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++ 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 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 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# 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 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
?>
<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
Time Complexity: O(1)
Auxiliary Space: O(1)