Probability of A winning the match when individual probabilities of hitting the target given (original) (raw)

Last Updated : 22 Jun, 2022

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

JavaScript

`

Time Complexity: O(1)
Auxiliary Space: O(1)