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Expected Number of Trials to get N Consecutive Heads
• Difficulty Level : Hard
• Last Updated : 08 Jul, 2020

Given a number N. The task is to find the expected number of times a coin must be flipped to get N heads consecutively.

Example:

Input: N = 2
Output: 6

Input: N = 5
Output: 62

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

Approach:
The key is to observe that if we see a tail between any consecutive N flip, it breaks the streak for continuous heads and we have to start over again for N consecutive head.

Let the expected number of trial be X to get N consecutive heads. Below are the possible Cases:

• Case 1: If, in the 1st trial, a tail occurs then it means that we have wasted one trial and we will have to do X more trial to get N consecutive head. The probability of this event is 1/2 and the total number of trial required to get N consecutive head is (X + count of the previous trial wasted).
• Case 2: If, in the 2nd trial, a tail occurs then it means that we have wasted our all previous trial and we will have to do X more trial to get N consecutive head. The probability of this event is 1/4 and the total number of trials required to get N consecutive flips is (X + count of previous trial wasted).
• Case 3:If, in the 3rd trial, a tail occurs then it means that we have wasted our all previous trial and we will have to do X more trial to get N. The probability of this event is 1/8 and the total number of trials required to get N consecutive flips is (X + count of the previous trial wasted). This will continue until we get N consecutive heads.
• Case N: Similarly, if in the Nth trial, a tail occurs, then it means that we have wasted our all previous trial and we will have to do X more trial to get N. The probability of this event is 1/2N and the total number of trials required to get N consecutive flips is (X + count of the previous trial wasted).

From the above cases, the summation of all probability gives will gives the count of trials for N consecutive heads. Mathematically:

X = (1/2)*(X+1) + (1/4)*(X+2) + (1/8)*(X+3)+. . .+(1/2N)*(X+N) + (1/2N)*N

Solving the above equation for X. We have:

```By opening the above expressions and arranging it we have:
X = X(1/2 + 1/4 + 1/8 + . . . . . . 1/2N)
+ (1/2 + 2/4 + 3/8 . . . . . . . + N/2N
+ N/2N)
```

The first part of the above equations form Geometric Progression and second part of the above equations forms an Arithmetico Geometric Sequence. Solving the above sequences separately we have:
For Geometric Sequence:

Sum of GP series = 1/2 + 1/4 + 1/8 + . . . . . . 1/2N
first term(a) is 1/2
common ratio(r) is 1/2
last term (nth term) is 1/2N which is also a * rN-1
Hence sum is given by:
Sum of GP series = (1/2)*( (1 – (1/2)N)/(1 – 1/2) ) using formula : (a * (1 – rN)) / (1 – r) since r < 1
Sum of GP series = (1 – (1/2)N)

Let S = Sum of Arithmetico Geometric Sequence:

=> S = (1/2 + 2/4 + 3/8 + . . . . . . N/2N) …….(1)

Multiplying By 2, we get
=> 2S = (1 + 2/2 + 3/4 + . . . . . . . + N/2N-1) …….(2)

Subtracting the equation(1) from the equation(2), we get
=> S = (1/2 + 1/4 + 1/8 + . . . . . . 1/2N-1) – N/2N
=> S = sum of GP series – N/2N
=> S = (2 – (1/2)N-1)) – N/2N

Using the sum of the GP series and Arithmetico Geometric Sequence:

=> X = X*(1 – (1/2)N) + (2 – (1/2)N-1) – N/2N + N/2N
=> X = X*(1 – (1/2)N) + (2 – (1/2)N-1)
=> X*((1/2)N) = (2 – (1/2)N-1)
=> X = 2N+1 – 2

Now the above formula for X gives the number of trials requires getting N consecutive heads.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach``#include "bits/stdc++.h"``using` `namespace` `std;`` ` `// Driver Code``int` `main()``{``    ``int` `N = 3;`` ` `    ``// Formula for number of trails for``    ``// N consecutive heads``    ``cout << ``pow``(2, N + 1) - 2;``    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``class` `GFG{`` ` `// Driver Code``public` `static` `void` `main(String[] args) ``{ ``    ``int` `N = ``3``;`` ` `    ``// Formula for number of trails for``    ``// N consecutive heads``    ``System.out.print(Math.pow(``2``, N + ``1``) - ``2``);``}``}`` ` `// This code is contributed ``// by shivanisinghss2110`

## Python3

 `# Python3 implementation of the above approach`` ` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``     ` `    ``N ``=` `3`` ` `    ``# Formula for number of trails for``    ``# N consecutive heads``    ``print``(``pow``(``2``, N ``+` `1``) ``-` `2``)`` ` `# This code is contributed by mohit kumar 29`

## C#

 `// C# implementation of the above approach``using` `System;``class` `GFG{`` ` `// Driver Code``public` `static` `void` `Main() ``{ ``    ``int` `N = 3;`` ` `    ``// Formula for number of trails for``    ``// N consecutive heads``    ``Console.Write(Math.Pow(2, N + 1) - 2);``}``}`` ` `// This code is contributed ``// by Code_Mech`
Output:
```14
```

Time Complexity: O(1)

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