An interview question: About Probability

Question

An interview question:

Given a function f(x) that 1/4 times returns 0, 3/4 times returns 1. Write a function g(x) using f(x) that 1/2 times returns 0, 1/2 times returns

Answer 1

Assuming

P(f[x] == 0) = 1/4
P(f[x] == 1) = 3/4

and requiring a function g[x] with the following assumptions

P(g[x] == 0) = 1/2
P(g[x] == 1) = 1/2

I believe the following definition of g[x] is sufficient (Mathematica)

g[x_] := If[f[x] + f[x + 1] == 1, 1, 0]

or, alternatively in C

int g(int x)
{
    return f(x) + f(x+1) == 1
           ? 1
           : 0;
}

This is based on the idea that invocations of {f[x], f[x+1]} would produce the following outcomes

{
  {0, 0},
  {0, 1},
  {1, 0},
  {1, 1}
}

Summing each of the outcomes we have

{
  0,
  1,
  1,
  2
}

where a sum of 1 represents 1/2 of the possible sum outcomes, with any other sum making up the other 1/2.

Edit. As bdk says - {0,0} is less likely than {1,1} because

1/4 * 1/4 < 3/4 * 3/4

However, I am confused myself because given the following definition for f[x] (Mathematica)

f[x_] := Mod[x, 4] > 0 /. {False -> 0, True -> 1}

or alternatively in C

int f(int x)
{
    return (x % 4) > 0
           ? 1
           : 0;
}

then the results obtained from executing f[x] and g[x] seem to have the expected distribution.

Table[f[x], {x, 0, 20}]
{0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0}

Table[g[x], {x, 0, 20}]
{1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1}