Assume I am counting [3-6] +1 and [T,J,Q,K] -1.

6 deck shoe, 4 decks exposed, and running count is +3.

I then get dealt 6-T, and dealer shows 7.

How would I go about deriving the expected number of cards remaining for each rank?

There are 2 decks remaining at the start of the hand. That means 8 of each rank.

Since we see a 6, 7, and T we can subtract one of each. Since the count is +3 we can subtract 3/4 of a card from 3, 4, 5, and 6. We have to add the +3 back into the other 9 cards so they each get +1/3.

That leaves us with expectations of:

8.33 deuces
7.25 threes
7.25 fours
7.25 fives
6.25 sixes
7.33 sevens
8.33 eights
8.33 nines
7.33 tens
8.33 jacks
8.33 queens
8.33 kings
8.33 aces

That gives us 101 expected cards or 2 decks minus the 3 we saw from those 2 decks. This works unless I mis-understood how you're using a term.

I agree that #s remaining for [A,2,8,9] are equal, and #s remaining for [3,4,5] are also equal. Also, to keep it simple I consider [T,J,Q,K] as the same rank, so I am only concerned about the total # remaining for [T,J,Q,K].

If your method were correct, # of 6s for a running count of +30 would be (8 - 1 - 30/4) = -0.5. That is impossible.

I think the solution requires a set of simultaneous equations for the following unknowns:
# for [A,2,8,9] = A
# for [3,4,5] = B
# for [6] = C
# for [7] = D
# for [T,J,Q,K] = E

The first equation is trivial, A+B+C+D+E = 101. The next equation involves the running count, so (72-B) + (24-C) - (96-E) = 3. After that I run into trouble. I would think A and D involve a binomial distribution, with A slightly more likely than D.

For the record, I ran a simulation of 50,000 shuffles, and I get this distribution:

A: 7.857
B: 7.465
C: 7.130
D: 7.524
E: 32.525

Looking at the above simulation (and another with 3 decks remaining), I believe I have 4 of the equations, 1 short.

4A + 3B + C + D + E = 101
3B + C - E = -3
B - C = 101/312
A - D = 101/312

Where:

# for [A,2,8,9] = 4A
# for [3,4,5] = 3B
# for [6] = C
# for [7] = D
# for [T,J,Q,K] = E

I am guessing on this equation, but it looks like it might be close enough:

A = 101/4 + 101/(4x312)

So, I get the following 5x5 matrix:

Inverting, and multiplying by the solution vector, I get:

This also behaves well with 153 cards remaining, compared with the sims.

The value of this is constructing a precise expectation of cards remaining for any count scenario and # of cards remaining. Playing strategy can then be determined with far greater accuracy than by simulation or general expectation of cards remaining. The important parameters are then:

-cards remaining
-running count
-size of shoe (# of cards, 6 decks = 312)
-suited-independent # of cards, typically 13 (Spanish is 12)

It also needs to be generalized for any possible set of cards dealt to the player and dealer, and counting algorithms.