How can you make a random number generator that chooses between the number 2 or 3. (not 1,2,3)

Question

Hi I am working on a monty hall generator and in part of my code I need it to generate random number 2 or 3. It cannot be 1,2,3 but the computer needs to select between 2 or 3. Thanks!

I have tried randomCarDoor = ( rand() % 3 ) + 1; but does not work.

randomCarDoor = ( rand() % 3 ) + 1;

It gives me the number 1,2,3 but I just want 2 and 3

Possible duplicate of [How to generate a random int in C?](https://stackoverflow.com/questions/822323/how-to-generate-a-random-int-in-c) — anatolyg, Jan 27 '19 at 22:36
If you're not worried about *bias*, use (the number of wanted results) as the operand of `%`. In your case that's `2` for the results `2` and `3`. So `rand()%2` gives `0` or `1`; and when you add two you get `2` or `3`. — pmg, Jan 27 '19 at 22:52
Random12 = ( rand() % 2 ) + 1; would this be how to set a random number from 1-2? — Shopimo Online, Jan 27 '19 at 23:12
@ShopimoOnline Yes, `rand() % 2` will be either `0` or `1`, so the result will be `1` or `2`. — Osiris, Jan 27 '19 at 23:28
@KerrekSB: That can suffer from a pathology in that linear congruential generators tend to draw odd -> even -> odd etc. — Bathsheba, Jan 28 '19 at 07:55

JuMoGar · Answer 1 · 2019-01-27T22:47:41.267

As @Kerrek SB suggest, your formula is: random() % 2 + 2:

random() % 2 ==> Gets [0 or 1]
[0 or 1] + 2 ==> Gets [2 or 3]

A functional code is:

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>



int main()
{
    // Declare variables
    int i;
    int randomNumber;

    // Set random seed
    srand(time(NULL));

    // Get 10 random numbers between 2 and 3
    for (i = 0; i < 10; ++i){
        randomNumber = rand() % 2 + 2;
        printf("Random %d: %d\n", i, randomNumber);
    }

    // End function
    return 0;
}

chqrlie · Accepted Answer · 2019-01-28T12:19:34.497

2

You can use the low order bit of the random value, but it is very risky as some pseudo-random number generators do not provide adequate dispersion on low order bits:

int two_or_three = 2 + rand() % 2;

A much better way is to use the magnitude of the random number which is specified as having a flat distribution:

int two_or_three = 2 + (rand() >= RAND_MAX / 2);

If you want numbers 1 and 3, here is a simple solution for any pair:

int random_value = (rand() < RAND_MAX / 2) ? 1 : 3;

edited Jan 28 '19 at 12:19

answered Jan 27 '19 at 22:39

chqrlie

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Your first way is awful! (Oops!) but the second approach much better. – Bathsheba Jan 28 '19 at 07:58
@Bathsheba: yes, I reworded the answer to make it more explicit. – chqrlie Jan 28 '19 at 12:20

Bathsheba · Answer 3 · 2019-01-28T08:00:42.480

This is not trivial using rand() since a linear congruential generator typically alternates between odd and even numbers.

So one of the worst things you can do it to use a formula based on rand() % 2.

In this particular case, I suggest you draw based on

n = rand();

and call it 1 if n < RAND_MAX / 2 and 3 otherwise, which you can do with

rand() < RAND_MAX / 2 ? 1 : 3

That might have adequate statistical properties - perhaps some unwanted autocorrelation - but probably no worse than rand() itself.

How can you make a random number generator that chooses between the number 2 or 3. (not 1,2,3)

3 Answers3