Probability of A winning the match when individual probabilities of hitting the target given

Given four integers a, b, c and d. Player A & B try to score a penalty. Probability of A shooting the target is a / b while probability of B shooting the target is c / d. The player who scores the penalty first wins. The task is to find the probability of A winning the match.

Examples:

Input: a = 1, b = 3, c = 1, d = 3
Output: 0.6



Input: a = 1, b = 2, c = 10, d = 11
Output: 0.52381

Approach: If we consider variables K = a / b as the probability of A shooting the target and R = (1 – (a / b)) * (1 – (c / d)) as the probability that A as well as B both missing the target.
Therefore, the solution forms a Geometric progression K * R0 + K * R1 + K * R2 + ….. whose sum is (K / 1 – R). After putting the values of K and R we get the formula as K * (1 / (1 – (1 – r) * (1 – k))).

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the probability of A winning
double getProbability(int a, int b, int c, int d)
{
  
    // p and q store the values
    // of fractions a / b and c / d
    double p = (double)a / (double)b;
    double q = (double)c / (double)d;
  
    // To store the winning probability of A
    double ans = p * (1 / (1 - (1 - q) * (1 - p)));
    return ans;
}
  
// Driver code
int main()
{
    int a = 1, b = 2, c = 10, d = 11;
    cout << getProbability(a, b, c, d);
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG 
{
  
// Function to return the probability
// of A winning
static double getProbability(int a, int b, 
                             int c, int d) 
{
  
    // p and q store the values
    // of fractions a / b and c / d
    double p = (double) a / (double) b;
    double q = (double) c / (double) d;
  
    // To store the winning probability of A
    double ans = p * (1 / (1 - (1 - q) * 
                               (1 - p)));
    return ans;
}
  
// Driver code
public static void main(String[] args)
{
    int a = 1, b = 2, c = 10, d = 11;
    System.out.printf("%.5f"
               getProbability(a, b, c, d));
}
}
  
// This code contributed by Rajput-Ji

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Python3

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# Python3 implementation of the approach 
  
# Function to return the probability
# of A winning 
def getProbability(a, b, c, d) : 
  
    # p and q store the values 
    # of fractions a / b and c / d 
    p = a / b;
    q = c / d;
      
    # To store the winning probability of A
    ans = p * (1 / (1 - (1 - q) * (1 - p)));
      
    return round(ans,5); 
  
# Driver code 
if __name__ == "__main__"
  
    a = 1; b = 2; c = 10; d = 11
    print(getProbability(a, b, c, d)); 
  
# This code is contributed by Ryuga

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C#

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// C# implementation of the approach 
using System;
  
class GFG
{
  
// Function to return the probability 
// of A winning 
public static double getProbability(int a, int b, 
                                    int c, int d)
{
  
    // p and q store the values 
    // of fractions a / b and c / d 
    double p = (double) a / (double) b;
    double q = (double) c / (double) d;
  
    // To store the winning probability of A 
    double ans = p * (1 / (1 - (1 - q) * 
                               (1 - p)));
    return ans;
}
  
// Driver code 
public static void Main(string[] args)
{
    int a = 1, b = 2, c = 10, d = 11;
    Console.Write("{0:F5}"
                   getProbability(a, b, c, d));
}
}
  
// This code is contributed by Shrikant13

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PHP

Output:

0.52381


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