Find probability that a player wins when probabilities of hitting the target are given

Given four integers p, q, r, and s. Two players are playing a game where both the players hit a target and the first player who hits the target wins the game. The probability of the first player hitting the target is p / q and that of the second player hitting the target is r / s. The task is to find the probability of the first player winning the game.

Examples: 

Input: p = 1, q = 4, r = 3, s = 4 
Output: 0.307692308

Input: p = 1, q = 2, r = 1, s = 2 
Output: 0.666666667 

Approach: The probability of the first player hitting the target is p / q and missing the target is 1 - p / q. 
The probability of the second player hitting the target is r / s and missing the target is 1 - r / s. 
Let the first player be x and the second player is y. 
So the total probability will be x won + (x lost * y lost * x won) + (x lost * y lost * x lost * y lost * x won) + ... so on. 
Because x can win at any turn, it's an infinite sequence. 
Let t = (1 - p / q) * (1 - r / s). Here t < 1 as p / q and r / s are always <1. 
So the series will become, p / q + (p / q) * t + (p / q) * t² + ... 
This is an infinite GP series with a common ratio of less than 1 and its sum will be (p / q) / (1 - t).

Below is the implementation of the above approach:  

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the probability of the winner
double find_probability(double p, double q,
                        double r, double s)
{

    double t = (1 - p / q) * (1 - r / s);

    double ans = (p / q) / (1 - t);

    return ans;
}

// Driver Code
int main()
{
    double p = 1, q = 2, r = 1, s = 2;

    // Will print 9 digits after the decimal point
    cout << fixed << setprecision(9)
         << find_probability(p, q, r, s);

    return 0;
}

// Java implementation of the approach
import java.util.*;
import java.text.DecimalFormat;

class solution
{

// Function to return the probability of the winner
static double find_probability(double p, double q,
                        double r, double s)
{

    double t = (1 - p / q) * (1 - r / s);

    double ans = (p / q) / (1 - t);

    return ans;
}

// Driver Code
public static void main(String args[])
{
    double p = 1, q = 2, r = 1, s = 2;

    // Will print 9 digits after the decimal point
    DecimalFormat dec = new DecimalFormat("#0.000000000");
    System.out.println(dec.format(find_probability(p, q, r, s)));
}
}
// This code is contributed by
// Surendra_Gangwar

# Python3 implementation of the approach 

# Function to return the probability
# of the winner 
def find_probability(p, q, r, s) :
    
    t = (1 - p / q) * (1 - r / s)

    ans = (p / q) / (1 - t); 

    return round(ans, 9)

# Driver Code 
if __name__ == "__main__" : 

    p, q, r, s = 1, 2, 1, 2

    # Will print 9 digits after 
    # the decimal point 
    print(find_probability(p, q, r, s)) 

# This code is contributed by Ryuga

// C# implementation of the approach
using System;

class GFG
{

// Function to return the probability of the winner
static double find_probability(double p, double q,
                        double r, double s)
{

    double t = (1 - p / q) * (1 - r / s);

    double ans = (p / q) / (1 - t);

    return ans;
}

// Driver Code
public static void Main()
{
    double p = 1, q = 2, r = 1, s = 2;
    Console.WriteLine(find_probability(p, q, r, s));
}
}

// This code is contributed by
// anuj_67..

<?php
// PHP implementation of the approach

// Function to return the probability
// of the winner
function find_probability($p, $q, $r, $s)
{
    $t = (1 - $p / $q) * (1 - $r / $s);

    $ans = ($p / $q) / (1 - $t);

    return $ans;
}

// Driver Code
$p = 1; $q = 2;
$r = 1; $s = 2;

// Will print 9 digits after 
// the decimal point
$res = find_probability($p, $q, $r, $s);
$update = number_format($res, 7);
echo $update;

// This code is contributed by Rajput-Ji
?>

<script>


// Javascript implementation of the approach

// Function to return the probability of the winner
function find_probability(p, q, r, s)
{

    var t = (1 - p / q) * (1 - r / s);

    var ans = (p / q) / (1 - t);

    return ans;
}

// Driver Code
var p = 1, q = 2, r = 1, s = 2;
// Will print 9 digits after the decimal point
document.write( find_probability(p, q, r, s).toFixed(9));


</script>

0.666666667

Time Complexity: O(1)

Auxiliary Space: O(1)