scobee 1: Floating Advantage and playing SAFE

[Don Schlesinger]

> 5/6 with a count of zero is really only
> comparable to 0/1, if 5/6 is zero before
> the round is dealt, or would have been zero
> at 5/6 if you subtract your cards and
> dealer's upcard after the round is dealt.
> In this case, the deflection from full deck
> composition, just from the cards in your
> hand, and dealer's upcard, is closer to 0/1
> basic than 0/6 basic. So my hunch is that 1
> deck basic would be more accurate here.

Interesting observation, and the definition would have to be one way or the other, for a fair comparison.

> I don't put this forth as mathematical
> proof. I understand the problem. I just
> suspect that the large deflection -- which
> is automatically reflected in 1d basic -- is
> a greater factor than the imperfect
> knowledge of composition at 5/6 with a count
> of zero.

Not sure. Still think no.

> I think MathProf is planning on putting
> something about a "counter's
> basic" in his book. I'll bet it would
> be possible to come up with a
> one-size-fits-all basic that's
> "better," from a RA point of view,
> than any EV maximizing basic that's tailored
> to the # of decks. Give us lazy people a
> hook to hang our hat on...

Yes, undoubtedly that could be done, as a sort of "best fit" exercise. But CBS is different. It involves taking into account a bet spread such that certain plays, like 16 v. 10, become a CBS stand, reflecting the greater amount of money on the table when the departure is made.

Hal Marcus already presented a fine paper on this topic at the last Gambling Conference in Las Vegas.

Don

[ET Fan]

After more thought, I think you're right. But it's a cloudy mirror. ;-) The clouding is caused by the imperfections in whatever count is being used. If a count had 100% PE and BC, then 5/6 at TC = x should have exactly the same advantage and strategy as 0/1 at TC = x.

Make sense?

ETF

> True basic does not change. The better way
> of putting it as you did, the correct play
> at a zero count with one deck remaining most
> likely is true IMHO. Also IMO indices would
> change to mirror SD indices.

[ET Fan]

If a count had 100% PE and BC and the distribution was perfectly normal, then 5+c/6 at TC = x should have exactly the same advantage and strategy as 0+c/1 at TC = x.

I added the "+c" because 0/6 can't be anything other than 0, and the statement is more general than that. I added the "perfectly normal" proviso because I wrote a program once for the bjmath workshop showing that even the average distribution isn't exactly the same for for all these cases. They are very, very close, though, because the binomial is an excellent approximation to the normal.

The objection could be made that no counting system is remotely close to 100% PE. I don't care. ;-) I could expand the definition of "counting system" (make it a count based on a 13 digit base 24 number) or talk about games similar to blackjack with a perfect PE.

I do think my "breakthrough insight" (breakthrough for me, anyhow), sheds some light on the problem, and that's the only intent.

If Don agrees with this insight, then I guess he has to explain why the cloud on the mirror bends the image towards a multi-deck BS (stretching the metaphor right to the breaking point).

ETF

> After more thought, I think you're right.
> But it's a cloudy mirror. ;-) The clouding
> is caused by the imperfections in whatever
> count is being used. If a count had 100% PE
> and BC, then 5/6 at TC = x should have
> exactly the same advantage and strategy as
> 0/1 at TC = x.

> Make sense?

> ETF

[ML]

Sure there will be variations of the 0 52 card stack which will have excess 7,8,9 and replacing aces with tens to equal out will change expectations and accuracy of departures too. So, for example, one could certainly construct a stack of 0 52 cards where doubling 9 v 2 would be suboptimal.

My premise remains, however, that the median deck when five decks are dealt from six and the count in the remaining cards is zero, is a "perfect" deck.

I wonder if expectation falls off on both sides of the "perfect" deck, however. It might well. I don't have time to do any sims right now but that might tell us something.

[Don Schlesinger]

> If Don agrees with this insight, then I
> guess he has to explain why the cloud on the
> mirror bends the image towards a multi-deck
> BS (stretching the metaphor right to the
> breaking point).

I've never given this too much thought, mainly because, as a practical point, anyone sophisticated enough to be counting and to be aware of the FA isn't likely to be playing the hands according to BS, as opposed to using indices, which render the whole discussion somewhat moot.

But, even with indices, you have the question of using "blended" ones, which average over several deck penetrations, as opposed to using slightly different ones for some plays, at different pen levels. That would apply here, as well, I would imagine.

Still, although the average edge at TC = 0 for 5/6 is very similar (if not identical) to 0/1, there is the nagging reality that, for 0/1, only a single deck composition is possible, and there are only, say, four aces in that remaining pack (with certainty), whereas for TC = 0, 5/6, there are God knows how many possible subsets remaining and any number from 0 to 24 aces remaining! That HAS to mean something insofar as BS is concerned.

Don

[ET Fan]

> Still, although the average edge at TC = 0
> for 5/6 is very similar (if not identical)
> to 0/1, there is the nagging reality that,
> for 0/1, only a single deck composition is
> possible, and there are only, say, four aces
> in that remaining pack (with certainty),
> whereas for TC = 0, 5/6, there are God knows
> how many possible subsets remaining and any
> number from 0 to 24 aces remaining! That HAS
> to mean something insofar as BS is
> concerned.

... but a perfect PE, if you can imagine such a thing, overcomes this objection.

See?

ETF

[ET Fan]

That the median (or what Epstein called the "most probable configuration") at 5/6, TC = 0 is the same as one complete deck (0/1), that's no guarantee that the expectation will be identical, nor that the perfect strategy will be the same.

Griffin goes through an example of this on p.183. The median strategy (ie. the strategy dictated by the count and the multivariate normal assumption) was stand for his example (and, of course, that's reflected in the full deck index), but complete CA at the 15 card level showed that hitting is better. Also see pages 26 and 99.

I agree that the expectation will be close, but not identical to 16 digits ala Cacarulo ;-) For that, you need a perfect BC = 1 -- and that's unattainable with a linear count.

An exception would be the insurance subgame. Insurance is perfectly linear, so there's one "perfect" index for any penetration.

ETF

> My premise remains, however, that the median
> deck when five decks are dealt from six and
> the count in the remaining cards is zero, is
> a "perfect" deck.

[ML]

I have posted sim results of basic strategy for eight decks and one deck on BJ21 Free Misc using SBA stacked decks of -10 and +10. These results confirm floating advantage tracks advantage for numbers of decks left and occur at extreme counts as well as in the middle. They further demonstrate the Cant observations here are inaccurate IMHO.

Also is a caution as to the accuracy of Cant posts in general.

[Don Schlesinger]

> These results confirm floating advantage
> tracks advantage for numbers of decks left
> and occur at extreme counts as well as in
> the middle.

I think that's what I explained to everyone, no?

>They further demonstrate the Cant observations here are inaccurate IMHO.

Now, there's a novelty!

Don

> Also is a caution as to the accuracy of Cant
> posts in general.