Expected Number of Trials to get N Consecutive Heads

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: 6Input: 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++
Java
Python3
C#
Javascript
// C++ implementation of the above approach#include "bits/stdc++.h"using namespace std; // Driver Codeint main(){    int N = 3;     // Formula for number of trails for    // N consecutive heads    cout << pow(2, N + 1) - 2;    return 0;}
// Java implementation of the above approachclass GFG{ // Driver Codepublic 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 implementation of the above approach # Driver codeif __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# implementation of the above approachusing System;class GFG{ // Driver Codepublic 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
<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>
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Time Complexity: O(1) 
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