Formula for proportion of entropy Let's say we have a probability distribution having 20 distinct outcomes. Then for that distribution the entropy is calculated is $2.5$ while the maximal possible entropy here is then of course $-\ln(\frac{1}{20}) \approx 3$.
How can I describe that $2.5$ is quite a high entropy given the number of possible outcomes? It is hard to get a feel for it just stating that it is $2.5$, my gut tells me it would be ok to simply divide the entropy by the maximum entropy possible resulting in a number between 0 and 1; in this case $\frac{2.5}{-\ln(\frac{1}{20})} \approx 0.83$. Is this a valid way of calculating it (since this is not a linear but logarithmic scale)? Has this been done before?

First of all, I would use log base 2 instead of natural log because it's easier to talk about its meaning as the number of yes/no questions on average to guess the value.
Given 20 choices, the maximum entropy distribution has entropy of 4.322 bits. While your distribution has 3.607 bits, which is 83% of the maximum possible value.
Of course you can normalize it, but I do not recommend normalizing it to be between 0 and 1 because it will lose its value. Without a context, it is not clear what this normalized entropy means.
