Entropy of a block of characters I have a question about the following statement about entropy:

If a source provides us with a sequence chosen from 4 symbols (say A, C, G, T), then the maximum average information per symbol is 2 bits. If the source provides blocks of 3 of these symbols, then the maximum average information is 6 bits per block.

In the first case, each symbol has probability 1/4 therefore the entropy is 2 bits on average -(4*1/4*log(1/4)).
The second case where we are sending blocks of 3 symbols is unclear. What is the meaning of sending block of 3 symbols? Are the symbols always different or could be the same? How do we calculate the entropy of group of symbols from a set?    
 A: By block, I assume they mean something like [AAA] or [ATC]: a set of several concatenated symbols.
We know, from the definition of joint entropy, that:
$$H(X,Y,Z) = - \sum_x \sum_y \sum_z P(x,y,z) \log_2(P(x,y,z))$$
so you could calculate it directly, assuming you knew the joint distributions for $X$, $Y$, and $Z$. 
Entropy is "subadditive", that is
$$\begin{align} H(X) + H(Y) \lt H(X,Y) \hspace{0.1in} \textrm{if }& X \textrm{ and $Y$ are dependent} \\
H(X) + H(Y) = H(X,Y) \hspace{0.1in} \textrm{if }& X \textrm{ and } Y \textrm{ are independent}
\end{align}$$
If you're willing to assume independence, the the entropy of a block is easy to calculate: it's the block length * the entropy of one element, so 3 symbols/block * 2 bits/symbol = 6 bits/block. If they are not independent or you do not know the joint distribution, then all you can say is that the entropy must be six bits or less. 
A: The symbols could be both different and the same (any of the 64 triplets: AAA, AAC, ..., TTT). If the MAI (maximal average information) of one symbol is 2 bits, the MAI of 3 symbols is 3*2=6 bits.  Another way to calculate the information is $ -(64*1/64*log_2(1/64))=6 $
