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📜  从有偏见的硬币中赚取公平的硬币

📅  最后修改于: 2021-04-28 00:02:13             🧑  作者: Mango

您将获得一个代表有偏硬币的函数foo()。调用foo()时，它以60％的概率返回0，以40％的概率返回1。编写一个新函数，以50％的概率返回0和1。您的函数应仅使用foo()，而不能使用其他库方法。

解决方案：
我们知道foo()以60％的概率返回0。我们如何确保以50％的概率返回0和1?
解决方案与此职位相似。如果我们可以通过某种方式以相等的概率获得两个案例，那么我们就完成了。我们两次调用foo()。两个调用都将以60％的概率返回0。因此，两次调用foo()会以相等的概率生成两对(0，1)和(1，0)。让我们看看如何。
(0，1)：两次调用foo()的结果为0后接1的概率= 0.6 * 0.4 = 0.24
(1，0)：从两次foo()调用中得到1后跟0的概率= 0.4 * 0.6 = 0.24
因此，这两种情况出现的可能性相等。想法是只考虑上述两种情况，一种情况下返回0，另一种情况下返回1。对于其他情况[(0，0)和(1，1)]，请重复进行直到遇到以上两种情况中的任何一种。

下面的程序描述了如何使用foo()以相等的概率返回0和1。

C++
#include 
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 
 
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

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