Generate integer from 0 to 1 with equal probability

Question

I am trying to get around this problem of my own making. I want to generate 0 or 1 with another function(gr0_4()) which generates random number from 0 to 4.

I am wondering if I can approach this way:

a). if gr0_4() = 0 or 1 then I will use 0

b). if gr0_4() = 2 or 3 then I will use 1

c). if gr0_4() = 4 I will repeat the a) and b) steps.

Is my understanding correct that the "a" and "b" step each has 50% probability of happening?

def gr0_1():
  while True:
    x = gr0_4()
      if x == 0 or x == 1:
        return 0
      elif x == 0 or x == 1:
        return 1

What if I want to use gr0_1() to create gr1_7() i.e. create number between 1 to 7 with equal probability?

Can I use below reasoning to create that function gr1_1().

As 7 consists of 3 bits. I can generate each bit with equal probability using gr0_1(). So I will call gr0_1() three times and based on that value I get, I will set/unset the corresponding bits to generate a number between 1 to 7 including the numbers 1 and 7. However I can get the number 0 but I don't want that so I will repeat the process again. Will the probability of each number generation between 1 to 7 will be 1/7 in that case also?

Some simple mathematical calculation will be nice to answer this. I tried to read up on rejection sampling but couldn't understand much.

Answer 1

Yes, this is exactly rejection sampling and it works in the way you think it works, assuming that your initial procedure gr_04() generates the numbers $0$–$4$ with equal probability (you say it's random but you don't say it's uniform).

If $X$ is distributed uniformly on $\{0, \dots, 4\}$, then $\Pr(x\in\{0,1\}) = \Pr(x\in\{2,3\}) = 2/5$. With probability $1/5$, you'll have to try again but, on your second attempt, the probability of getting $0$ or $1$ is still equal to the probability of getting $2$ or $3$. The number of attempts you have to make before you get something in $\{0, \dots, 3\}$ is just a geometric random variable with parameter $p=4/5$ and this has expectation $(1-p)/p = 1/4$ so, on average, you'll have to make $5/4$ calls to gr_04(), including the one that succeeds.

The analysis of your second example is similar: your procedure generates each possible answer with the same probability and each attempt succeeds with probability $7/8$, so you'll need an average of $8/7$ calls to gr_01(), which means an average of $(8/7)(5/4)=10/7$ calls to gr_04().