Maximum number of dots after throwing a dice N times

Given a dice with m-faces. The first face of the dice contains a dot, the second one contains two dots, and so on, the m-th face contains m dots. Each face appears with a probability . Our task is to calculate the expected maximum number of dots after tossing the dice  times.

Examples:

Input: 2 2 
Output: 1.750000000000
Here the dice includes {1, 2}. 
So, the sample space of throwing the dice two times = 
{(1, 2), (1, 1), (2, 1), (2, 2)} 
For (1, 2)–> maximum=2 
For (1, 1)–> maximum=1 
For (2, 2)–> maximum=2 
For (2, 1)–> maximum=2 
The probability of each outcome is 0.25,
that is, expectation equals to 
(2+1+2+2)*(0.25) = 7/4 = 1.750000000000

Input: 6 3 
Output: 4.958333333333  

Approach: 
The key observation in this problem is that no. of times a number can occur a maximum of times depending upon its previous number. 
For i-th number, it will be . 
Take m = 6, n = 2 as an instance. 
Total numbers with a maximum =6 are equal to . 
The total numbers with a maximum, 5 are equal to . 
Similarly, we can find out for 4,3,2, and 1. 
6 6 6 6 6 6 
5 5 5 5 5 6 
4 4 4 4 5 6 
3 3 3 4 5 6 
2 2 3 4 5 6 
1 2 3 4 5 6 
Enumerate the maximum number, the distribution will be an n-dimensional super-cube with m-length-side. Each layer will be a large cube minus a smaller cube. 
So, our answer will be the sum of all i-th elements from 1 to m given by:  

Calculating may cause overflow, so we could move the divisor into the sum and calculate instead. 

C++

// CPP program for above implementation
#include <bits/stdc++.h>
using namespace std;

// Function find the maximum expectation
double expect(double m, double n)
{
    double ans = 0.0, i;


       for (i = m; i; i--)
        // formula to find the maximum number and
        // sum of maximum numbers
        ans += (pow(i / m, n) - pow((i - 1) / m, n)) * i;

    return ans;
}

// Driver code
int main()
{
    double m = 6, n = 3;
    cout << expect(m, n);

 return 0;
}

Java

// Java program for above implementation
class GFG
{
// Function find the maximum expectation
static double expect(double m, double n)
{
    double ans = 0.0, i;

    for (i = m; i > 0; i--)

        // formula to find the maximum number
        // and sum of maximum numbers
        ans += (Math.pow(i / m, n) -
                Math.pow((i - 1) / m, n)) * i;

    return ans;
}

// Driver code
public static void main(String[] args)
{
    double m = 6, n = 3;
    System.out.println(String.format("%.5f",
                             expect(m, n)));
}
}

// This code is contributed by mits

Python3

# Python3 program for finding maximum
# number of dots after throwing a
# dice N times.

# Function to find the maximum
# expectation
def expect(m,n) :

    ans = 0.0
    i = m
    while (i):

        # formula to find the maximum
        # number and
        # sum of maximum numbers
        ans += (pow(i / m, n) - pow((i-1) / m, n)) * i
        i -= 1

    return ans

# Driver code
if __name__ == "__main__" :

    # multiple assignments
    m,n = 6,3

    # function calling
    print(expect(m,n))

C#

// C# program for above implementation
using System;

class GFG
{
// Function find the maximum expectation
static double expect(double m, double n)
{
    double ans = 0.0, i;

    for (i = m; i > 0; i--)

        // formula to find the maximum number
        // and sum of maximum numbers
        ans += (Math.Pow(i / m, n) -
                Math.Pow((i - 1) / m, n)) * i;

    return ans;
}

// Driver code
public static void Main()
{
    double m = 6, n = 3;
    Console.WriteLine(expect(m, n));
}
}

// This code is contributed
// by Akanksha Rai

PHP

<?php
// PHP program for above implementation

// Function find the maximum expectation
function expect($m, $n)
{
    $ans = 0.0;

    for ($i = $m; $i; $i--)

        // formula to find the maximum number
        // and sum of maximum numbers
        $ans += (pow($i / $m, $n) -
                 pow(($i - 1) / $m, $n)) * $i;

    return $ans;
}

// Driver code
$m = 6;
$n = 3;
echo expect($m, $n);

// This code is contributed by ChitraNayal
?>

Javascript

<script>
// Javascript program for above implementation

    // Function find the maximum expectation
    function expect(m,n)
    {
        let ans = 0.0, i;

        for (i = m; i > 0; i--)

        // formula to find the maximum number
        // and sum of maximum numbers
            ans += (Math.pow(i / m, n) -
                Math.pow((i - 1) / m, n)) * i;

        return ans;
    }

    // Driver code
    let m = 6, n = 3;

    document.write(expect(m, n).toFixed(5))



// This code is contributed by avanitrachhadiya2155
</script>

Output:

4.95833

Time Complexity: O(m * log n), where m and n are given inputs.
Auxiliary Space: O(1), as constant space is used.