ReRefuting ML

This link was originally provided by T-Hopper to a discussion by Don Schlesinger and Brett Harris about the floating advantage, that was first discussed here and then became part of a thread on bjmath.com: the link from T-Hopper. The followup sections on bjmath.com have been retitled, but involve a very similar post to ML's.

Both the Harris/Schlesinger and ML posts made the very same mistake in reasoning, but let's dispell one point raised by ML first: there is no one way or convention on styles for proofs. ML's demand I follow his style is a demand that I follow his error in fact.

Both posts took the true count prediction of edge as fully representative of the edge in the subsets of an initial pack. This is simply not the case as differing populations are involved.

The true count predicts the edge for all possible packs that can have a given true count. For every true count, at any given specific number of cards left in a pack, there is a most probable pack composition. The mean edge of that composition is the mean of the edges of every possible such pack composition.

This relates to Theorem II as given by ML, which can be better stated as: the mean of a given pack is also the mean of the means of every subset of that pack.

But the compositions of a pack have a different distribution and composition, where Theorem II applies not just for the true count, and all possible packs with that TC, but to the subsets that each subsequent subset is drawn from. You are dealing with, instead of the set of all packs that can have a given true count, with a given number of cards left, the set of all subsets of prior subsets that can have a given true count with a given number of cards left.

That is an entirely different population with an entirely different set of applicable means. The true count is a less accurate measure of the edge than the actual composition of such subsets. Thus these are two different measures for predicting basic strategy edge that are topomorphic to each other as per standard Graham-Stokes equation tools for evaluating these different measures.

The bow effect results from the binomial theorem predictable decrease in the probability of subsets having means relating to the entire spectrum of possible true counts sets, relative to the set involving the available spectrum from the actual card subsets.

The best resolution of this is to look again at Theorems I and II, from ML. Theorem I is that the mean basic strategy edge is invariant with penetration in that any penetration, for beginning a round, can be transformed into any other penetration by simply cutting the starting pack to bring that point into play.

That is in conflict with Theorem II, as presented by ML with the assumption that the population of all possible true count subsets has the same means as the population of true count subsets that is available from the subsets of the original pack (from now on subsets_(TCG), and subsets_(TCS) respectfully).

The mean edges of subsets_(TCG) increase in that the mean of each TC, at each penetration point, for subsets_(TCG), does show an increase in edge, in that the mean composition is smaller in each case. But the composition of subsets_(TCS) shows a change from compositions from that mean composition of subsets_(TCG). And their edges thus differ as well.

Were ML's viewpoint to be true, Theorem I could not be true. Theorem I is true however if the mean edge of subsets_(TCS) is equal to the mean edge of the the largest subset_(TCG) of the original pack, where the original pack is the largest such subset for neutral pack subsets (ie for basic strategy the largest neutral pack subset is the starting pack). That requires some mechanism that is within the observed characteristics of the bow effect and results in the mean of subsets_TCS being less than the mean of subsets_TCG).

The alternative is the contradiction that rounds begun at various penetration points have equal means, if you examine the cards used to complete each round, and unequal means if you consider that each round is drawn from smaller and smaller subsets of the pack overall.

That mechanism is found in the general demonstration that the mean of subsets_(TCS) is less than the mean of subsets_(TCG) in ML's "proof" post. It is also found by simulation of the effects of skewing away from mean compositions using hi-lo as a simple example count. Trading aces for tens, and tens for aces, results in lower overall mean edges subsets_(TCS) for various true counts, but higher edges in true count ranges near zero, for packs composed of such skews.

The resolution is to be found in my postulate of a Theorem III, which I gave below in another thread, which is:

Theorem III:
within an original pack of cards the probability of the appearence of subsets with mean compositons, and thus mean expectations, is reduced, and the probability of non-mean compostions enhanced, which have generally reduced expecations, such to reduce the expectations of such subsets to the same values as the similarly measured largest possible subsets of the original pack of cards.

The exception is the rare and statistically insignificant random occurences of subsets with mean compositions. This is of the same significance as the specific results of the binomial distribution vanishing at the limit which forms the normal distribution.

Since from ML it has become a matter of easy disposal of alternative ideas that don't match certain demands of certain posters for certain style formats I will leave a formal proof of this to others.

In summary:every form of claiming that the edge in blackjack, with a given true count, is the same as if the number of cards left were the number of cards started with, involves a conflict between obviously true Theorems I and II, except to postulate The Theorem III that I have presented

A further summary is my postulated Theorem III.

Still another point for summary is that with increased penetration, there is always a skewing away from the means of the original pack or larger pack subsets. This is true for both observed true counts and actual compositions. This is an obvious corollary of the binomial theorem.

This post refutes all of the Cant objections which I understand to the recent proof posted here concerning the floating advantage.

The best I can understand, these are the Cantian objections not remaining and remaining.

Not remaining:

1. There are no errors of math in the body of the proof. He says the math is misapplied but that is not an objection to the math. That has to be an objection to the theorems and postulate because the theorems and postulate were inserted in the proof in the identical form they were defined.

2. That the notation E(i,n) stands for one expectation of one arbitrary grouping of n cards which happens to also have the attribute of an i true count. I think we agree a specific E(i,n) can be skewed in any manner, disproportionate numbers of aces as to tens,
disproportionate numbers of five�s to twos, disproportionate numbers of neutral cards or whatever. I think we agree the expectation of each specific E(i,n) more than likely varies from another specific E(i,n) (which is why the proof like all similar proofs used the concept of an undefined variable as its basis of proof. There is no other concept that could possibly have been used in a proof of this sort). E(i,n) is not an average of any sort but a value.

3. That Theorem I is correct. At one time he agreed Theorem II was correct but backed off when he finally understood what E(i,n) meant.

Theorem II does incorporate means which is probably how Cant got confused. The notation f(sub i,m) * E(i,m) is an averaging process and E(i,m) the way things are defined is the mean of the expectation which of course is an expectation, a number with an undefined value. The theorem was written that way rather than incorporating another ddefinitional element into the mix and confusing matters. I could have used notation like f(sub p,i,m) * E(p,i,m) to show each possible permutation of dispersal and added terms but that is unnecessarily confusing because Theorem II is the fundamental equation used in all blackjack mathematical calculations. The average of the values of all possible subsets after all combinations are removed have the same value as the original set of cards. The form used to make this calculation is the summation of all f*ev for all the true counts observed where ev is a mean used for convenience. That is simply the way it is done in blackjack. When we say there is a floating advantage we are saying the mean expectation of a specific true count is larger than the starting expectation for that true count. It cannot be one number because there has to be dispersal. That is simply the way things are calculated in blackjack because of dispersal and the arbitrary way people before us made blackjack calculations. It is the same calculation. But one has to realize the invariant starting value for n cards is an undefined value as are the mean expectation for various true counts values. A number has no concept of how it got to be that value, it only knows it has a value. Mathematical proofs incorporating undefined values are mathematical proofs using numbers.

As to the statement Theorem II is invalid because it is inconsistent with Theorem I is absurd. In fact, the statements are identical and with more rigor the second could have been derived from the first but I skipped that step because it would have been simply proving the fundamental blackjack calculation technique which can stand by itself.

If value 1 equal value 2 a valid equation exists. Theorem II is a valid equation and has been accepted as the fundamental equation since blackjack calculations have been made. It supplied the starting point for the proof.

I think I see where Cant is confused and perhaps others are so I will discuss the equivalency and the confusion. He says that because the overall expectation is identical at any point in a stack of n cards which have not been messed with is invariant, the expectation of the summation of all the true counts must add up to the same at any depth. Okay, so far so good. From this point he says that if there is a floating advantage in the middle, the sides must make up for the middle, the original Schlesinger concept of a balloon which I have disagreed with before has to occur. He says that phenomenon occurs because bizarre mixtures away from the mean don�t obey the same rules and make things come out even.

What is being overlooked in saying such a thing is all that is dealt with in Theorem II is one *number*, E(i,n). That that number corresponds to an arbitrary grouping of cards of i true count is meaningless except for labeling purposes. The expectation is one value, one number describing edge, which is determined by the unique set of cards which comprise it. That the sum of the frequency * Ev of all the subsets add to the same number does not impose any limitation of the value of Ev at specific true counts of the subsets comprising the whole.

The last sentence of the preceding paragraph is an important concept so I will explain it as completely as I can. We start from an original value (number if you will), E(i,n). We all know true counts will diverge both ways from that original number, sometimes the true count at m will be greater, sometimes the same, and sometimes less. The expectation of each of the subsets will almost certainly be different from the expectation of the original subset, some higher and some lower. The expectation of subsets of higher true count will be higher, etc. The value representing the expectation of the mean of all possible subsets of size m which happen to count out as i+j, E(i+j,m) will be higher than the value of E(i,n) but that does not impose on it the additional restraint that it be identical to E(i+j,n). That restraint is not at all necessary for the value of the summation of f*Ev of all the possible true counts of subsets of cards of number m to equal the value of E(i,n). E(i+j,n) is irrelevant because it is in a different universe. Whatever the value of the expectation in a different universe might be does not affect the valuations in the universe being looked at. Cant is imposing the restriction that either (1) the expectations of the mean of all true counts of E(i,n) must be equal to the mean of all true counts of E(i,m) or (2) that at some values of true counts E((i+,-x),m) must be less than E((i+,-x),n) to make up the difference in expectation at central distributions when whatever happens to other distributions of not E(i,n) are not in the universe being examined. The sum equals the whole of its parts and could not care the least about something which is not one of its parts. Every mean expectation of each true count of subsets of size m may be larger than each mean expectation of each true count of all possibilities of sets of size n (it is my firm belief they are) and still sum when multiplied by frequency to the value of E(i.n) because, once all the not E(i,n) are discarded from the universe, they are no longer relevant.

Just one more point. Cant has not explicitly said it but he has rejected the mathematical expression used to describe the bow effect. He has rejected it as being only associated with a �perfect� deck of cards and that skewed permutations counterbalance it. He says it applies only to the mean of perfect cards, that is it only applies when a casino deals two decks and the remainder of fifty two cards is a perfect deck and since that is a highly unlikely occurrence the skewed permutations overcome the bow effect demonstrated only in cases of perfect removal. He asserts that with no evidence to back him up. In fact he asserts that despite contrary evidence. If I start with any particular set of cards of size n, again E(i,m) that particular set of cards, however skewed from �perfect� cards will have subsets of size n and the mean of all of the subsets will have the identical skew of the original cards. The cards do not know if aces are proportionally more frequent than tens or twos proportionately more frequent than five�s. All the cards know is the mean subset will have the same skew as the original set. If the skew is a perfect deck, the mean subset will be a perfect deck. There is no reason to believe or assert the bow effect which exists for the mean subsets of perfect cards would not exist for the mean subset of any skewed distribution. The bow effect is caused by the things Griffin said it is caused by, more pushes at high counts and less success at low counts. Just because E(i,m) might be skewed, E(i+10,m) will definitely have relatively more pushes at high counts than E(i,n) and E(i-10) will have relative less success at winning doubles and things like that than E(i,n). There is no reason at all to assert differently. How could it exist for �perfect� starting point and not exist for a skewed starting point? It is based on what happens with relative abundance�s of large cards or relative abundance�s of small cards and should not be affected by starting points except to the extent the original expectation was affected.

In any event, the proposition was made as a postulate and, as expressed and as used in the proof, skewed distributions at the beginning point, values of E(i.n) of a specific skewed stack of cards are taken into account in the proof because E(i,n) is not an average but instead is an undefined value. The language (E(i+1,m) - E(i.m)) < (E(i,m) - E(i-1.m)) explicitly incorporated that concept and if that was Cant�s objection, he should have said that and saved thousands of words. That it was not said indicates he had no idea what was going on just as he mistook an undifined value for a mean.

The only thing used as a starting point is the fundamental of blackjack calculations, Theorem II, expressed with E(i,n) as an undefined value which is certainly the only thing proper for a proof of this sort. Each mathematical step of the proof follows from the previous step. Step 4 is mathematically correct if the postulate is correct. The proof is mathematically correct and Cant has not really disputed that. The word �misapplication� was used but that did not attack the validity of the math. Therefore the proof is correct, the only gray area left being the accuracy of the postulate. If the accuracy of the postulate is being questioned, Cant certainly described his objection imprecisely by saying it was a misapplication. At its worst, a correct application of an invalid postulate occurred in the proof. My assertion is the postulate, as stated in mathematical terms when the starting point is any of the possible unique distributions of n cards (the undefined value of the expectation of that unique distribution) is accurate for the reasons I have given in this post.

Cant can quarrel with the postulate to his heart�s content. Did I not say it was a postulate and did I not make that clear when my final statement was *If the bow effect......�?

"but he has rejected the mathematical expression used to describe the bow effect. He has rejected it as being only associated with a �perfect� "

Not true at all. I said that the bow effect occurs due to the growing difference in measures of edge between the actual composition and the true count. I said that the floating of the basic strategy edge only occurs with the perfectly mean decks, and is insignificant.

As to the statement Theorem II is invalid because it is inconsistent with Theorem I is absurd.

I never made this claim either. What I have consistantly said was that Theorem II should be applied to the compositions of each subset as they occur. This would then reflect how prior distributions affect the following subsets within each pack.

This is typical of ML and we can wait until the nearest black hole boils over due to Hawking radiation before ML stops creating in his own mind views that are not mine and exist only in his imagination. While ML keeps on battling his imaginary Clarke, and keeps getting bizzare endorsements, from the others who keep this going, the plain fact remains as I have stated:

ML has demonstrated the bow effect without needing to apply Graham-Stokes to the topomorphism of the true count and the actual composition, but he links this to a false hypothesis of claiming that the mean prediction of the true count is universaly applicable, when we know true counts are less accurate as penetration increases. He also equates the distribution assumed by the true count, which is from the population of all possible packs having that true count at a specific penetration, and equates this with the population of pack subsets that have that same true count at a specific penetration, which is another population, and where both have predictably different means. That is how he switches populations and means By switching in such non-representative means ML's then uses the rise in the edge for such mean compositions to claim that the basic expectation grows with penetration generally.

If you read what I actually posted above you know how ML's claim is contradictory and false!

But responding to ML when he keeps blathering on in response to his imaginary version of myself is pointless and is to missprison his lying about my statements.

Just who the hell do you think you are fooling by this missrepresentation ML?

Nothing at all has been said about the proof in this post nor has anything at all relevant to the proof has been said in any other post.

If you say something that deals with the proof and the later explanations, you can say it in a few words. This part is wrong because (less than twenty words).

Since you do not and have not, each element of the proof has remained uncriticised as written. Tangible criticisms are invited but they have not yet appeared.

Once again--you have proven in the above you don't really read my posts but make up what you think I said:

In step 4 Theorem II is not properly applied to the pack subsets.

That would be sufficient for an honest poster wishing to have an error pointed out.

The precise error is actually derived from the conventions you insist on using: you cannot distinguish that two different populations: (1)the entire range of possible subsets that have a given true count, and a given number of cards left, and (2) the subsets that are possible from a given pack, and the prior observed subsets of that given pack, for the same true count and nubmer of cards left.

By making that equate (DS claims equate cannot be used as a noun but he is not too computer literate either--a past controvery of similar sort to this one) by how E(i,t) is used in its most general way, in your proof, your proof then falsely claims that the perfect mean compositions for each true count are representative and that thus, by reason of such perfectly mean compositionsshowing a general increase in edge with penetration, the basic strategy edge increases with penetration. It does not! (use of html coding for emphasis to avoid further misquotes).

Instead it remains invarient and the bow effect occurs as a predictable curve of the true counts prediction of edge versus the composition's prediction of edge. The balloon analogy given by Schlesinger is totally correct, but to be precisely stated this has to be stated as I have. You without truly understanding this have given another demonstration that does not involve the Graham-Stokes equation and the property of topomorphy that is related to it. Your accidental, apparently, contribution is still to be saluted!