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

The problem with your algorithm is that it repeats itself with high probability. My code:

function g(x) = {
    var s = f(x) + f(x) + f(x); 
    // s = 0, probability:  1/64
    // s = 1, probability:  9/64
    // s = 2, probability: 27/64
    // s = 3, probability: 27/64
    if (s == 2) return 0;
    if (s == 3) return 1;

    return g(x); // probability to go into recursion = 10/64, with only 1 additional f(x) calculation
}

I've measured average number of times f(x) was calculated for your algorithm and for mine. For yours f(x) was calculated around 5.3 times per one g(x) calculation. With my algorithm this number reduced to around 3.5. The same is true for other answers so far since they are actually the same algorithm as you said.

P.S.: your definition doesn't mention 'random' at the moment, but probably it is assumed. See my other answer.

Answer 2

This is much like the Monty Hall paradox.

In general.

Public Class Form1

    'the general case
    '
    'twiceThis = 2 is 1 in four chance of 0
    'twiceThis = 3 is 1 in six chance of 0
    '
    'twiceThis = x is 1 in 2x chance of 0

    Const twiceThis As Integer = 7
    Const numOf As Integer = twiceThis * 2

    Private Sub Button1_Click(ByVal sender As System.Object, _
                              ByVal e As System.EventArgs) Handles Button1.Click

        Const tries As Integer = 1000
        y = New List(Of Integer)

        Dim ct0 As Integer = 0
        Dim ct1 As Integer = 0
        Debug.WriteLine("")
        ''show all possible values of fx
        'For x As Integer = 1 To numOf
        '    Debug.WriteLine(fx)
        'Next

        'test that gx returns 50% 0's and 50% 1's
        Dim stpw As New Stopwatch
        stpw.Start()
        For x As Integer = 1 To tries
            Dim g_x As Integer = gx()
            'Debug.WriteLine(g_x.ToString) 'used to verify that gx returns 0 or 1 randomly
            If g_x = 0 Then ct0 += 1 Else ct1 += 1
        Next
        stpw.Stop()
        'the results
        Debug.WriteLine((ct0 / tries).ToString("p1"))
        Debug.WriteLine((ct1 / tries).ToString("p1"))
        Debug.WriteLine((stpw.ElapsedTicks / tries).ToString("n0"))

    End Sub

    Dim prng As New Random
    Dim y As New List(Of Integer)

    Private Function fx() As Integer

        '1 in numOf chance of zero being returned
        If y.Count = 0 Then
            'reload y
            y.Add(0) 'fx has only one zero value
            Do
                y.Add(1) 'the rest are ones
            Loop While y.Count < numOf
        End If
        'return a random value 
        Dim idx As Integer = prng.Next(y.Count)
        Dim rv As Integer = y(idx)
        y.RemoveAt(idx) 'remove the value selected
        Return rv

    End Function

    Private Function gx() As Integer

        'a function g(x) using f(x) that 50% of the time returns 0
        '                           that 50% of the time returns 1
        Dim rv As Integer = 0
        For x As Integer = 1 To twiceThis
            fx()
        Next
        For x As Integer = 1 To twiceThis
            rv += fx()
        Next
        If rv = twiceThis Then Return 1 Else Return 0

    End Function
End Class

Answer 3

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}

Answer 4

If you call f(x) twice in a row, the following outcomes are possible (assuming that successive calls to f(x) are independent, identically distributed trials):

00 (probability 1/4 * 1/4)
01 (probability 1/4 * 3/4)  
10 (probability 3/4 * 1/4)  
11 (probability 3/4 * 3/4)

01 and 10 occur with equal probability. So iterate until you get one of those cases, then return 0 or 1 appropriately:

do
  a=f(x); b=f(x);
while (a == b);

return a;

It might be tempting to call f(x) only once per iteration and keep track of the two most recent values, but that won't work. Suppose the very first roll is 1, with probability 3/4. You'd loop until the first 0, then return 1 (with probability 3/4).