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 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 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 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 <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; ` `} ` |

*chevron_right*

*filter_none*

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.

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