A basic data compression course would have sufficed too, but as you say, you don’t have any knowledge of the basics of information theory or data compression. You sound like an over-eager yet ignorant grad student who insists he has found flaws in the professor’s lecture material.
This isn't how you calculate Shannon limit fwiw. If all symbols are of equal frequency simply take the total number of different symbols and simply log2(different_symbol_count). That's it.
So say you had 60 different permutations. Each of those is a symbol in entropy encoding. Each individual symbol takes log2(60) bits to store using arithmetic encoding. Which is exactly the statement and correct calculation you had for calculating the Shannon limit in the very next line :). As in the Shannon limit for 60 different symbols is absolutely 5.9bits not 9bits
Instead you have some weird calculation here that seems to take individual probabilities and add them back up again to give 9bits. That's very far off and incorrect.
This seems to do something similar: https://en.wikipedia.org/wiki/Context-adaptive_binary_arithm...