Multi-deck SMs

If the segments that used to be associated with a deck behind the cutcard show a higher count when seen, than the the segments that used to be associated with a prior deck, you can be certain the probability of higher cards ACTUALLY being in the deck cutoff is higher;

Let's try to break this down. "the segments that used to be associated with a deck behind the cutcard". Just as an example, we say that the first deck in the shoe and the deck behind the cut card are two segments that had been shuffled together. They were subsequently broken apart after being married. Half of this two deck section ended up behind the cut card and half remained in the front of the shoe. Let's also say that the deck in the back of the shoe, just prior to the cut card, has the relationship with the second deck in the shoe. You are saying that if the first deck in the shoe shows a higher count (meaning more low cards dealt) than the second deck of the shoe, then it means that there is a greater probability that there are more high cards behind the cut card than there are in the deck prior to the cut card.

I disagree. This infers that each of the two deck segments, 1&6, and 2&5 were balanced with each other before the shuffle. The excess small cards that you see dealt in the first deck of the shoe indicate a richness of high cards distributed throughout the entire remaining five decks. They do not indicate, or give a greater probability of, those excess high cards being among the deck with which this deck was shuffled.

What you are essentially saying is this. Take any six deck shoe and remove two sections of two decks (104 cards) each and put them next to each other on the table. From each of these two deck sections count the first 52 cards. You are saying that which ever of these two stacks has the higher running count for the first 52 cards has a greater chance of having an excess of high cards among cards 53-104. I see no logical reason for this to be so.

You are forgetting how you have other simultanius information from seeing the way that other segments ended. I assume you are familiar, in differential equations, with examining the Wronskian of a differential system, to see if the elements have an algebraic solution (??).

I know that sounds really complicated but it is simpler and more general to discuss that way than talking about "suppose you have segment A...." further. You can show that you can have a system with an excess of unknowns, in this case two segments, and missing one unknown (you havn't counted the two segments yet), and still have an valid inference on the simple difference in projected total points between that ending segment and the segment behind the cut-card. It is an excercise in one problem in the revision of, How to Ace the Rest of Calculus, 2001 edition (boy did they do a good job this time).

Once again I point out before proceding that you are forgeting all the information you have by continuing to count along.

Now all it takes is some information about the original segments and what they total. And that information is staring everyone in the face from the prior shoe: how many cards sit on the top of the cutcard where the dealer inserted it into the discard rack? You can guage the number of cards, and just ask the table who ended-up with the joker last round? I doubt that there is much heat in that? You can ask did she have to pull alot of cards past the joker to beat you guys last go round?

The number of cards is enough to make an inference on the difference between the last cards used and the original plugs, the number of count points that is. Now can you see where there is a partial solution available---and I appologise about leaving this in general terms but I can't confuse this the way some of your buds might want to.

The trouble is that once you get the point you are probably going to be better at explaining it than I can.

And even going back to Hall's postings on shuffle tracking there is suprising profit from knowing that high cards are simply more or less likely to be available for play or beind the cut-card.

There are other clues to getting a partial solution as well that admittedly involve some math tricks. But most important of all they allow some extra information to come for shuffle tracking the next shoe. Normally in shuffle tracking you don't care about the shuffle point except as having a deeper shuffle point gives you more information that does not have to be averaged over the entire region of the plugs (insert your own term if you wish). Well some use of this can have you at least having a good guess past the cut card each shoe.

Even during the first shoe, with the prior shoe sight unseen:
(a) A Tracker can probably note several shuffles where there is definite solutions to the total points in the last deck and deck that is behind the cutcard. Experienced trackers will note these are usually shuffles where some segments have more riffles than others.
(b) (This is why I mentioned the new How to Ace Math books, which are great in their December 2001 revisions) An indeterminent system is not always indeterminent for all variables~segments toi solve.
(c) A memory of cards flashed during the shuffle can be used to apply cut control lite to appropriate segments when they are dealt.
(d) The same can result in partial algebraic solutions.
(e) Even the cards used used to complete the final round in the last shoe can be used---just ask the table "how many cards did she need to draw to beat you guys in the last shoe?" as in the pro's pseudo count system---can be enough to partially solve for some segments.