intuition behind discrepancy in expected number of coin tosses Yesterday evening I read a very interesting article on prime numbers which contained the following paragraph on coin tosses:

If Alice tosses a coin until she sees a head followed by a tail, and
  Bob tosses a coin until he sees two heads in a row, then on average,
  Alice will require four tosses while Bob will require six tosses (try
  this at home!), even though head-tail and head-head have an equal
  chance of appearing after two coin tosses.

I immediately tried to look into this question. Using the formula provided by André Nicolas, it can be shown that the expected number of coin tosses we need before getting $n$ consecutive heads is given by:
\begin{equation}
e_n = \sum_{k=1}^n\frac{1}{2^k}(e_n+k)+\frac{n}{2^n}
\end{equation}
For $n=2$, the expected number of heads($e_2$) is 6. Analytically, this makes sense once you become familiar with the above equation. Now, what I find interesting is that, as mentioned in the article, the probability of obtaining two heads is the same as the probability of obtaining a head followed by a tail:
\begin{equation}
P(HH)=P(HT)=\frac{1}{4}
\end{equation}
However, the expected number of coin tosses required to get the $HT$ pattern is 4 not 6. I still find this quite counter-intuitive. In fact, 
I ran a simulation using the following python code:
import numpy as np

head_tail = np.zeros(10000)
two_heads = np.zeros(10000)

for i in range(10000):
    z = np.random.randint(2, size=100)

    for j in range(100):
        if z[j] == 1 and z[j+1] == 0:

            head_tail[i] = j+2

            break

    for j in range(100):

        if z[j] == 1 and z[j+1] == 1:

            two_heads[i] = j+2

            break

And I noticed that their distributions behaved very differently:

It's by no means intuitive to me that the behaviour of these two distributions should be very different and yet they are remarkably different. Is there an intuitive reason why this must be the case?

Alice and Bob toss their coins until they see their first head.   The expected tosses until this happens is the same for each.   Then things are different.
Alice just needs to continue until she finally sees a tail.
However, if Bob doesn't immediately see a head on the next throw, he needs to start over from the beginning.
Thus $\mathsf E(A) ~{= \mathsf E(H)+\mathsf E(T)\\ = 2+2\\= 4}\\ \mathsf E(B)~{=\mathsf E(H)+1+\tfrac12\mathsf E(B)\\ =3+\tfrac 12\mathsf E(B)\\=6}$ 
