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.
#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 |
#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;
} |
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 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 |
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 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 ?> |
<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.