Technically yes. But the method is by far strong enough that this isn't an issue. This is sort of always the issue with calculating entropy. We say that password has less entropy than 8(A>Ni'[. But that is baking in assumptions about the search space. If password is a randomly generated string of lower, upper, numbers and symbols it is just as secure as the latter. (80^8^ ≈ 10^15^ candidates) but if password was generated as just lowercase characters it is far less secure (26^8^ ≈ 10^11^ candidates) but if it was a random dictionary word it is not very secure at all (≈ 10^5^ candidates) and if it was chosen as one of the most popular passwords it is even less secure. How can one password have different entropy?

The answer is basically it matters how the attacker searches. But in practice the attacker will search the more likely smaller sets first, then expand to the larger. So the added time to search the smaller sets is effectively negligible.

What may be more useful is the "worst case" entropy. Basically the attacker knows exactly what set you picked. For the password case that is 1 because I just picked the most common password. For the rolling method described above it is 6^5^6^ ≈ 10^23^ because even if they know the word list they don't know the rolls. You may be able to go slightly higher by building your own word list, but the gains will probably be fairly small and you would likely get far more value just by rolling one more word on the existing list than spending the time to generate your own.