Make a fair coin from a biased coin

You are given a function foo() that represents a biased coin. When foo() is called, it returns 0 with 60% probability, and 1 with 40% probability. Write a new function that returns 0 and 1 with a 50% probability each. Your function should use only foo(), no other library method.

Solution:
We know foo() returns 0 with 60% probability. How can we ensure that 0 and 1 are returned with a 50% probability? 
The solution is similar to this post. If we can somehow get two cases with equal probability, then we are done. We call foo() two times. Both calls will return 0 with a 60% probability. So the two pairs (0, 1) and (1, 0) will be generated with equal probability from two calls of foo(). Let us see how.
(0, 1): The probability to get 0 followed by 1 from two calls of foo() = 0.6 * 0.4 = 0.24 
(1, 0): The probability to get 1 followed by 0 from two calls of foo() = 0.4 * 0.6 = 0.24
So the two cases appear with equal probability. The idea is to return consider only the above two cases, return 0 in one case, return 1 in other case. For other cases [(0, 0) and (1, 1)], recur until you end up in any of the above two cases. 

The below program depicts how we can use foo() to return 0 and 1 with equal probability.  

C++

#include <bits/stdc++.h>
using namespace std;

int foo() // given method that returns 0
          // with 60% probability and 1 with 40%
{
    // some code here
}

// returns both 0 and 1 with 50% probability
int my_fun()
{
    int val1 = foo();
    int val2 = foo();
    if (val1 == 0 && val2 == 1)
        return 0; // Will reach here with
                  // 0.24 probability
    if (val1 == 1 && val2 == 0)
        return 1; // Will reach here with
                  // 0.24 probability
    return my_fun(); // will reach here with
                     // (1 - 0.24 - 0.24) probability
}

// Driver Code
int main()
{
    cout << my_fun();
    return 0;
}

// This is code is contributed
// by rathbhupendra

C

#include <stdio.h>

int foo() // given method that returns 0 with 60%
          // probability and 1 with 40%
{
    // some code here
}

// returns both 0 and 1 with 50% probability
int my_fun()
{
    int val1 = foo();
    int val2 = foo();
    if (val1 == 0 && val2 == 1)
        return 0; // Will reach here with 0.24 probability
    if (val1 == 1 && val2 == 0)
        return 1; // // Will reach here with 0.24
                  // probability
    return my_fun(); // will reach here with (1 - 0.24 -
                     // 0.24) probability
}

int main()
{
    printf("%d ", my_fun());
    return 0;
}

Java

import java.io.*;

class GFG {

    // Given method that returns 0
    // with 60% probability and 1 with 40%
    static int foo()
    {
        // some code here
    }

    // Returns both 0 and 1 with 50% probability
    static int my_fun()
    {
        int val1 = foo();
        int val2 = foo();
        if (val1 == 0 && val2 == 1)

            return 0; // Will reach here with
                      // 0.24 probability

        if (val1 == 1 && val2 == 0)

            return 1; // Will reach here with
                      // 0.24 probability

        return my_fun(); // will reach here with
                         // (1 - 0.24 - 0.24) probability
    }

    // Driver Code
    public static void main(String[] args)
    {
        System.out.println(my_fun());
    }
}

// This code is contributed by ShubhamCoder

Python3

# Python3 program for the
# above approach
def foo():

    # Some code here
    pass

# Returns both 0 and 1
# with 50% probability
def my_fun():

    val1, val2 = foo(), foo()

    if val1 ^ val2:

        # Will reach here with
        # (0.24 + 0.24) probability
        return val1

    # Will reach here with
    # (1 - 0.24 - 0.24) probability
    return my_fun()

# Driver Code
if __name__ == '__main__':
    print(my_fun())

# This code is contributed by sgshah2

C#

using System;

class GFG {

    // given method that returns 0
    // with 60% probability and 1 with 40%
    static int foo()
    {
        // some code here
    }

    // returns both 0 and 1 with 50% probability
    static int my_fun()
    {
        int val1 = foo();
        int val2 = foo();
        if (val1 == 0 && val2 == 1)
            return 0; // Will reach here with
                      // 0.24 probability
        if (val1 == 1 && val2 == 0)
            return 1; // Will reach here with
                      // 0.24 probability
        return my_fun(); // will reach here with
                         // (1 - 0.24 - 0.24) probability
    }

    // Driver Code
    static public void Main() { Console.Write(my_fun()); }
}

// This is code is contributed
// by ShubhamCoder

PHP

<?php

function foo()  // given method that returns 0
                // with 60% probability and 1 with 40%
{
    // some code here
}

// returns both 0 and 1 with 50% probability
function my_fun()
{
    $val1 = foo();
    $val2 = foo();
    if ($val1 == 0 && $val2 == 1)
        return 0; // Will reach here with
                  // 0.24 probability
    if ($val1 == 1 && $val2 == 0)
        return 1; // Will reach here with
                  // 0.24 probability
    return my_fun(); // will reach here with
                     // (1 - 0.24 - 0.24) probability
}

// Driver Code
echo my_fun();

// This is code is contributed
// by Akanksha Rai
?>

Javascript

<script>

// Given method that returns 0 with
// 60% probability and 1 with 40%
function foo()
{
    // Some code here
}

// Returns both 0 and 1 with
// 50% probability
function my_fun()
{
    var val1 = foo();
    var val2 = foo();

    if (val1 == 0 && val2 == 1)
        return 0; // Will reach here with
                  // 0.24 probability

    if (val1 == 1 && val2 == 0)
        return 1; // Will reach here with
                  // 0.24 probability

    return my_fun(); // Will reach here with
                     // (1 - 0.24 - 0.24) probability
}

// Driver Code
document.write(my_fun());

// This code is contributed by noob2000

</script>

Time Complexity: O(1)

Auxiliary Space: O(1)

References: 
http://en.wikipedia.org/wiki/Fair_coin#Fair_results_from_a_biased_coin
This article is compiled by Shashank Sinha and reviewed by GeeksforGeeks team. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.