Studying the patent for 7,137,627

Ringer · Jun 1, 2007

In studying the patent for the primary shuffling device (CSM) used at my local casino (I'd rather not say which one at this time), I have found some points of interest for those of you that desire to beat these confounded things.

Let me first pose my hypothesis:
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I am working on a running theory about CSM's in blackjack. It is my hypothesis that card counting is equally effective on a CSM so long as you figure: TC = RC / # of Decks

After much research we know that most CSM's have between 15 and 20 compartments to which each card gets shuffled individually (one at a time) into a compartment at random. Then one of those compartments is chosen randomly for the next hand of blackjack.

I believe though that we could say if two or more compartments are chosen in a row with low cards, the remaining compartments must have a higher concentration of high cards.

So here's my question about the "odds".

If we delt out 15 piles of cards with 5 decks onto a table, then removed the two piles with the most amount of low cards, redistributed those piles one card at a time (randomly) between the 15 piles again, what are the odds that the next pile chosen will have a high concentration of high cards? Also, how are the odds of this affected over the number of hands delt?

I believe the odd's calculated for this could lead to a breakthrough on how to count on the CSM. It might require a new counting scheme, but it could also just equal out to nothing. I am prepared for either result, the rewards are quite worth the time IMO.
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Anyway, moving on to the patent.

Section 6 of the topic "Device And Method for Continuously Shuffling And Monitoring Cards", line 29-41 states:

"According to the present invention, the operation of the apparatus is continuous. That is, once the apparatus is turned on, any group of cards loaded into the card receiver will be entirely procesed into one or more groups of random cards in the compartments. The software assigns an identity to each card and then directs each identified card to a randomly selected compartment by operating the elevator motor to position that randomly selected compartment to receive the card. The cards are unloaded in groups from the compartments, a compartment at a time, as the need for cards is sensed by the apparatus. Thus, instead of stopping play to shuffle or reshuffle cards, a dealer always has shuffled cards available for distribution to players."

What I have put in bold is extremely important to us. The cards come out one full compartment at a time. The cards are not individually selected one at a time randomly from any of the varying compartments, instead one compartment at a time is selected. So, how many cards are in each compartment?

Section 17, line 40:

"Most preferably, the microprocessor is programmed to skip compartments having seven or fewer cards to maintain reasonable shuffling speed."

Reading further into the 4 deck model (although not for our 5 decks it still gives us some good numbers to work with):

Section 17 line 53:

"Maximum number of cards/compartment: variable between 10-14"

And

Line 59:

"Number of cards in the second card receiver to trigger unloading of a compartment: variable between 6-10"

What we can take from this is that each time the "shoe" has 6-10 cards in it, then a compartment will be emptied into the shoe. This puts the varying rate of cards waiting to be played between 6 and 24.

This changes the number of cards still stored in the compartments to a reduced amount. For those of you working on "latency" of the machine, this could be quite helpful since you have a latency of both the cards being placed back into the system as well as waiting for the dealer.

So just after a hand has been delt. Lets assume 6 players with an average of 3 cards per person, that is 21 cards just drawn from the machine. Assuming the best of circumstances there will be 24 more cards in the shoe giving us a total of 45 cards not in circulation prior to the hand just delt being recycled.

With 5 decks that means 21 cards are being reshuffled amongst a total of 215 cards.

All that info I'm sure you "latency" guru's might find helpful.

But I'm going to be biased a bit toward my theory for just a moment.

Since the compartments are basically storage area, I find it no different than if the dealer simply took the 21 cards played and randomly inserted them into a stack of 215 cards. This is not sufficient to break up a grouping of 10's and Aces. To take the situation to it's most potentially benefitial point, lets say 21 low cards were just played and in wait another 24 low cards are in the shoe. That would increase the likelyhood of grouped tens and Aces even further making our insertion of 21 low cards even less impactufull. (There are now only 55 low cards spread out within 215 cards raising the density of high cards remaining to just under 75%!

Because a grouping is inevitable to occur, the predictability might change slightly from that of a cut card shoe, but I suggest it is no different than keeping the odds based on a all decks still included in the CSM since all decks (minus the cards awaiting to be played in the shoe) are in the stack.

Even better than a cut card shoe, here we have the benefit of ALL cards eventually being played out over time. There is no way for the deck to be "reshuffled entirely" when a high count still exists as there is in a cut card shoe. It will take a longer period of time for all cards to eventually be played because of the nature of the CSM. That does not negate the fact that the card will still be played!

If there are 15 compartments, 2 have been emptied and thus cannot be chosen to be put into the shoe, and we witness two hands in a row with a majority of low cards, we can safely say that either all 13 remaining compartments will have a slightly higher concentration of high cards or two compartments will have an extremely high concentration of high cards.

Either way, once the low cards have been played and our card count goes up we will have an advantage over the house, either in the long run over the next 13 hands or the short term with a high density of high cards delt right away. The third option is of course that we maintain an equality with the house until the final two of the 13 compartments are chosen which means unfortunately by that time most of the low cards that have been delt will now be fully back into circulation. There is still a very low impact on those compartments with high cards since once they reach a maximum amount(10-14) the machine cannot put any more cards into the compartment.

I'm sure some of you can find a flaw in my logic. I will be happy to discuss any of them at length until we can all possibly come to a conclusion either yay or nay. I do also realize that for a lot of this I am using the optimal conditions. Obviously it's extremely unlikely to see 21 low cards in one hand since most of the players will have to take more than one hit, esspecially the dealer. But still, the more low cards we see played, the higher the density of high cards left in the stack. Whether they all get placed one at a time back into the deck or not, the statistical advantage is still there.

bigbjfan · Jun 2, 2007

Ringer,

I did the majority of my research on the CSM about a year ago. I would count and keep track of the number of rounds until the 5 decks were close to exhausted based on the number of players and the average of 2.5 hands per. I think we are agreeing on how we think these CSM's work. My post at that time is here if you are interested in viewing:

http://www.blackjackinfo.com/bb/showthread.php?t=2629

Ringer · Jun 2, 2007

bigbjfan said:

I have read every thread here that had CSM in the thread. (gotta love the search function) Your thread was of interest to me as it seems to follow my observations. The trouble with both of our comments was the lack of real data and time that we could put into it.

Sonny refuted both what you were saying and what I was saying with basically the same suggestion, that the cards placed into the machine from the previous hand were again immediately available to the next hand. The patent itself says the shoe contains no less than 6 cards.. at which point if it gets that low the machine will randomly select a compartment and dispense that compartment into the shoe (or que if you will) for use in the next hand. If the shoe is already full with 24 cards (this particular machines maximum in the shoe) then there is no way for any of the cards just delt to be played in the next hand. The cards that have just been played are placed one at a time into random compartments, but if no compartments are emptied into the shoe then it's impossible for those cards to be played.

Anyway, I"m not here to just refute what Sonny says as I think he might be able to bring in some good insight.

I think if you were to keep the cards in piles instead of an elevator and compartments, the CSM doesn't seem quite as unbeatable. It's just doing a simple set of functions that we could do ourselves without the machine. It's not stacking the odds against you through some mathematical algorithm, it's just sorting out the cards just used, back into the piles we have laid out on the table.

Surely that system is quite beatable.

halcyon1234 · Jun 2, 2007

And I'm going to take you up on that. What follows is a lengthy post that I spent a good deal of the night working on. It's easy enough to say "you can't do this on a CSM". But I wanted to do the math and come up with the figures anw know for sure one way or the other.

This is an indepth dissection of what you are proposing. It takes into account the techniques you are theorizing, the CSM technical information revealed, and the statistics of Blackjack. If you are serious about persuing your system, read the entire post. Either it will dissuade you, or show you where the serious flaws are in your reasoning and put you on a different path that no one's come up with before.

First, the hypothesis:

Ringer said:

By definition, this is always true:

Let N = number of decks used
Let d = number of decks dealt

TC(d) = RC / (U - d)

A CSM is TC(d) where d always = 0. This isn't a re-defining of TC(d). It is just limiting the domain of TC(d) to d=0.

Ringer said:

Imagine it this way. Below are the composition of imaginary decks. They contain one of three cards: H, L or N.
H represents a significant majority of high cards, and is worth 1.
L represents a significant majority of low cards, and is worth -1.
N represents a significant majority of normal cards, and is worth 0.

A new, undealt deck is composted of HLN, and is worth 1 + 0 + (-1) = 0.

If we take 5 of these decks, we get:

Deck - Composition
1 - HLN = 0
2 - HLN = 0
3 - HLN = 0
4 - HLN = 0
5 - HLN = 0
= 0

Now after a some playing, there are two decks that have a high concentration of low cards. This must mean that the rest of the decks contain that deck's neutral and high cards. A scenario:

Deck - Composition
1 - HNL = 0
2 - HNN = 1
3 - HNN = 1
4 - HLL = -1
5 - HLL = -1
= 0

So we're going to take decks 4 and 5, and randomly redistribute them across decks 1, 2 and 3, evenly, so that we get 3 decks of 5 widgets. There are 2 H and 4 L to relocate. Each deck MUST get at least 1 L, leaving over 2H and 1L

1 - HNLL = -1
2 - HNNL = 0
3 - HNNL = 0
x - HHL = 1
= 0

There are 6 ways of distributing them across 3 decks (using a lower case h to illustrate this)

hHL
HhL

hLH
HLh

LHh
LhH

Since decks 2 and 3 are the same, the first 4 are the same as each other, and the last two are the same as each other. This means:

HHL occurs 4/6 times (2/3)
LHH occurs 2/6 times (1/3)

Scenario 1.1 (2/3 of the time)
1 - HNLLH = 0
2 - HNNLH = +1
3 - HNNLL = -1
= 0

Scenarion 1.2 (1/3 of the time)
1 - HNLLL = -2
2 - HNNLH = +1
3 - HNNLH = +1
= 0

So we have 3 scenearios, each with 3 decks, meaning 9 possibilites. Each are equally likely to occur:

Deck 1.1.1 = 0
Deck 1.1.2 = +1
Deck 1.1.3 = -1

Deck 1.1.1 = 0
Deck 1.1.2 = +1
Deck 1.1.3 = -1

Deck 1.2.1 = -2
Deck 1.2.2 = +1
Deck 1.2.3 = +1

= 0

There are 4 cases of +1, and the odds of selecting any of them are 1/9
There are 0 cases of +2, and the odds of selecting it is 0/9
There are 0 cases of +3, and the odds of selecting it is 0/9

There are 1 cases of 0, and the odds of selection it is 1/9

There are 2 cases of -1, and the odds of selecting any of them are 1/9
There are 1 cases of -2, and the odds of selecting any of them are 1/9
There are 0 cases of -3, and the odds of selecting any of them are 0/9

4*1*(1/9) + 0*2*(0/9) + 0*3*(0/9)
+ 0*0*(1/9)
+ 2*(-1)*(1/9) + 1*(-2)*(1/9) + 0*(-3)*(1/9)

= 4*1*(1/9) + 2*(-1)*(1/9) + 1*(-2)*(1/9)

= 4*(1/9) + (-2)*(1/9) + (-2)*(1/9)

= (4/9) + (-2/9) + (-2/9)

= 0

OR

You get a positive 4/9
You get a neutral 1/9
You get a negative 4/9
= 9/9 = 1

The numbers in this scenario can be shuffled, more decks added, or anything else you desire-- and you'll get the same conclusion. Yes, indeed, there will be some high-count "pockets". But, by the very definition of count, there will be a balancing number of low-count pockets. And the odds of drawing any one of those pockets are exactly the same. And the player-edge gained by drawing a high-pocket is taken away by the player-edge lost to drawing a low packet.

The only reason why card counting works, is because the composition of the deck changes, and the odds of drawing a "packet" of high cards becomes disproportional to the odds of drawing a packet of low cards.

The flaw you are running into is:

You do not know the composition of the undealt decks. This means you cannot say they are balanced (HLN) or unbalanced (HHL, etc). If an L comes out, you can accurately assume that there is now an unbalance of H remaining in the CSM. But when that L gets fed back into the CSM, you do not know if the L was put into a balanced deck, or an unbalanced deck.

This about it this way:

Let TD be the total number of undealt decks
Let LD be the number of unbalanced-low decks
Let HD be the number of unbalanced-high decks
Let ND be the number balanced decks

If an L comes out and is put back in a way as to create an LD, there must be an HD because:

TD = ND + LD + HD

(ND/TD) + (LD/TD) + (HD/TD) will always equal 1.

Conclusion:

Because there are always a balanced number of cards remaining in a CSM, and because cards are divided into compartments, and because each compartment has an equal probability of being selected, there is no way to get a desired composition a disproportional amount of time.

In order to break the machine:

1) The machine would have to favour one (or more) compartments and deal from it more often than the others
2) You would have to know the method it uses to distribute cards that are fed-in
3) You would have to recognize exactly which cards are put into that compartment, and count them
4) You would have to know when the machine is about to deal from that compartment (which you can only know by recognizing cards going in)

You would then have to watch a table a do this:
1) Count the hand.
2) Track the cards going into your favoured compartment (their exact value and order) by applying the machine's shuffle method (see #2 above) to the ordered discards.
3) Once the cards you recognize come out of the CSM, you know that the machine is dealing from the favoured compartment, and that it is now empty of all unknown cards. (You cannot play if there are any unknown cards)

Now that you know the favoured compartment is of a good composition, you can play. And it will only be worth playing if the player edge of the favoured compartment, times the likelyhood of it coming up, is greater than the house edge of the rest of the CSM combined (times the likelyhood of a non-favoured compartment coming up).

In other words:

Let
x = number of compartments
N = total number of decks inside CSM
n = number of decks inside the favoured compartment
u = percentage of time the fav compartment is selected
RC = Running count of cards placed into the compartment only.
E = player edge per count.

Notes:
n: This can be a fraction or decimal, if the compartment doesn't hold a whole deck. Ideally, it is N/x.

u: 100% of the time = only that compartment is dealt = 1. A perfectly balanced CSM has u = 1/x (ie: each of 15 compartments has 1/15 chance of being chosen)

RC: It is important to only track the RC in this manner, because you are only interested in that one compartment. You want to know when it's "good". The true count of that compartment is then RC/n, while the true count of the rest of the CSM is -RC/(N-n). 0 = RC/n - RC/(N-n)

E: We'll assume the universally accepted 0.5% (0.005)

Thus, your edge becomes:

Edge =
[E * ((RC/n) - 1) * u] - [E * ((-RC/(N-n)) -1) * (1 - u)]

(The edge of the "deck" in the favoured compartment, times the probability that it will come up, minus the edge of the rest of the shoe, times the probability that the favoured won't come up)

(Note: I tried to factor this down, but the number of variables makes it a bit complex)

The above formula is different than just a plain old CSM, because u=0 doesn't mean no edge. In "random distribution", you don't know which of the remaining piles has a clump, if any. In this "one section tracking" method, you are able to tell when one particular section has a clump, because you are tracking one section for sure.

*IF* you are able to do this-- track the composition of one compartment, and you know how often it comes up, then you can gain an edge. Here's some numbers:

Assume:
15 compartments
5 decks
therefore, 0.3 decks / compartment

If you can track that one compartment, then your edge is:
E(RC, u)
0<u<1 (because a compartment's range is "never selected" to "always selected")
-(n*52)<RC<(n*52) (since, there can only be so many cards in the favoured section)

(For simplicity sake, we're rounding 17.3333 to 17, and refer to 1/x as Q)

And yes, a negative count inside the section is valuable, because it means a positive count OUTSIDE, and if the edge outside is greater than the chance of the favoured section being selected, it's worth it.

So, this means your edge during certain circumstances is:

*E(-17, 0) = 1.36% (A section full of low cards is never selected)
*E(-17, Q) = -0.57% (A section full of low cards is equally selected)
*E(-17, 1) = -27.50% (A section full of low cards is always selected)

*E(0, 0) = -0.50% (A neutral section is never selected)
*E(0, Q) = -0.57% (A neutral section is equally selected)
*E(0, 0) = -1.50% (A neutral section is always selected)

*E(17, 0) = -2.36% (A positive section is never selected)
*E(17, Q) = -0.57% (A positive section is equally selected)
*E(17, 1) = 24.50% (A positive section is always selected)

*nb: I will gladly upload my Excel spreadsheet so my figures and calculations can be doublechecked

So the highest edge you could get is 24.5%, by betting when the RC in the shoe is high, and you are guarenteed to get it.

If there was only a slight bias-- say the favoured section is twice as likely to be selected ((1/x) * 2), if that section is full of high cards, you are only a 1.22% favour. This is the same as straight counting and betting at TC +3.

So there you go. *IF* you can perfectly track the way the machine shuffles and you can *PERFECTLY* predict when a favourable section will be used, then you will get at most 24.50%.

*IF* you can perfectly track the way the machine shuffles, and you have to play every hand because of an untrackable but quantifiable bias, you get around 1.22% at most.

*IF* you can figure out which compartment is going to be used, but you don't know the count of it, you are at -0.5%, which is exactly the same as just playing without any system.

One thing I can't quite calculate would be if you could sequence the shuffle. That is, if you know not only which cards go into Section X, but also in the exact order, then you could gain a huge advantage, if you could control the table. You'd know exactly when a Blackjack was coming, exactly what the dealer hand was, etc. It could theoretically be a near 100% advantage. BUT in order to do this, you need to track every card in the discad tray (which you need to do for a section track), track every section, know exactly which section will deal which card, and be able to figure out exactly the best way to play and bet-- all in the 5 seconds or so between hands. AND nothing can change-- the machine can't change its shuffle pattern, drop a card, get a deck changed, nothing. To be perfectly honest, if you had this kind of brain, you'd already be a mutli-billionaire somehow.

As for the patents:

There's no information as to how big of a group is taken from the compartment, where it's taken from, how many compartments are used, in what order, or when the "need" threshold is.

No, it says they come out in groups. It doesn't say an entire compartment. It could be half a compartment, a full compartment, 10 cards, or some random number each time.

That is useful if you can track when a section is low on cards, because then it won't be selected and it will increase the odds of a "good" section being selected. But just by a small percent.

Is this a constant set at boot time, or a variable that changes on the fly? Without knowing that, you can't accurately track a section because you'll never know if card 11 goes into your section or not. Another piece of information needed to calculate the algorithm in your head.

Two issues:
1) Ambiguous wording. Will it unload a whole compartment, or like it said before, a "group of cards".
2) You can't see the tray, and you can't see how many cards are added to the tray. Therefore, you won't know when new cards are added, or how many until the next grab.

The number of cards in play is unimportant. It's the order/composition of the cards being put back into the compartments, and the odds of a good compartment being selected (by biased selection, or by you knowing the order of compartments)

Sure it is. If all 21 cards were High, and the dealer has 15 spots to insert those cards into, then each spot will get, on average, 1.4 high cards. Each spot has a 1/15 chance of being hit, 21 times. There may be some clumping, not consistently.

Yes, a grouping is inevitable to occur-- statistically speaking. It's also statistically LOW that a large clumping will occur, and thus (as I've shown many times above), the odds balance out. Clumping is valuable when it occurs, but its rarer occurance lowers its value-- because of the high-number of non-clumps. Since you can't predict the clumping, you cannot use it.

CONTINUED IN NEXT MESSAGE

halcyon1234 · Jun 2, 2007

I still don't think you get the concept of the shuffle. A count on a shoe isn't reset because all the cards have been played. A count on a shoe is reset because the deck has been shuffled, and there is once again an equal probability of any one of those cards coming up.

Except that the machines first shuffles the discards before spitting out new cards. Even taking into account the patent from above, think about it like this:

* Dealer pulls out 21 cards (out of 208, for a 4 deck CSM)
* Let's assume it empties 2 compartments by pure chance.
* All are low cards.
* RC of the machine before the discards are put back is +21
* Machine queues up 6-10 cards (as per the patent)
* The dealer puts the 21 cards back into the discard tray. Let's say that they are evenly distributed, so that means the two empty trays now have (liberally) 2 cards each, which is below the "deal out" threshold.
* Those 6-10 cards are worth TC = RC/(4 decks - 21 cards) = RC/~3.5 = 6.
* 6 seems good, but ONLY for those first 6-10 cards. Even if it is 10 cards, if there's any more than 4 players, the players after the 4th seat will not benifit from it, because:
* The machine shuffles those 21 cards back into the CSM, resetting the entire CSM to a count of 0.
* After the dealer yanks out enough cards to drop "cards in tray" to the "spit out more" threshold. The tray fills up with cards of an unknown quantity.
* You reset your count to 0, and count up the hand played. That is your new count.
* You then add 4, since there are 2 trays with 2 low cards that won't be used, due to threshold.
* If the combined "that hand" Running Count PLUS 4 is greater than 3.5, you have a TC > 1, you can bet accordingly.
* After that shuffle, do the same thing. Except this time, the 2 previously empty trays have 2 low cards each, and 2 UNKNOWN cards. So you must use a count adjustment of 4 again.
* After the third shuffle, statisitically, the 2 presviously empty trays MUST have enough cards in them to be selected. Because those cards are now available to be "shuffled", you can't count them. Your next count cannot be offset.

For all that work, all you get is a game where, at best, you get a count of, at most, +6 for ONE hand (less if there's more decks), POSSIBLY a high count for MAYBE 2 more. And all this hinges on knowing when the trays are empty, which is knowledge you don't have since they're hidden inside the machine.

Without that knowledge, you can utilize the "6 cards in tray" knowledge. But you must sit at 1st base for maximum effect. If there are any more than 2 players, then the "high count" cards in the tray will run out before the hand is done. (2 players = 6 cards, theoretically you can other player can stop on 2 cards, dealer busts).

For maximum efficiency, you must be the only player at the table, playing heads up against the dealer, since that way both you and the dealer get high-count cards, AND the dealer hits from high-count cards. But even that won't work because then there are only 4-9 cards being played per hand, and even if they are all high cards, the best you get is a TC of 1-2.

TC 1 is 0%, TC 2 is 0.5%! You would have to jump from a $5 bet to table max to get ANY advantage-- and even then, I doubt there's a $5 CSM table with a high enough max to give you a proper spread.

You COULD play all 7 spots to get more cards onto the table, and only max-bet your 1st base hand when the time is right-- but then you'll get, at best, TC 6, which is only a 2.5% advantage. You wouldn't be able to spread enough to cover the $30 @ -0.5% you'll be playing in the other six spots. (And you won't get 2.5%, since the high cards will run out before you're 1st hand gets its 2nd card).

Since there is at most 31 cards being dealt, what you are playing is a 4 deck game with just over 15% pen! Any more than 4 decks, and-- well, I'll leave that figure up to you to calculate.

Yes. The main flaw, from everything of you're that I've read, is that you are not properly using the running count. Yes, the running count is a way to track cards that have come out of the shoe/csm/deck-- but once they go back in, if you cannot account for their EXACT position, they become unknowns, and must be taken away from your count.

What you are proposing is called shuffletracking. It is done on non-CSM tables. When a large concentration of high cards comes out within the first deck of the shoe, a player "watches" that bunch of cards through a particular type of shuffle, makes some RC adjustments, then that player cuts that section of the deck to the front. To a casual observer, it looks like the deck starts at RC 0. To them, they know the next X cards *DO* have Y number of high cards, and can change their initial RC to it. That is what you're trying to do with a CSM, but it won't work because you cannot observe or manipulate the shuffle.

zengrifter · Jun 2, 2007

This is likely a wild goose chase, BUT its of interest, nonetheless. In my interview I describe how I beat seemingly unbeatable 1st-generation CSMs, so the patent could reveal some vulnerability. Though, I suspect some very smart minds have already done this. zg

Ringer · Jun 2, 2007

halcyon, thank you so very much for taking such great care and time to respond with extremely valuable information. I am going to have to take quite some time to fully discern it and re-postulate my theory.

Your efforts certainly do not go un-appreciated.

Ringer · Jun 4, 2007

After much contemplation and studying I fully understand the redistribution of the cards leads to a zeroing out of the remaining decks.... according to your distribution.

There still is a small issue of empty compartments that would be recieving low cards that could allow for a slight advantage to the remaining compartments, but only by a factor of +2 per emptied compartment throughout the entire 5 decks, assuming 2 cards get sent to the empty compartment and all were low cards.

I simply wrote out the following with the rules as such... take one row and redistribute the numbers among the same 6 rows. You must take a row that has at least 5 numbers in it:

-1 -1 -1 -1 -1 = -5
+1 +1 +1 +1 +1 = +5
0 0 0 0 0 0 = 0
0 0 0 0 0 0 = 0
+1 +1 +1 +1 +1 = +5
-1 -1 -1 -1 -1 = -5

Assuming the first row is taken and redistributed close to evenly we get:

-1 = -1
+1 +1 +1 +1 +1 -1 = +4
0 0 0 0 0 0 -1 = -1
0 0 0 0 0 0 = 0
+1 +1 +1 +1 +1 -1 = +4
-1 -1 -1 -1 -1 -1 = -6

Row one is not an acceptable row to choose from since it does not meet our criteria of 5 numbers. Because of this we only have a slight edge since the total count remaining of rows 2-6 is only +1. If we're lucky though we get the last row chosen now.

-1 -1 = -2
+1 +1 +1 +1 +1 -1 -1 = +3
0 0 0 0 0 0 -1 -1 = -2
0 0 0 0 0 0 -1 = -1
+1 +1 +1 +1 +1 -1 = +4
-1 = -1

Again we are still not permitted to take row 1 and now we cannot take row 6. Our chances of getting a positive row are 50% and the low counts aren't that dramatic.. but if we get one of the high counts, they are dramatic.

Granted, the chances of low card compartments selected in a row is not very high (1 in 169 if there are 15 compartments but 2 always remain unselectable). But if it is observed that they come out, we do have an edge so long as we know there are compartments that cannot be chosen.

If we were to hit the "trifecta" of opportunity and have our third row selected for the next redistribution (about 1 in 2197), then we're in statistical heaven.

-1 -1 0 -1 = -3
+1 +1 +1 +1 +1 -1 -1 0 -1 = +2
0 = 0
0 0 0 0 0 0 -1 0 = -1
+1 +1 +1 +1 +1 -1 0 = +4
-1 0 = -1

We now have a 2/3 chance a positive count row will be selected and the remaining low count row is about the house edge.

I'm not disputing your post in any way. I just think the sugnificance of compartments not selectable needs to be factored in as I emphasized here.

Again, my logic could be wrong... and I'm starting to believe a running count won't be nearly as effective as simple observation and anticipation. A slow running positive count doesn't really mean anything other than probably a slow running negative count soon to come. But a fast one where two or three hands in a row come up low cards could be quite sugnificant. It would explain why not long after we see runs of low cards, in general we see runs of high cards on a CSM.

I am open to changing the original hypothesis, so long as we find a hypothesis that is beneficial. :cool2:

halcyon1234 · Jun 5, 2007

The situation you are describing, where a clump of low cards is partially distributed into an non-selectable compartment, is advantageous because more high cards are available to be dealt than low cards. The CSM's true count is up. However, the conditions for this occuring are:

1) A compartment is empty
2) A large amount of low cards is going to be "shuffled" back into the CSM

In order for you to take advantage of this situation, you must be able to prove that 1) is true. Without intimiate knowledge of the state of each compartment, and knowledge of the CSM's compartment chosing algorithm, you cannot prove that 1) is true. You won't be able to tell if a compartment is empty/near enough to empty that it won't recieve enough cards to be beyond the selectable threshold.

In actual play, without that knowledge, rather than an empty compartment, you end up having another row. That row balances out with the rest of the shoe, but has an unknown number of units inside of it. Thus, when the 6 -1s are redistrobuted, they may land in that compartment, and they may be selected.

With perfect knowledge of the state of the compartment, you can use the formula I gave you in my last message. In this case:

u (the chance that the favoured compartment will be chosen) = 0. This is because it is impossible for that compartment to be selected.
RC (the running count inside that compartment) is equal to the threshold of selectability of that comaprtment, -1. Based on the patents you posted, the highest that number can be seems to be 7. That would put the RC at -6.

Hence, the best we can hope for is a 4 deck CSM with 10 compartments and a threshold of 7. Therefore, E(-6, 0) is 0.33%.

That means that if you knew for sure that a compartment was empty, and if a large group of low-count cards was about to be shuffled, your MAXIMUM possible advantage is 0.33%. That is less that just playing a shoe that is TC +1.5. And this advantage goes down with each deck you add, with each compartment you add, and as the threshold decreases.

I would suspect that in order to take perfect advantage of it, you'd have to set it up, then spread to 7 hands of table max (possibly having a partner table-max backbet each one) to beat the money you'd lose playing a table at min bet to set this up.

For the record, if you were on an average $5-$200 table, and table-maxed with back bet each spot, you'd have $2800 on the table. With a 0.33% advantage, this means you'd earn $9.24. Playing three spots at table min (the fewest hands I'd play if I was trying to force a -6 count in a single hand), at 100 hands an hour, you're playing $500 per hour at a 0.5% disadvantage. This means you are losing $2.50 per hour. The -6 with an empty compartment situation would have to come up at least once every ~4 hours to be advantageous. It would have to come up twice per hour to pay you more than a minimum wage job, or once every 50 hands. We won't even talk of the variance.

If you had a table full of perfect-BS playing ploppies who would let you backbet them, you would need to spend the $2.50 / hour playing the tables themselves, but you cut in half the amount you can bet (since you are the backbetter), and you only make $4.14 per situation.

So the hypothosis that "a CSM that has just dealt a large amount of low-count cards will be advantageous" is incomplete. The hypothosis should read:

"A CSM that has just deal a quantifiable number of low-count cards will be advantageous, iff there exists one or more compartments ineligable to deal the next that can accept those cards. This CSM will remain in an advantageous state until such time as all ineligable compartments are once again eligable to be dealt from."

The pre-condition for this statement is that you can prove the "there exists one or more compartments..." part. I propose that if you are able to do the computations on the fly, in your head, in under 5 seconds, than you are worth far more than the $27/hour you *might* get if your situation came up three times an hour.

In fact, I propose thati f you could do these computations on the fly, in your head, in under 5 seconds, and still wanted to beat the crap out of a CSM, you could easily track BLACKJACKS through the CSM, backbet first base, and make a hell of a lot more.

What it comes down to is this: The effort to produce the effect you are describing is massively disproportional to the reward you wil receieve from it.

My counterproposal to you is this: You're smart. You know how to play. Make sure your BS is perfect, you can hi-lo a deck in 30 seconds, remember to reset the RC to 0 after each shuffle, learn the Illustrious 18, and print out a copy of the Security Report that Zengrifter posted in the "News" forum. Work out your unit size for an optimal ROR. Then, when you go to Vegas, hit the non-CSM tables and take some money.

Studying the patent for 7,137,627

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