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Clarke Cant on Floating Advantage
I understand that you are interested in this "deep math" subject but don't want a "deep math" explanation, so I would not really bother myself with it. How many hours do you play a year x bankroll x count system?
To me, floating advantage has caused more trouble than Sammy Bean Ladden has. The TC count represents a simplified version of your advantage either + or - and tells us when to play hands differently and change bet size within the realm of the table heat we could expect. That is what it is supposed to do. If you want to knock yourself out, read the following. Have you seen the movie A Beautiful Mind?? ;>
===============================================================
From: clarke_cant
Date: Thu Sep 26, 2002 10:46 pm
Subject: Before I forger just to archive
Since Puiu is driving the PBers nuts on bj21.com I figure I better
archive here--in case my laptop crashes too--a copy of my final
floating advantage paper. I still await the review Steepen is setting
up.
Here is the slighly revised version of my paper to use, of course not
all typos have been lassoed either:
Wongy on bjmath.com wanted to see a version of this without any
partition theory being included. This would be just another boring
idea for me except that elimination of partition theory mention leads
to a proof, in the proof of Postulate IIb, that penetration
absolutely cannot change any basic strategy expectation.
Theorem I: That any pack of remaining or original cards can be cut
such that any penetration can be transformed into any other.
Theorem II: That the average mean path of a random walk is a straight
line from origin to destination. Inclusive of this would be all
Euclidian postulates for a straight line.
Postulates:
I: The otherwise proven Jalib's True Count Theorem in a simpler proof
here, that any observed true count tends to remain at the same value
until all cards are drawn.
II: That actual mean basic strategy edge does not change with
penetration within a given pack or subpack.
III: That the difference between actual and true count predicted edge
(with basic strategy) increases with penetration.
IV: That more extreme True Counts progressively overestimate edge,
and less extreme True Counts progressively underestimate edge with
penetration.
Axioms:
I: True count =running count/cards left unseen in the pack.
II:Any pack composition subset can be precisely defined by a
combination of balanced rank versus rank, or ranks versus ranks,
balanced counts
III: All balanced counts, and some unbalanced counts, end at a
running count of zero.
IV: The difference between any actual edge and single, or multiple,
true count prediction of edge can be positive or negative.
V: Mean changes in expected value for a remaining pack subset are
proportional to the true counts used to measure them.
Postulate I: By Axiom III for any running count, RC(j) at j cards
left, RC(0)=0, for zero cards left.
Applying Theorem II for all j and k not equal to zero, with k less
than j, but j and k greater than zero, RC(k)=(k/j)*RC(j).
Applying Axiom I True count TRC(j)=n=RC(j)/j, TRC(k)=m=RC(k)/k.
(j/k)=(n/m) j*m/k=n j*m=k*n n/j= m/k
QED
Once you observe a true count the average true count does not change
no matter how many other cards are seen, for the boundaries set for j
and k above, in that the true count is the slope of the line of the
mean tendency of the running count to approach zero with no cards
left.
Postualate IIa: Proven by Theorem I with challenges answered in Cases
I thru IV. The basic strategy edge for any given round is independent
of the penetration in that any round starting point can be
transformed by cutting the pack into any other. The basic strategy
edge is the mean of all rounds played, and their original or new
position does not change the edge for any round.
Postulate III:
Step 1:A complete pack has the ijmplied balanced counts: +4 for every
non ten ranked card, and -9 for every ten ranked card, and serperate
balanced counts -12 for every rank, and +1 for all other ranks. by
the True Count Therem, or TCT for short, the mean composition for
this complete set of counts, after drawing 13 cards, is for the true
count of them all to remain the same as at the start, or zero. This
composition is the most predicted: removing one each of every rank
aces thru 9s, but removing 4 ten ranked cards. (See PIIb and Case I
however down below).
Step 2: At 13 cards drawn this most predicted mean pack is, by
inspection, only a tiny fraction of the total number of compostions
possible from drawing any 13 cards. At each level of penetration via
Axiom II, a new series of counts is needed is otherwise needed (the
10 seperate counts in Step 1 is different from the usual situations--
see below) to track the difference between the main count(s)'s
predictions and the actual composition. The solution in Step 1 is
trivial for comparing the difference between most predicted
compositions, and actual compostions, in that this set of counts
directly gives you the actual composition, but the result that each
most predicted pack subset is such a small fraction remains at least
down to the 3 cards left level--well past the smallest size needed to
complete a round of cards.
Step 3: At 3 cards left there will a constalation of balanced counts
added to the main count (see Axiom II) such that there will be some
card drawn, of the 3 left, that keeps the most number of such true
counts the same. But the 3 cards have the same probability of being
drawn. Thus the number of possible compositions that results in other
than the most predicted pack always excedes the number of
compositions that are most predicted.
In all such cases compositions are comparable to two dice totals in
craps, where the most likely total is 7, but the most likely set of
totals is some number other than 7.
Step 4: By Step 3 with just one rank versus another rank, used as a
balancd count, this property can be extended to drawing just one
card, to the 3 cards left point, which is far smaller that the number
of cards left to have a complete round.
Step 5: An algebraic simplificaton occurs when all such counts that
are added via Axiom II, to the main count(s), have a zero count, in
that then the composition is exactly that predicted by the main count
(s).
Step 6: The probability that any round will result in the need for a
new side count to precisely define the remaining subset pack will not
change for any round, via Theorem I. But with deeper penetration
every new count needed via Axiom II will have higher initial true
counts via Axiom I, and involve more difference between the actual
remaining pack, via Axiom V (with Axiom IV also applying), and the
subset pack predicted by all previously applied counts.
QED
Penetration increases the difference between the edge predicted by
the main count(s) and the exact pack composition. This difference,
and need to add new counts, occurs with a frequency that is
proportional to the number of cards sampled from the start of the
orginal pack of cards too, as explained in step 6. Overall the
presense of extreme counts, added to the main counts to define the
actual pack, as opposed to the most probable pack indicated by the
main count(s), provides a mechanism by which the expected value
associated with the most probable pack, for a given penetration
point, is reduced by such exteme true counts also being included, to
lower the actual mean expected value to that of the original starting
pack. The added counts are tending to have higher initial true
counts, and extreme true counts have negative bow effects.
Postulate III is proven without needing to modify Postulate II. See
Cases I thru IV below to refute arguments otherwise.
And the mechanism discussed above by which differences between the
main count(s)'s prediction of edge, and actual edges, occur negates
any postulate that the rise in the value of true counts in middle
ranges implies any sort of rise in overall basic strategy edge for
all true counts, as penetration increases.
I understand that you are interested in this "deep math" subject but don't want a "deep math" explanation, so I would not really bother myself with it. How many hours do you play a year x bankroll x count system?
To me, floating advantage has caused more trouble than Sammy Bean Ladden has. The TC count represents a simplified version of your advantage either + or - and tells us when to play hands differently and change bet size within the realm of the table heat we could expect. That is what it is supposed to do. If you want to knock yourself out, read the following. Have you seen the movie A Beautiful Mind?? ;>
===============================================================
From: clarke_cant
Date: Thu Sep 26, 2002 10:46 pm
Subject: Before I forger just to archive
Since Puiu is driving the PBers nuts on bj21.com I figure I better
archive here--in case my laptop crashes too--a copy of my final
floating advantage paper. I still await the review Steepen is setting
up.
Here is the slighly revised version of my paper to use, of course not
all typos have been lassoed either:
Wongy on bjmath.com wanted to see a version of this without any
partition theory being included. This would be just another boring
idea for me except that elimination of partition theory mention leads
to a proof, in the proof of Postulate IIb, that penetration
absolutely cannot change any basic strategy expectation.
Theorem I: That any pack of remaining or original cards can be cut
such that any penetration can be transformed into any other.
Theorem II: That the average mean path of a random walk is a straight
line from origin to destination. Inclusive of this would be all
Euclidian postulates for a straight line.
Postulates:
I: The otherwise proven Jalib's True Count Theorem in a simpler proof
here, that any observed true count tends to remain at the same value
until all cards are drawn.
II: That actual mean basic strategy edge does not change with
penetration within a given pack or subpack.
III: That the difference between actual and true count predicted edge
(with basic strategy) increases with penetration.
IV: That more extreme True Counts progressively overestimate edge,
and less extreme True Counts progressively underestimate edge with
penetration.
Axioms:
I: True count =running count/cards left unseen in the pack.
II:Any pack composition subset can be precisely defined by a
combination of balanced rank versus rank, or ranks versus ranks,
balanced counts
III: All balanced counts, and some unbalanced counts, end at a
running count of zero.
IV: The difference between any actual edge and single, or multiple,
true count prediction of edge can be positive or negative.
V: Mean changes in expected value for a remaining pack subset are
proportional to the true counts used to measure them.
Postulate I: By Axiom III for any running count, RC(j) at j cards
left, RC(0)=0, for zero cards left.
Applying Theorem II for all j and k not equal to zero, with k less
than j, but j and k greater than zero, RC(k)=(k/j)*RC(j).
Applying Axiom I True count TRC(j)=n=RC(j)/j, TRC(k)=m=RC(k)/k.
(j/k)=(n/m) j*m/k=n j*m=k*n n/j= m/k
QED
Once you observe a true count the average true count does not change
no matter how many other cards are seen, for the boundaries set for j
and k above, in that the true count is the slope of the line of the
mean tendency of the running count to approach zero with no cards
left.
Postualate IIa: Proven by Theorem I with challenges answered in Cases
I thru IV. The basic strategy edge for any given round is independent
of the penetration in that any round starting point can be
transformed by cutting the pack into any other. The basic strategy
edge is the mean of all rounds played, and their original or new
position does not change the edge for any round.
Postulate III:
Step 1:A complete pack has the ijmplied balanced counts: +4 for every
non ten ranked card, and -9 for every ten ranked card, and serperate
balanced counts -12 for every rank, and +1 for all other ranks. by
the True Count Therem, or TCT for short, the mean composition for
this complete set of counts, after drawing 13 cards, is for the true
count of them all to remain the same as at the start, or zero. This
composition is the most predicted: removing one each of every rank
aces thru 9s, but removing 4 ten ranked cards. (See PIIb and Case I
however down below).
Step 2: At 13 cards drawn this most predicted mean pack is, by
inspection, only a tiny fraction of the total number of compostions
possible from drawing any 13 cards. At each level of penetration via
Axiom II, a new series of counts is needed is otherwise needed (the
10 seperate counts in Step 1 is different from the usual situations--
see below) to track the difference between the main count(s)'s
predictions and the actual composition. The solution in Step 1 is
trivial for comparing the difference between most predicted
compositions, and actual compostions, in that this set of counts
directly gives you the actual composition, but the result that each
most predicted pack subset is such a small fraction remains at least
down to the 3 cards left level--well past the smallest size needed to
complete a round of cards.
Step 3: At 3 cards left there will a constalation of balanced counts
added to the main count (see Axiom II) such that there will be some
card drawn, of the 3 left, that keeps the most number of such true
counts the same. But the 3 cards have the same probability of being
drawn. Thus the number of possible compositions that results in other
than the most predicted pack always excedes the number of
compositions that are most predicted.
In all such cases compositions are comparable to two dice totals in
craps, where the most likely total is 7, but the most likely set of
totals is some number other than 7.
Step 4: By Step 3 with just one rank versus another rank, used as a
balancd count, this property can be extended to drawing just one
card, to the 3 cards left point, which is far smaller that the number
of cards left to have a complete round.
Step 5: An algebraic simplificaton occurs when all such counts that
are added via Axiom II, to the main count(s), have a zero count, in
that then the composition is exactly that predicted by the main count
(s).
Step 6: The probability that any round will result in the need for a
new side count to precisely define the remaining subset pack will not
change for any round, via Theorem I. But with deeper penetration
every new count needed via Axiom II will have higher initial true
counts via Axiom I, and involve more difference between the actual
remaining pack, via Axiom V (with Axiom IV also applying), and the
subset pack predicted by all previously applied counts.
QED
Penetration increases the difference between the edge predicted by
the main count(s) and the exact pack composition. This difference,
and need to add new counts, occurs with a frequency that is
proportional to the number of cards sampled from the start of the
orginal pack of cards too, as explained in step 6. Overall the
presense of extreme counts, added to the main counts to define the
actual pack, as opposed to the most probable pack indicated by the
main count(s), provides a mechanism by which the expected value
associated with the most probable pack, for a given penetration
point, is reduced by such exteme true counts also being included, to
lower the actual mean expected value to that of the original starting
pack. The added counts are tending to have higher initial true
counts, and extreme true counts have negative bow effects.
Postulate III is proven without needing to modify Postulate II. See
Cases I thru IV below to refute arguments otherwise.
And the mechanism discussed above by which differences between the
main count(s)'s prediction of edge, and actual edges, occur negates
any postulate that the rise in the value of true counts in middle
ranges implies any sort of rise in overall basic strategy edge for
all true counts, as penetration increases.
Re: Clarke Cant on Floating Advantage cont.
Postulate IIb: A cards left penetration count ends in zero and is
subject to the TCT as well. Anytime you specify the start of a round
you specify a penetration point for a shuffled pack of cards.
Penetration cannot have any effect on a strategy that is optimal for
a full pack of cards in that penetration not only is subject to the
TCT. and on average does not change, but the true count of the
penetration, cards left, number absolutely never changes. Axiom V
applies meaning that the effect of setting a boundary for the start
of a round of cards does not change anywhere with penetration.
Penetration by itself cannot change basic strategy edge or basic
strategy recomendations.
Postulate IV: A balanced rank count of all tens, versus all other
ranks is well accepted to result in an underestimate of edge at its
extremes, in that such a count has all pushes at its all tens
extreme, which is all 20 to 20 pushes with the dealer.
A balanced rank count of all aces versus all other ranks will result,
at its all aces extreme, in a player loss in that most rules do not
allow resplitting or rehitting split aces. Thus the ace count
underperforms at its extremes.
A balanced count of a single low card and all others will
underperform predicted edge when it is at its all low card extreme in
that the player will spit and stand on lower totals than the dealer,
with the exception of all 7s, where the player and dealer will push
with 21 totals, the player having split to 4 hands.
A neutral count, especially with more practical counts, implies a
pack that resembles a pack that started with less decks, as
penetration increases.
In all of the above, such bow effects increase with penetration.
As per Axiom II, every practical count that is intended for actual
use has to be a combination of the above, and exhibit the sum of such
characteristics.
Thus the general characteristics of the bow effect have been proven
without any Postulate (I,III,IV) being shown to necessarily modify
Postulate II.
The following Cases refute challenges to the above statement:
CaseI: The most expected composition at any penetration level (though
any penetration level also means predictions involve fractions of
cards compositions and not just integer numbers of cards) matches the
most expected composition, at that level, for a true count =0.
Doesn't this imply that since the edge for a True Count of
0, "floats" upward with penetration, that the basic strategy edge
does as well?
Answer: A TC=0 is a statement that excludes other TCs, where some
have postive bow effects and the extreme TCs have negative bow
effects. Case I does not modify Theorem II in that the sum of all of
the excluded bow effects at those other TCs is allowed to balance the
postive bow effects at TC=0. Imbalance must be demonstrated to
require this modification.
Case II: if basic strategy edge does not change over all TCs, doesn't
the behavior of all likely TCs, as penetration increases, still
require some rise in basic strategy edge?
Answer: Bow effects, as proven in Postulate III, included any
balanced count, and can be postive or negative as per Axiom IV. The
flaw in the Case II objection is that the effects of extreme counts
is being excluded on their probability alone and not the product of
their probability and effect when observed, and it is ignored how
such effects can involve a main count near middle ranges, and
necisarily included side counts, as per Axiom IV, that may be well
into such negative bow effects ranges. Case II is a pure attempt to
evade Axiom IV, by attempting to exclude more extreme count ranges.
Case III: Simulations still appear to show increases in basic
strategy edge with increased penetration. What gives?
Answer: Simulation results are typically reported with several layers
of rounding in how observed true counts are grouped with similar
groups of observed true counts at different penetrations. Such
reports, or reporting modules of code frozen within the popular
simulators, ignore how the true count is a discrete number, and how
the average true counts that are possible, within any rounding range,
creap upward as penetration increases.
Cut card effects are present also that are near universally
overcorrected, and because cut card effects increase with
penetration, such adjustments result in the appearence of more edge
with more penetration for basic strategy playing decisions also.
Case III: is a simple example of equating apples to oranges in two
areas.
Case IV: Sometimes a composition will be formed at random that is the
same as starting out with less decks of cards initially. Gotcha?
Answer: This is a highly improbable event, but it is a limited
exception. There is otherwise a collection of true counts, required
by Axiom II, that have conserved true counts, where there are
pertibations in composition and edge that are only possible from
those subsets originating with a larger initial pack. But this
exception is also overvalued in a reverse of Case II fallacies.
In all of the above the idea of a strong floating advantage is a
stubborn one in that the fallacies decribed in Cases I thru IV often
are combined. They arrise simply because most people are too trusting
or too compartmentalized in their thinking to reason through them.
Notes:
Griffin omitted any sort of Axiom IV in his discussion, in Theory of
Blackjack, of a regression function operating with penetration for
the changes in compostion as a pack depletes.
ML, in his paper posted on bj21.com and bjmath.com, made intersection
of means errors, as shown in Case I and Case II, by omitting any sort
of Axiom IV, in step 4 of his "proof" claiming that the bow effect
required modification of what is given here as Postulate II.
Don Schlesinger, in Blackjack Attack, both editions, used a squeezed
balloon analogy, for his version of step 4(ML)/Postulate III(myself)
that is based upon the Graham-Stokes result for the topomorphic
properties of the true count prediction of edge and the actual edge.
His version of error, in claiming this need to modify Theorem III, is
different than ML's. His derivation misses how what is set here as
Postulate II is a boundary condition for the Graham-Stokes result to
be applied and is not modified.
Those who accept simulation reports rather blindly usually make Case
III errors.
This started an an alternative proof of the True Count Theorem, that
was requested by Rob McCarvey, but the alternative proof of the TCT,
allowed a new simple proof of ML's step 4, that did not involve his
Case I and Case II errors, and proceded as a necessity from that. I
appologise for making a new post on this topic, after everyone has
been justifiable bored by it, in past posts by myself and ML, but a
friend's request and a need for completion compelled me. Woggy's
request to, "avoid the science fiction" that results from including
partition theory, led to adding Axiom V, to the proof that more edge
variance is involved with deeper penetration, but this also led to
the proof of Postulate IIb, that should forever bury the idea that
penetration modifies basic strategy in any way. I hope there are no
missunderstandings for doing this.
Postulate IIb: A cards left penetration count ends in zero and is
subject to the TCT as well. Anytime you specify the start of a round
you specify a penetration point for a shuffled pack of cards.
Penetration cannot have any effect on a strategy that is optimal for
a full pack of cards in that penetration not only is subject to the
TCT. and on average does not change, but the true count of the
penetration, cards left, number absolutely never changes. Axiom V
applies meaning that the effect of setting a boundary for the start
of a round of cards does not change anywhere with penetration.
Penetration by itself cannot change basic strategy edge or basic
strategy recomendations.
Postulate IV: A balanced rank count of all tens, versus all other
ranks is well accepted to result in an underestimate of edge at its
extremes, in that such a count has all pushes at its all tens
extreme, which is all 20 to 20 pushes with the dealer.
A balanced rank count of all aces versus all other ranks will result,
at its all aces extreme, in a player loss in that most rules do not
allow resplitting or rehitting split aces. Thus the ace count
underperforms at its extremes.
A balanced count of a single low card and all others will
underperform predicted edge when it is at its all low card extreme in
that the player will spit and stand on lower totals than the dealer,
with the exception of all 7s, where the player and dealer will push
with 21 totals, the player having split to 4 hands.
A neutral count, especially with more practical counts, implies a
pack that resembles a pack that started with less decks, as
penetration increases.
In all of the above, such bow effects increase with penetration.
As per Axiom II, every practical count that is intended for actual
use has to be a combination of the above, and exhibit the sum of such
characteristics.
Thus the general characteristics of the bow effect have been proven
without any Postulate (I,III,IV) being shown to necessarily modify
Postulate II.
The following Cases refute challenges to the above statement:
CaseI: The most expected composition at any penetration level (though
any penetration level also means predictions involve fractions of
cards compositions and not just integer numbers of cards) matches the
most expected composition, at that level, for a true count =0.
Doesn't this imply that since the edge for a True Count of
0, "floats" upward with penetration, that the basic strategy edge
does as well?
Answer: A TC=0 is a statement that excludes other TCs, where some
have postive bow effects and the extreme TCs have negative bow
effects. Case I does not modify Theorem II in that the sum of all of
the excluded bow effects at those other TCs is allowed to balance the
postive bow effects at TC=0. Imbalance must be demonstrated to
require this modification.
Case II: if basic strategy edge does not change over all TCs, doesn't
the behavior of all likely TCs, as penetration increases, still
require some rise in basic strategy edge?
Answer: Bow effects, as proven in Postulate III, included any
balanced count, and can be postive or negative as per Axiom IV. The
flaw in the Case II objection is that the effects of extreme counts
is being excluded on their probability alone and not the product of
their probability and effect when observed, and it is ignored how
such effects can involve a main count near middle ranges, and
necisarily included side counts, as per Axiom IV, that may be well
into such negative bow effects ranges. Case II is a pure attempt to
evade Axiom IV, by attempting to exclude more extreme count ranges.
Case III: Simulations still appear to show increases in basic
strategy edge with increased penetration. What gives?
Answer: Simulation results are typically reported with several layers
of rounding in how observed true counts are grouped with similar
groups of observed true counts at different penetrations. Such
reports, or reporting modules of code frozen within the popular
simulators, ignore how the true count is a discrete number, and how
the average true counts that are possible, within any rounding range,
creap upward as penetration increases.
Cut card effects are present also that are near universally
overcorrected, and because cut card effects increase with
penetration, such adjustments result in the appearence of more edge
with more penetration for basic strategy playing decisions also.
Case III: is a simple example of equating apples to oranges in two
areas.
Case IV: Sometimes a composition will be formed at random that is the
same as starting out with less decks of cards initially. Gotcha?
Answer: This is a highly improbable event, but it is a limited
exception. There is otherwise a collection of true counts, required
by Axiom II, that have conserved true counts, where there are
pertibations in composition and edge that are only possible from
those subsets originating with a larger initial pack. But this
exception is also overvalued in a reverse of Case II fallacies.
In all of the above the idea of a strong floating advantage is a
stubborn one in that the fallacies decribed in Cases I thru IV often
are combined. They arrise simply because most people are too trusting
or too compartmentalized in their thinking to reason through them.
Notes:
Griffin omitted any sort of Axiom IV in his discussion, in Theory of
Blackjack, of a regression function operating with penetration for
the changes in compostion as a pack depletes.
ML, in his paper posted on bj21.com and bjmath.com, made intersection
of means errors, as shown in Case I and Case II, by omitting any sort
of Axiom IV, in step 4 of his "proof" claiming that the bow effect
required modification of what is given here as Postulate II.
Don Schlesinger, in Blackjack Attack, both editions, used a squeezed
balloon analogy, for his version of step 4(ML)/Postulate III(myself)
that is based upon the Graham-Stokes result for the topomorphic
properties of the true count prediction of edge and the actual edge.
His version of error, in claiming this need to modify Theorem III, is
different than ML's. His derivation misses how what is set here as
Postulate II is a boundary condition for the Graham-Stokes result to
be applied and is not modified.
Those who accept simulation reports rather blindly usually make Case
III errors.
This started an an alternative proof of the True Count Theorem, that
was requested by Rob McCarvey, but the alternative proof of the TCT,
allowed a new simple proof of ML's step 4, that did not involve his
Case I and Case II errors, and proceded as a necessity from that. I
appologise for making a new post on this topic, after everyone has
been justifiable bored by it, in past posts by myself and ML, but a
friend's request and a need for completion compelled me. Woggy's
request to, "avoid the science fiction" that results from including
partition theory, led to adding Axiom V, to the proof that more edge
variance is involved with deeper penetration, but this also led to
the proof of Postulate IIb, that should forever bury the idea that
penetration modifies basic strategy in any way. I hope there are no
missunderstandings for doing this.
the post you copied here from Clarke...
... was specifically NOT about floating-advantage per'se, it was in response to YOUR confusion about whether/not the TC tends to remain constant or return to zero.
As long as you are in the mode of mis-applying Cant, repost some of his several admonishments against your proposed and current Totaliterian Protocol CCCafe moderation. zg
... was specifically NOT about floating-advantage per'se, it was in response to YOUR confusion about whether/not the TC tends to remain constant or return to zero.
As long as you are in the mode of mis-applying Cant, repost some of his several admonishments against your proposed and current Totaliterian Protocol CCCafe moderation. zg
Re: Clarke Cant on Floating Advantage
Hello Rob,Like ltc i do not really understand the ins and out of the FA.to me it means that if you start counting a 6 deck(Ev -.5%) and say after 1 deck is dealt and the tC is +2 your advantage is even with the house is .5% or thereabouts).But then as the deck is dealt down and say the pen is quite good and you get to 5 decks in the discard tray,1 deck left and the TC is +2 then you should have a bigger bet out than before ,because now you can consider playing against 1 deck with the same rules,is this it?
Hello Rob,Like ltc i do not really understand the ins and out of the FA.to me it means that if you start counting a 6 deck(Ev -.5%) and say after 1 deck is dealt and the tC is +2 your advantage is even with the house is .5% or thereabouts).But then as the deck is dealt down and say the pen is quite good and you get to 5 decks in the discard tray,1 deck left and the TC is +2 then you should have a bigger bet out than before ,because now you can consider playing against 1 deck with the same rules,is this it?
yes, essentially...
...for a given TC, rules being equal, as the #decks deplete from say 8Ds to 2Ds to 0.5Ds the house or player edge tends to recapitulate - thus with a TC=0 count of the top of 6D h17 DAS the housEdge is relatively high (-.69% approx?) that same shoe dealt down to the 0.33D level with a TC=0 now yields a player advantage of (+0.50 approx?).
This is something that was emphasized by LRevere and later forgotten untill DSchlesinger revived it in debates with ASnyder and PGriffin in BJForum circa '85 (see BJAttack). Notwithstanding Clarke Cant rebukes the FA and the net effect of a theoretical FA is negligable in any event. zg
...for a given TC, rules being equal, as the #decks deplete from say 8Ds to 2Ds to 0.5Ds the house or player edge tends to recapitulate - thus with a TC=0 count of the top of 6D h17 DAS the housEdge is relatively high (-.69% approx?) that same shoe dealt down to the 0.33D level with a TC=0 now yields a player advantage of (+0.50 approx?).
This is something that was emphasized by LRevere and later forgotten untill DSchlesinger revived it in debates with ASnyder and PGriffin in BJForum circa '85 (see BJAttack). Notwithstanding Clarke Cant rebukes the FA and the net effect of a theoretical FA is negligable in any event. zg
Re: yes, essentially...
This is something that was emphasized by LRevere and later forgotten untill DSchlesinger revived it in debates with ASnyder and PGriffin in BJForum circa '85 (see BJAttack). Notwithstanding Clarke Cant rebukes the FA and the net effect of a theoretical FA is negligable in any event. zg
how do you mean the net effect is negligable, if you get into a deep pen should you not use it?
Password:
This is something that was emphasized by LRevere and later forgotten untill DSchlesinger revived it in debates with ASnyder and PGriffin in BJForum circa '85 (see BJAttack). Notwithstanding Clarke Cant rebukes the FA and the net effect of a theoretical FA is negligable in any event. zg
how do you mean the net effect is negligable, if you get into a deep pen should you not use it?
Password:
Re: yes, essentially...
>>how do you mean the net effect is negligable, if you get into a deep pen should you not use it? <<
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Well, I will say that I DO use it, but the longterm use-benifit is minimal... and there is still some scientific debate over whether it even exists - if it does, it ain't worth much. zg
>>how do you mean the net effect is negligable, if you get into a deep pen should you not use it? <<
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Well, I will say that I DO use it, but the longterm use-benifit is minimal... and there is still some scientific debate over whether it even exists - if it does, it ain't worth much. zg
Re: Clarke Cant on Floating Advantage
"Hello Rob,
Like LTC i do not really understand the ins and out of the FA.to me it means that if you start counting a 6 deck(Ev -.5%) and say after 1 deck is dealt and the tC is +2 your advantage is even with the house is .5% or thereabouts).But then as the deck is dealt down and say the pen is quite good and you get to 5 decks in the discard tray,1 deck left and the TC is +2 then you should have a bigger bet out than before ,because now you can consider playing against 1 deck with the same rules,is this it?"
I think this over simplifies it, but you are getting some of the idea, which is great. I think the Mayor can give a better explanation since he has a degree in higher math?? I really don't think that is what you want.
Example:
The QBack fires the ball down the field to be caught. He gets the trajectory right, put the stabilizing spin on the ball, and lets it go. A rifle bullet works much the same way. There is a ton of math involved, but the ball goes to its target. A shorter distance (2 decks left) can make for a better shot. The more math you understand does not necessarily make you a better shot. If you have a correct TC, have your indices down, have your bankroll loaded, you are doing fine.
"Hello Rob,
Like LTC i do not really understand the ins and out of the FA.to me it means that if you start counting a 6 deck(Ev -.5%) and say after 1 deck is dealt and the tC is +2 your advantage is even with the house is .5% or thereabouts).But then as the deck is dealt down and say the pen is quite good and you get to 5 decks in the discard tray,1 deck left and the TC is +2 then you should have a bigger bet out than before ,because now you can consider playing against 1 deck with the same rules,is this it?"
I think this over simplifies it, but you are getting some of the idea, which is great. I think the Mayor can give a better explanation since he has a degree in higher math?? I really don't think that is what you want.
Example:
The QBack fires the ball down the field to be caught. He gets the trajectory right, put the stabilizing spin on the ball, and lets it go. A rifle bullet works much the same way. There is a ton of math involved, but the ball goes to its target. A shorter distance (2 decks left) can make for a better shot. The more math you understand does not necessarily make you a better shot. If you have a correct TC, have your indices down, have your bankroll loaded, you are doing fine.
Re: Clarke Cant on Floating Advantage
... was specifically NOT about floating-advantage per'se, it was in response to YOUR confusion about whether/not the TC tends to remain constant or return to zero.
I think you are confused.
"Since Puiu is driving the PBers nuts on bj21.com I figure I better
archive here--in case my laptop crashes too--a copy of my final
***floating advantage paper***
Why do we bet more into high counts? Because we expect them to try to return to zero. In other words, start spitting 10's and Aces out. This is why tracking is so important - we can track spots that are neutral, ie, stick to 0 23%, and track dead spots, ie neg EV, and track plus EV spots, and what they are shuffled with.
No one has to worry about having ontopic posts go down. Clarke was making sure that the "mod squad" knew his higher math was part of advantage play, and not some psycho-bable astro physics drug induced delerium.
Try to be more constructive over here instead of wasting your time dogin me. It's old, it's tired, it's history. Go for higher ground.
Balanced count start at 0
6 deck
TC 0 23.2877%
TC 1 13.1418%
TC 2 9.6193%
TC 3 7.0801 %
TC 4 5.4209 %
TC 5 4.0655 %
TC 6 3.1428 %
TC 7 2.3829 %
TC 8 1.8886 %
TC 9 1.4364 %
TC 10 1.1376%
Would end at 0 if game brought to completion
... was specifically NOT about floating-advantage per'se, it was in response to YOUR confusion about whether/not the TC tends to remain constant or return to zero.
I think you are confused.
"Since Puiu is driving the PBers nuts on bj21.com I figure I better
archive here--in case my laptop crashes too--a copy of my final
***floating advantage paper***
Why do we bet more into high counts? Because we expect them to try to return to zero. In other words, start spitting 10's and Aces out. This is why tracking is so important - we can track spots that are neutral, ie, stick to 0 23%, and track dead spots, ie neg EV, and track plus EV spots, and what they are shuffled with.
No one has to worry about having ontopic posts go down. Clarke was making sure that the "mod squad" knew his higher math was part of advantage play, and not some psycho-bable astro physics drug induced delerium.
Try to be more constructive over here instead of wasting your time dogin me. It's old, it's tired, it's history. Go for higher ground.
Balanced count start at 0
6 deck
TC 0 23.2877%
TC 1 13.1418%
TC 2 9.6193%
TC 3 7.0801 %
TC 4 5.4209 %
TC 5 4.0655 %
TC 6 3.1428 %
TC 7 2.3829 %
TC 8 1.8886 %
TC 9 1.4364 %
TC 10 1.1376%
Would end at 0 if game brought to completion
Re: Clarke Cant on Floating Advantage
Hey i dont know about all the math.but if i am playing a 4d and then get down to 1d left,dont you think as if you are playing against 1d regardless.I thought this before i even heard of this theory,it feels right,it seems right.
My thinking is that if i start off playing a 4d at 0TC and then 3 D are depleted and then you are still at 0TC,then on average every rank should be depleted equally leaving a 1 Deck with same rules left,but with a better EV for the player.
Is this such a great concept seems intuitive to me?
Or is it counter intuitive,such as NRS formulea(havent a clue about this formulea but know it is not what you would expect!)
Hey i dont know about all the math.but if i am playing a 4d and then get down to 1d left,dont you think as if you are playing against 1d regardless.I thought this before i even heard of this theory,it feels right,it seems right.
My thinking is that if i start off playing a 4d at 0TC and then 3 D are depleted and then you are still at 0TC,then on average every rank should be depleted equally leaving a 1 Deck with same rules left,but with a better EV for the player.
Is this such a great concept seems intuitive to me?
Or is it counter intuitive,such as NRS formulea(havent a clue about this formulea but know it is not what you would expect!)
Re: yes, essentially...
6 Deck off the top edge .54, same rules deal 5 decks now 1 deck remaining true count is 0...house edge is now .19 (the exact edge of a 1 deck game with count of zero) That is the floating advantage.
For practical considerations it means that in neutral counts, deeper in the pack, you can use a bit more betting cover with less cost or bet slightly bigger slightly faster in moderate counts.
6 Deck off the top edge .54, same rules deal 5 decks now 1 deck remaining true count is 0...house edge is now .19 (the exact edge of a 1 deck game with count of zero) That is the floating advantage.
For practical considerations it means that in neutral counts, deeper in the pack, you can use a bit more betting cover with less cost or bet slightly bigger slightly faster in moderate counts.
True-count theorem
The true-count theorem tells us that we expect the true count to stay exactly where it is, not return towards zero. Rather, we expect the _running_ count to return towards zero. It is a common misperception that the true count tends towards zero (the first time I considered the matter I made the same error), so I think it is important to distinguish carefully between tendencies of the running count and tendencies of true counts.
Example: 2 decks remaining, running count = 4, true count = 2. What is our 'best guess' of the running count once only 1 deck remains? Since there are 2 more big than small cards remaining per deck, we expect the running count to fall by 2 over the next deck. Thus, we expect a running count of 2 once only 1 deck remains. So what do we expect the true count to be? True count equals running count divided by decks remaining, which we expect to be (4 - 2)/1 = 2. To sum up, the running count is expected to return towards zero (down from 4 to 2), whereas the true count is expected to remain exactly where it is (at 2).
The basic intuition behind this result is simple. While we expect the running count to return to zero (ie, become smaller in absolute terms), we also know that the number of decks remaining will also become smaller. Consider that:
t = r/d
where t = true count, r = running count and d = decks remaining. As the shoe is depleted, we expect |r| to fall, which acts to reduce |t|, but at the same time we expect d to fall, which acts to increase |t|. The two effects offset each other exactly provided more than one card remains to be dealt. (The true count is undefined once no cards remain; ie, r/d = 0/0, which is undefined.)
The true-count theorem tells us that we expect the true count to stay exactly where it is, not return towards zero. Rather, we expect the _running_ count to return towards zero. It is a common misperception that the true count tends towards zero (the first time I considered the matter I made the same error), so I think it is important to distinguish carefully between tendencies of the running count and tendencies of true counts.
Example: 2 decks remaining, running count = 4, true count = 2. What is our 'best guess' of the running count once only 1 deck remains? Since there are 2 more big than small cards remaining per deck, we expect the running count to fall by 2 over the next deck. Thus, we expect a running count of 2 once only 1 deck remains. So what do we expect the true count to be? True count equals running count divided by decks remaining, which we expect to be (4 - 2)/1 = 2. To sum up, the running count is expected to return towards zero (down from 4 to 2), whereas the true count is expected to remain exactly where it is (at 2).
The basic intuition behind this result is simple. While we expect the running count to return to zero (ie, become smaller in absolute terms), we also know that the number of decks remaining will also become smaller. Consider that:
t = r/d
where t = true count, r = running count and d = decks remaining. As the shoe is depleted, we expect |r| to fall, which acts to reduce |t|, but at the same time we expect d to fall, which acts to increase |t|. The two effects offset each other exactly provided more than one card remains to be dealt. (The true count is undefined once no cards remain; ie, r/d = 0/0, which is undefined.)
I think you have a Misconception, Robert
You said: Why do we bet more into high counts? Because we expect them to try to return to zero.
This is false, we expect the runnin gcount to return to 0, but the TC to stay the same. Think of a line that begins at a high point and returns to the x axis. Its slope (the TC) is the same, but it still is returning to 0 (RC).
I have seen you make this same mistake elsewhere, you might give this one some serious thought.
--Mayor
You said: Why do we bet more into high counts? Because we expect them to try to return to zero.
This is false, we expect the runnin gcount to return to 0, but the TC to stay the same. Think of a line that begins at a high point and returns to the x axis. Its slope (the TC) is the same, but it still is returning to 0 (RC).
I have seen you make this same mistake elsewhere, you might give this one some serious thought.
--Mayor
Re: Clarke Cant on Floating Advantage
Thanks for saying that I can explain this, but I can't very well. The truth is, on this one I leave it to DS and MathProf to do the explaining. This is one of those parts of playing that is so microscopic and academic, that I have never bothered. I know that it has brought DS and Clarke to near blows.
--Mayor
Thanks for saying that I can explain this, but I can't very well. The truth is, on this one I leave it to DS and MathProf to do the explaining. This is one of those parts of playing that is so microscopic and academic, that I have never bothered. I know that it has brought DS and Clarke to near blows.
--Mayor
Re: I think you have a Misconception, Robert
"You said: Why do we bet more into high counts? Because we expect them to try to return to zero.
"This is false, we expect the running count to return to 0, but the TC to stay the same. Think of a line that begins at a high point and returns to the x axis. Its slope (the TC) is the same, but it still is returning to 0 (RC).
I understand this. Just as we expect the count to stay close to 0 23% of the time, we also know that it may stay constant (23%) for periods of time at +1TC or +2TC or -3TC etc. These are neutral areas of the decks. This is why tracking is soooo important to multi-deck players, and the reason why they are facing more and more CSM's.
An example would look like this where each | is an RC
6 left 0|||||1|||||2|||||3
5 left 0||||1||||2||||3
4 left 0|||1|||2|||3
3 left 0||1||2||3
2 left 0|1|2|3
1 left 0123
Of course it will take longer for the TC to change when there are many decks left, and the RC to match the TC when we are at one deck left. It would be like a speedometer that only told you that you were doing 40 or 50 or 60.
"I have seen you make this same mistake elsewhere, you might give this one some serious thought.
I understand these concepts, and that when we go from 60 to 59 mph we haven't hit 50 yet, but since there is a stop sign up ahead we will be back to 0, or be starting from 0 again, unless you want to blow the next light......that's what trackers are doing by avoiding neutral and neg EV areas, if not pounding plus EV areas.
And thank you for being so kind and patient with me. I respond much better to your positive atmosphere. I probably didn't stop long enough on some of these steping stones as I traversed the AP River, so please forgive me for being in such a hurry to get to the other side. Patient I am not....with myself that is.
"You said: Why do we bet more into high counts? Because we expect them to try to return to zero.
"This is false, we expect the running count to return to 0, but the TC to stay the same. Think of a line that begins at a high point and returns to the x axis. Its slope (the TC) is the same, but it still is returning to 0 (RC).
I understand this. Just as we expect the count to stay close to 0 23% of the time, we also know that it may stay constant (23%) for periods of time at +1TC or +2TC or -3TC etc. These are neutral areas of the decks. This is why tracking is soooo important to multi-deck players, and the reason why they are facing more and more CSM's.
An example would look like this where each | is an RC
6 left 0|||||1|||||2|||||3
5 left 0||||1||||2||||3
4 left 0|||1|||2|||3
3 left 0||1||2||3
2 left 0|1|2|3
1 left 0123
Of course it will take longer for the TC to change when there are many decks left, and the RC to match the TC when we are at one deck left. It would be like a speedometer that only told you that you were doing 40 or 50 or 60.
"I have seen you make this same mistake elsewhere, you might give this one some serious thought.
I understand these concepts, and that when we go from 60 to 59 mph we haven't hit 50 yet, but since there is a stop sign up ahead we will be back to 0, or be starting from 0 again, unless you want to blow the next light......that's what trackers are doing by avoiding neutral and neg EV areas, if not pounding plus EV areas.
And thank you for being so kind and patient with me. I respond much better to your positive atmosphere. I probably didn't stop long enough on some of these steping stones as I traversed the AP River, so please forgive me for being in such a hurry to get to the other side. Patient I am not....with myself that is.