facility

std::random_device rd;
std::uniform_int_distribution<int> distribution(1, 100);
std::mt19937 engine(rd()); // Mersenne twister MT19937

int value=distribution(engine);
if(value > threshold) ...

like this, then say all above 75 is true, and all below is false, or whatever threshold you want

Unlike rand you can actually control the properties of the number generation, also I don't believe rand as used in other answers (using modulo) even presents a uniform distribution, which is bad.

Answer 3

std::random_device or boost::random if std::random_device is not implemented by your C++ compiler, using a boost::random you can use bernoulli_distribution to generate a random bool value!

Answer 4

#include <cstdlib>
#include <iostream>
using namespace std;

bool yesOrNo(float probabilityOfYes) {
  return rand()%100 < (probabilityOfYes * 100);
}

int main() {
    srand((unsigned)time(0)); 
    cout<<yesOrNo(0.897);
}

if you call yesOrNo(0.75) it will return true 75% of the time.