I really think this is too rude to post here, Refuting ML's proof post

So it is posted on the yahoo.com CCC also. I appologise to the PBers for this, but it was simply too complex in more nice versions. The labels are more inline with ML's originals and ML is still to be congratulated for finding a non-Graham-Stokes proof of the bow effect even as his proof is flawed by the problems I give below:

I tried to develope a nice version of this post and could not. It
simply got too complex in nice versions. Perhaps Ted Forester or T-
Hopper can put it gently and still have it make sense. I either have
to base this on ML's recent post on bj21.com and evidence of clone
prevaricvation or rewrite everything.

First the two significant Theorems of ML's post:

Theorem I

Basic Strategy expectation is invariant for the original size of the
shuffled original pack of cards--and the rules etc.--with
penetration. Proof: any penetration can be transformed into any other
by simply cutting the pack.

Theorem II

The mean for all possible subsets of a given pack is the same as the
mean of that given pack.

ML took these two --and then strangely added a third not in his
original post in a reply to me--and demonstrated the bow effect
without needing to appeal to the Graham-Stokes formula. From that
formula the true count and actual edge cover the same range of
possible pack combinations but the true count has less accuracy. The
integral of their differences must thus involve equal areas of less
and more predicted edge.

ML took a different and unique approach of stepwise summation of the
edge differences.

In either case the bow effect curve of the true count versus actual
edge is just the required curve needed for the true count
underestimating edges with TCs near zero, and overestimating the edge
at more extreme TCs, and getting more bent with deeper penetration.

But the application by ML of Theorem II is incorrect and even
dishonest. Theorem II applies to the actual subsets and not the
perfect mean subsets of each True Count or other pack composition. ML
did not apply this theorem continuously.

The proper equation, for how such edges average out is: E(m,i)=edge
for the perfectly mean pack(s) m; E(m,a)=edge for actual packs:

E(m,a)=average summa subsets E(m,a)

ML's version is to substitute E(m,i)=average summa subsets E(m,a)

ML gives a very good, non-Graham-Stokes, proof for how E(m,i)>average
E(m,a), which points to dishonesty in the following (this is where I
had to get blunt to stay simple and clear) in substituting E(m,i) for
E(m,a), in that this is only true for pack compositions where
composition E(m,a)= composition E(m,i), which is very rare except for
the original pack in that the original full pack is the only
composition E(m,i) certain to occur.

The rarity of composition E(m,i)=composition E(m,a) is a binomial
distribution for predicting the probability that a subset will be
some perfect mean density. For blackjack this involves a reduction in
probability of E(m,i) by approximately 13!/10! for every level of
penetration where a perfect mean is possible. Thus every possibility
of a perfect mean is reduced in probability by 1/1716, at every such
point.

Don Schlesinger used a deep knowledge of the Graham-Stokes equation
for years, on Wall Street, to weed out bad predictive models for
market trading--presumably sucessfully. ML was able to demonstrate
the bow effect without G-S being invoked. They know what they are
doing. It is simply too far-feteched for such skilled mathematicians
to fail to apply Theorem II continuously. Every subset has to be
considered the same as the original pack set, where the claim that
edges average summa E(m,i)=edges average summan E(m,a) simply smells.

The distribution of subserts is a distribution from prior actual
subsets and cannot and does not involve substitution of perfect mean
subsets at the current penetration levels.

Donald Schlesinger knows this. Seek out a publication from Morgan-
Stanely, variously titled over the years, two past titles are, Where
Your Portfolio Is, and Where Do You Go From Now?, which were largerly
written by Don for private clients. That publication advised private
clients in how to decide on changing their strategy goals, and to
only include past results if they involved using the same strategies
in the past. ML and DS are violating these principles when they jump
from current situations to ideal starting points for a distribution.
It simply adds to the evidence that they know better than their
current claims.

The ML post is thus flawed and dishonest, despite its success in
achieving a seperate derivation of the bow effect, by changing
distributions in this way.

Message 14891

Of course I used the expectation exactly the same way expectation is used in all blackjack calculations. We find a figure, either by combinatorial analysis or by simulation which is the expectation of all the possible configurations for a certain true count. You can call it a mean, average, overall, whatever, expectation but it remains an expectation. We do not have two overall expectations for one true count, one being a "perfect" configuration and the other bizarre configurations. There is one figure which is the expectation between +1 and +2 which contains data for a +1.00 count and a +1.99 count (if flooring) which are different or for a two removed or a five removed which are also different. There may be strange configurations where a +1 count for, say double deck, is not favorable for the player and that is part of the data comprising expectation for the +1 true count.

Clarke, do you not understand the power of a mathematical proof is that what I called E(i,n) is a *number*. It has a value. And in blackjack calculations it always has one value, the mean or whatever one wishes to call it. The power is that i can vary from -infinity to plus infinity and the statement remains true if the proof is valid.

If the bow effect is true, it exists for SBA data which, of course, uses all possibilities to come up with a one number expectation (assuming the sim is long enough to be able to be relied) for the aggragate expectation at each specific true count. If it does not exist, the SBA output would not show the existence. Now I made no definitive statement as to the existence of the bow effect. I postulated it. If you say there is no bow effect in actuality because the overall expectancies for each specific true count do not demonstrate such a thing then I would say the proof proved nothing for the real world but, since the value of the expectation was used as the value of expectation is invariably used in conjunction with blackjack, one value incorporating all permutations and different expectations which might be present in cards consisting of a specific true count, your objections so far have not mathematically supported. Your objections remain unsupported double talk.

Those expectations center around the compositions. The Proof you gave has these valid elements:

It demonstrates how E(m,i) (expectation for perfect means) excedes the average E(m,a) (expectation for all other compositions). This is exactly how come the expectation for compositions at other than the perfect mean falls back to the overall basic strategy expectation.

It demonstrates how the bow curve is driven into its shape by having to accomodate an overall fixed basic strategy edge and an increasingly less linear True Count prediction of edge.

It does so without going to depend on the Graham-Stokes equation etc.

It however does not show that the perfect mean at each penetration is the mean about which subsequent means are distributed. It shows the opposite in fact, even as it shows that a floating of the basic strategy edge occurs when such perfect means actually occur, and that this limited floating must be true if the bow effect is true.

Once again your error(?) is in not applying Theorem II continuously. Every actual deck composition is the parent of all subsequent deck compositions. Only one--I used the edge E(_,_) in my post as the subscript for the compositions involved-- composition sub_E(m,i) is certain to occur, the complete pack, and only one composition sub_E(m,i) for each true count composed pack for a fixed initial size pack is at the mean for all such true counts.

Now for the nasty parts which I tried to tone down--sorry PBers. The following is far more nasty than it seems though I will try to tone it down.

Your switching of means is well recognized to be a fraudulent exercise in statistics. It is totally refuted in some of DS' own writings for Morgan-Stanely in that he participated in some writings about how to make sure that one does not ever fail to isolate distributions between different trading models used. That should be reviewed if anyone is an M-S customer.

The distribution of means is not about the perfect mean at each penetration level, but it is a distribution about the mean of the prior deck composition. To say otherwise is the same as to assign different distributions of fluctuations to different trading strategys!

If the expectations for true counts as commonly dealt with in blackjack as a number incorporating the average win rate for all combinations of cards constituting a specific true count and as defined in my proof as E(i,n) do not demonstrate a bow effect, there is no bow effect to be seen at all. I postulated a bow effect because it is seen in data and postulated by Griffin. I cannot understand your continuing to say I defined a term differently than the way it was defined. An expectation for a true count and a depth is the expectation for a true count and a depth. What else could it be? I continue to feel your repeated mistatements of my definitions are your typical doubletalk. Now if you want to say part of the data which constitutes an expectation would show a true count while other parts do not, meaning expectations as always used in blackjack calculations before do not demonstrate the bow effect, you may. My postulate is the expectation as everyone has always used it before demonstrates a bow effect.

More double talk occurs when you say "The distribution of means is not about the perfect mean at each penetration level, but it is a distribution about the mean of the prior deck composition" as a criticism of the proof. The proof incorporates that concept by using the normal distribution around true count i at the reduced number of cards. Implying the proof does not incorporate that demonstrates you do not understand the proof or are trying to create a false impression.

Since expectations

The proof incorporates that concept by using the normal distribution around true count i at the reduced number of cards. Implying the proof does not incorporate that demonstrates you do not understand the proof or are trying to create a false impression.

You are distributing about the true count with a perfect mean pack composition for that true count. The distribution of the perfect mean pack for that true count thus replaces the available actual distribution from the prior pack compositions.

For the rest I tried to stay within your labels. The above is how, unless you are being disingenuous, you are substituting means. This is why I took a passage from one of the Morgan-Stanely publications that DS apparently helped write, in a post below, "switching means is fraud." It is more pointing toward lying, I am sorry to say, to have to point this out.

Your resoning follows this paraphrase: because we are dealing with the distributions of the true count, we then take the mean of that true count at a given penetration level. Thus the mean for the TCs is applied that is nearly certain to be different from the actual mean. This mean is not representative for the means that are possible from prior means, even though it is representative for all true counts at that level of penetration, because they involve two totally different distributions. The mean of all such perfect mean true count packs is not the same as the mean that is possible from coming from prior pack composition means.

The mean for all of the true counts at a given penetration level, is simply NOT the same mean as the mean that matches the distribution of means, that is set by prior pack compositions , unless your pack composition has fallen on some such perfect mean.

The expectation for a true count is the expectation. That one value for each true count is shown, along with frequency, in every blackjack statistic ever shown. The effect of your statement is to sayevery simulation ever run is wrong because it shows one value for expectation and that value is wrong. E(i,m) or E(i-10,m) or whatever is one value, stated in the abstract to comply with the formalities of a legal proof. How in the world can you say that value does not incorporate what it incorporates. It is the value of your chances of winning or losing at a specific true count. If the values of winning and losing demonstrate a bow effect, which is the postulate, a floating advantage exists. I am not using a wrong mean or substituting means. I am using a value which defines winning and losing possibilities at specific true counts. What else could be used?

The expectation for a true count is the expectation.

Totally wrong. There is one prediction from the actual composition and another from the true count. The bow effect is from the predictable distortion in the TC as it tries to predict the actual composition expectation.

I am not using a wrong mean or substituting means.

You are! The prediction given by the true count, is based on the mean deck for that true count, and predicts all such mean edges for all such possible true counts, at each level of penetration. That mean differs from the actual mean edge given by the actual composition.

One mean is constrained by the mean of the previous composition. The mean you give for the true count is just that: for the mean pack with that true count.

Theorem II deals only with the means of the actual compositions. The mean for the full pack is the mean of the mean of every pack that can be drawn from it. The true count mean differs from this mean. Each round involves a new subset pack, where every such new subset pack has it own distribution of means for new rounds.

Increasingly the mean of each subset, diverts from the mean of each observed true count at that penetration point. That is what the bow effect is.

The true count distribution of means is from the perfect mean pack at each penetration point for that true count. The means of each actual subset are from a distribution of the prior actual subsets. Those distributions are totally different within a given starting pack.

The average of all subsets, when subsets are grouped together from the same starting shoe, averages the edge of the starting shoe--which is Theorem I restated. This differs from the distribution of the true count.

Example: 12 aces through 9s and 48 tens, is a perfect mean for both a zero hi-lo TC=0 and a 3 deck pack. But it is not them most probable mean for a zero hi-lo TC, at 3 decks, that started with a 4 deck pack that had 18 aces, 16 2 through 9s and 62 tens. The most probable 3 deck pack would then actually be: 14 aces, 12 2 through 9s, and 46 tens, if drawn from this 4 deck pack.

But you would be assuming that the TC=0 most probable pack, 12 aces through 48 tens, applied which differs. That is how you endup substituting means in this distribution. The actual edges follow a distribution from the original full pack, while the true counts substitute another distribution.

The divergence increases with penetration such that:

The actual edges average the full pack edge, while

the true count predictions follow the edge for perfectly mean true count packs and averages. This is the largest size pack for a given TC in that the same metric applies that limits distribution to smaller possible packs for the same TC, only if they are actually observed, the same way it limits the basic strategy edge with penetration.

the average edge for actual packs is less than the edge for the perfect mean packs.

That is the bow-effect restated.

If you are not able to see that there are two distributions involved you must be lying or nuts your choice?

1. Either Theorem.

2. The mathematical validity of each step given the Theorems and the Postulate.

You have done neither. All of the words you have thrown out (none of which are verified by anything objective, they are just words) have touched those two things.

You can quarrel with the Postulate the bow effect exists across the board as I have said from the first. I think the postulate is true but never asserted anything except belief and gave reasons for my belief. If, as I believe after torturing myself to read your garbled thought processes, you are saying the bow effect only exists in "perfect" decks and does not in others, then you should have said that simply enough to be understood. But that does not attack the validity of the proof, only the validity of the Postulate.

Perhaps you do not understand E(i,n) does not stand for predictions. It does not incorporate Chapter 10 or any other prediction. It is simply the overall value of the expectation of true count i and number of cards n. Starting from that point, the expectation of all possible subsets of size m when m is smaller than n can be sorted by true counts and they were in the proof. Nothing is left out, no configuration whatsoever. If the bow effect applies to all configurations as the Postulate states, the rest follows.

Please, in less than twenty words or some number in that area state where the quarrel with the a theorem or a postulate exists rather than redifining previously defined items. That can be done very easily and in about twenty words for any mathematical proof. That is why mathematical proofs contain definitions, Theorems, Corollaries, and Postulates and numbered lines.

Theorem I is that the basic strategy edge is invarient.
Theorem II is that the mean of any given pack, is also the mean of the means of every subset of that pack.

The corolary of Theorem II is that it applies to every subset as they are encountered. Any subsets actual mean expectation is the mean for applying Theorem II to all further subsets that develope. This has the further corolary that the mean for every measured true count thus encountered is not applicable unless such subsets actually have what I have called a perfect mean composition. That is highly rare!

The divergence that developes is the bow effect, where the application of the Graham-Stokes equation, dealing with two measures that have the same range and means over their range requires that the intergral measure of their differences result in that measure summing to zero. Because the errors of the true count are more extreme, with more extreme counts, the plot of actual versus true count measures of edge has the bow effect curve.

What you have proved is:

An alternative to invoking Graham-Stokes, to define how this curve is the required curve.

A proof of how the pack compositions that diverge from the perfect mean compositions has, averaged over various true count ranges, less edge than the perfectly mean pack, thus showing the mechanism by which basic strategy remains invarient. At some TCs the edge increases with penetration, those near zero, but it also falls at the more extreme TCs with penetration.

You have not proved that the overall basic strategy edge as measured by the true count, rises with penetration, except for showing how the bow-effect requires some floating for the limited situations where the mean edge of the pack subsets, matches the mean edge of perfectly mean packs.

Now the alternative to your postulate of this is simple:

When forced to select which of all possible perfectly mean subsets, best matches the mean of all the means of all encountered subsets, the choice is overwhelmingly the subset which is the original starting pack or the largest mean pack for a given true count, given that that pack is of smaller size than the starting pack. The proof is found via applying a binomial distribution estimate of the probability of such perfectly mean subsets being formed at random from the starting pack, as an estimate of the probability of the mean edge of a subset pack having a mean that matches that of a perfectly mean subset.

In the proof is inaccurate. If the theorems are accurate and the steps are accurate, the proof is accurate.

Further state when and how in the proofs, in the theorem, in the postulate, or in the body of the proof I shifted any mean as you have claimed I have.

Before you respond you should carefully consider what E(i,n) really means because you have not caught that. It is a theoretical expression of the expectation of stacks of cards of size m that when counted have a true count but, as is mathematically consistent with the use of such a term it stands for one configuration which has the true count i and not the average of all the possible stacks which have the true count i or the mean stack. As such, the value of E(i,n) when consisting of different permutations of cards might vary slightly. Any person with much mathematical knowledge would have recognized a proof always deals with showing the same thing occurs, whatever the circumstance. That is why I could not understand what you were talking about when you said I shifted means. E(i,n) is not an average, it is a representative. Expressing this representative theoretically allows it to be applied to each instance there are n cards with a true count of i but your saying it was a mean when mathematics considers things expressed this way a representative rather than a mean demonstrates your total ignorance of elementary mathematical concepts. It is elementary a proof deals with unspecified values and since the value is not specified and relations hold true despite the nonspecification of values, then the statement is true. As such, the proof has nothing to do with the average or means of expectation. And when this proof showed an unspecified value of E(i,m) is larger than an unspecified value of E(i,m), the proof is complete if the theorems are accurate and the math is accurate. The math is accurate. If you think any conversion or any step or the proof by induction is inaccurate, it will take you less than twenty words to point out the supposed inaccuracy. You will not find an error of math and if you do not post a supposed error in math within twenty-four hours I will take it concurred there is no mathematical error in the proof.

That is enough for now. Let us get by this step.

... is in assuming CC will somehow, someday, run out of words. Or do you sincerely believe he is using logic, to the best of his ability, in a sane, rational way to try to dispute and/or confirm your theorem!? Do you have any evidence for that belief, because I haven't seen any.

[ML:] The expectation for a true count is the expectation.

[CC:] Totally wrong.

What is there to discuss after this? If anyone wants to know how the supposed "debate" turned out, just supply this text. It's critical to the discussion, and it's in context. Any rational person reading this can figure out what's going on.

ETF

Step 4 in the application of Theorem II, in that actual edge is related to the actual compositions, where the true count relates to the ideal compositions for each true count. I have expanded on that and condensed on that. I have restated it and....

The result is that due to the fact that the true count is not a perfect evaluation of the actual edges of the subsets that occur, the measures diverge, and the bow effect as I have outlined is the result. The correlation coefficient is the measure of the convergence over the full range involved of...though in this context it should perhaps be modified to be the convergence between a perfect linear estimate etc.

You thus obviously, except for being so blind, have two difference measures, and distributions of those measures, of the edge for each subset compositon. Those measures differ predictably in that they come to the same overall invarient mean which is the basic strategy mean for the entire pack, but diverge with deeper penetration. Thus the value of a true count point, in evaluating your edge is, toward a zero true count: higher, toward a more extreme true count: lower, with deeper penetration, even while the mean value of a true count point does not change.

Your "proof" post takes the mean for every true count, at a given penetration, from evaluating what I have called perfectly mean packs, and uses that to evaluate the mean for diagnosed decks, whose actual means follow the actual composition such that actual composition skews the actual mean, with depletion, with respect to the true count and this is how you substitute means. You then used the mean for all possible packs with a given true count, which is unrepresentative within a given starting pack, to claim that the mean edge for a neutral deck and for a given true count floats with penetration. This occurs only when an actual perfectly mean pack is encountered.

At no time has E(i,n) as I defined it and as such terms have to be used and understood in a mathematical proof stood for anything except a undifined value which expresses the expectation for each of the situations where the true count of a stack of n cards is the number i. I have said that over and over and you, apparently because you are so dificient in math, have kept believing the term stood for an average expectancy of i with n cards. Never did. Doesn't now.
Neither in the Postulate is a mean used and anyone with a tiny bit of knowledge of math and mathematical proofs would have thought that. That is why I have had such difficulty understanding what you were saying. You have not understood the first thing as to why the proof was valid because you were saying the undefined values were something they were not, means. Then you said I was changing means and that was deceptive. That was horribly and totally wrong because I had never used means in the first place. I did use an averaging process correctly in that step and when you said mean, I presumed you were talking about that because my math notation could not have, except by mistake, have been termed means.

If you had understood elementary mathematical notation as used in proofs, your objection would have been the postulate was wrong and that could have been said in a few words. You should understand the postulate as written in no way should have been mistaken as dealing with means either. Again, the terms are undefined values expressing relationships.

It is my opinion the corollary correctly defines relationships.

Now, since you clearly do not understand the concept of andefined value as used in mathematical proofs, I am going to give you a short course.

Now the term E(i,n) represents the expectation of each combination of n cards which when counted have a true count of i. So it could represent a 0 true count where tens have been drawn disproportionately from aces so the expectation might be higher than another combination which has a disproportionately high number of fives. That should have been understood from the start. That is one of the reason why the expectation is an undefined value (the other reason is the i is also undefined and can stand for any true count). Using any combination as the starting point, subsets of that combination are described by Theorem I and sorted by true count and the frequency of those counts by Theorem II to have the expectancy of the subsets of the original combination equal the expectancy of the combination E(i,n). Thus E(i,m) and other true counts are formed from the original combination, not some kind of average. One would then expect the expectation of E(i,m) if i=0 of the starting combination with a surplusage of aces to be larger than the combination having a surplusage of fives described above. Your objection, then, that different combinations are not accounted for in the proof is just one of inability to understand the elementary proof concept of undefined values.

And clearly you do not understand the import of the postulate which you say are averages. As should have been understood immediately by anyone with minimal math knowledge and as stated and as used in the proof, the postulate states that for a divergence of true counts from a set starting point (and that is the only way such divergence can be detected at all i.e, the only way it occurs) the bow effect is present. Again, averaging is being used and it describes the relationship of expectations bowing from any arbitrary starting combination of cards.

It appears you are saying the corollary is invalid because the bow evvect occurs only for some kind of mean expectations for perfect decks of cards. There is no evidence for such a proposition. Just because I start out with E(i,n) which might be overrich in something and end up with f(sub i-j, m)*E(1-j,m) (i-j being the lowest count encountered and 1+j being the largest count encountered)+.... + f(sub i+j,m)*E(i,m)...etc. does not mean the same things which cause the bow effect would not be happening, pushing twenties at high counts, making and losing more doubles at low counts. There is no reason to think the thing happens only at mean expectations of all true counts. In fact, however, that it does happen at mean expectations of all true counts is pretty strong evidence it happens all combinations of true counts of subsets resulting from taking subsets of a larger stack of cards of any arbitrary combination.

So I continue to assert the postulate, as stated, accurately describes what does happen. If your assertion is skewed starting combinations have not been taken into account in the proof, they most certainly have. Your ignorance has made it impossible to understand skewed and other distributions are implicitely there under the fundamentals of mathematical proofs.

If your assertion is the bow effect does not occur for subsets of m cards taken from n cards, you simply have no evidence to back that up.

So taking you through this one more time, E(whatever) is the mean of nothing. It is an abstract expression describing the specifics of the uncountable combinations which can result in stacks of cards. But it only stands for one combination at a time. It stands for an ace rich original stack of true count zero at the same time it stands for a five rich original stack of true count zero and how the expectancy of the subsets of each of these equal the expectancy of the original stack. It stands for sorting the original stack into each subset but nothing is averaged. Each is independent but the proof shows each independent occurrence results in E(i,m) being greater than E(i,n). That is the power of mathematical proofs.

If I had used the notation dealing with means of all possible true counts that would have been a necessary part of definitions and when I did not the notation unambiguously described to anyone with elementary knowledge what was being done.

E(i,n) is not an average, it is a representative

You use it as such!

You use the mean for all possible decks with a given true count, as the mean at each occurance. Yet within a given starting pack not all subsets are possible. The mean of the actual subsets diverges from the prediction given by the true count.

It is elementary a proof deals with unspecified values and since the value is not specified and relations hold true despite the nonspecification of values, then the statement is true.

This is not the case when you are dealing with two different sets of populations. One set is the subsets of a given starting pack (1), which happen to have given true counts, another is the set of all possible subsets for a given true count and number of decks left (2). The mean of (1) is constrained by Theorem II, within the starting pack, while (2) has the mean of the perfectly mean subsets for a given true count and number of decks left. (1) comes from a starting pack of a given size, which has probabilities of (2) that are much reduced by normal binomial estimates of their probability. Thus your representative is NOT representative.

As such, the value of E(i,n) when consisting of different permutations of cards might vary slightly. (The following sentence was just another normal clone insult for daring to question the "collective wisdom.")

They vary predictably such that the overall basic strategy edge for all subsets of a given pack size, has the same basic strategy mean as the original given pack size. The binimial reduction in the probability of perfectly mean packs also reduces the mean of the packs that occur--with bow effects and etc. ---which is how Theorem I is expressed for the subsets of the original pack size. This is the mechanism, whereby Theorem I occurs for all rounds, even though each round comes from a smaller subset. This is the only way Theorem I, easily proven by being able to cut the pack such that any point for starting a round is transformable to any other, is compatable with the fact that each round comes from a smaller subset of the original pack. Thus your conditions imply the following theorem:

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.

not been consistant in using yourself.

In the post I made Still another attempt to get through to ML, I have tried to isolate my objection further. If you think that you can dispose of my points just because I had not used your notation throughout, and thus these objections don't exist, neither have you.

My Postulated Theorem III is fully implied by Theorems I and II. And that you have switched in means from different actual populations is well proven.

Now your posts have gotten predictably more and more nasty and strange in thta now you claim not using your terms allows you to dispose of my objections etc. This is typical clone Blackjack Attacking and not discussion.

I am saying the notation I used would be understood by any person with minimal mathematical knowledge as not indicating a mean but as indicating an occurrence. That is simply the way mathematical proofs are done. They cannot be done any other way. No person with minimal mathematical knowledge would understand the notation any other way. It is appalling you post as a person with mathematical knowledge and did not understand this standard mathematical notation.

Notation has nothing to do with the thing. You say my notation meant I was using means when my notation, used as mathematicians use such notation have absolutely no relation to means.

I repeat. Such notation stands for one value. That values can vary depending on circumstances is not a question. If a >b and c >d then a+c >b+d when a,b,c,d are integers. That forms the basis of a mathematical proof. The value of the numbers are unimportant. They are certainly not averages. But they relate mathematically, one to the other so a+c >b+d can certainly be proven.

How can you not comprehend this really simple thing? In fact the proof includes the concept of mean and I used different notation to express that concept. A mathematical proof vetted by a competent mathemetician which uses one notation for a mean surely would not have used different notation for another mean. When I meant mean I said it.

Of course my notation represented each possibility of a true count of i in n cards. I used other notation for different true counts so the thing was not overly complicated.

There is no way the notation should have been mistaken as representing a mean. That you misunderstood is no excuse. Two languages are being spoken here, The language of mathematics and the language of Cant. The establishment of the conventions I used was done so two different languages would not be spoken one to the other.

OKAY. If you want to you can go from here after the short course. E(i,n) is NOT a mean. If I did the first thing in the world to suggest it was a mean I made a mistake. It is a representative undefined value as similar terms are used in the proof representing in the abstract EACH combination of n cards which demonstrate a true count of i. From these multidudinous starting points, everything flows from each starting point. If that is assumed, the proof then makes all your objections irrelevant. We were talking different languages, me talking mathematics and you talking Cant.

Speaking the language of mathematics, now what are your objections?