# Probability that the pieces of a broken stick form a n sided polygon

We have a stick of length L. The stick got broken at (n-1) randomly chosen points (lengths of parts can be non-integer or floating point numbers also) so we get n parts. We need to find the probability that these n pieces can form a n sided polygon. **Examples:**

Input : L = 5 n = 3 Output : 0.25 We need to cut rope of length 5 into three parts.

First we need to find the condition when n lengths can form a n sided polygon. Let us consider a triangle, we know for a triangle the length of the largest side must be smaller than the sum of the lengths of other sides. Similarly for a n sided polygon the length of the largest side must be less than the sum of the other (n-1) sides. Why? Suppose we break the stick into two equal halves. We further break one of the halves into (n-1) parts. We can never place them such that they make a closed polygon. (Actually the best we can do is to make 2 parallel lines). So we just need to find the probability that no part has the length greater than or equal to L/2.

Now we need to work on the probability. There are many ways to calculate the required probability we will use a geometric approach. Consider a circle of perimeter L. We place n points on the perimeter. The probability that they lie on the same semicircle is . Please refer this link for more information, let us denote it by P(E).

This probability is actually same as breaking the stick such that at least one part is of length L/2. But we want just the complement of this event hence our answer is

## C++

`// CPP program to find probability that` `// a rope of length L when broken into` `// n parts form a polygon.` `#include<iostream>` `using` `namespace` `std;` `double` `printProbability(unsigned L, unsigned n)` `{` ` ` `unsigned p = (1 << (n-1));` ` ` `return` `1.0 - ((` `double` `)n) / ((` `double` `)p);` `}` `int` `main()` `{` ` ` `unsigned n = 3, L = 5;` ` ` `cout << printProbability(L, n);` ` ` `return` `0;` `}` |

## Java

`// Java program to find probability that` `// a rope of length L when broken into` `// n parts form a polygon.` `public` `class` `GFG {` ` ` ` ` `static` `double` `printProbability(` `int` `L, ` `int` `n)` ` ` `{` ` ` `int` `p = (` `1` `<< (n-` `1` `));` ` ` `return` `1.0` `- ((` `double` `)n) / ((` `double` `)p);` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `int` `n = ` `3` `, L = ` `5` `;` ` ` `System.out.println(printProbability(L, n));` ` ` ` ` `}` ` ` `// This Code is contributed by ANKITRAI1` `}` |

## Python3

`# Python3 program to find probability that` `# a rope of length L when broken into` `# n parts form a polygon.` `def` `printProbability(L, n):` ` ` `p ` `=` `(` `1` `<< (n` `-` `1` `))` ` ` `return` `1.0` `-` `(` `float` `(n) ` `/` `float` `(p) )` `if` `__name__` `=` `=` `'__main__'` `:` ` ` `n ` `=` `3` ` ` `L ` `=` `5` ` ` `print` `(printProbability(L, n))` `# this code is contributed by ash264` |

## C#

`// C# program to find probability ` `// that a rope of length L when` `// broken into n parts form a polygon.` `using System;` `class` `GFG` `{` `static` `double` `printProbability(` `int` `L, ` `int` `n)` `{` ` ` `int` `p = (` `1` `<< (n - ` `1` `));` ` ` `return` `1.0` `- ((` `double` `)n) /` ` ` `((` `double` `)p);` `}` `// Driver code` `public` `static` `void` `Main()` `{` ` ` `int` `n = ` `3` `, L = ` `5` `;` ` ` `Console.WriteLine(printProbability(L, n));` `}` `}` `// This code is contributed` `// by inder_verma` |

## PHP

`<?php` `// PHP program to find probability that` `// a rope of length L when broken into` `// n parts form a polygon.` `function` `printProbability(` `$L` `, ` `$n` `)` `{` ` ` `$p` `= (1 << (` `$n` `- 1));` ` ` `return` `1.0 - (` `$n` `) / (` `$p` `);` `}` `// Driver Code` `$n` `= 3; ` `$L` `= 5;` `echo` `printProbability(` `$L` `, ` `$n` `);` `// This code is contributed` `// by Abby_akku` `?>` |

## Javascript

`<script>` `// Javascript program to find probability` `// that a rope of length L when broken into` `// n parts form a polygon.` `// Function to find out the` `// number of that vertices` `function` `printProbability(L, n)` `{` ` ` `var` `p = (1 << (n - 1));` ` ` `return` `1.0 - n/ p;` `}` `// Driver code` `var` `n = 3, L = 5;` `document.write(printProbability(L, n));` ` ` `// This code is contributed by Kirti` `</script>` |

**Output:**

0.25

**Time Complexity:** O(1)**Auxiliary Space:** O(1)