Wonging Count

[blackjack avenger]

So We Agree

Okay finally got some time to read this thread, and here are some clarifications:

A) The True Count as i always say is not "truely" a count, it is a density, a ratio of two quantities, a positive TC will indicate to you that at any given moment there is a higher probability of drawing a high card, the rest is history.

B) As far as the True Count theorem, IT DOES MATTER whether the cards are seen or not seen . The True Count Theorem states:

"The expected true count after any number of cards are revealed and removed from any deck composition is the same as before the cards were removed, for any balanced count, provided you do not run out of cards."

C)Unseen cards are unseen cards, it doesnt matter if you failed to see them because they got burnt, you had to go to take a piss, you were staring at the cocktail waitress, or whether they are behind the cut card or discard tray. All what unseen cards do is reduce the effective penetration.

We actually don't have to see any cards to know the TC of a shoe, it starts at TC0 and ends at TC0. Does anyone dispute this? However, for our discussion, yes you do need to see some of the cards to establish a working TC. However, once cards are revealed and removed we now know what the TC is going forward or even backward. Does anyone dispute this? Also, because of the TC theorem, we know on "average" the value distribution of the remaining cards, so we know at any time the "average" whether we are present or not. This has been known for well over 10 yrs.

A form of shuffle tracking as an example:
6 decks
We know a 1 deck slug has 10 extra high cards
It is mixed with an unknown 1 deck slug
because of the TC theorem
we know on "average" the remainig 5 decks are deficient by 10 high cards
or an "average" of 2 per deck.
We mix the known slug with the average of the other deck being deficient 2 high cards.
10-2=8 extra high cards in the new 2 deck slug
Now to those who have been saying you don't know the exact value of those 1 or 2 decks in my previous examples, the same could be said for this ST example, yet it is an accepted form of ST.
Another interesting application of the TC theorem which hopefully will help some to understand what I have been saying.

Now I am sure someone will comment that there is a better way to ST then my example, which is fine. My example shows that the TC theorem is used in practical settings.

I once had an article that talked about if one is in disagreement with many others then there is a high probability the one is wrong. It appears that KC, Icountintrack and Automonkey and others don't agree with me. I am also quite certain their individual math education and background are higher then mine. However, as this has gone on and on I am in fact more convinced that using the TC theorem in this manor is simple and quite valid.

:joker::whip:

 

[Automatic Monkey]

Unseen Card Monty!

We actually don't have to see any cards to know the TC of a shoe, it starts at TC0 and ends at TC0. Does anyone dispute this? However, for our discussion, yes you do need to see some of the cards to establish a working TC. However, once cards are revealed and removed we now know what the TC is going forward or even backward. Does anyone dispute this? Also, because of the TC theorem, we know on "average" the value distribution of the remaining cards, so we know at any time the "average" whether we are present or not. This has been known for well over 10 yrs.

A form of shuffle tracking as an example:
6 decks
We know a 1 deck slug has 10 extra high cards
It is mixed with an unknown 1 deck slug
because of the TC theorem
we know on "average" the remainig 5 decks are deficient by 10 high cards
or an "average" of 2 per deck.
We mix the known slug with the average of the other deck being deficient 2 high cards.
10-2=8 extra high cards in the new 2 deck slug
Now to those who have been saying you don't know the exact value of those 1 or 2 decks in my previous examples, the same could be said for this ST example, yet it is an accepted form of ST.
Another interesting application of the TC theorem which hopefully will help some to understand what I have been saying.

Now I am sure someone will comment that there is a better way to ST then my example, which is fine. My example shows that the TC theorem is used in practical settings.

I once had an article that talked about if one is in disagreement with many others then there is a high probability the one is wrong. It appears that KC, Icountintrack and Automonkey and others don't agree with me. I am also quite certain their individual math education and background are higher then mine. However, as this has gone on and on I am in fact more convinced that using the TC theorem in this manor is simple and quite valid.

:joker::whip:

It is simple and valid, and is better than nothing when seeking an advantage. No one's challenging that fact.

But doing it "the other way" complements the TCT and does not violate it. All unseen cards can be treated the same and if so, all seen cards must be treated the same. All this other method does is allow you to weight the cards you see when you return to be the same as the ones you saw before you left. Thus when you are betting the first round to be dealt upon your return, both methods give the same answer.

Thinking about this, the math is the same as the famous "Monty Hall problem" which is counterintuitive and the smarter you are the more confusing it is. It gets me all the time, same for the push-on-22 rule in BJ Switch. All problems caused by mixing observed data with theoretical conclusions.

 

[blackjack avenger]

Not Quite

It is simple and valid, and is better than nothing when seeking an advantage. No one's challenging that fact.

But doing it "the other way" complements the TCT and does not violate it. All unseen cards can be treated the same and if so, all seen cards must be treated the same. All this other method does is allow you to weight the cards you see when you return to be the same as the ones you saw before you left. Thus when you are betting the first round to be dealt upon your return, both methods give the same answer.

Both methods give the same answer for the next hand only, after that using the TC Theorem method gives you a smaller divisor in calculating TC for the remaining decks. This is quite powerful, even if based on an average TC for "some" of the decks played. The TC Theorem gives us some info vs no info.

Aren't unbalanced counts based on an "average" TC across the whole shoe. Yet, another application of the TC Theorem. I am advocating using an average for a section of a shoe.

:joker::whip:

 

[k_c]

We actually don't have to see any cards to know the TC of a shoe, it starts at TC0 and ends at TC0. Does anyone dispute this? However, for our discussion, yes you do need to see some of the cards to establish a working TC. However, once cards are revealed and removed we now know what the TC is going forward or even backward. Does anyone dispute this? Also, because of the TC theorem, we know on "average" the value distribution of the remaining cards, so we know at any time the "average" whether we are present or not. This has been known for well over 10 yrs.

A form of shuffle tracking as an example:
6 decks
We know a 1 deck slug has 10 extra high cards
It is mixed with an unknown 1 deck slug
because of the TC theorem
we know on "average" the remainig 5 decks are deficient by 10 high cards
or an "average" of 2 per deck.
We mix the known slug with the average of the other deck being deficient 2 high cards.
10-2=8 extra high cards in the new 2 deck slug
Now to those who have been saying you don't know the exact value of those 1 or 2 decks in my previous examples, the same could be said for this ST example, yet it is an accepted form of ST.
Another interesting application of the TC theorem which hopefully will help some to understand what I have been saying.

Now I am sure someone will comment that there is a better way to ST then my example, which is fine. My example shows that the TC theorem is used in practical settings.

I once had an article that talked about if one is in disagreement with many others then there is a high probability the one is wrong. It appears that KC, Icountintrack and Automonkey and others don't agree with me. I am also quite certain their individual math education and background are higher then mine. However, as this has gone on and on I am in fact more convinced that using the TC theorem in this manor is simple and quite valid.

:joker::whip:

I'll try one more time to show where the error in your logic is. If this doesn't do it I gotta give up.

Assume the first 50 cards out of a 6 decks shoe are all (2-6) so at that point Hi-Lo RC = +50 and Hi-Lo TC = +10.

Player now leaves and comes back when 106 cards remain. 156 additional cards have been dealt which player doesn't see but he assumes current RC = +20 with 106 cards remaining to be dealt because on average this will be the case.

Actual RC could be anything in the range -70 to +106.

Assume unbeknownst to player actual RC = +106 with 106 cards remaining although he assumes RC = +20. Actual shoe consists of nothing but high cards so as cards continue to be dealt and counted with an assumed RC = +20 then with 42 cards remaining player's RC = -44. Impossible! RC can't be more negative than the number of cards remaining. By the end of the shoe player's RC will be -86 with 0 cards remaining. However RC must be 0 at shoe's end for a balanced count.

Relating to unseen cards accounts for all cases. Interpolating using average expected RC accounts exactly for one and only one case, which is when RC is exactly what it is expected to be at the point player resumes play.

 

[sagefr0g]

For those of you who have played a lot of shoes, what generally happens when there is a positive shoe?

The count goes up to say tc2 as an example and the rc drops but the tc remains the same, this is generally how it goes and is the tc theorem in action. The tc stays the same while the rc drops.

Example:
1 out of 6 dealt and rc is 10 so tc2 play 1 deck
2 out of 6 dealt and rc is 8 so tc2 play 1 deck
3 out of 6 dealt and rc is 6 so tc2 play 1 deck
4 out of 6 dealt and rc is 4 so tc2 play 1 deck
etc.
This is what happens on average and is the basis of the tc theorem

Now let's put the burden of proof on the other camp:
If one misses a deck of play, why would they pretend the cards are behind the cut card when in fact they are not? They are indeed in the discard tray and since they are in the discard tray they have an effect on the remaining rc because as the tc theorem tells us the rc drops while the tc remains the same. So all one has to do is drop the rc enough to equal the same tc as when they left. Now moving forward we have as a divisor for the tc only the cards left to be played. Those who act as if those cars were not dealt and mentally put them behind the cut card now have to add those cards into their divisor and will underbet moving forward. The tc theorem allows us to have some information on the unplayed missed cards vs not using the tc theorem and pretending we have no information.

to sum up:
tc theorm some information on unseen cards
put behind cut card ignores information on unseen cards
tc theorem smaller divisor for tc moving forward
put behind cut card has larger divisor moving forward

The smaller divisor moving forward I would think is the trump
:joker::whip:

just nit picking here i guess far as the part bolded in blue.
like for your example the rc would be presumed to drop, but not in general apparently, as there appears to be a misprint in a portion of the article on the true count theorem where Jalib writes about the running count.
http://www.blackjackinfo.com/bb/showpost.php?p=14998&postcount=21
compare in the link below:
http://www.bjmath.com/bjmath/counting/tcproof.htm (Archive copy)
apparently whether the rc rises or falls depends upon if the tc was negative or positive, sorta thing.
also the following statement would seem to alert one to perhaps making too many presumptions regarding the running count when considering the true count theorem, maybe?
The Running Count
-----------------
The theorem applies only to the true count, not to the running count. The running count does not obey the same laws of as the true count.

also kind of interesting how Jalib uses some different terminology in his corollaries compared to his original statement of the true count theorem.

Theorem: the expected value of the true count after a card is revealed and removed from any deck composition is exactly the same as before the card was removed, for any balanced count, provided you do not run out of cards.
............................
Theorem:
The expected true count after a card is revealed and removed from any deck composition is the same as before the card was removed, for any balanced count, provided you do not run out of cards.

Corollary:
The expected true count after any number of cards are revealed and removed from any deck composition is the same as before the cards were removed, for any balanced count, provided you do not run out of cards.

Corollary:
The expected true count after a round is the same as before the round, for any balanced count, provided you do not run out of cards.

again, pretty much just nit picking here.:devil::whip:

 

[rrwoods]

using the TC Theorem method gives you a smaller divisor in calculating TC for the remaining decks. This is quite powerful

Ugh. It's not powerful, it's deceptive.

Whatever. I'm done here. ICNT said it better than anyone could.

 

[blackjack avenger]

Never Going to Happen

I'll try one more time to show where the error in your logic is. If this doesn't do it I gotta give up.

Assume the first 50 cards out of a 6 decks shoe are all (2-6) so at that point Hi-Lo RC = +50 and Hi-Lo TC = +10.

Player now leaves and comes back when 106 cards remain. 156 additional cards have been dealt which player doesn't see but he assumes current RC = +20 with 106 cards remaining to be dealt because on average this will be the case.

Actual RC could be anything in the range -70 to +106.

Assume unbeknownst to player actual RC = +106 with 106 cards remaining although he assumes RC = +20. Actual shoe consists of nothing but high cards so as cards continue to be dealt and counted with an assumed RC = +20 then with 42 cards remaining player's RC = -44. Impossible! RC can't be more negative than the number of cards remaining. By the end of the shoe player's RC will be -86 with 0 cards remaining. However RC must be 0 at shoe's end for a balanced count.

Relating to unseen cards accounts for all cases. Interpolating using average expected RC accounts exactly for one and only one case, which is when RC is exactly what it is expected to be at the point player resumes play.

I agree with your math and what you have stated.

However,
Your example is so rare as to be shall I say "real world" impossible.
What is most likely to happen?
The average will play out and the TC theorem will prove itself, what are the next likely things to happen? the cards are close enough to the average that little will be lost.

To deny the TC theorem then you are denying some of the procedures used in shuffle tracking that I pointed out, and you are also denying the basis of unbalanced counts.

As far as error rate. We are surrounded by error and estimates and averages. All of the counts we use have a built in error in regard to BC, PE, IC. The only counts that would truly have no error is if one uses effect of removal counts. BS charts are based on averages of hands, in shoes one rarely advises use of composition dependent BS. We fudge and estimate in many different ways.

Would you use an extremely unlikey ordering of cards or distribution to nullify BS? I don't think so.

So while we are talking error
I ran a 6 deck sim on what the error would be if you overbet by 1tc and tc2 on positive counts, if the average of the Tc theorem had you error and overbet
tc1 overbet SCORE went from 42 to 41 and ror went from 13% to 22%
tc2 overbet SCORE went from 42 to 38 and ror went from 13% to 32%
Now, I think most of those heavily involved in our discussion would recommend betting a small fraction of kelly, so the higher ror is less of an issue if one plays a reasonabl fraction of kelly. As you can see the overbet if over by 1tc is not that painful. However, the tc2 overbet does get a little painful considering SCORE but remember. On average the TC theorem will work just fine.

Now, let's not forget that we will only emply the tc theorem in this rare occurence for a few hands so the real world effect will be extremely negligible. To sagefrog I can guarantee that if someone does this on occasion the earth won't open up and swallow them.

I tend to agree with KC. I don't think any minds have been changed, nor much more can be said. Both sides I think have valid points and both are correct. I think everyone would recommend to try and not miss cards, but if you do miss cards you have 2 different methods you can use moving forward.

I am going to try to limit my replies moving forward to an "average" of 1 a day. I will try my best! LOL

:joker::whip:

 

[blackjack avenger]

Address the Shuffle Tracking Example

No one has commented on how the TC theorem is used in shuffle tracking? It shows the TC Theorem has real world application. I guess one cannot use an average of unseen decks in shuffle tracking? We know that is not the case.

:joker::whip:

 

[blackjack avenger]

I Agree 100%

just nit picking here i guess far as the part bolded in blue.
like for your example the rc would be presumed to drop, but not in general apparently, as there appears to be a misprint in a portion of the article on the true count theorem where Jalib writes about the running count.
http://www.blackjackinfo.com/bb/showpost.php?p=14998&postcount=21
compare in the link below:
http://www.bjmath.com/bjmath/counting/tcproof.htm (Archive copy)
apparently whether the rc rises or falls depends upon if the tc was negative or positive, sorta thing.
also the following statement would seem to alert one to perhaps making too many presumptions regarding the running count when considering the true count theorem, maybe?
The Running Count
-----------------
The theorem applies only to the true count, not to the running count. The running count does not obey the same laws of as the true count.

also kind of interesting how Jalib uses some different terminology in his corollaries compared to his original statement of the true count theorem.

Theorem: the expected value of the true count after a card is revealed and removed from any deck composition is exactly the same as before the card was removed, for any balanced count, provided you do not run out of cards.
............................
Theorem:
The expected true count after a card is revealed and removed from any deck composition is the same as before the card was removed, for any balanced count, provided you do not run out of cards.

Corollary:
The expected true count after any number of cards are revealed and removed from any deck composition is the same as before the cards were removed, for any balanced count, provided you do not run out of cards.

Corollary:
The expected true count after a round is the same as before the round, for any balanced count, provided you do not run out of cards.

again, pretty much just nit picking here.:devil::whip:

I agree with everything said above, and think I have stated same.
I think the other side would also agree with above.

The difference is I give the TC theorem more value then the others. I think it has application in the real world, in shuffle tracking, in the creation of unbalanced counts and what do to if one misses some cards.

I guess the other side denies the above?

One thing about friendly debate, I have read the legion of responses to me and tried to answer all questions. It would appear the other side either does not read all or only picks out parts they can disagree with. There has been no comments on the use of the TC theorem or using averages in the creation of unbalanced counts or shuffle tracking. Some on the opposite side swear by shuffle tracking which in some forms using an average of unknown cards.

:joker::whip:

 

[k_c]

I agree with your math and what you have stated.

However,
Your example is so rare as to be shall I say "real world" impossible.
What is most likely to happen?
The average will play out and the TC theorem will prove itself, what are the next likely things to happen? the cards are close enough to the average that little will be lost.

To deny the TC theorem then you are denying some of the procedures used in shuffle tracking that I pointed out, and you are also denying the basis of unbalanced counts.

As far as error rate. We are surrounded by error and estimates and averages. All of the counts we use have a built in error in regard to BC, PE, IC. The only counts that would truly have no error is if one uses effect of removal counts. BS charts are based on averages of hands, in shoes one rarely advises use of composition dependent BS. We fudge and estimate in many different ways.

Would you use an extremely unlikey ordering of cards or distribution to nullify BS? I don't think so.

So while we are talking error
I ran a 6 deck sim on what the error would be if you overbet by 1tc and tc2 on positive counts, if the average of the Tc theorem had you error and overbet
tc1 overbet SCORE went from 42 to 41 and ror went from 13% to 22%
tc2 overbet SCORE went from 42 to 38 and ror went from 13% to 32%
Now, I think most of those heavily involved in our discussion would recommend betting a small fraction of kelly, so the higher ror is less of an issue if one plays a reasonabl fraction of kelly. As you can see the overbet if over by 1tc is not that painful. However, the tc2 overbet does get a little painful considering SCORE but remember. On average the TC theorem will work just fine.

Now, let's not forget that we will only emply the tc theorem in this rare occurence for a few hands so the real world effect will be extremely negligible. To sagefrog I can guarantee that if someone does this on occasion the earth won't open up and swallow them.

I tend to agree with KC. I don't think any minds have been changed, nor much more can be said. Both sides I think have valid points and both are correct. I think everyone would recommend to try and not miss cards, but if you do miss cards you have 2 different methods you can use moving forward.

I am going to try to limit my replies moving forward to an "average" of 1 a day. I will try my best! LOL

:joker::whip:

Fair enough. I tried to throw together a simple example to emphasize the problem but I probably could have done a better job. Actually I think the full shoe example expressed the problem better but that didn't seem to resonate with you.

Anyway I'll just leave it at that. :devil:

 

[London Colin]

Not sure if this will help or hinder...

I had to read through this thread a number of times before I felt I understood both what was being argued and where the discrepancy lies between the two points of view. (But then I'm more easily confused than most.)

I failed to come up with a purely theoretical response, but I think I may be able to offer a useful thought experiment to clarify things...

Imagine three players -

Player A sees every card dealt from the shoe and bets according to the precise TC.

Player B plays the first deck, takes a break, then plays the last deck before the shuffle. He treats the cards he missed during the break as unseen in the same way as those behind the cut card, adjusting the TC accordingly as he sees further cards removed.

Player C takes the same break as Player B, but he reasons that since the missed cards will, on average, have an even distribution, he can play out the last deck based on that assumption. That is, he gives the same weight to the cards he sees removed in the last deck as Player A does, dividing by the number of decks behind the cut card to get his TC.

[Edit: should have talked of cards remaining to be dealt, rather than behind the cut card in the above two paragraphs.]


So does Player C gain additional EV, compared to Player B (at the cost of raised variance)? If I've undersood correctly, that is blackjack avenger's premise.

It's a seductive argument, but what would be the consequences if it were true?

Suppose, instead of sitting out the middle portion of the shoe, players B and C both decide that they will only play the first two decks, then quit until the next shoe. This is entirely equivalent. In both cases they see only two decks. And if it is valid to assume an even distribution of the cards missed during the break, then it is equally valid to assume the same even distribution in the cards that will remain undealt when they quit the shoe after two decks.

What does this change mean for the two players? -

Player B, by playing the first two decks just as any counter (such as Player A) would, is playing a game with a reduced penetration (say 2 decks from 6), which is exactly what he was doing before, when he took a break in the middle, beacause he was using all the unseen cards (both missed and behind the cut card) as his divisor when calculating the TC.

Player C, by comparison, is now behaving very strangely. He plays the first deck normally, then arbitrarily decides to give increased weight to the cards he sees removed in the second deck, simply beacuse he plans to leave.

That clearly cannot make sense. Moreover, if it did, it should always be done, even if you don't plan to leave early.


Like blackjack avenger, I do find it a bit of a paradox that we can apparently use the assumption of an average, even distribution of the cards as the basis for something like an unbalanced counting methodology, but not for other purposes.

It's a confusing business!

 

[blackjack avenger]

Thanks for Adding

Player C, by comparison, is now behaving very strangely. He plays the first deck normally, then arbitrarily decides to give increased weight to the cards he sees removed in the second deck, simply beacuse he plans to leave.

That clearly cannot make sense. Moreover, if it did, it should always be done, even if you don't plan to leave early.

Like blackjack avenger, I do find it a bit of a paradox that we can apparently use the assumption of an average, even distribution of the cards as the basis for something like an unbalanced counting methodology, but not for other purposes.

Apparently I enjoy a good debate more then others.

I don't follow your thoughts on player C.

We also use cards we don't have knowledge of; even though we did see them, in some shuffle tracking strategies. When known and unknows slugs are mixed. As I tried to point out.

:joker::whip:

 

[iCountNTrack]

Apparently I enjoy a good debate more then others.

I don't follow your thoughts on player C.

We also use cards we don't have knowledge of; even though we did see them, in some shuffle tracking strategies. When known and unknows slugs are mixed. As I tried to point out.

:joker::whip:

The NRS Theory is based on using an average BUT NOT a True Count average, a Slug Count (basically a running count), furthermore the True Count theorem is not whatsoever used in the development of the theory.

 

[kewljason]

Apparently I enjoy a good debate more then others.

I don't follow your thoughts on player C.

We also use cards we don't have knowledge of; even though we did see them, in some shuffle tracking strategies. When known and unknows slugs are mixed. As I tried to point out.

:joker::whip:

I am enjoying this debate. I am just not participating because it seems irrelivent to me. (shuffle tracking aside, because thats a different debate)

I can guarantee I am not leaving the table during a good count. No Phone calls answered, no bathroom breaks. I'd rather have urine running down my leg than give up a juicy count. If the count is negative enough that I walk away, I am not returning to that shoe, unless I have actual evidence that there is reason to do so, like walking by and seeing a dozen or so small cards on the felt. So while the TCT is one that I understand, I see no real practical use for it.

 

[rrwoods]

<snip> This is entirely equivalent.

Listen to this man. He knows what he speaks.

 

[London Colin]

I don't follow your thoughts on player C.

We also use cards we don't have knowledge of; even though we did see them, in some shuffle tracking strategies. When known and unknows slugs are mixed. As I tried to point out.

I'm afraid I probably can't explain my thoughts on player C any more clearly than my first attempt. (That was supposed to be my best effort at a clear explanation! ) [But see my slight edit in the original post, in case that helps.]

The basic idea is -

If you accept that the two cases (a break midway vs quitting after two decks) are equivalent, and you accept that player C's methodology makes no sense in the second case, then you must also accept that it makes no sense in the first case.)

One further thought: If you sit down at a shoe that is already halfway completed, with no knowledge of any of the cards that have been dealt so far, so that on average the TC must be assumed to be 0, are you suggesting that you should calculate your TC for the remainder of the shoe based only on the number of undealt cards?

If not, how is that any different to re-entering a shoe which you left with a known count?

And if you are in fact suggesting that, it would imply that a large advantage could be gained simply by entering each shoe towards the end, effectively turning a 6-deck shoe into, say 2 decks! Alas, that is not my understanding of the reality.

 

[Automatic Monkey]

I am enjoying this debate. I am just not participating because it seems irrelivent to me. (shuffle tracking aside, because thats a different debate)

I can guarantee I am not leaving the table during a good count. No Phone calls answered, no bathroom breaks. I'd rather have urine running down my leg than give up a juicy count. If the count is negative enough that I walk away, I am not returning to that shoe, unless I have actual evidence that there is reason to do so, like walking by and seeing a dozen or so small cards on the felt. So while the TCT is one that I understand, I see no real practical use for it.

Generally I use the principle when I've walked away from a shoe (in a generally play-all situation) and have to return to it and don't want to wait it out. I've used it when I had to miss a hand because I dropped chips on the floor and had to retrieve them. Most of the TCT value is in developing theory- if whatever you're considering doesn't square with the TCT don't do it.

But I feel the same way you do about leaving good counts and I've never done it voluntarily. I think I'd draw the line at fouling myself because you can get banned or arrested for that. If I'm going to do that, it's going to be for a high EV gambit and intentional.

 

[kewljason]

I think I'd draw the line at fouling myself because you can get banned or arrested for that.

Ok, that was a little exageration for effect. (most likely) :laugh:

 

[k_c]

I had to read through this thread a number of times before I felt I understood both what was being argued and where the discrepancy lies between the two points of view. (But then I'm more easily confused than most.)

I failed to come up with a purely theoretical response, but I think I may be able to offer a useful thought experiment to clarify things...

Imagine three players -

Player A sees every card dealt from the shoe and bets according to the precise TC.

Player B plays the first deck, takes a break, then plays the last deck before the shuffle. He treats the cards he missed during the break as unseen in the same way as those behind the cut card, adjusting the TC accordingly as he sees further cards removed.

Player C takes the same break as Player B, but he reasons that since the missed cards will, on average, have an even distribution, he can play out the last deck based on that assumption. That is, he gives the same weight to the cards he sees removed in the last deck as Player A does, dividing by the number of decks behind the cut card to get his TC.

[Edit: should have talked of cards remaining to be dealt, rather than behind the cut card in the above two paragraphs.]


So does Player C gain additional EV, compared to Player B (at the cost of raised variance)? If I've undersood correctly, that is blackjack avenger's premise.

It's a seductive argument, but what would be the consequences if it were true?

Suppose, instead of sitting out the middle portion of the shoe, players B and C both decide that they will only play the first two decks, then quit until the next shoe. This is entirely equivalent. In both cases they see only two decks. And if it is valid to assume an even distribution of the cards missed during the break, then it is equally valid to assume the same even distribution in the cards that will remain undealt when they quit the shoe after two decks.

What does this change mean for the two players? -

Player B, by playing the first two decks just as any counter (such as Player A) would, is playing a game with a reduced penetration (say 2 decks from 6), which is exactly what he was doing before, when he took a break in the middle, beacause he was using all the unseen cards (both missed and behind the cut card) as his divisor when calculating the TC.

Player C, by comparison, is now behaving very strangely. He plays the first deck normally, then arbitrarily decides to give increased weight to the cards he sees removed in the second deck, simply beacuse he plans to leave.

That clearly cannot make sense. Moreover, if it did, it should always be done, even if you don't plan to leave early.


Like blackjack avenger, I do find it a bit of a paradox that we can apparently use the assumption of an average, even distribution of the cards as the basis for something like an unbalanced counting methodology, but not for other purposes.

It's a confusing business!

Today I have more time to try to come up with a coherent example. It parallels my example where I started with a full shoe. That example showed the the discrepancy in the 2 methods but blackjack_avenger was not satisfied that it did.

This is for purposes of debate. I am impartial to what method anyone chooses. I am just out to let the numbers speak for themselves. I think that in a lot of cases blackjack players tend to ignore the truth because the truth gets in the way of the ability to speculate.

Starting out with the original example and then building upon it to parallel my full shoe example yields the following data:

1) Assume a Hi-Lo counter sees 2 decks of a 6 deck shoe and knows RC at that point equals +16, which is a TC of +4

2) Hi-Lo counter leaves and returns when 52 cards remain to be dealt and does not see the 156 cards that were dealt in the interim.

3) At this point dealer promises to deal 2 more rounds. The first round uses 26 cards and in that round 3 more low cards than high are dealt, so existing RC is increased by 3.

4) Method 1 says that RC = 16-3*4+3 since 3 decks have been dealt at an average RC of 4 per deck and then RC has increased by 3 in the round that was seen upon resuming play. Method 1 also says that number of decks is the actual number of decks remaining to be dealt, which is .5. Method 1 TC = +7/.5 = +14

5) Method 2 says that an unseen card is an unseen card regardless of its physical location, so method 2 RC = 16+3 and number of decks remaining = 4-.5. Method 2 TC = 19/3.5 = ~+5.4.

For my money method 2 is correct because it assumes nothing. In essence player is just counting normally and requires no additional faith.

Method 1 is correct in one case. That is where RC does indeed drop by 12 for the 3 unseen decks. What is the probability of that happening? Unfortunately I have no way of getting an exact figure for this. However, I can tell you that although this is the most likely change in RC this will happen no more than 10% of the time and most likely less. Evidence of this is that if 156 cards are dealt from a full 6 deck shoe then probability of Hi-Lo running count being equal to 0 at that point = ~5.13% even though it is the most likely RC at that point.

Obviously the methods do not agree but anyone is free to choose what they want to employ.

 

[London Colin]

For my money method 2 is correct because it assumes nothing. In essence player is just counting normally and requires no additional faith.

Method 1 is correct in one case. That is where RC does indeed drop by 12 for the 3 unseen decks. What is the probability of that happening? Unfortunately I have no way of getting an exact figure for this. However, I can tell you that although this is the most likely change in RC this will happen no more than 10% of the time and most likely less. Evidence of this is that if 156 cards are dealt from a full 6 deck shoe then probability of Hi-Lo running count being equal to 0 at that point = ~5.13% even though it is the most likely RC at that point.

Obviously the methods do not agree but anyone is free to choose what they want to employ.

I'd be a little more dogmatic than that. Method 1 is simply wrong.

What I tried to show with my example is that there is no need to wait for missed cards in order to employ method 1. The unseen cards that have yet to be dealt (or lie behind the cut card) are just as amenable to the assumption that the RC changes evenly throughout as any block of missed cards would be.

So if method 1 has any merit, then you should be free to swap steps 2 and 3 in your example -

After two decks have been dealt and the TC is +4, make the assumption that there is a block of 156 cards (or any size you wish) which will come after the next two rounds to be played and will have an averge RC of +4 per deck.

Therefore play the next round, having not missed any cards, using the calculations you gave in step 4. (Or some variation thereof, depending on just how large a chunk of the undealt shoe you arbitrarily decide to treat in the manner of missed cards.)

So the implication is that we are at liberty to apply any weight we choose to the removal of a card, dividing the RC by anything we like, since any number of the unseen cards can be eliminated from our calculations by assuming an even distribution.

All the above is obviously nonsense, but it is a logical consequence of accepting the validity of method 1.