Wonging Count

[k_c]

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.

I agree with you but hey, it's hard to decline instant free pen. :laugh::laugh:

 

[blackjack avenger]

Terminology

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.

As soon as you say:
The average of the unseen slug
The average of the known slug
when mixed
you are applying, I will say aspects of the TC Theorem

The NRS talks about how high or low cards are dispersed through a section or the shoe. Where do you think that idea of dispersed comes from? or what supports these thoughts?

The TC Theorem

Just because the words "TC theorem" are not used does not mean it's not part of the foundation.

In shuffle tracking you use incomplete information on cards not known, just as I am suggesting.
:joker::whip:

 

[Bojack1]

In shuffle tracking you use incomplete information on cards not known, just as I am suggesting.
:joker::whip:

Suggest all you want, but good shuffletrackers do not track in this manner. This is a generic, flawed, variance through the roof, easy to sell method. Tracking in the manner in which you suggest kills more bankrolls than it builds.

 

[blackjack avenger]

Heretics

You guys are forcing me to post more, I can't do 1 a day when there are 7 against! LOL
Don't you guys have dates on Fri nights LOL, or at least play! LOL

You guys do realize you are the ones talking heresy, you are all denying the TC theorem and its applications. You all see it, but can't take the next step.

KC brings up variance as if there is no variance in any bets we make. What is the long term variance in employing my technique rarely during a career? Not very much. I can easily bring up an example which greatly supports my position. Especially considering my method employs a smaller TC divisor moving forward.

London Colin has not quite understood what I am saying, and it has ran for several posts. I am not saying this!
You walk up to a table 2 decks before shuffle, you play 1 deck and have rc10, so then one can say the last deck is TC 10.
THIS IS WRONG, WRONG, WRONG, WRONG!!!!!!!!!!!!!!!

TC Theorem in use every day.
You have a RC of 10 with 2 decks left, TC is 2 and we know the high cards are there because of the TC theorem. The cards are randomly distributed after we take out known cards.

Some here would advise using easy counts, rounded indices etc. but in this debate they want to use the harder method. When you use the unseen method, you have to remember those unseen, missed cards every time you make your TC adjustment moving forward. Anyone want to consider the cost of forgetting to include the missed cards? With my method you make one math calculation and you only have to consider the unseen cards moving forward. So it appears I also have the ease of use argument! This also makes the answer subjective in nature. Those of you whom are proponents of keeping things simple I should have now won over.

It has been said several times, how bad it is to miss cards. We all agree!
If one has left a negative shoe, it's better to start a new one.
The TC probably has to be rather high before you consider coming back because you are facing poor penetration, but the higher the TC the more inclined one should be to NOT LEAVE in the first place.

:joker::whip:

 

[blackjack avenger]

Truly Disagree

Suggest all you want, but good shuffletrackers do not track in this manner. This is a generic, flawed, variance through the roof, easy to sell method. Tracking in the manner in which you suggest kills more bankrolls than it builds.

There is ST that employs complete knowledge
There is also ST that employs incomplete knowledge
Are you saying an ST team would not act if they missed the first hand? Had no real knowledge of that slug? We know if they had the rest of the shoe known they would act. I also imagine they could give you an idea of what that first hand actually was.

Ahhhhh?
Maybe a breakthrough?
One is willing to bet bigger because they have cut X extra cards in Y slug. Often we do not know the exact value of Y slug.

I am saying be willing to bet because we know X number of cards on average are in a section of the shoe we missed.

If I offer any of you a +TC section of a shoe, but you only get to play 1 part of it, you would all do it. Wait, that is what we do all the time!

Some of you guys scream about penetration is everything, the unseen, unknown method offers poor penetration. My method offers deeper penetration. Is penetration king or not? Would you rather play deeper penetration with a weak count (lower betting correlation, less accurate count) or poor penetration with a strong count (higher Betting correlation count, more accurate count)
I will say it again, employing the tc theorem gives some information vs no information with the unseen method. Those who scream penetration is everything I should have now won over.

No one has addressed that averages are used in unbalnced counts? The Tc theorem seems to be totally accepted in this context. Guess what, sometimes when you emply an unbalanced count you are overbetting or underbetting. Those indices are sometimes incorrect when used

Just about every bet we make is on incomplete/unknown knowledge, we bet into those TCs hoping those good cards come out. Every TC bet we make we are hoping the TC theory holds true. Well it does.

 

[assume_R]

Just a thought

What seems to be at the heart of the debate is this:

Let's say we are 4 decks into an 8 deck shoe with a RC of +16 (TC = +4) and we miss the next deck.

BlackjackAvenger is saying that on average, that next deck will drop the RC by 4. He knows that the TC theorem says that on average, 4 big cards will come out. If this situation were to happy millions of times, the TC Theorem says that the unseen slug will drop the RC by 4. Whether or not they did doesn't matter to him, because on average the RC will drop by 4. So he'll just play as if they did. (with a RC of +12 when he comes back, and continuing on his merry way!)

 

[London Colin]

London Colin has not quite understood what I am saying, and it has ran for several posts. I am not saying this!
You walk up to a table 2 decks before shuffle, you play 1 deck and have rc10, so then one can say the last deck is TC 10.
THIS IS WRONG, WRONG, WRONG, WRONG!!!!!!!!!!!!!!!

I know that's not what you are intending to say. That's the whole point!
It is a direct consequence of what you are saying, which surely should make you stop and think.

A implies B.
B is nonsense, therefore A is nonsense.
It's not rocket science.

 

[blackjack avenger]

Saved by the TC Theorem

I know that's not what you are intending to say. That's the whole point!
It is a direct consequence of what you are saying, which surely should make you stop and think.

A implies B.
B is nonsense, therefore A is nonsense.
It's not rocket science.

the TC theorem is not nonsense
accepted for over 10 yrs.
it's actually common sense
You are making an assumption, I am not making

if you walk up to a table
2 decks left
you play one deck
then rc 10
what then is your TC?
Its tc2
RC spread out over the 5 decks you did not see
consistent with what I have been saying.

Now the incorrect method
2 decks left
you play one deck
then rc 10
what then is your TC?
Its not tc10
because you are not using the tc theory which tells us the excess good cards will on average be spread among all unsees cards. London Colin I hope you see that the TC theory keeps us from playing this situation incorrectly.
:joker::whip:

 

[blackjack avenger]

Nice

say we are 4 decks into an 8 deck shoe with a RC of +16 (TC = +4) and we miss the next deck.

BlackjackAvenger is saying that on average, that next deck will drop the RC by 4. He knows that the TC theorem says that on average, 4 big cards will come out. If this situation were to happy millions of times, the TC Theorem says that the unseen slug will drop the RC by 4. Whether or not they did doesn't matter to him, because on average the RC will drop by 4. So he'll just play as if they did. (with a RC of +12 when he comes back, and continuing on his merry way!)

Well Put
:joker::whip:

 

[London Colin]

Back to the beginning

if you walk up to a table
2 decks left
you play one deck
then rc 10
what then is your TC?
Its tc2
RC spread out over the 5 decks you did not see
consistent with what I have been saying.

RC = 10.
5 decks unseen.
TC = 10/5 = 2.

No one, throughout these many pages, has been arguing anything different.

The TC stays the same while the RC changes
So
if you leave a shoe at TCx
when you return it is still TCx

you need to adjust the RC to give you the appropriate TCx

example
you leave at 4 out of 8 decks, running count is 8 TC is 2
you return at 6 out of 8 decks, TC is 2 and you have to assume a running count of 4 because of tc theorem

there will be wild variance but on average this is what it will be
:joker::whip:

with my method you make the one RC adjustment then you play as normal. A one step process.

With Ferrets method you have to remember to add that one deck behind the cut card every time you convert RC to TC

I think mine is easer, but it's subjective

It's the above methodology that gives rise to all the absurdities that I have pointed out, if you will only follow the logic through.

Yes, it gives the correct TC at the moment you re-enter the shoe, as has been said. But thereafter it causes you to overvalue every further card that you see, treating an 8-deck shoe with 2 decks missed as if it has magically become a 6-deck shoe.

That in itself sounds like only a minor complaint, but it is symptomatic of a much deeper problem, which I laid out in my very first post in this thread.

In response to that post, you said you didn't quite follow what I was saying about player C. By all means ask specific questions about that, and I will try to elucidate. But otherwise, I really can't think of anything else to say!

 

[assume_R]

Player C - London Collin

That in itself sounds like only a minor complaint, but it is symptomatic of a much deeper problem, which I laid out in my very first post in this thread.

In response to that post, you said you didn't quite follow what I was saying about player C. By all means ask specific questions about that, and I will try to elucidate. But otherwise, I really can't think of anything else to say!

So player C would play 1 deck, and then have a RC of +14 (7 unseen decks, TC of +2). He would reason that the RC for a random deck later (i.e. deck 4), the TC should be +2 if that deck was seen. So 2 extra big cards would come out on average, and the RC would be +12 (with 6 unseen decks). So for his next deck (which is the last one he's playing), he could use a RC of +12 with 6 decks left (even though he only played 1).

So an extension would be if he presumed that in 5 unseen decks, the RC would drop by 10. So he would have a RC of +4, with 2 decks left. He would then play his next deck (which is the last one he's playing) as if it was a RC of +4 with 2 decks left. This would make the TC calculation more inaccurate, because it's making assumptions that in the unseen decks, the RC drops what it "should", "on average", drop.

Just because this leads to a less accurate TC calculation, is it incorrect, though? Because on average, truly, the RC will indeed drop by 2 per deck (with a TC of +2).

Correct? Is this what you're saying, LC?

 

[rrwoods]

London Colin has not quite understood what I am saying, and it has ran for several posts. I am not saying this!
You walk up to a table 2 decks before shuffle, you play 1 deck and have rc10, so then one can say the last deck is TC 10.
THIS IS WRONG, WRONG, WRONG, WRONG!!!!!!!!!!!!!!!

holy crap are you even reading his posts

 

[rrwoods]

Just because this leads to a less accurate TC calculation, is it incorrect, though? Because on average, truly, the RC will indeed drop by 2 per deck (with a TC of +2).

Yes, it is incorrect. Entirely because of the consequences it has. If the unseen cards were, say, actually behind the cut card instead of between two segments of play, you could make the same argument, essentially giving yourself whatever pen you wanted. Clearly this is not correct.

 

[London Colin]

It's worse than that

So player C would play 1 deck, and then have a RC of +14 (7 unseen decks, TC of +2). He would reason that the RC for a random deck later (i.e. deck 4), the TC should be +2 if that deck was seen. So 2 extra big cards would come out on average, and the RC would be +12 (with 6 unseen decks). So for his next deck (which is the last one he's playing), he could use a RC of +12 with 6 decks left (even though he only played 1).

That's the gist of it. But for direct equivalence with the case where 4 decks are missed while taking a break, player C could assume that any random block of 4 decks that he will not be seeing after he quits has the 2 extra big cards per deck average. Thus he would be playing the second deck as if it were RC +6 with 3 decks unseen, just as he would have been after taking the 4-deck break. (Assuming an 8-deck shoe, as per your example).

But there's nothing special about 4 decks, why not assume that all the decks you have yet to see have this arrangement? Why limit yourself to whole decks?

Where does it end? You could play the next quarter deck as if all other unseen cards have the RC evenly distributed. Why not? If it is valid in one case, it should be valid in all cases.

So an extension would be if he presumed that in 5 unseen decks, the RC would drop by 10. So he would have a RC of +4, with 2 decks left. He would then play his next deck (which is the last one he's playing) as if it was a RC of +4 with 2 decks left. This would make the TC calculation more inaccurate, because it's making assumptions that in the unseen decks, the RC drops what it "should", "on average", drop.

Just because this leads to a less accurate TC calculation, is it incorrect, though? Because on average, truly, the RC will indeed drop by 2 per deck (with a TC of +2).

Correct? Is this what you're saying, LC?

I probably should have read that bit before I wrote my previous pararaph. :grin: So yes indeed, that is what I have been saying.

And it clearly must be incorrect, because you could arrive at any answer you want, just by changing your choice of how large a chunk of the shoe to apply your 2-per-deck assumption to.

As for the theoretical underpinnings about why it is incorrect, the True Count Theorem, and so on, I'm afraid that's where I start to get out of my depth.

My way to get a handle on all this was simply to imagine that blackjack avenger's approach might be valid, and then to think through the consequences if that were so. (No theorems were involved)

 

[k_c]

Yes, it is incorrect. Entirely because of the consequences it has. If the unseen cards were, say, actually behind the cut card instead of between two segments of play, you could make the same argument, essentially giving yourself whatever pen you wanted. Clearly this is not correct.

rrwoods gets it. Pen is dependent on unseen cards and not cards remaining to be dealt. Most of the time these are the same thing but in this discussion they are not.

As London Colin has implied the consequence of making pen dependent on cards remaining to be dealt makes pen an arbitrary number.

Unfortunately there is no such thing as free instant pen.

 

[blackjack avenger]

Legions Against Me LOL

I have read everything; more then once, but it's a lot for me to answer. So let's try this thought experiment:
In both examples we will deal to the last card
And WE ARE USING THE SAME SHOE

TREAT AS UNSEEN METHOD:
6 deck shoe
dealt 1
rc10, tc2
miss 2 decks
upon return
3 decks remain + 2 decks unseen
rc10, tc2
the divisor moving forward is 5 decks
We deal down to the last card
what do we end with?
The first deck we played, the 2 we missed and the 3 we played
rc0 tc0 as the TC theorem tells us and simple observation.

TC THEOREM METHOD:
6 deck shoe
dealt 1
rc10, tc2
miss 2 decks
upon return
3 decks remain
employ tc theorem
We lose on average 2 cards per deck we missed so
rc 6 tc 2
the divisor moving forward is 3 decks (different from above example)
We deal down to the last card
what do we end with?
If you take the first deck, 2 decks we missed and the 3 we played you will end up with
rc0, tc0 as the tc theorm tells us and simple observation.

Now do both examples 10 billion times and what do we expect?

On average the TC theorem will hold, just like the average of any other sim.
Using the TC theorem allowed us to employ a smaller divisor going forward, we will make more money because of the deeper penetration.

Now back in the real world we have a cut card so we don't get to fully employ the complete knowledge of treating all the cards as unknown and then dealing down to the last card. With either method we take the chance of betting into a TC and the good cards are not where they are suppose to be, they could be out of play.

:joker::whip:

 

[rrwoods]

So, BJA, when you walk up to a shoe in progress, and there's four of eight dealt, do you just divide by four?

That's what you're saying you should do. The TC started at zero. Four decks later, the TC is zero, so the RC is also zero. Divide by four, right?

 

[blackjack avenger]

Lost in Translation

So, BJA, when you walk up to a shoe in progress, and there's four of eight dealt, do you just divide by four?

That's what you're saying you should do. The TC started at zero. Four decks later, the TC is zero, so the RC is also zero. Divide by four, right?

No, your statements are incorrect. I don't think I ever said you ignore cards. I account for all cards in the TC theorem method.

In your example you would have to take the rc and divide by all unseen decks in this case.
:joker::whip:

 

[bigplayer]

So if the TC used to be -2.5, and on average it didn't change, and I still have 4 unseen decks when I get back from the break, the RC will still be -10 on average, right?

Either way, if you have 4/8 played with RC -10 and you come back with 6/8 played you can adjust the RC to -5 and the TC is still -2.5 with 2 decks left or leave the RC as -10 and count the two decks left and two you didn't see as unseen...either way the TC is still -2.5

 

[blackjack avenger]

TC Theorem Example

I think some of you don't grasp the TC Theorem. This will be an attempt to help correct that:

6 decks
you watch the first deck
rc10
where are those 10 extra cards?
there are on average 2 in each of the next 5 decks
employ TC theorem
tc2 divisor 5
play normal

you miss the first deck
you play the second deck
rc10
where are those 10 extra cards?
there are on average 2 in the first deck and 2 in each of the 4 decks you have yet to play
employ TC theorem
rc8
tc2 divisor 4
play normal

you miss the first 2 decks
you play the third deck
rc10
where are those 10 extra cards?
they are on average 2 in the first 2 decks you missed and an average of 2 in the next 3 decks you have yet to play.
employ tc theorem
rc6
tc2 divisor 3
play normal

you miss the first 3 decks
you play the 4th deck
rc10
where are those 10 extra cards?
there on average 2 in each of the first 3 decks you missed and an average of 2 in each of the next 2 decks you have yet to play.
rc4
employ tc theorem
tc2 divisor 2
play normal

you miss the first 4 decks
you play the 5 deck
rc10
where are those 10 extra cards?
there are on average 2 in each of the first 4 decks you missed and an average of 2 in the final deck you have yet to play.
employ tc theorem
rc2
tc2 divisor 1
play normal

Notice with the TC theorem in each instance I have an idea of what I missed and what I have moving forward. If you give me a running count at any point in the shoe I can tell you what the TC is for the entire shoe because I TAKE THE ENTIRE SHOE INTO CONSIDERATION.

It's important everyone sees when using the TC theory:
The TC stays the same
The RC drops

Now, I believe many will say, "Why not wait until the last deck of every shoe". One gets the advantage of a small divisor? The answer is you don't get as much of an advantage as you think. Your RC on that deck played is reduced because you are taking all cards into consideration, those you have not seen. Also, you will only be playing 1 deck, the cut card is soon to come. One still suffers a penalty from not seeing all unplayed cards.
:joker::whip: