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Make a fair coin from a biased coin

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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.  

 



Last Updated : 10 Sep, 2021
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