Wonging Count
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blackjack avenger
Well-Known Member
Not So Much
if you are playing any +TC in any game on "average" the good cards will be equally distributed amongst the remaining cards. One is just playing the "average"
I offer you these rules
you have an advantage off the top
you have to flat bet entire pack or you can only play 1 hand with a shuffle
would you play both scenarios?
the answer is yes, depending on advantage, variance, bank
If you don't see it yet, I think you just don't quite understand the TC theorem and it's potential application in this; hopefully, rare situation.
With straight counting you don't know what cards will come out, you are playing with incomplete information, this is not much different. With straight counting one places that bet, hoping the TC theorem is correct and some of those good cards come out.rrwoods said:
if you are playing any +TC in any game on "average" the good cards will be equally distributed amongst the remaining cards. One is just playing the "average"
I offer you these rules
you have an advantage off the top
you have to flat bet entire pack or you can only play 1 hand with a shuffle
would you play both scenarios?
the answer is yes, depending on advantage, variance, bank
If you don't see it yet, I think you just don't quite understand the TC theorem and it's potential application in this; hopefully, rare situation.
blackjack avenger
Well-Known Member
At Least Some Credit
:joker::whip:
I said it LOLassume_R said:
:joker::whip:
blackjack avenger
Well-Known Member
I Have Never Gotten a Wart From a Frog
If you were to apply the put the unseen cards behind the cut card plan you would place a Tc2 bet instead of tc4, again ho hum not much difference for 1 hand.
Has anyone ever let a winning bet ride if the count dropped from TC4 to perhaps tc3 or even tc2? If you have done that then this should be less of an issue.
:joker::whip:
So applying the TC theorem the RC has dropped from 16 to 8. The TC applying the TC theorem is still tc4. It's a positive expectation hand. If you make one bet early on before you leave you are hoping the TC theorem is true. As you make that last bet are you also not hoping the TC theorem is true? Also, in your scenario it's only 1 round in hopefully a rare situation, not going to make or break you.sagefr0g said:what your saying is far as i know correct theory.
thing is certain questions make one wonder about all this.
like ok, say your two decks into a six deck shoe and the tc=+4
also say at tc=+4 is when you make your max bet.
but say for whatever reason you must temporarily get up from your seat and are not able to monitor the next two decks worth of play.
and say there is a round worth of play left before the shuffle and you can sit back in and make that play.
are you gonna lay out your max bet?:whip:
If you were to apply the put the unseen cards behind the cut card plan you would place a Tc2 bet instead of tc4, again ho hum not much difference for 1 hand.
Has anyone ever let a winning bet ride if the count dropped from TC4 to perhaps tc3 or even tc2? If you have done that then this should be less of an issue.
:joker::whip:
Firstly, I'd like to credit blackjack avenger for the insightful analogy lol.blackjack avenger said:
Secondly, how can there be 2 different answers? Won't 1 of these 2 situations (either a TC +4 bet or a TC +2 bet in this example) yield, on average, a higher EV? What's correct?blackjack avenger said:
blackjack avenger
Well-Known Member
Don't Do This at Home, or All The Time
In both; hopefully, rare situations you will face the same cards on the following hands.
They are both the same positive situations.
So the bigger bet will yield the higher EV, there will be more variance.
Applying the TC theorem can probably be considered more aggressive. If one wanted to bet a bit on the conservative side there is nothing wrong with that. However, not I
Of course I do bet a small fraction of kelly
I am the avenger
:joker::whip:
The rareness of the situation makes either method valid. You will probably have lost more in the hands missed then on what is the correct thing to do with the cards that are left.assume_R said:
In both; hopefully, rare situations you will face the same cards on the following hands.
They are both the same positive situations.
So the bigger bet will yield the higher EV, there will be more variance.
Applying the TC theorem can probably be considered more aggressive. If one wanted to bet a bit on the conservative side there is nothing wrong with that. However, not I
Of course I do bet a small fraction of kelly
I am the avenger
:joker::whip:
sagefr0g
Well-Known Member
i remember Renzey writes about this sort of stuff in his Blackjack Bluebook II, errhh or at least sort of. where he talks about how multiple deck is sorta sluggish as far as the volatility of the true count. he even goes so far as to say at some point with a TC high enough one could just stop counting all together and bet some appropriate amount the rest of the way through the shoe, that this would end up being some small advantage, sorta thing. and alternatively at some point if the TC was low enough at some point in a shoe that it would be worthwhile abandoning the shoe. well, i hope i'm describing what i read fairly accurately. i think Blackjack Attack touches on this sorta thing as well, when referencing wonging.blackjack avenger said:
whatever, again questions arise. like what would relying on the TC theorem mean as far the optimality of your betting? kind of throws the idea of proportional betting out the window, no?
what about risk of ruin, i know you mentioned how variance would be heightened, so your ROR would take a hit from this sort of stuff, no?
now i'm also remembering how Renzey described such stuff would have high variance. so one wonders, how much higher would the variance be, and if ROR is raised, how much would it be raised.
the other thing you mention is how this would likely be a rare situation, hopefully. just me maybe, but that's interesting, because rarity is a hallmark of advantage, far as i've seen.
but again just me maybe, but the rarity of certain advantages sometimes has me taking pause for thought far as the concern of do i wanna make this play or not, sorta thing. pretty much because of the tag along variance and the extreme of the rarity, sorta thing.
i also remember QFIT writing about how in some sense a novice sort of player who only plays rarely, say on vacations, sorta thing, might be better off just playing basic strategy and flat betting than trying to play a card counting strategy.
whatever, avenger, you seem to be one that has an interest, far as Kelly stuff goes, for sizing and resizing bets, i think perhaps according to damage done to ones bankroll, or health of ones bankroll, sorta thing.
how about resizing bets according to rarity and potential variance, sorta thing?
sometimes it just seems certain situations are just begging to take us on that horrific extreme roller coaster ride.:yikes::whip:
.
This is wrong.blackjack avenger said:
If TC = +4 at 4/6 decks remaining then RC = +16. RC should decrease on average by 4 per remaining SEEN deck. 4 decks remain UNSEEN.
If 2 decks are dealt but are UNSEEN then with your method you are saying that you haven't actually seen the cards but if you did the average RC will have dropped by 8.
However since the cards are actually UNSEEN then RC doesn't drop at all and still = +16 with 4 UNSEEN decks remaining to be dealt. It's the same as just cutting the unseen 4 decks in half and playing the second half instead of the first.
Either method ends up with TC = +4.
So if I leave with 4/8, and the TC = +4. Then I come back at 6/8. I know the TC should still be +4. So I estimate that the RC is probably +8, and there are 2 decks left. You're saying this is incorrect logic? That I have to keep the RC at +16, and just divide by the total unseen decks in all cases?k_c said:
Yes to me it is preferable to divide known RC by unseen decks rather than to assume RC has dropped by the average so you can instead divide by the actual number of remaining decks by estimating what is in the discard tray.assume_R said:
One way is based on an assumption while the other just relies on what is known.
Automatic Monkey
Banned
For the first hand, there is no difference, the TC you will be using is still +4 in either case.assume_R said:
Let's say it's a deeply dealt shoe and you are now at the last hand, 7/8 decks dealt out. The RC of the cards you've seen dealt since you've been back is -2, 2 extra 10's.
Using the first method of estimating the new RC: you started with an estimated RC of +8, your RC is now +6, there is 1 deck left so your TC is +6
Using the second method of keeping the observed RC constant: your RC was +16, it is now +14, there are 3 unobserved decks, your TC is +4.7.
The second is more accurate because seeing that extra deck dealt out after you returned to the table gives you more real-world information about the cards that were dealt while you were gone that the TCT can not. The TCT tells you (accurately) that the expected TC of the decks you missed was +4, but because the TC of the deck you saw upon your return was -2, there were probably more high cards in the missed portion than you expected, and your RC should now be adjusted down for that fact.
blackjack avenger
Well-Known Member
Unseen Does not Mean Nothing
the TC theorem gives us some information on the cards we missed while away. One can use the tc theorem info or not
All the cards we bet into are unseen.k_c said:
the TC theorem gives us some information on the cards we missed while away. One can use the tc theorem info or not
blackjack avenger
Well-Known Member
One Man Stands Alone LOL
A weakness of treating the unseen cards as completely unknown (ignoring the TC theorm) is moving forward you have a much larger divisor caused by considering all cards unseen and (completely unknowable, which is not true)
Here is perhaps another way to look at the TC theorem method. If you are at tc2 and instead of playing the next hand from the next cards you played the hand from 2 decks deaper in the shoe, the dealer literally took the 2 decks out of the center of the shoe, would you do it? sure! Yet, those cards left in the shoe are alllll unseen.
It's subjective; neither is wrong, which method one choose's:
Using the tc theorem is more aggressive, and accurate because using some knowledge I think is preferable to pretending one has no knowledge.
Putting the unsees cards behind the cut card is more conservative and one is choosing not to use the knowledge provided by the tc theorem
A weakness of treating the unseen cards as completely unknown (ignoring the TC theorm) is moving forward you have a much larger divisor caused by considering all cards unseen and (completely unknowable, which is not true)
Here is perhaps another way to look at the TC theorem method. If you are at tc2 and instead of playing the next hand from the next cards you played the hand from 2 decks deaper in the shoe, the dealer literally took the 2 decks out of the center of the shoe, would you do it? sure! Yet, those cards left in the shoe are alllll unseen.
It's subjective; neither is wrong, which method one choose's:
Using the tc theorem is more aggressive, and accurate because using some knowledge I think is preferable to pretending one has no knowledge.
Putting the unsees cards behind the cut card is more conservative and one is choosing not to use the knowledge provided by the tc theorem
rrwoods
Well-Known Member
What makes you think you have any more knowledge about cards in the discard tray that you haven't seen as compared to cards actually behind the cut card? Why are they any different?blackjack avenger said:
The fact that they have been dealt does not matter. You have exactly the same information about them as you do about the ones behind the cut. You can estimate their composition based on the true count, but you can't completely infer that composition, which is what "using the TC theorem" presumes to do.
blackjack avenger
Well-Known Member
rrwoods said:
blackjack avenger
Well-Known Member
What Does Experience Show Us?
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:
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:
You're not pretending the unseen cards are behind the cut card. They are simply unseen cards. Yes they are in the discard tray but why would it be better to assume what they should be on average when they can simply be treated as unseen, which is exactly what they are.blackjack avenger said:
As an example start with a full six deck shoe and assume player leaves and comes back after 5 decks have been dealt and dealer is game to deal 2 more rounds. Let's say at the end of the first round running count turns out to be +3 with 1/2 deck remaining to be dealt. Would you be more comfortable calling the TC +6 (3/.5) or +.545 (3/5.5)? The principle is the same.
With method 1 you are speculating that the unseen cards are exactly what they should be on average.
With method 2 you are just simply treating them as unseen cards.
Smaller divisor is speculative in nature. Watch out for overbetting.blackjack avenger said:
blackjack avenger
Well-Known Member
Nothing has Changed
3rc. If we are again distracted for a quarter deck and return the TC is still
.545 thanks to the tc theorem and all we need to do is pick up where we left off, approx rc2.85? To use your example to further show my point, if dealt down to the last hand on average we will end up with tc0 as the final big cards come out. Perfect info? No, but useful, yes.
To sum up as briefly as possible:
Once you have a TC, on average it remains the same as hands are played thanks to the tc theorem. Whether the played cards are seen or not does not matter.
:joker::whip:
I don't see where your example nullifies the tc theorem. The answer is .545. One does not get to employ the tc theorem because one has no knowledge of any cards. I can tell you where that .545tc average is, it is spread evenly on average through the 5.5 decks. So for the 5.5 decks we have .545tck_c said:
3rc. If we are again distracted for a quarter deck and return the TC is still
.545 thanks to the tc theorem and all we need to do is pick up where we left off, approx rc2.85? To use your example to further show my point, if dealt down to the last hand on average we will end up with tc0 as the final big cards come out. Perfect info? No, but useful, yes.
To sum up as briefly as possible:
Once you have a TC, on average it remains the same as hands are played thanks to the tc theorem. Whether the played cards are seen or not does not matter.
:joker::whip:
iCountNTrack
Well-Known Member
blackjack avenger said:
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.
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