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

  • Last Updated : 02 Jun, 2021

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++




// 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;
}

Java




// 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

Python3




# 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

C#




// 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

PHP




<?php
// PHP implementation of the approach
 
// Function to return the probability
// of A winning
function 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,6);
}
 
// Driver code
$a = 1;
$b = 2;
$c = 10;
$d = 11;
echo getProbability($a, $b, $c, $d);
 
// This code is contributed by chandan_jnu
?>

Javascript




<script>
 
// JavaScript implementation of the approach   
 
// Function to return the probability
// of A winning
    function getProbability(a , b , c , d) {
 
        // p and q store the values
        // of fractions a / b and c / d
        var p =  a /  b;
        var q =  c /  d;
 
        // To store the winning probability of A
        var ans = p * (1 / (1 - (1 - q) * (1 - p)));
        return ans;
    }
 
    // Driver code
     
        var a = 1, b = 2, c = 10, d = 11;
        document.write( getProbability(a, b, c, d).toFixed(5));
 
// This code contributed by aashish1995
 
</script>
Output: 
0.52381

 




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