Note 1

This value was :  0.68
We could wonder what this value would be when we had this same 2222 distribution while having only four boxes A, B, C, D, and thus having the distribution

Well, let us compute the corresponding H value :

The distrubution is evidently an equilibrium distribution.
P_eq[A] = P_eq[B] = P_eq[C] = P_eq[D] = 2/8 .

Box A.
P[A, t] = 2/8 .
P_eq[A] = 2/8 .
P[A,t] / P_eq[A] = (2/8) / (2/8) = 1 .
log ( P[A,t] / P_eq[A] ) = log 1 = 0 .
P[A,t] . { log ( P[A,t] / P_eq[A] ) } = (2/8)(0) = 0 .

The same goes for the remaining boxes (B, C, and D).
So the H value for this distribution is
0 + 0 + 0 + 0 = 0 .

Let us compute it :

Also this distribution is evidently an equilibrium distribution.
P_eq[A] = P_eq[B] = P_eq[C] = P_eq[D] = P_eq[E] = P_eq[F] = P_eq[G] = P_eq[H] = 2/16 .

Box A.
P[A, t] = 2/16 .
P_eq[A] = 2/16 .
P[A,t] / P_eq[A] = (2/16) / (2/16) = 1 .
log ( P[A,t] / P_eq[A] ) = log 1 = 0 .
P[A,t] . { log ( P[A,t] / P_eq[A] ) } = (2/16)(0) = 0 .

The same goes for the remaining boxes (B, C, D, E, F, G, and H).
So the H value for this distribution is
0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 = 0 .

So while the distribution  2 2 2 2 0 0 0 0  corresponds to a H value larger than zero (namely 0.68) because it is a non-equilibrium distribution, the distributions  2 2 2 2  and  2 2 2 2 2 2 2 2  have zero H values, because they are equilibrium distributions.