[ed: no, that's not right. With two cards, you can have red-back or black-red. I guess that'd translate to N bits for 2 n cards, but it's not quite trivial...
For four, we have: rrbb, rbrb, rbbr, bbrr, brbr, brrb etc]
I think I'm misunderstanding you - but say you have 2n cards, n red, n black. If you shuffle them, you have 2n random bits, read from first to last? (this is assuming it's feasible to actually randomly shuffle cards).
No, in this application cards are all distinct, not merely half red and half black. There are n! permutation of a deck of n cards, and a uniformly chosen random permutation contains entropy equal to the base 2 logarithm of n!.
Both choosing a random permutation given a supply of random bits and, as noted in the parent post, extracting random bits from a given random permutation are nontrivial computational problems leading to interesting algorithms; shuffling cards or tiles is a good practical shortcut to obtain a random permutation more directly.
For four, we have: rrbb, rbrb, rbbr, bbrr, brbr, brrb etc]
I think I'm misunderstanding you - but say you have 2n cards, n red, n black. If you shuffle them, you have 2n random bits, read from first to last? (this is assuming it's feasible to actually randomly shuffle cards).