Question

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Expand a random range from 1-5 to 1-7

I understood the solution using rejection sampling i.e

public static int rand7() {
    while (true) {
       int num = 5 * (rand5() - 1) + (rand5() - 1);
       if (num < 21) return (num % 7 + 1);
    }
}

but I am thinking of another solution, i.e rand5() is called 7 times and result is divided by 5, but I am not sure whether this is correct. Please let me know if is or isn’t.

public static int rand7() {    
    int num = rand5()+rand5()+rand5()+rand5()+rand5()+rand5()+rand5();
    return num/5;
}

EDIT: It looks like the probability of generating 1 is (1/5)^7 but to generate 2 it is 7*(1/5)^7. It is uneven so it is not going to work.

Answer 1

While that other solution will generate a random number in the range 1..7, it will do so with a non-uniform probability distribution, i.e. the number 3 will be a lot more likely than the number 1.

In contrast, the rejection sampling approach will return all numbers in the range 1..7 with equal probability.