Wonging Count

[k_c]

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:

I really don't know why I'm responding because there is probably nothing I can say that will change your thinking but....

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

This is the consequence of your logic:

If true why not just cut to the chase. After 1 deck RC = +10, decks remaining = 5, TC = +2, any subsequent change in RC is divided by a little less than 5. Now tell yourself "I'm skipping 4 decks so RC = +2, decks remaining = 1, TC = +2, any subsequent change in RC is divided by a liitle less than 1." But why wait for those 4 skipped decks to be actually dealt? If you simply cut the last of the remaining 5 decks to the top then that's the same as putting 4 decks in the discard tray. All you need to do is quit after the next deck is played. Any 1 of those 5 decks will on average be the same. It doesn't matter which 4 of the 5 decks were skipped. Voila, instant pen to 1 deck.

But wait; you can do even better. Why not skip 4.5 decks and get instant pen to .5 decks so now you can multiply by 2 instead of dividing by 5?

But wait; you can do even better. Why not skip 4 2/3 decks and get instant pen to 1/3 deck so now you can multiply by 3 instead of dividing by 5?

The consequence of your logic is that you can define your own pen on the very next round by just telling yourself, "I'm just playing for X more cards and skipping the rest." This is the same thing as first skipping (remaining cards - X) and then playing the final X number of cards.

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?

That is a completely valid observation.

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.

You are not accepting the consequence of your own logic.

 

[London Colin]

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

With the result that if the next card to be dealt is a ten, you will infer that the TC has dropped from tc2 to tc1, and if it is a small card you will infer that it has risen from tc2 to tc3.

(And that's just for the first card! The effect gets magnified as the remaining cards are depleted and your divisor gets even smaller.)

Whereas if you make the normal calculation of rc / unseen cards, the tc will have barely changed at all. You've only seen 53 out of 312 cards, yet you are inferring huge swings in the TC, as if this were a 2 deck game!

So what are the consequences of this?

In this example, suppose +2 is the point at which you first assume an advantage and raise your bets, at +3 you raise your bets even higher, and at +1 you go back to 1 unit.

As you play out the remaining cards, you will be missing opportunities to raise your bet, because you think the TC has fallen below +2, when it has not, you will be overbetting when you do have an advantage, and you will even be raising your bets when you do not in fact have any advantage.

 

[blackjack avenger]

I Account For All Cards

What is the TC theorm saying?
TC stays the same
RC drops
So the TC theorem accounts for the missed cards, it tells you what cards you missed on average. So by inference you have counted them. So you divisor can become smaller.

So if I return later
what has happend?
TC stays the same
rc drops

Missed cards and the TC are accounted for without us even seeing them, if the TC has to be the same, then by inference we can establish the running count

When you count don't you know on average the distribution of the cards moving forward? Using the same logic you can know what they were going backwards.

the missed cards are accounted for; despite your protests to the contray, because the rc drops. So yes as one continues they have a smaller divisor because the cards we missed ARE ACCOUNTED FOR.

Perhaps better terminology is the TC theorem allows one to have an approximation of missed cards by inference.
The inference is based on the TC theory which has been accepted for over 10 yrs and I think most of you agree with. If we know a TC we can know an RC
:joker::whip:

 

[blackjack avenger]

TC Theorem, Not Perfect Fix

6 decks 1 cut
tc1 is the beginning of positive expectation.

you know the first deck dealt is rc5, tc1 for rest of shoe
you leave for 3 decks
employ TC theorem
come back with 2 decks to go, tc1 so rc must be on average 2
play normal starting with 2 deck divisor

You guys are claiming this is a good situaion? That I should just jump forward at any time for "better penetration"
This is horrible, I missed 3 decks of positive expectation play. The farther ahead I jump the more good hands I miss and the fewer good hands I have left.
I only have 2 decks left to play. One deck of actual play due to 1 deck cut.

However, if I did play all the way through instead of leaving I would be on average; because of the tc theorm, facing exactly the same situation. A small divisor with few hands to play ending with same rc 2 and same tc 1. If I end up in the same exact place on average wether I played all or not, how can it be wrong?

The tc theorem allows me to pseudo count by inference what those cards were I did miss, but I suffer the loss of positive expectation ev because I did not play the missed cards. This is a horrible way to play. However I prefer counting the unseen decks by inference vs not at all. Not considering the unseen cards you have tc 1, rc 5, divisor 5 decks upon returning.

Let's play the TC theorem through and see what happens:
there is 1 deck to play; then the cut card, then the deck behind cut card.
tc 1 rc 2
on average 1 extra high card comes out during the last deck of play.
cut card comes out
On average what was behind the cut card?
1 extra high card
now if we were allowed to count the deck behind the cut card and the 3 we missed, what would we find for the 4 decks on average? RC4 and if we were allowed to play those cards to the last card, what would we find? rc0 tc0

probably the main points are:
jumping to the last few hands in order to employ a small divisor is not good because you missed out on playing the cards you inferred using the tc theorem. The farther ahead one jumps the worse this is. Also, I hope obviously I would like to infer a TC as little as possible.

I also believed I showed there is on average no great distortion of play moving forward using the TC theorem. If one brings up what if X or Y cards come out, well we never know if the big cards are out of play or behind the cut card. We can always be betting into garbage.
:joker::whip:

 

[London Colin]

:bomb:

 

[iCountNTrack]

Final Attempt

Reading this thread gives me a headache, BA i think you are getting to a point where you dont want to accept that you might be wrong on this one, overlooking all the examples and reasoning that have been graciously given to you. I am going to try to explain one more time why your logic is flawed, by going back to basics.

I want to first start with the counter-logical aspect of your approach which seemingly discriminate between the types of unseen cards??!!! A pack of unseen cards is a pack of unseen cards, it DOES NOT matter WHY or HOW we failed to see them. It doesnt matter whether you didnt seem them because in the house of hell the dealer burns 10 cards after every round, or because you left the table because the long lost love of your life walked by you. It doesnt matter whether they are in the discard tray or behind the cut-card.
This line of thinking is blasphemic because it goes along the line of thinking that states that a pack of cards picked from the beginning of the shoe is less random than one picked from the middle of the shoe!!

I am not going to enumerate all the erroneous applications of the true count theorem that you have mentioned in your posts, however i will focus on a very important detail. The True count theorem states the following

"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."

The True Count theorem WILL NOT apply if the cards are not REVEALED and removed, in fact in the proof of the True Count Theorem the word "COUNTED" is also added, "REVEALED, COUNTED, and REMOVED"
I have reformatted the proof given by Mike Hall long time ago to better explain what i am talking about. It clearly indicates that the cards have to be REVEALED, COUNTED, REMOVED.
REMOVED alone does not cut it, and the TC theorem does not apply.

 

[iCountNTrack]

:bomb:

 

[blackjack avenger]

Can't Infer Anything, Can't Shuffle Track

6 deck
cut 3
straight shuffle

Apparently you can't shuffle track this because we cannot infer anything we can't see an count.

I can shuffle track it because I can infer what can't be seen
:joker::whip:

 

[blackjack avenger]

Simplicity to Prove My Point

6 deck cut 1.5
first deck rc 10, tc2

Now stop counting but drop the RC 2 for every deck you play.

OR

Now stop counting and one can play every remaining hand with a tc2 bet.


BOTH EXAMPLES ARE HORRIBLE WAYS TO PLAY, but they can be done. In both examples on average every hand moving forward is tc2. In both examples I am in effect missing cards, by not counting them. I am just playing through even though not counting.

They both are simple ways to play, again still horrible.

For those who are saying you "can't" do this, just don't see it. They don't understand the TC Theorem.

The above is basically what I have been saying one can do if they miss cards.

I have to apologize to everyone for not coming up with this clear, easy to follow example pages ago that shows what is possible using the TC Theorem.

:joker::whip:

 

[blackjack avenger]

One Can Infer

you walk up to a table, first hand
cards are on the table
you scan the table
get the RC
first hand is 2 card 20
the second hand has no cards, but 2 green chips and in front of them are three green chips. The player is also acting quite happy
the dealer is playing out the third hand
One can infer that the second hand was a bj and subtract the RC by 2
Yep, I did not see the cards.
Sure, the bet could be from the previous hand, but if you are wrong the earth won't swallow you.

Another example
Especially if a high count
A player has 13 and hits, breaks
you miss the hit card
you can infer that it was a big card and use that info if you wish. Now, the break card could have been a 9, but if you are wrong the earth won't swallow you. I belive Wong says you can do this? Don't hold me to it.

Yet, 2 more examples of how one can use incomplete information.

This post with the previous one I hope shows one can reasonably act with incomplete information.
:joker::whip:

 

[iCountNTrack]

BA, it is rather clear to me that you are not even reading the responses given to you, and you ar ejust posting the same things over and over again like JSTAT. I invite you to carefully reread the entire thread perhaps you will be enlightened.

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