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Probability that a random pair chosen from an array (a[i], a[j]) has the maximum sum
• Last Updated : 23 Oct, 2019

Given an array arr[] of N integers, the task is to find the probability of getting the maximum sum pair (arr[i], arr[j]) from the array when a random pair is chosen.

Examples:

Input: arr[] = {3, 3, 3, 3}
Output: 1
All the pairs will give the maximum sum i.e. 6.

Input: arr[] = {1, 1, 1, 2, 2, 2}
Output: 0.2
Only the pairs (2, 2), (2, 2) and (2, 2) will give
the maximum sum out of 15 pairs.
3 / 15 = 0.2

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Run two nested loops to get the sum for every single pair, keep the maximum sum for any pair and its count (i.e. the number of pairs that give this sum). Now, the probability of getting this sum will be (count / totalPairs) where totalPairs = (n * (n – 1)) / 2.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the probability ` `// of getting the maximum pair sum ` `// when a random pair is chosen ` `// from the given array ` `float` `findProb(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``// Initialize the maximum sum, its count ` `    ``// and the count of total pairs ` `    ``long` `maxSum = INT_MIN, maxCount = 0, totalPairs = 0; ` ` `  `    ``// For every single pair ` `    ``for` `(``int` `i = 0; i < n - 1; i++) { ` `        ``for` `(``int` `j = i + 1; j < n; j++) { ` ` `  `            ``// Get the sum of the current pair ` `            ``int` `sum = arr[i] + arr[j]; ` ` `  `            ``// If the sum is equal to the current ` `            ``// maximum sum so far ` `            ``if` `(sum == maxSum) { ` ` `  `                ``// Increment its count ` `                ``maxCount++; ` `            ``} ` ` `  `            ``// If the sum is greater than ` `            ``// the current maximum ` `            ``else` `if` `(sum > maxSum) { ` ` `  `                ``// Update the current maximum and ` `                ``// re-initialize the count to 1 ` `                ``maxSum = sum; ` `                ``maxCount = 1; ` `            ``} ` ` `  `            ``totalPairs++; ` `        ``} ` `    ``} ` ` `  `    ``// Find the required probability ` `    ``float` `prob = (``float``)maxCount / (``float``)totalPairs; ` `    ``return` `prob; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 1, 1, 2, 2, 2 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``); ` ` `  `    ``cout << findProb(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to return the probability ` `// of getting the maximum pair sum ` `// when a random pair is chosen ` `// from the given array ` `static` `float` `findProb(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``// Initialize the maximum sum, its count ` `    ``// and the count of total pairs ` `    ``long` `maxSum = Integer.MIN_VALUE, ` `         ``maxCount = ``0``, totalPairs = ``0``; ` ` `  `    ``// For every single pair ` `    ``for` `(``int` `i = ``0``; i < n - ``1``; i++)  ` `    ``{ ` `        ``for` `(``int` `j = i + ``1``; j < n; j++)  ` `        ``{ ` ` `  `            ``// Get the sum of the current pair ` `            ``int` `sum = arr[i] + arr[j]; ` ` `  `            ``// If the sum is equal to the current ` `            ``// maximum sum so far ` `            ``if` `(sum == maxSum) ` `            ``{ ` ` `  `                ``// Increment its count ` `                ``maxCount++; ` `            ``} ` ` `  `            ``// If the sum is greater than ` `            ``// the current maximum ` `            ``else` `if` `(sum > maxSum)  ` `            ``{ ` ` `  `                ``// Update the current maximum and ` `                ``// re-initialize the count to 1 ` `                ``maxSum = sum; ` `                ``maxCount = ``1``; ` `            ``} ` ` `  `            ``totalPairs++; ` `        ``} ` `    ``} ` ` `  `    ``// Find the required probability ` `    ``float` `prob = (``float``)maxCount /  ` `                 ``(``float``)totalPairs; ` `    ``return` `prob; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``int` `arr[] = { ``1``, ``1``, ``1``, ``2``, ``2``, ``2` `}; ` `    ``int` `n = arr.length; ` ` `  `    ``System.out.println(findProb(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## Python3

 `# Python3 implementation of the approach  ` `import` `sys ` ` `  `# Function to return the probability  ` `# of getting the maximum pair sum  ` `# when a random pair is chosen  ` `# from the given array  ` `def` `findProb(arr, n) : ` ` `  `    ``# Initialize the maximum sum, its count  ` `    ``# and the count of total pairs  ` `    ``maxSum ``=` `-``(sys.maxsize ``-` `1``); ` `    ``maxCount ``=` `0``; ` `    ``totalPairs ``=` `0``;  ` ` `  `    ``# For every single pair  ` `    ``for` `i ``in` `range``(n ``-` `1``) : ` `        ``for` `j ``in` `range``(i ``+` `1``, n) : ` `             `  `            ``# Get the sum of the current pair  ` `            ``sum` `=` `arr[i] ``+` `arr[j]; ` `             `  `            ``# If the sum is equal to the curren ` `            ``# maximum sum so far ` `            ``if` `(``sum` `=``=` `maxSum) : ` `                 `  `                ``# Increment its count ` `                ``maxCount ``+``=` `1``;  ` ` `  `            ``# If the sum is greater than  ` `            ``# the current maximum  ` `            ``elif` `(``sum` `> maxSum) : ` ` `  `                ``# Update the current maximum and  ` `                ``# re-initialize the count to 1  ` `                ``maxSum ``=` `sum``;  ` `                ``maxCount ``=` `1``;  ` ` `  `            ``totalPairs ``+``=` `1``;  ` ` `  `    ``# Find the required probability  ` `    ``prob ``=` `maxCount ``/` `totalPairs;  ` `     `  `    ``return` `prob;  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``arr ``=` `[ ``1``, ``1``, ``1``, ``2``, ``2``, ``2` `];  ` `    ``n ``=` `len``(arr); ` `     `  `    ``print``(findProb(arr, n));  ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of above approach ` `using` `System; ` `     `  `class` `GFG ` `{ ` ` `  `// Function to return the probability ` `// of getting the maximum pair sum ` `// when a random pair is chosen ` `// from the given array ` `static` `float` `findProb(``int` `[]arr, ``int` `n) ` `{ ` ` `  `    ``// Initialize the maximum sum, its count ` `    ``// and the count of total pairs ` `    ``long` `maxSum = ``int``.MinValue, ` `        ``maxCount = 0, totalPairs = 0; ` ` `  `    ``// For every single pair ` `    ``for` `(``int` `i = 0; i < n - 1; i++)  ` `    ``{ ` `        ``for` `(``int` `j = i + 1; j < n; j++)  ` `        ``{ ` ` `  `            ``// Get the sum of the current pair ` `            ``int` `sum = arr[i] + arr[j]; ` ` `  `            ``// If the sum is equal to the current ` `            ``// maximum sum so far ` `            ``if` `(sum == maxSum) ` `            ``{ ` ` `  `                ``// Increment its count ` `                ``maxCount++; ` `            ``} ` ` `  `            ``// If the sum is greater than ` `            ``// the current maximum ` `            ``else` `if` `(sum > maxSum)  ` `            ``{ ` ` `  `                ``// Update the current maximum and ` `                ``// re-initialize the count to 1 ` `                ``maxSum = sum; ` `                ``maxCount = 1; ` `            ``} ` ` `  `            ``totalPairs++; ` `        ``} ` `    ``} ` ` `  `    ``// Find the required probability ` `    ``float` `prob = (``float``)maxCount /  ` `                 ``(``float``)totalPairs; ` `    ``return` `prob; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String []args) ` `{ ` `    ``int` `[]arr = { 1, 1, 1, 2, 2, 2 }; ` `    ``int` `n = arr.Length; ` ` `  `    ``Console.WriteLine(findProb(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```0.2
```

Time Complexity: O(n2)

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