Manipulating odds with JS when using Math.random ()?

Question

Manipulating odds with JS when using Math.random ()?

When using this function Math.floor((Math.random() * 10) + 1) i get a random number from 1 to 10. We assume that each number has a 10% chance so we have a 50% chance for odd or even, but what if I wanted to manipulate those chances so that the function is more likely to give me an odd number, for example 70% chance of getting an odd number and 30% chance of getting an even number.

is it possible to do this?
What would be the logic applied and how can I do it?

4

javascript random

Author: Leo Letto, 2017-04-04

Source

4 answers

score 2 · Answer 1

The algorithm below consists of two parts:

numTipo is generated: its value will be 0 if the return of a random generation from 1 to 10 is greater than 8, or 1 if negative; and
numFinal simply contains a value between 2 and 100 that will always be even , where numTipo is subsequently added.

Adding an odd value to an even value will always generate an odd value. Any other case of addition (even + even, odd + odd) generates a number pair.

This algorithm forces the generation of values whose average distribution is 70% odd and 30% even.

function geraNum() {

  var numTipo = Math.floor((Math.random() * 10) + 1) > 7 ? 0 : 1;
  var numFinal = Math.floor((Math.random() * 50) + 1) *2 + numTipo;

  return numFinal;
}

function geraSerie() {

    var contaPar = 0;
    var contaImpar = 0;

    for (i = 0; i < 1000; i++) { 
        
        if (geraNum() % 2 == 0) {
            contaPar++;
        } else {
            contaImpar++;
        }
    }
    
    console.log("Pares: " + contaPar + ", Ímpares: " + contaImpar);    
}

geraSerie();
geraSerie();
geraSerie();
geraSerie();
geraSerie();

score 0 · Answer 2

The name itself already Says ramdom(), that is, Returns a random number. If you want to change the chances of the returned number being more even or being more odd, you can create a simple function to check the return of the Code:

let myNumber = Math.floor((Math.random() * 10) + 1);
if (myNumber % 2 == 0) { 
  // par = faça alguma coisa
} else {
  // ímpar = faça alguma coisa
}

However, since it is a random return, only with the if's you can manipulate the result.

Another Way Is You create your own function (Ex: Math.myRandomFunction()) and in it you create your logic yourself, storing information and forcing the return to be more even or be more odd.

score 0 · Answer 3

It would be more interesting for you to say which problem exactly you want to solve. But, according to your example, you can first make a draw to determine whether the number you want is even or odd. Then, home in case, call a function that generates a number only in the pattern you want. For example:

if (x > 7) {
    return gerarNumeroPar();
} else {
    return gerarNumeroImpar();
}

This function can be done by selecting from a previously generated list or even by drawing successive times until you hit with what you need to generate.

score -2 · Answer 4

I did not understand very well, but if the goal is to be more conducive to a result, it can be done this way:

if(valor > 7){
    // 30% de chance de cair aqui 
}else{
    // 70% de chance de cair aqui 
}