On Locking the Wonging Thread

[blackjack avenger]

Working Count Dance

or pullus is latin for chicken:laugh:

a 6 deck shoe:

deck 1 rc2
deck 2 rc-3
deck 3 rc1
deck 4 rc9
deck 5 rc-4
deck 6 rc-5
the sum of above is 0

If I know the count of any 5 decks I know the count of the last deck, no matter which deck is the last deck. This is typical counting where we infer the rest of the shoe from a known working count.

add together any 2 decks, I can tell you the sum of the remaing 4 decks, no matter which ones you choose. They don't have to be next to each other. I can also tell you the average of each deck of the 4 deck group.

Here is the big one
give me the count of 2 decks
I don't need to know another 2 decks, I can be in the bathroom
I can tell you the average count of the final 2 decks and if that average is positive you can play it. Does not matter which decks you choose
An example:
deck 4 + deck 2 = rc6
the rc for the remaining 4 decks is -6
the average rc for each remaining deck is -1.5
Now because the average count per deck is -1.5 meaning on average 1.5 extra good cards per deck we can play the final 2 decks.
We can also make this leap, to play the final 2 decks, we don't have to count them. Because we know on average the final 2 decks are playable. The rc on average at the beginning of the final 2 decks is 3 and the divisor is 2 for tc conversion.

To borrow from Shuffle Tracking:
I can mix any 2 decks and tell you their sum. From this I can tell you the sum of the remaining 4 decks and the average of each deck.

One can plug in any rc's for the 6 decks as long as the sum is 0 and can do all the above.

Can we all finally agree that given a working count one can infer the rest of the shoe with simple math?

A horrible way to play, don't do it, but can be done. It's better to see every card and count it.
:joker::whip:

A quick edit to answer the 0 question by KC
first deck is rc0 the average of rest of shoe is rc0 correct?
any deck is rc0, from anywhere, the average of rest of shoe is rc0 correct?
now let's consider 3 decks:
deck 1 rc 0
deck 2 rc 1
deck 3 rc -1
given deck 1 I can tell you sum of deck 2 and 3, I can give you average of each
given deck 3 I can tell you sum of decks 1 and 2, I can give you average of each
When we count normally a 0 doesn't ruin are inference moving forward, this is no different.
In each example I am using all the cards.
In your examples you are not

 

[QFIT]

The rc on average at the beginning of the final 2 decks is 3 and the divisor is 2 for tc conversion.

NO. The divisor is 4 because the count of only half of the unseen cards is known. This is an important concept in NRS. It is also why tracking without using NRS can give poor results.

 

[blackjack avenger]

No Panacea

NO. The divisor is 4 because the count of only half of the unseen cards is known. This is an important concept in NRS. It is also why tracking without using NRS can give poor results.

In a shoe we can infer the count of all; or some, unseen cards, if we know the count of any. Don't you agree?
However,
I can see where application of the NRS formula would improve things.
I was not comfortable with the line you reference.
I almost deleted it.
:joker::whip:

 

[QFIT]

In a shoe we can infer the count of all unseen cards, if we know the count of any. Don't you agree?

Of course. But, the TC divisor is all unseen cards. Not some of them.

 

[assume_R]

6. We have eliminated 7 decks worth of cards and even though we haven't seen them, knowing what their average RC is allows us to continue counting as if there are only 52 cards that remain.
7. In the next 26 cards RC increases by 4 so RC = +4, decks remaining = 1/2, TC = +8. This applies to first 52 card slug, last 52 card slug, or any slug in between.

The above is the consequence of assuming that unseen cards can be accounted for by assuming they have been removed according to an expected average. The (erroneous) conclusion was that an increase of RC by +4 in the first 26 cards of an 8 deck shoe results in a TC of +8.

Okay, I follow this, and could you elucidate why you think it's erroneous to assume the RC will, on average, drop back to 0 in the second half of the first deck?

Right now I am of the opinion that it is a very very inaccurate method for estimating the advantage (lots of variance), but why is it incorrect? I.E. why are you saying that, on average, the RC won't drop by 8 in the second half of the first deck? What's the expected (mean) drop of the RC in the second half of the first deck if it's not 8?

 

[QFIT]

It doesn't matter. It's like walking up to an 8 deck shoe, stating that the last seven decks on average have an RC of zero (true) and then playing the first deck as if it is a single-deck game using only that deck for TC calcs. Won't work.

 

[London Colin]

why are you saying that, on average, the RC won't drop by 8 in the second half of the first deck? What's the expected (mean) drop of the RC in the second half of the first deck if it's not 8?

I think you meant to say 4 there. The actual value would be 0.266667, for what it's worth. (RC = +4, divided by 15 unseen half decks.)

Here's the basic contradiction -

You start out with 8 unseen decks, and can therefore assert that the expected TC for any group of cards (e.g., each of the 8 decks) is 0.

You then start playing one of the decks under the assumption that this average holds true, and yet at the same time try to take advantage of the fluctuations from this average, the fact that it probably won't hold true, by counting the cards.

But the fluctuations relate to the entire shoe, not just the one deck (or whatever sized chunk you arbitrarily decide to focus on).

With each individual card that is revealed, you have new information about the average composition of all the remaining unseen cards. Your old average doesn't apply any more. It's a completely fluid, dynamic process.

At any stage all you know is the RC and the number of unseen cards.

 

[iCountNTrack]

At any stage all you know is the RC and the number of unseen cards.

I like it, i will also want to add that, all the information we have (when we are card counting) at any stage about the seen cards is contained in the Running Count, and about unseen cards is their number.

 

[kewljason]

I am truely stunned that this discussion is still going on...even bypassing a locked thread into a new one. :laugh: I know you guys are all math nerds, and I say that with respect....I wish I had a fraction of your mathematical knowledge. As I have stated, to me the true count therem is almost irrelivent. It only comes into play when you have left a shoe and wish to return. If you left a positive count shoe, well then I want to smack you upside your head. If you left a negative count shoe, the chances are slim that the count this shoe will turn positive enough to get excited about. No need to get back in. Your efforts are better spent, finding a new game. If there is not a new game nearby, just wait for the new shoe! :laugh: To me this discussion is like discussing the best index plays for negative true counts of -5 or more. It's irrelivent. There is a far better option. But please, continue on. It's entertaining. :laugh:

 

[k_c]

This thread needs an example that is so simple that anyone can see it. I'll try but I'm not going to hold my breath waiting for success.

At the core of the problem is the desire to treat unseen cards as if they have already been played resulting in a reduced number of cards at a RC that has been computed from an initial TC by interpolating what would be expected on average if the unseen cards had actually been seen.

Starting with a full shoe if we deal, SEE, and count cards until one card remains we know with 100% certainty that the card is high, medium, or low. RC will be +1, 0, or -1. TC will be +52, 0, or -52.

Starting with a full shoe if we deal and DON"T SEE the cards until one card remains we don't know any more than when we started. What can be said about the last card in this case? Since we have seen no cards all we know is probability of high = 5/13, probability of medium = 3/13, probability of low = 5/13. probability TC +52 = 5/13, probability TC 0 = 3/13, probability TC -52 = 5/13. Since probability high = probability low average RC = 0, average TC = 0. The average RC case is not the most likely case. If we say that we can predict the last card because we assume that RC will be the average value then we'll be right 3/13 of the time. We'll be wrong 10/13 of the time and our supposed TC of 0 will be way off. The only thing we know for sure about that last card is that trying to determine what it is is identical to the problem of trying to determine what it is by picking a card from a full shoe.

The bottom line is this "problem" is extremely simple and is rightly summed up by London Colin

At any stage all you know is the RC and the number of unseen cards.

Any convoluted effort to get around the above will not work.

 

[mathman]

Use your head

Why do you think casino's went from one deck to two, from two decks to four, from four decks to six and from six decks to eight?

If the Casino's did this just to increase the number of hands played per hour only then why do the "tougher" casino's place the cut card two decks or more into a shoe?

Why does a competent AP pay strong attention to pen?

Enough said...

 

[assume_R]

I think you meant to say 4 there. The actual value would be 0.266667, for what it's worth. (RC = +4, divided by 15 unseen half decks.)

Here's the basic contradiction -

You start out with 8 unseen decks, and can therefore assert that the expected TC for any group of cards (e.g., each of the 8 decks) is 0.

You then start playing one of the decks under the assumption that this average holds true, and yet at the same time try to take advantage of the fluctuations from this average, the fact that it probably won't hold true, by counting the cards.

But the fluctuations relate to the entire shoe, not just the one deck (or whatever sized chunk you arbitrarily decide to focus on).

With each individual card that is revealed, you have new information about the average composition of all the remaining unseen cards. Your old average doesn't apply any more. It's a completely fluid, dynamic process.

At any stage all you know is the RC and the number of unseen cards.

This thread needs an example that is so simple that anyone can see it. I'll try but I'm not going to hold my breath waiting for success.

At the core of the problem is the desire to treat unseen cards as if they have already been played resulting in a reduced number of cards at a RC that has been computed from an initial TC by interpolating what would be expected on average if the unseen cards had actually been seen.

Starting with a full shoe if we deal, SEE, and count cards until one card remains we know with 100% certainty that the card is high, medium, or low. RC will be +1, 0, or -1. TC will be +52, 0, or -52.

Starting with a full shoe if we deal and DON"T SEE the cards until one card remains we don't know any more than when we started. What can be said about the last card in this case? Since we have seen no cards all we know is probability of high = 5/13, probability of medium = 3/13, probability of low = 5/13. probability TC +52 = 5/13, probability TC 0 = 3/13, probability TC -52 = 5/13. Since probability high = probability low average RC = 0, average TC = 0. The average RC case is not the most likely case. If we say that we can predict the last card because we assume that RC will be the average value then we'll be right 3/13 of the time. We'll be wrong 10/13 of the time and our supposed TC of 0 will be way off. The only thing we know for sure about that last card is that trying to determine what it is is identical to the problem of trying to determine what it is by picking a card from a full shoe.

I want to thank you guys for these last 2 posts, as they explain to me why the other method is incorrect. The pieces that made the most sense to me (if they are indeed correct they seem to refute the "TC Theorem method") I made in red.

[Title = "Use Your Head"] Why do you think casino's went from one deck to two, from two decks to four, from four decks to six and from six decks to eight?

If the Casino's did this just to increase the number of hands played per hour only then why do the "tougher" casino's place the cut card two decks or more into a shoe?

Why does a competent AP pay strong attention to pen?

Enough said...

With all due respect, I have been using my head, and that wasn't enough said. I have presented cogent arguments and questions with good mathematical basis, and have asked specific things about what I didn't understand. Just saying that the casino chooses more decks for a reason, or asking me why I pay attention to penetration isn't pertinent to the discussion. Anyway, no ill will towards you, but I just felt like that was a bit snarky.

 

[QFIT]

The pieces that made the most sense to me (if they are indeed correct they seem to refute the "TC Theorem method") I made in red.

Red is good since the TC Theorem is a red herring in these posts.

Seriously, nothing in this thread relates to the TC theorem. It is based on unseen cards. When you close your eyes, or go away for awhile, nothing has happened relevant to the TC Theorem. The number of unseen cards has not budged. Cards that you miss are like trees falling in a deserted forest. They don't count. What is the difference if unseen cards are behind the cut card or in front of it but you didn't see them? Or the dealer burned two decks?

 

[assume_R]

Seriously, nothing in this thread relates to the TC theorem. It is based on unseen cards. When you close your eyes, or go away for awhile, nothing has happened relevant to the TC Theorem.

Okay, could you give a practical or theoretical example in which the TC theorem is used? Shuffle tracking as bja has said, perhaps?

 

[psyduck]

Assuming RC is evenly distributed among all the unseen cards is a way of estimating, but with potentially large errors in my opinion. I personally always treat the unseen cards as the cards behind the cut card.

For a 6 deck shoe with one deck cut out, if we play the first deck and the TC is 0, we left the table and came back when 2 of the 6 decks are left. At this point, assuming the TC is still 0 will most likely be wrong because the frequency of TC = 0 decreases as we play deeper into the shoe.

In short, BA's estimation is one way of treating the shoe, but I will not use it because of the potential large errors.

 

[QFIT]

Okay, could you give a practical or theoretical example in which the TC theorem is used? Shuffle tracking as bja has said, perhaps?

TC Theorem is what it is. Kudos to Abdul Jalib for his wonderful work. But, unrelated to this thread. It has been used more to disprove poorly conceived systems than to create new systems. Which does not make it less valuable. It is in the back of the mind of those that build concepts like NRS.

As the Qu'ran says "You shall have your religion and I shall have my religion" [109:006]. My "religion" is based on math.

norm

 

[Automatic Monkey]

TC Theorem is what it is. Kudos to Abdul Jalib for his wonderful work. But, unrelated to this thread. It has been used more to disprove poorly conceived systems than to create new systems. Which does not make it less valuable. It is in the back of the mind of those that build concepts like NRS.

Maybe what still needs to be said is that TCT gives you useful information after you have determined a true count and have no additional knowledge of the number of unseen cards. If you walk away from a true count, and discover when you return that the dealer has thrown a newspaper over the shoe and the discards the TCT gives you the most accurate information you're going to get. If you know the number of cards that have been dealt in your absence, you have additional data that the TCT was not derived to take into account.

As the Qu'ran says "You shall have your religion and I shall have my religion" [109:006]. My "religion" is based on math.

norm

Not the best analogy, because it also says to kill the unbelievers, and because the postulates of math are more difficult to prove than those of religion. More people have seen God than have seen the ultimate fate of two parallel lines.

 

[QFIT]

Not the best analogy, because it also says to kill the unbelievers, and because the postulates of math are more difficult to prove than those of religion.

As does the Old Testament. Which is why my religion is math.

 

[zoomie]

Lock It

When our friends Norm and AM start snapping at each other, it is definitely time to LOCK THE THREAD.

 

[QFIT]

When our friends Norm and AM start snapping at each other, it is definitely time to LOCK THE THREAD.