ML's single mean demand

Every true count and penetration combination has associated with them a mean composition, and a mean expectation for that composition. ML simply demands that one has to use that mean and that mean alone for discussions about the mean for a given true count.

There are no such requirements. The use of such mean compositions is simply a useful practice for reducing computations needed to find the mean predicted by a given true count.

It is not a requirement to use this mean and never has been. The actual bow effect is simply the result of the totally predictable difference between the mean expected value for the mean composition associated with a given true count and penetration, and the actual compositions and their expected values.

There in fact is no reversion to such means either. As I detailed in my Blame Woggy.... proof, each composition becomes the starting point for all further distributions.

What results is ML adding to this absurd definition requirement an equally absurd demand that this is the only definition for real mathematicians to use and that one is not allowed to post any response unless one uses this same definition.

Again: there are no such requirements! The ruler used is not the thing being measured. They have different means associated with them.

But once you go down this absurd trail with ML, you have to replace a simple difference in measures with the further absurdity that the basic strategy edge, rather than the edge associated with the actual compositions, floats instead. And that is a simple replay of an ultimate absurdity: that somehow measurement of a population changes the mean expected value of any remainder of that population. That is exactly what is involved when you claim that penetration alone changes expected values, which is what you are saying when you say that basic strategy edge changes with penetration!

Hi Clark, just one question for you.

You use the terminology: "mean composition." I don't see how that is defined. The "mean" is an arithmetic term, and a deck is more of an object. Please give a detailed definition of this for me. Please don't include anything else in your response, I am just interested in the answer to this one question. You know you have a tendency to ramble and be misunderstood, so I am requesting you be very focused in your response.

When I was going for my Ph.D., for one of the questions in my oral I had was to explain such and such concept (I think it was Lebesque integration) to a 2-nd grade student. Could you do the same here? Explain the "mean composition" as if I was in 2nd grade.

I am very curious about this concept. It has a lot of possible applications, as it implies that some sort of metric applies to the category of "decks". I really want to know what you are talking about.

Thanks

Eliot

Mean Composition is a useful device for analysis of the packs that can have a given true count, or for showing what is the average composition that can result from a prior composition.

The traditional explanation is that one composes (verb) a pack according to the exact long term probabilities for drawing each card. An example would be a hi-lo pack, given the running count and cards left. You would compose such a deck with the probabilities of drawing out of the initial pack 2 thru 6 being equal, and the probabilities of drawing a Ten, Jack, Queen, King, and Ace being equal. So you would draw out equal numbers of 2s thru 6s and Tens thru Aces. Should you do this for a given true count, you would do so for an aimed for difference between the number of low and high cards, and number of cards left, but would do so with the same equal probability included. For multi-level counts you do this by having the same probabilities for drawing like valued cards (that have the same sign), but where the value aimed at includes the different values given.

The alternative explanation I used in my proof was that the True Count Theorem tendency could directly be used to show how the mean composition for drawing 13 cards, would be for 4 tens, and one each ace thru 9, to be drawn. For various true counts this would involve the composition, for a given true count, that does not require adding new side counts to describe the actual composition.

The results are the same, but my alternative actually is more powerful.

What ML takes as a requirement is simply a computationaly useful assumption, that the mean composition for a given true count has a mean expectation that is equal to the mean expectation of all such possible true count packs. He needs to read Chapter 7, Theory of Blackjack, especially the subheading: An Exercise in Futility. While it has proven useful to use mean compositions in these ways, it is by no means proven that they are representative. They often are not.

The rest is in my proof!

Clarke, I'm a pretty smart guy, but I still don't get it. You know I was a professor of mathematics, so I should be able to understand what you are talking about. But there doesn't seem to be a linear ordering, or any other metric you are associating that leads to this concept naturally.

Perhaps you can give me an example. That is, give me an exact deck composition, TC, and counting system such that the particular deck represents the mean. No, give me three examples, that way I will see the pattern. Please, lay out the 52 cards explicitly, so I can see what you mean by the mean.

Thanks.

--Eliot

Hilo for a true count of +6, and one deck left would have 3 extra Aces and Tens grouped together, and 3 less 2 thru 6 grouped together.

Thus you would have as a mean pack 3.4 2s....3.4 6s, 4 7s...4 9s, 4.6 10s, 4.6 Jacks, 4.6 Queens, 4.6 Kings, or a total of 18.4 of various tens, 4.6 Aces. That would be to satisfy having 3 less cards amoung the low cards 2 thru 6, and 3 more cards amoung the aces and tens, while having the modifications for every denomination being equal.

If you follow my modified method in my proof you would have a goal of a +6 hi-lo true count and no other count being needed to define the actual pack composition. That would involve having no other relationships occuring other than the card values given by hi-lo. If there are thus no counts such as an ace side count involved, and you retain the equal probability assumptions...

The example of a mean pack for a hi-lo true count of -6, with one deck left would simply be: 4.6 each 2 thru 6, 4 7 thru 9, 13.6 tens (removing the differences in tens ranks I used in the first example for illustration) and 3.4 aces.

Other examples should come to mind.

Back to my orignal point in this thread --sorry to go off the topic of your question -- the mean expectation, for the first example pack for a +6 hi-lo true count, is certainly close to the mean expectation of all possible +6 hi-lo true counts with one deck left, but this is not the mean that you will see for actual packs given different numbers of starting decks. The distribution spread for starting with 6 decks initially is different from the distribution spread for starting with 2 decks initially, for this +6 hi-lo true count with one deck left. ML takes the usefulness of the mean composition as a requirement again, to..... But the different distributions involve different degrees to which the actual compositions pull away from the mean composition, such that one has mean compositions that would involve a floating of basic strategy edge, being exactly canceled by such variant compositions. I explained that better in my proof than I can here by itself.

You have really just tried to creat a continuous version of a deck -- but since a deck is discrete (as is the process of playing), simply adding a decimal point or two does nothing to change this. You might as well just say you have 10 decks with 37 aces rather than saying you have one deck with 3.7 aces. I think your attempt to create a continuous form of blackjack is intelligent and interesting, but I am not sure this is enough.

I actually thought of a similar thing several years ago, where we play with a real number line from 0.5 to 11. 5, rather than a deck, with a probability denisity function determining the likelyhood of a number in any given interval being chosen (as usual, the integral of the distribution function). This might be a more reasonable approach to what you are trying to accomplish here, and actually leads to some real mathematics.

--Eliot

Your ignorance of the form and terminology of mathematical proofs or just sheer mendacity makes you continue to misrepresent the terminology. Witness:

Cant: Every true count and penetration combination has associated with them a mean composition, and a mean expectation for that composition.

ML: Absolutely deceptive. It is true every true count and penetration has associated with them (sic) a mean composition but it is mathematical error to say there is a mean expectation for that composition. It would have been accurate if you had said "an expectation for that composition. There is only one sample, the mean composition so there is no mean value, but one value. There you go switching means.

Cant: ML simply demands that one has to use that mean and that mean alone for discussions about the mean for a given true count.

Totally incorrect. The proof used the terminology of expected value of the expectation of subsets of specific true counts at specific depths from an original specific set of cards greater in number than the cards contained in the subsets, not the expectation of the mean composition of the subsets. IMHO the values would be identical at least to all significant figures, but that is not relevant to the fact you continue to say I did not use expected values when the proof clearly demonstrates that is what was used. Saying otherwise is a misrepresentation.

Cant: There are no such requirements.

ML: Agreed. That is why I used expected value rather than the value you say I used and as pointed out above that is not a mean value but instead a set, definite value.

Cant: The use of such mean compositions is simply a useful practice for reducing computations needed to find the mean predicted by a given true count.

ML: Agreed. But that is irrelevant when expected values are being used.

Cant: It is not a requirement to use this mean and never has been.

ML: And I did not do that.

Cant: The actual bow effect is simply the result of the totally predictable difference between the mean expected value for the mean composition associated with a given true count and penetration, and the actual compositions and their expected values.

ML: Disagree but that has nothing to do with the matters being discussed. Again there is not a "mean" expected value of the mean composition but one value. Further, it was to take into account all possibilities and not just the value of the mean composition I used the terminology E(i,m) which included all compositions which might occur and as you say, if there is a bow effect, all the different compositions have to be taken into account. So in your saying the bow effect occurs when all compositions are taken into account as E(i,n) unambiguously does, I take that as an endorsement of my Postulate which stated precisely that point.

Cant: There in fact is no reversion to such means either. As I detailed in my Blame Woggy.... proof, each composition becomes the starting point for all further distributions.

ML: True, there is no reversion to means. That was my reason for using expected values rather than the value of the mean composition in my proof.

Cant: What results is ML adding to this absurd definition requirement

ML: Cant is now saying using expected values is an absurd thing to use in a mathematical proof. Perhaps he is calling Dr. Thorp an absurd person because the proofs in the paper Cant cited uses expected values over and over. I can think of no proofs dealing with gambling type probability topics which do not use expected value. Perhaps Cant would point one out (he won't because he is lying here).

Cant: an equally absurd demand that this is the only definition for real mathematicians to use and that one is not allowed to post any response unless one uses this same definition.

ML: No. I say you cannot attack a mathematical proof by saying an expected value expressed in the proof is not an expected value at all. A proof can only be talked about in the terms of the proof. The propounder of the proof can define. I expressly used expected values and I have the right as the propounder to do that. If Cant had said the Postulate or the Theorems did not apply to expected values, that would be a valid disagreement. Saying the terms were not expected values but, instead, something else is not a valid disagreement. They cannot be something else because I did not define them as something else and unambiguously defined them as expected values.

Cant: Again: there are no such requirements!

ML: There most definitely are. Once the values were defined as expected values, they are expected values.

Cant: The ruler used is not the thing being measured.

ML: For the life of me I cannot see how that statement relates to anything.

Cant: They have different means associated with them.

ML: Expected values don't and nothing except expected values were used. One of the things you are talking about does not even have a mean but has, instead, a value.

Cant: But once you go down this absurd trail with ML,

ML: The absurd trail of using expected values in a proof blazed by Thorp and often trod by Griffin?

Cant: you have to replace a simple difference in measures

ML: What did I replace? I used expected values of expectation throughout

Cant: with the further absurdity that the basic strategy edge, rather than the edge associated with the actual compositions, floats instead.

ML: That statement is absolutely undecypherable. There are many reasons to believe, including the proof, the expected values of expectations for specific true counts of subsets of m size are larger than the expectations for the same true counts of sets of n size which leads one to believe the E(i,m) is larger than E(i,n) when n>m. One reason to believe that is the expectation (not expected value) of one deck before any cards are dealt is larger than the expectation of a stack of eight decks before any cards are dealt. That is comparing a "perfect" set with a median subset of the set. One can think about creating a "nonperfect" -1 eight deck stack by replacing any four large cards with any four small cards and looking at the expectation and then replacing an eight with a small card in the one deck stack and one can say with some certainty the expectation for the 416 cards would be smaller than the expectation for the 52 cards. However replacement is effected in each stack to preserve a -1 count in each stack, I would say it is reasonable to conclude the expectation for the eight deck stack would be smaller than the expectation for the one deck stack because the one deck stack has a .5+ running start. And it is reasonable to conclude the same for +10 or whatever. That the expectation for the same count for the 52 card stack remains higher, however the composition which established the same count seems more than reasonable and that is consistent with the floating edge proposition. The 52 deck stacks have a .5+ head start which ain't shabby.

Cant: And that is a simple replay of an ultimate absurdity: that somehow measurement of a population changes the mean expected value of any remainder of that population.

ML: It is absurd of Cant to make a statement I ever said such a thing. He will be unable to respond and point out a place where I said such a thing. He will not respond to this request because I certainly never have said such a thing.

Cant: That is exactly what is involved when you claim that penetration alone changes expected values,

ML: Penetration does. If it did not the floating advantage would not exist yet it does.

Cant: which is what you are saying when you say that basic strategy edge changes with penetration!

ML: And, there in a nutshell is your ignorance exposed to the world. The statement the expectation for identical true counts change with penetration does not say in any way overall expectation changes for a specific number of cards drawn from a certain larger stack. Saying fifty-two cards drawn from a 104 card stack which, by coincidence, has the same true count as the 104 card stack have a higher expectation than the original stack is not the least bit inconsistent with saying all the possible 52 card subsets which can be drawn from the 104 card stack have the same expectation as the original 104 card stack. In fact, Cant not recognizing both might very well be true demonstrates how really ignorant Cant is of mathematics and about the fundamentals of blackjack mathematics while posing as conversant with the works of the masters.

The reason both might well be true and there are no inconsistencies is a very simple concept. Any stack of cards has a value for expectation and can be divided into subsets having an average expected equal to the set. The subsets have true counts which spread and if you take the expected value of the expectation of all the true counts involved and average them, some (larger than the starting true count for sure) will have expected values larger than the original expectation and some will have expected values of expectation smaller. What the individual magnitude of each might be is not important so long as the average expected value of all add up to the starting value. Cant has talked about imposing restrictions but it is imposing a restriction to say the expected values of the samples with a true count identical to the true count of the starting set have to equal the value of the starting expectation. The expected values may be larger if the average expected values of all the other true counts are smaller, i.e., steps 4 and 5 of the proof.

And my proof adequately explains how such a thing can happen.

Cant cant be trusted when he talks math. He knows no math.

There is a bow effect but no overall floating advantage. The extreme ranges in any true count involve negative bow effects, while the middle ranges tend to be positive. Go back to my proof readers and you see the entire flaw in ML's logic(?). As each composition occurs different from the mean composition, or the mean composition of the constelation of counts needed to precisely describe pack composition previously, you add another balanced count, or overall count that ends in zero. All of those counts are subject to the True Count Theorem, including penetration itself.

Each such count involves a prior composition that affects further distributions, but the TCT itself is sufficient to describe such effects. Every variation from the mean composition indicated by the main count(s) involves conservation of some added true count. As those added counts become more extreme in current values, they have their own negative bow effects, pulling down the overall edge and exactly canceling the effects of the mean expectation for the mean compositions tending to become higher.

For infinite deck approximation analysis. What I have tried in vain to do is point ML (OK not ML who I think knows better but readers of his screams about what is and is not math) to how the tendencies, of the mean compositions, to grow in edge as the pack size decreases, are canceled by the effects of compositions that occur that can have only originated with the original pack.

You are correct in that doing this by only the TCT and a simple random walk theorem leads to an entirely new class of Gauss-Lagrange solutions and proofs. I have to confess I find some of the editing boring to complete this -hint- and that my purpose is simply to stop this attack series from ML's and the ridiculous conclusions that have been brewed forth from the entire floating advantage debate since 1985. I also point you to my direct answer to ML.

I'm a 2nd grader.

What is Clarke Cant saying?

He's in a world of his own.
Not much any of us can do about it.
Sad thing is, he found ML as a catalyst; or rather: they found one another.

For a moment, Clarke, you were actually beginning to make sense (one sure sign of the Apocolypse). Glad to see you back to your old self! ;-)

Eliot, sincere congratulations are in order. To the best of my knowledge, this is the first time anyone got CC to write down ANY accurate mathematics (his example of a "mean pack" at +6 on 12 September 02, 2:32 p.m). I hope you don't think Cant invented this system (to wit: "If you follow my modified method in my proof..."). This is the best linear estimate of the composition of a deck induced by a given count, first described, AFAIK, by Griffin in ToB. Please see the Appendix to Chapter 7 for a justification and an example. It is based on the "multivariate normal assumption," he develops in earlier chapters, and by no means is application limited to infinite decks. It really forms the whole justification for adding and subtracting Effects Of Removal (EORs).

As an exercise, you might like to find the insurance index for your favorite # of decks by 1) removing the ace upcard from the complete shoe, 2) carefully applying the principle of proportional deflection, and 3) Finding which index deflects the tens up to one-third of the remaining shoe. It's tricky, but if you do it correctly, for all practical purposes you get exactly the same index a program will generate.

ETF

I was using this to dispell ML's single mean garbage.

This might be a more reasonable approach to what you are trying to accomplish here, and actually leads to some real mathematics.

In physics it is noted, especially in the new paper by C. Ewing, in the Gravity Research Foundation Archives, (Hawking and Kelley are given credit as coauthors) that we see a baseline reality of 4 dimensions (3 +time) due to Fermat's Last Theorem, in that all geometry above 3 dimensions is differential geometry. FLT states that every equation involving powers higher than 3 requires complex number solutions. Complex numbers in dimensions do not interact directly with other dimensions, except in other differential equation relationships. Thus the 3 basic dimensions are percieved as the basic 3.

But it would be nice if there were similar higher dimensional extensions of standard normal statistics. Statistics is bascially derived in only 2 dimensions. The best extension is the Graham-Stokes solution in that it deals with different measures of a population space, by placing them as vectors on a 2 dimensional surface manifold to that space.

My Blame Woggy little proof can be used to derive the Central Limit Theorem, and all the other basic Theorems in Statistics, but its basis includes higher dimensions. Each new sidecount can be considered a new dimensional measure of the population space of the blackjack pack of cards. Things like the realtionship of different endpoints from a mean can be added, like in one CLT proof the Mayor worked on.

I just did this to show how when ML appeals to their being only one definition of the mean expectation for a pack, he is drawing that definition from the mean composition, and is thus ignoring the effects of more extreme compositions being included, from the pack orginating from a larger initial pack, that by involving more extreme true count ranges (for such side counts predictably), they exactly cancel the rise in the expected value of the mean composition as the pack size decreases.

I did not intend or suspect the impact or power of this proof that the Mayor has hinted at.

I have no idea wha Clarke is saying. But, for now, I think I understand what he means by a "mean composition" --

I have this weird feeling about Clarke, that he might actually be saying something incredibly important, and I should really fight through his words to the content. But in truth, I have never been able to get more than 2 sentences into anything he writes without my head spinning.

Best advice for Clarke: come over for a beer and an evening of discussion, and I will attempt to rephrase your thoughts in a way others can understand.

--Eliot

"Sad thing is, he found ML as a catalyst;
or rather: they found one another."

You didn't see me asking for an explanation of ML's posts, now, did you? Guess why. I'll tell you why. Even when I disagree with ML's posts, I can understand what he says.

But I never disagree with Clarke Cant's posts! Guess why.