I haven’t run any simulations or seen any from a source I would trust (i.e. the Wizard), so my opinion that a CSM cannot be counted is based solely on logical conclusions not based on precise math.
Assume that in a six-deck CSM with all the discards (“muck” in the above sited article) placed back in the CSM immediately at the end of every round, the true count at the beginning of each round is zero. You have no knowledge of any previously-used cards when you place your bet – no cards have been shown for this round – they’re all in the CSM.
If there is a twelve or sixteen card “buffer” such that you can know for sure that the last twelve or sixteen cards from the previous round are not yet reentered into the shuffle, then whatever the running count is, you must divide that count by six, since you only know about sixteen cards from the entire six-deck shoe. (Technically, divide by 5.692308, since that’s the precise number of decks remaining in the CSM after sixteen cards have been shown.
5.692308 is a constant in this play, since the number of cards no longer in play will always be no more than sixteen. Assume that the dealer keeps more than one round out of the CSM and that he then mucks 20 or 30 cards, even 52 cards, and the lowest the divisor is going to get should be no lower than five.
The running count would need to be +20 before the true count would reach even +4. That running count of +20 would need to be reached by counting as few as sixteen cards.
That’s the thought experiment that I figure makes counting a CSM a waste of time. Maybe someone can think about it differently and present a scenario which sounds like it makes sense.
Someone else ought to be able to do some simulations that assume a “window” of 12, 16 or 52 cards in a six-deck shoe and present some real math that tells us.
Until that happens, I’m staying away from CSMs.