Shannon entropy of a fair dice

To recover entropy, you have to consider a sequence of dice throws, and ask how many questions per roll you need in an optimal strategy, in the limit that the number of rolls goes to infinity. Note that each question can cover all the rolls, for example for two rolls, you could ask at some point: “Are the results in $\{16,21,22,23\}$?” (where the first digit denotes the first throw, and the second digit denotes the second throw).

I'm too lazy to do it for 36 possibilities, therefore here a simpler example: Consider a die for which each roll gives only one of three results with equal probability. Then the entropy is about $1.58496$.

For one toss, the optimal strategy is simply to ask “was it $1$?” followed by ”was it $2$?”, which on average gives $5/3 = 1.66$ questions.

For two tosses, an optimal strategy would be to first ask “was it one of $\{11,12,13,21\}$?” (where the first digit gives the result of the first toss, and the second digit the result of the second toss). If the answer is “yes”, then use two questions to single out one of the four results. Otherwise, ask “was the first toss a $2$?”, if yes then it was one of $22$ or $23$, and one question is sufficient to determine that. In the remaining case you know the first toss was $3$ and know nothing about the second, so you employ the one-toss strategy to determine the second toss.

This strategy needs on average $29/9=3.2222$ questions, or $1.61111$ questions per toss. Which is already much better, and indeed only $1.65\,\%$ worse that the value given by the entropy.

Note that the average number of questions of the single-toss optimal strategy can differ dramatically from the entropy. For this, consider the toss of a biased coin. The entropy of this can be made arbitrary low by making the coin sufficiently biased. But obviously there's no way you can get the result of a coin toss with less than one question.

In your setting, the Shannon entropy is "just" a lower bound for an entropy of any decision tree (including optimal ones). These don't have to coincide. To get closer to what the Shannon entropy is, imagine an optimal decision tree identifying outcomes of throwing a dice $N$ times with some large $N$ (assuming independence). The larger $N$ is, the smaller (yet nonnegative) is the difference between the "averaged" (i.e. divided by $N$) entropy of this "compound" decision tree and the Shannon entropy of the dice. (It resembles a background of arithmetic coding).

If you have $1$ die, there are $6$ possible outcomes. Label them 0 through 5 and express as a binary number. This takes $\lceil\log_2{6}\rceil = 3$ bits. You can always determine the 1 die with 3 questions, just ask about each bit in turn.

If you have $10$ dice, then there are $6^{10}$ possible outcomes. Label them 0 through $6^{10}-1$ and express as a binary number. This takes $\lceil\log_2{6^{10}}\rceil = \lceil10\log_2{6}\rceil = 26$ bits. You can always determine the 10 dice with 26 questions, just ask about each bit in turn. The average is 26 questions / 10 dice = 2.6.

If you have $100$ dice, then there are $6^{100}$ possible outcomes. Label them 0 through $6^{100}-1$ and express as a binary number. This takes $\lceil\log_2{6^{100}}\rceil = \lceil100\log_2{6}\rceil = 259$ bits. You can always determine the 100 dice with 259 questions, just ask about each bit in turn. The average is 259 questions / 100 dice = 2.59.

If you have $1000$ dice, then there are $6^{1000}$ possible outcomes. Label them 0 through $6^{1000}-1$ and express as a binary number. This takes $\lceil\log_2{6^{1000}}\rceil = \lceil1000\log_2{6}\rceil = 2585$ bits. You can always determine the 1000 dice with 2585 questions, just ask about each bit in turn. The average is 2585 questions / 1000 dice = 2.585.

Each order of magnitude gets you one more digit, converging toward the Shannon entropy.

On the other hand, with the decision tree in your example, you would not converge towards splitting the outcome space in half with each question. The first question $d_1 \in \{1,2,3\}$? does, but then there is waste if you have to ask two questions to determine 3 remaining outcomes. The second question (given a yes to the first), could be is either $d_1 = 1$ or $d_1 = 2$ and $d_2 \in \{1,2,3\}$?, which does split the outcomes space in half for multiple dice. Now you are forced to ask 3 questions to get the first die, but have gained information about the following dice. The strategy of enumerating and encoding the outcomes as above is just an extension of this idea. It doesn't pay off for a low number of dice, but does for many.