Robo needed a simpler proof of Abdul Jalib's True Count Theorem and one damn thing lead to another...
Theorem I: the mean path for any random walk is a straight line from origin to destination.
Theorem II: Chandahar's partition theorem and 3 of its corallaries:
__(a) For cutting any segment c=a+b, the greatest product a*b is where a=b.
__(b) Applying this to nspace of n dimensions (which is why this is called Chandahar's Theorem as it would be too easy just to do this in one dimension as in (a) above)
__(c)That the effects of random cutting is to lead to equal partitions, but maximum variation within each partition. (see notes later on Theorem II(c) for an explanation)
Theorem III: That any penetration, within any pack or subpack can be transformed into any other penetration, leading to no changes in mean edge within that pack or subpack.
Postulates:
I: The otherwise proven Jalib's True Count Theorem in a simpler proof here.
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 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.
Postulate I: By Axiom III for any running count, RC(j) at j cards left, RC(0)=0, for zero cards left.
Applying Theorem I 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 RC(j)=TRC(j)/j, TRC(j)=RC(j)/j, TRC(k)=RC(k)/k
by simple algebra the average TRC(k)=average TRC(j)
Once you observe a true count the average true count does not change no matter how many other cards are seen, for the boudaries set for j and k above,
Postualate II: Proven by Theorem III with challenges answered in Cases I thru IV.
Postulate III: At every penetration level and running count (or true count see Axiom I) at that level, there is a most probable pack composition given only those two parameters, but that most probable compostion is a very small fraction of the possible compositions that can have those parameters. The proof is binimial and beyond this paper's purpose, but can be demonstrated by comparing the probability of a full pack coming to a subset that has the same card rank densities at the first opportunity. This requires drawing 13 cards, where exactly 4 of those cards are ten count ranks (ranks for ten count cards do not matter in blackjack) and only one each of card ranks ace to 9 are drawn in 13 draws. The exercise is left to the interested reader, but the ratio of most probable compositions to all possible copositions is similar, for any true count and penetration.
The difference between the most probable subset and an actually seen subset, of pack compositions, can be mapped by adding a new set of balanced rank versus rank counts, as per Axiom II.
But each time that a new "sidecount" of this sort is added, its true count is also subject to the True Count Theorem (reproven in Postulate I), such that the current value of all such previously used "side counts," from all prior additions of such counts, redefines what is the most probable composition.
But the improbability of an actual pack being the same as the most probable pack stills holds with each new observation. With each increase in penetration, which occurs with each sampling observataion, a new set of "side counts" is massively probable to be needed. This requires that as penetration increases there is an increasing more complex and more modifying set of such side counts needed to define the actual composition. Theorem II(c) and Axiom II apply also, in that there is a tendency toward the most extreme compositions.
Thus the difference between your main true count, or main set of true count's, prediction(s) of edge, grows, and grows more complex to evaluate, as penetration increases.
QED
Postulate III is proven without needing to modify Postulate II. See Cases I thru IV below to refute arguments otherwise.
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 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 porduct 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. But it is overvalued in a reverse of Case II fallacys.
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 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.
All of the above are provable without Theorem II(c), but Theorem II(c) is included, sometimes called Chandahar's Paradoxical Corollary, to clarify how the addition of new "side counts" involves a tendency toward maximum variation in new pack subsets.
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. I hope there are no missunderstandings for doing this.
Note on Theorem II(c):
If you cut a segment randomly you tend to cut more into the largest subsegments than the smaller ones. This involves a tendency toward eventually equal sized segments. In n dimension however, each slice is a hyperspatial manifold. Where each cut involves complex segment boundaries, they involve higher possibilities of extreme subsets that enlarge as the power of the number of dimensions involved. Each use of Axiom II adds a dimension to evaluate the actual composition of the blackjack remaining pack subset, while every change, in every true count that has been previously added to evaluate that pack and its subsets, is a new slice in that hyperspace.
