Confusion about Typical Set concept The typical set is defined as the sequences which have a probability in $[2^{-nH(X) - \epsilon}, 2^{-nH(X) + \epsilon}]$. These are the sequences which give the average informativeness but certainly do not include the most probable sequences. For instance, for a Bernoulli distribution generates binary sequences with a probability .9 for 0's, the most probable sequence is all 0's but it is not included in the typical set.  However, the probability of having a sequence that belongs to the typical set goes to 1 as n (the number of elements generated or the length of the sequence) goes to infinity per AEP theorem. I understand the mathematical proof but this sounds counter-intuitive. How could a very long sequence generated by a Bernoulli process with .9 probability for the 0's to not contain with high probability the sequence of all zeros?
 A: There is one most probable sequence, the all-zeroes sequence. While other observed sequences have smaller probability than the all-zeroes sequence, there are so many of them that the probability that the observed sequence is some member (we don't care which one) of the typical set is very close to $1$, and indeed approaches $1$ as the sequence length increases. In contrast, the most probable sequence of length $n$ has probability $(0.9)^n$, and this probability is approaching $0$ as $n$ increases.  Read your last sentence carefully.

How could a very long sequence generated by a Bernoulli process with .9 probability for the 0's to not contain with high probability the sequence of all zeros?

The sequence (of length $n$) that is being generated cannot contain the sequence of all zeroes;  it is either exactly the sequence of $n$ zeroes, or it is not. The probability that the sequence being generated is exactly the sequence of $n$ zeroes is
$$P(000\cdots 000) = (0.9)^n$$  which is very small for large $n$.  The sequence being generated will certainly have lots of zeroes in it -- typically (pun intended) the sequence will likely be approximately $90\%$ zeroes and will be a member of the typical set. What it is very unlikely to be is the sequence of $n$ zeroes.  Don't be misled by LSN, the law of small numbers, which says that if you try it for $n=2$, then $P(00) = 0.81$ and the typical set better contain $00$ or else....  Try it for $n=1000$ when the most likely all-zeros sequence has probability $\approx 10^{-46}$ while the sequences in the typical set have even smaller probability than $\approx 10^{-46}$ but there are so many sequences (most of them having $90\%$ zeroes) in the typical set that the probability that the observed sequence is some member of the typical set (we don't care which one) is close to $1$.
