Find probability that a player wins when probabilities of hitting the target are given
Last Updated :
23 Jun, 2022
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++
#include <bits/stdc++.h>
using namespace std;
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;
}
int main()
{
double p = 1, q = 2, r = 1, s = 2;
cout << fixed << setprecision(9)
<< find_probability(p, q, r, s);
return 0;
}
|
Java
import java.util.*;
import java.text.DecimalFormat;
class solution
{
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;
}
public static void main(String args[])
{
double p = 1 , q = 2 , r = 1 , s = 2 ;
DecimalFormat dec = new DecimalFormat( "#0.000000000" );
System.out.println(dec.format(find_probability(p, q, r, s)));
}
}
|
Python3
def find_probability(p, q, r, s) :
t = ( 1 - p / q) * ( 1 - r / s)
ans = (p / q) / ( 1 - t);
return round (ans, 9 )
if __name__ = = "__main__" :
p, q, r, s = 1 , 2 , 1 , 2
print (find_probability(p, q, r, s))
|
C#
using System;
class GFG
{
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;
}
public static void Main()
{
double p = 1, q = 2, r = 1, s = 2;
Console.WriteLine(find_probability(p, q, r, s));
}
}
|
PHP
<?php
function find_probability( $p , $q , $r , $s )
{
$t = (1 - $p / $q ) * (1 - $r / $s );
$ans = ( $p / $q ) / (1 - $t );
return $ans ;
}
$p = 1; $q = 2;
$r = 1; $s = 2;
$res = find_probability( $p , $q , $r , $s );
$update = number_format( $res , 7);
echo $update ;
?>
|
Javascript
<script>
function find_probability(p, q, r, s)
{
var t = (1 - p / q) * (1 - r / s);
var ans = (p / q) / (1 - t);
return ans;
}
var p = 1, q = 2, r = 1, s = 2;
document.write( find_probability(p, q, r, s).toFixed(9));
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
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