# Expected Number of Trials to get N Consecutive Heads

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)
+ (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 seperately we have:
For Geometric Sequence:

Sum of GP series = 1/2 + 1/4 + 1/8 + . . . . . . 1/2N-1
common ratio(r) is 1/2
Hence sum is given by:
Sum of GP series = (1/2)*( (1 – (1/2)N-1)/(1 – 1/2) )
Sum of GP series = (1 – (1/2)N-1)

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 = (1/2)N-1) – N/2N

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

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

## 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 `

Output:

```14
```

Time Complexity: O(1)

GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details

My Personal Notes arrow_drop_up If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : mohit kumar 29