# Probability of getting a sum on throwing 2 Dices N times

Given the sum. The task is to find out the probability of occurring that sum on the thrown of the two dice N times.
Probability is defined as the favorable numbers of outcomes upon total numbers of the outcome. Probability always lies between 0 and 1.

Examples:

```Input: sum = 11, times = 1
Output: 2 / 36
favorable outcomes = (5, 6) and (6, 5) i.e 2
Total outcomes = (1, 1), (1, 2), (1, 3)...(6, 6) i.e 36
Probability = (2 / 36)

Input: sum = 7, times = 7
Output: 1 / 279936
```

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

Formula:-

The probability of occurring sum on throwing 2 dices N times = (favorable/total) ^ N

Approach:-

First of All, Calculates the probability of Occurring that sum on thrown of 2 dice 1 times.
Let say it Probability1.
Now, to calculate the Probability of occurring that sum on thrown of 2 dice N times be:
Probability2 = (Probability1) ^ N. i.e Probability1 raise to power N

Below is the implementation of above approach:

## C++

 `// C++ implementation of above approach ` `#include ` `using` `namespace` `std; ` ` `  `// function that calculates Probability. ` `int` `Probability(``int` `sum, ``int` `times) ` `{ ` ` `  `    ``float` `favorable = 0.0, total = 36.0; ` `    ``long` `int` `probability = 0; ` ` `  `    ``// To calculate favorable outcomes ` `    ``// in thrown of 2 dices 1 times. ` `    ``for` `(``int` `i = 1; i <= 6; i++) { ` `        ``for` `(``int` `j = 1; j <= 6; j++) { ` `            ``if` `((i + j) == sum) ` `                ``favorable++; ` `        ``} ` `    ``} ` ` `  `    ``int` `gcd1 = __gcd((``int``)favorable, (``int``)total); ` ` `  `    ``// Reduce to simplest Form. ` `    ``favorable = favorable / (``float``)gcd1; ` `    ``total = total / (``float``)gcd1; ` ` `  `    ``// Probability of occurring sum on 2 dice N times. ` `    ``probability = ``pow``(total, times); ` ` `  `    ``return` `probability; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `sum = 7, times = 7; ` ` `  `    ``cout << ``"1"` `         ``<< ``"/"` `<< Probability(sum, times); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of above approach ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` `// Recursive function to return ` `// gcd of a and b  ` `static` `int` `__gcd(``int` `a, ``int` `b)  ` `{  ` `    ``// Everything divides 0  ` `    ``if` `(a == ``0``)  ` `    ``return` `b;  ` `    ``if` `(b == ``0``)  ` `    ``return` `a;  ` `     `  `    ``// base case  ` `    ``if` `(a == b)  ` `        ``return` `a;  ` `     `  `    ``// a is greater  ` `    ``if` `(a > b)  ` `        ``return` `__gcd(a - b, b);  ` `    ``return` `__gcd(a, b - a);  ` `} ` ` `  `// function that calculates  ` `// Probability. ` `static` `long` `Probability(``int` `sum, ` `                        ``int` `times) ` `{ ` ` `  `    ``float` `favorable = ``0``, total = ``36``; ` `    ``long` `probability = ``0``; ` ` `  `    ``// To calculate favorable outcomes ` `    ``// in thrown of 2 dices 1 times. ` `    ``for` `(``int` `i = ``1``; i <= ``6``; i++)  ` `    ``{ ` `        ``for` `(``int` `j = ``1``; j <= ``6``; j++) ` `        ``{ ` `            ``if` `((i + j) == sum) ` `                ``favorable++; ` `        ``} ` `    ``} ` ` `  `    ``int` `gcd1 = __gcd((``int``)favorable,  ` `                     ``(``int``)total); ` ` `  `    ``// Reduce to simplest Form. ` `    ``favorable = favorable / (``float``)gcd1; ` `    ``total = total / (``float``)gcd1; ` ` `  `    ``// Probability of occurring  ` `    ``// sum on 2 dice N times. ` `    ``probability = (``long``)Math.pow(total, times); ` ` `  `    ``return` `probability; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main (String[] args)  ` `{ ` `    ``int` `sum = ``7``, times = ``7``; ` `     `  `    ``System.out.println( ``"1"` `+ ``"/"` `+  ` `          ``Probability(sum, times)); ` `} ` `} ` ` `  `// This code is contributed  ` `// by inder_verma `

## Python 3

 `# Python 3 implementation of above approach ` ` `  `# from math import everything ` `from` `math ``import` `*` ` `  `# function that calculates Probability. ` `def` `Probability(``sum``, times) : ` `    ``favorable, total, probability ``=` `0.0``, ``36.0``, ``0` ` `  `    ``# To calculate favorable outcomes  ` `    ``# in thrown of 2 dices 1 times.  ` `    ``for` `i ``in` `range``(``7``) : ` `        ``for` `j ``in` `range``(``7``) : ` `            ``if` `((i ``+` `j) ``=``=` `sum``) : ` `                ``favorable ``+``=` `1` ` `  `    ``gcd1 ``=` `gcd(``int``(favorable), ``int``(total)) ` ` `  `    ``# Reduce to simplest Form.  ` `    ``favorable ``=` `favorable ``/` `gcd1 ` `    ``total ``=` `total ``/` `gcd1 ` ` `  `    ``# Probability of occurring sum on 2 dice N times. ` `    ``probability ``=` `pow``(total, times) ` ` `  `    ``return` `int``(probability) ` ` `  ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"` `: ` ` `  `    ``sum``, times ``=` `7``, ``7` ` `  `    ``print``(``"1"``,``"/"``,Probability(``sum``, times)) ` ` `  ` `  `# This code is contributed by ANKITRAI1 `

## C#

 `// C# implementation of above approach ` ` `  `class` `GFG  ` `{ ` `// Recursive function to return ` `// gcd of a and b  ` `static` `int` `__gcd(``int` `a, ``int` `b)  ` `{  ` `    ``// Everything divides 0  ` `    ``if` `(a == 0)  ` `    ``return` `b;  ` `    ``if` `(b == 0)  ` `    ``return` `a;  ` `     `  `    ``// base case  ` `    ``if` `(a == b)  ` `        ``return` `a;  ` `     `  `    ``// a is greater  ` `    ``if` `(a > b)  ` `        ``return` `__gcd(a - b, b);  ` `    ``return` `__gcd(a, b - a);  ` `} ` ` `  `// function that calculates  ` `// Probability. ` `static` `long` `Probability(``int` `sum, ` `                        ``int` `times) ` `{ ` ` `  `    ``float` `favorable = 0, total = 36; ` `    ``long` `probability = 0; ` ` `  `    ``// To calculate favorable outcomes ` `    ``// in thrown of 2 dices 1 times. ` `    ``for` `(``int` `i = 1; i <= 6; i++)  ` `    ``{ ` `        ``for` `(``int` `j = 1; j <= 6; j++) ` `        ``{ ` `            ``if` `((i + j) == sum) ` `                ``favorable++; ` `        ``} ` `    ``} ` ` `  `    ``int` `gcd1 = __gcd((``int``)favorable,  ` `                    ``(``int``)total); ` ` `  `    ``// Reduce to simplest Form. ` `    ``favorable = favorable / (``float``)gcd1; ` `    ``total = total / (``float``)gcd1; ` ` `  `    ``// Probability of occurring  ` `    ``// sum on 2 dice N times. ` `    ``probability = (``long``)System.Math.Pow(total, times); ` ` `  `    ``return` `probability; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main()  ` `{ ` `    ``int` `sum = 7, times = 7; ` `     `  `    ``System.Console.WriteLine( ``"1"` `+ ``"/"` `+  ` `        ``Probability(sum, times)); ` `} ` `} ` ` `  `// This code is contributed  ` `// by mits `

## PHP

 ` `

Output:

```1/279936
```

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