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# Maximum number of dots after throwing a dice N times

• Last Updated : 19 May, 2021

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 ``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

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## Javascript

 ``
Output:
`4.95833`

Time Complexity: O(m)

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