Skip to content
Related Articles

Related Articles

Improve Article
Expected Number of Trials to get N Consecutive Heads
  • Difficulty Level : Hard
  • Last Updated : 17 May, 2021

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 
 

 

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
 

For Arithmetico Geometric Sequence
 

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

Javascript




<script>
 
// Javascript implementation of the above approach
 
// Driver Code
     
    let N = 3;
   
    // Formula for number of trails for
    // N consecutive heads
    document.write(Math.pow(2, N + 1) - 2);
       
</script>
Output: 
14

 

Time Complexity: O(1)
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live




My Personal Notes arrow_drop_up
Recommended Articles
Page :