A coin is tossed 100 times. How many instances of at least 5 heads in a row do we expect to see?

Question

No overlaps. We are counting runs of at least 5, whereby for example a run of 6 does not count as 2 runs of 5.

I have received an answer to this question from someone with a PhD in Statistics, yet their theoretical answer does not agree with my code simulation.

Theoretically, the answer would be

$ \frac{96}{32} - \frac{95}{64} + \frac{94}{128} - \frac{93}{256} + \frac{92}{512} - ...$

since we expect $\frac{96}{32}=3$ instances of 5 heads in a row (if we allow overlap), and if we apply the Inclusion-Exclusion Principle, we can correct for double counts of runs of 6, 7, 8, etc...

The problem is, this answer is approx 2, but my coded simulation always results in about 1.5:

import numpy as np
import pandas as pd
def random_list(length):
    random_list = np.zeros(length) #list of "length" zeros
    for i in range(len(random_list)):
        random_value = np.random.random() #instantiate random value for each i
        if random_value > 0.5: #for approximately half of the random values
            random_list[i] = 1
        else:
            random_list[i] = 0
return random_list #random_list is now a random list of zeros and ones


def count_ones(array):
    runs = 0
    i = 0
    while i < (len(array) - 1): #iterate over each index in list
        if array[i] == 1: #find a value of one
            j = i + 1 #the next value is j
            while array[j] == 1: #iterate over indices until we hit a zero or end of array
                if j == (len(array)-1):
                    break #break out of loop if we are at the end of the list
                j += 1
            k = j #we now have either the first zero after a list of ones, or we are at the end of the list
            ones = k - i #how many ones in a row
            if ones >= 5:
                runs += 1 #count this as a run of 5
        else: #if array[i] == 0
            k = i # necessary so that code after if/else conditional runs
        i = k + 1 # loop will iterate over index after k
return runs


def average_runs(trials):
    results = np.zeros(trials)
    for i in range(trials): #do it a large number of times
        array = random_list(100)
        runs = count_ones(array)
        results[i] = runs
    average = sum(results)/len(results) #take the average
return average


average_runs(1000)

Can anyone explain why the simulation and theory do not agree?

Answer 1

There is no triple count here, because you cannot have $A_i$ and $A_{i+2}$ without $A_{i+1}$ (where $A_i$ means 5 consecutive heads starting in round $i$). So the answer is the expected number of runs of 5 consecutive (possibly with overlaps) minus the expected number of overlaps (which is exactly the expected number of runs of 6 consecutive): $$\frac{96}{32}-\frac{95}{64}\approx 1.52$$

Answer 2

The infinite sum seems like overkill to me.

A toss "starts a run" of five or more consecutive heads if it and the next four coins are heads, and the previous coin, if any, was tails (otherwise it is just part of a longer run).

So the very first toss has a $\frac{1}{2^5} = \frac{1}{32}$ chance of starting a run, the last four cannot start a run at all, and the remaining 95 have a $\frac{1}{2^6} = \frac{1}{64}$ chance of starting a run. Any run must start in exactly one place, so this covers all the possibilities. Total: 97/64 = 1.515625.

Answer 3

Here is another way to see that their "theoretical answer" is wrong (and Christophe Boilley's is correct, although verifying that our two expressions are exactly equal is not so easy).

What is the expected number of runs of exactly $k$? There are $101-k$ places this can start. In all but two of these, the probability of having a run of exactly $k$ starting there is $(\frac12)^{k+2}$, since we need $k$ heads with a tail either side. The two exceptions are when the run starts at the first coin or ends at the 100th coin, and these have probability $(\frac12)^{k+1}$. Thus the overall expected number of runs of exactly $k$ is $(103-k)\times(\frac12)^{k+2}$, for any $k\leq 99$. (For $k=100$ the probability is $(\frac12)^{100}$; what goes wrong with the above is that the "two exceptions" coincide.

Thus the total expectation of the number of runs of at least $5$ is $$\frac{98}{2^7}+\frac{97}{2^8}+\cdots+\frac{5}{2^{100}}+\frac{4}{2^{101}}+\frac{1}{2^{100}}\approx 1.516.$$