# Can't understanding information entropy and data compression, for beginners

I watched a very well done video about information entropy, and I thought I got the concept in my mind but then I asked my self a question that denied every certainty I had. The video if you are interested: https://www.youtube.com/watch?v=2s3aJfRr9gE&t

My question is about data compression, image these 2 strings:

String1: $\mathbf{AAAABBBB}$ and String2: $\mathbf{ABABABAB}$

In both string I have the same number of $\mathbf{A}$ and $\mathbf{B}$ so the chance of getting an $\mathbf{A}$ is the same of the chance of getting a $\mathbf{B}$ in both strings.

With only 2 symbols I can make a Huffman encoding of 1 bit each. This means every symbol has "weight 1"

I will call $\mathbf{N_1} := number\,\, of\,\, bits\,\, for\,\, encoding\,\, A$

And $\mathbf{N_2} := number\,\, of\,\, bits\,\, for\,\, encoding\,\, B$

And $\mathbf{P(N_i)} := The\,\,probablity\,\,of\,\,this\,\,simboly$

In this case $\mathbf{N_1=N_2=1}$

So the entropy is calculated in this way: $$\mathbf{ H=\sum_{i=1}^2 i = N_i\,P(N_i)}$$

So for both strings the result is $\mathbf{N_1\,P(N_1) + N_2\,P(N_2) = 1\cdot0,5 + 1\cdot0,5 = 1}$

But string1 is clearly more easy to compress with RLE, I can say: I have 4 $\mathbf{A}$ and 4 $\mathbf{B}$ -> 4A4B.

So if I encode $\mathbf{A}$ with 1 and $\mathbf{B}$ with 0 the result will be 4140, also if the number of $\mathbf{A}$ and $\mathbf{B}$ is very large but symmetric I can say $\mathbf{n1n0}$ and $\mathbf{n}$ can be every big number I can image.

I can't make the same with string2.

So WHY if they have the same entropy it seems that string1 is more easy to compress than string2? Where is my fault? Thx :)