Zenfighter: Side counting Aces. Dreams of perfection.

[Zenfighter]

A side count of aces for the Hi Opt I.

We will work all this thoroughly, with the aid of our new EoR?s tables from BJA3 Appendix D.
We?ll assume here the following rules: s17, das, spl3 and spa1.

Average absolute effect of the Hi Opt I monitored cards: 3,4,5,6 and Ts

Looking at Table D17 (betting correlations) for the above-mentioned rules we have:

Avg. eff. = (.4339 + .5680 + .7274 + .4118 + (4 * .5121))/ 8

Avg. eff. = 0.5237

Enter Griffin: just a bit less than that of the ace, [. 5794] the most important uncounted cards. It therefore seems reasonable to regard an excess ace in the deck as meriting a temporary readjusting of the running count (for betting purposes only) by plus one point

With roughly calculations we can make a quick inference:

(.5794/.5237)* 100 ? 100 = 10.64% more for the ace than the average effect we got above. That is:

Adjusted index (Ai) = -1.1064 approximately.

Striving for accuracy with the aid of Griffin.

Ai = [(13* Ei * Sum Ci^2)/(12 * I) where

Ei = EoR for the ace

Ci = tag values for the Hi Opt I

I = inner product (tag values times EoR?s)

So we have then:

Ei = -0.5794

Sum Ci = 1^2 + 1^2 + 1^2 + 1^2 + 4 * [(-1) ^2] = 8

I = 1* .4339 + 1*. 5680 + 1* .7274 + 1* .4118 + 4 * (-1) * (-.5121) = 4.1895

Finally plugging all the figures inside we get:

Ai = (13 *(-.5794) * 8) /(12 * 4.1895) and solving we get:

Ai = -1.1986 as the correct adjusted index, so

Ai = -1.2


Adjusting our running count with the side count of Aces. Theory.

1) Find the excess or deficiency of Aces at the selected card level. E.g. ? deck remains
and you have seen two aces gone, thus your pack is one Ace deficient.

2) For each card deficient that our pack has we add the value of Ai to the RC.

3) For each card in excess that our pack has we subtract the value of Ai from the RC.

4) Finally we divide by the number of decks remaining to find the TC and place our bet.


An example

26 cards already gone, RC = 2 and 3 Aces played.

Here the pack is one Ace poor, therefore our Ai = -1.2

RC = 2 - 1.2 = 0.8 and our

TC = 0.8 / (1/2) = 1.6

Lets say you?re spreading $25 to $100 with TC 1 = $50, TC2= $75 and TC=>3 $100

How much you would bet here? Answer: $50 ($200 for a black chip player)

Same example with the traditional Ai = -1

RC = 2-1 = 1

TC = 1/(1/2) = 2

How much you bet here? Answer $ 75 ($300 for a black chip player)

One form of avoiding the worries here, is to think, that using the traditional index (-1), the times the player is over betting any given TC will compensate those times he is under betting it, and so, in the long run the expectation will remain the same. Go ahead and do what you want, but side counting aces is not for everybody, or at least not everybody can do it exactly as it is.


Side count adjusted indices for different ace-neutralized counts

 

 

System	       Adj.index	Rounded	     Common 

 

						 

Hi Opt I	-1.1986		-1.2		-1 

						 

Gordon		-1.2075		-1.2		-1 

						 

Hi Opt II	-2.1859		-2.2		-2 

						 

Omega		-2.2873		-2.3		-2 

						 

UAPC		-3.2740		-3.3		-3 

 

 

To be continued. :-)

Zenfighter

[Zenfighter]

Correct the Hi Opt II as follows:

 

Ai = -2.14422 		-2.1 

 

The rest are double checked and are all fine.

Enjoy

Zenfighter

[John Lewis]

"Enter Griffin: just a bit less than that of the ace, [. 5794] the most important uncounted cards. It therefore seems reasonable to regard an excess ace in the deck as meriting a temporary readjusting of the running count (for betting purposes only) by plus one point"

I understand that this does not apply to hi lo, correct?

It's my understanding that even if one keeps a side count of aces in hi lo and detects an excess of aces (e.g., no aces dealt by 1/2 deck of SD) that a betting adjustment outside of the standard hi lo count is unwarranted. Somewhat counterintuitive, but correct.

Using your EOR tables from BJA 3 paperback I note that the betting EOR of ace removal (H17, NDAS, SPA1, SPL3) is -0.5311, and the effect of ten removal is -0.5470. Thus, though virtually identical in EOR, the ten is even slightly more advantageous to the player than the ace (H17). Very interesting.

Thanks, JL

[Zenfighter]

I understand that this does not apply to hi lo, correct?

The Hilo average absolute effect for the above mentioned rules:

Avg eff. = (.5794 +. 3809+ .4339 +. 5680 +. 7274 +. 4118 + 4 * .5121)/10 = .51498 = .5150

Here the difference between the average and the EoR?s of the corresponding ace is a bit wider than for the HiOpt I example.

It's my understanding that even if one keeps a side count of aces in hi lo and detects an excess of aces (e.g., no aces dealt by 1/2 deck of SD) that a betting adjustment outside of the standard hi lo count is unwarranted. Somewhat counterintuitive, but correct.

For ace-reckoned counts, keeping a side count of aces points mainly at increasing somehow the playing efficiency of the hands and not with betting purposes. But If I trust my memory, I remember Cacarulo stating that a slightly improvement in BC is also possible. Here you?ll need him for an added clarification. Not the main reason to side count aces with the Hilo count, anyway.

Have a look at the Main Page. Quite interesting thread regarding this. As usual, for SD it?s a lot easier than for multiple decks.

Hope this helps.

Zenfighter

[John Lewis]

"Have a look at the Main Page. Quite interesting thread regarding this."

Thank you for pointing this out.

I found the posts there very interesting, and posted comments of my own.

John

[Cacarulo]

> But If I trust my memory,
> I remember Cacarulo stating that a slightly
> improvement in BC is also possible. Here
> you?ll need him for an added clarification.
> Not the main reason to side count aces with
> the Hilo count, anyway.

Your memory is good

Yes, for betting purposes there is some gain (only in S17 games) in using an Ace-Side count but it is really not worth it. The best card to use for this is the seven which increases the BC from 0.9648 to 0.9804 (S17,DAS). The Ace side is better used for playing purposes.

Here is how the BC is increased in Hi-Lo:

Ace = -1        ==> BC = 0.9648 (S17,DAS) 

Ace = -1.15435  ==> BC = 0.9657 (S17,DAS) (optimal) 

 

Ace = -1        ==> BC = 0.9652 (S17,NDAS) 

Ace = -1.19708  ==> BC = 0.9667 (S17,NDAS) (optimal)

Now, using a Seven-Side count:

Seven = 0       ==> BC = 0.9648 (S17,DAS) 

Seven = 0.59384 ==> BC = 0.9804 (S17,DAS) (optimal) 

Seven = 0.5     ==> BC = 0.9800 (S17,DAS) 

Seven = 1       ==> BC = 0.9737 (S17,DAS) 

 

Seven = 0       ==> BC = 0.9652 (S17,NDAS) 

Seven = 0.61225 ==> BC = 0.9818 (S17,NDAS) (optimal) 

Seven = 0.5     ==> BC = 0.9813 (S17,NDAS) 

Seven = 1       ==> BC = 0.9757 (S17,NDAS)

For insurance purposes:

Ace = -1        ==> IC = 0.7885 (1D) 

Ace =  0.76923  ==> IC = 0.8920 (1D) (optimal) 

Ace =  0        ==> IC = 0.8717 (1D) 

Ace =  1        ==> IC = 0.8903 (1D)  

 

Ace = -1        ==> IC = 0.7647 (6D) 

Ace =  0.76923  ==> IC = 0.8929 (6D) (optimal) 

Ace =  0        ==> IC = 0.8674 (6D) 

Ace =  1        ==> IC = 0.8908 (6D)

Hope this helps.

Sincerely,
Cac

[Don Schlesinger]

From everything I'm reading below, it is clear that, in fact, we should count the ace as +1 instead of -1 to increase Hi-Lo insurance efficiency. That would mean reversing the "error" by multiplying deficient or excess aces by 2, as Wong suggested all along.

Don

> For insurance purposes:
> Ace = -1 ==> IC = 0.7885 (1D)
> Ace = 0.76923 ==> IC = 0.8920 (1D)
> (optimal)
> Ace = 0 ==> IC = 0.8717 (1D)
> Ace = 1 ==> IC = 0.8903 (1D)
> Ace = -1 ==> IC = 0.7647 (6D)
> Ace = 0.76923 ==> IC = 0.8929 (6D)
> (optimal)
> Ace = 0 ==> IC = 0.8674 (6D)
> Ace = 1 ==> IC = 0.8908 (6D)

[Cacarulo]

[Zenfighter]

 

Ace = -1        ==> IC = 0.7647 (6D)  

Ace =  0.76923  ==> IC = 0.8929 (6D) (optimal)  

Ace =  0        ==> IC = 0.8674 (6D)  

Ace =  1        ==> IC = 0.8908 (6D) 

 

In order to arrive at the .8674 figures Cacarulo has used here an unbalanced side count:

0 1 1 1 1 1 0 0 0 ?1

So he start with a IRC = - 24

With this side count the true counted index = -1 (see Main)

Obviously if you incorporate the aces too:

1 1 1 1 1 1 0 0 0 ?1 your side insurance correlation will pop up in efficiency to an astronomical .89 as the result of adding both correlation the primary and the ace indicator count with Griffin?s multiple correlation formula (page 62 TOB)

This is equivalent to keeping this side count apart and buying insurance whenever the index = -5 You would have to start here at an IRC = - 48 if you were to use this unbalanced count incorporating the aces inside.

Having an insurance counter to work for us we could touch heaven obviously:

1 1 1 1 1 1 1 1 1 ?2 with a 100% insurance correlation.

The problem is that Cardkountr is using only his Hilo count .7647 while estimating his side count of aces as a proportion to the amount of others. This is an ace indicator count, what means that his IC is a meagre (for 6 dks) .19245

So he is entitled only to estimate a gain as:

IC = sqr (.7647^2 + .1925^2) = .7886

were he using the traditional (like is the case for all the rest of the playing hands, were the side count = 1) adjustment.

What is still difficult for me to believe, is that this counter just subtracting one point more, is convinced he can match the perfection of his counter partner who started using the above-mentioned count since the beginning namely 111111000-1 with an unbalanced side count.

I?m full of doubts, Don. So I prefer to remain skeptical.

Can someone out there simulate this exactly? (The traditional method, I mean)

Sincerely

Zenfighter

[Cacarulo]

AND ONLY IF you are able to calculate an exact number of "excess aces". Of course, nobody does that so that's why the modern way is more accurate. Besides, I think the modern way is much more easier.

Example: Hi-Lo, RC = +10. Three decks remain unseen. 10 aces have been seen so 14 aces remain unseen.

PC = -1 1 1 1 1 1 0 0 0 -1 

SC =  1 0 0 0 0 0 0 0 0  0

PC (Hi-Lo) = +10
SC = -24 + 10 = -14

TC = [PC+2*SC]/DR = (10+2*(-14))/3 = -18/3 = -6
Since the index = -5 we don't buy insurance. This is the optimal method but we can use a simpler one:

TC = [PC+SC]/DR = (10+(-14))/3 = -4/3 = -1.33
Since the index = -1 we don't buy insurance.

Let's use now the traditional method:

1) Three decks remain; they should contain 12 aces;

2) But, they contain 14 aces, or two in excess;

3) For insurance purposes, aces are the same as any other low card; they should have been counted as +1, but hi-lo counts them as -1, thereby producing an error of two for each excess or deficient ace;

4) The error here is 2 x 2 = 4, which we subtract from the +10 running count;

5) The ace-adjusted running count is now +6, which when divided by three decks remaining gives +2;

6) Hi-lo ace-adjusted insurance index is +3, for all decks, so we don't insure.

The problem with this method appears when the number of decks remaining is not an integer. In this case you need to calculate a fractional number of "excess or deficient" number of aces.

Example: Hi-Lo, RC = +10. 3 3/4 decks remain unseen. 10 aces have been seen so 14 aces remain unseen.

Hope this helps.

Sincerely,
Cac

[Zenfighter]

[Don Schlesinger]

> 6) Hi-lo ace-adjusted insurance index is +3,
> for all decks, so we don't insure.

I've used this method forever and find it extremely simple.

> The problem with this method appears when
> the number of decks remaining is not an
> integer.

No, not at all. Same as always calculating the TC when the decks remaining is not an integer. We expect one ace every quarter-deck; so what's the problem?

> In this case you need to calculate
> a fractional number of "excess or
> deficient" number of aces.

No, sorry, not true.

> Example: Hi-Lo, RC = +10. 3 3/4 decks remain
> unseen. 10 aces have been seen so 14 aces
> remain unseen.

We've seen 2 1/4 decks. That should have been 9 aces (no fractions!). We saw 10 aces, so there is one deficient in the decks remaining. Add 2 to the RC, producing +12. Divide by 3 3/4, yielding 3.2, so insure. In this case, all you have to see is that the division produces a value greater than 3, which is instanteously discernible.

Don

[Cacarulo]

> 6) Hi-lo ace-adjusted insurance index is +3,

> I've used this method forever and find it
> extremely simple.

> No, not at all. Same as always calculating
> the TC when the decks remaining is not an
> integer. We expect one ace every
> quarter-deck; so what's the problem?

> No, sorry, not true.

> We've seen 2 1/4 decks. That should have
> been 9 aces (no fractions!). We saw 10 aces,
> so there is one deficient in the decks
> remaining. Add 2 to the RC, producing +12.
> Divide by 3 3/4, yielding 3.2, so insure. In
> this case, all you have to see is that the
> division produces a value greater than 3,
> which is instanteously discernible.

Say we've seen 2 1/8 decks. In this case 8.5 aces should be the number. Of course, we can approximate it to 9 (or to 8) but in order to be exactly the same method we should use 8.5.
See what I mean?

Sincerely,
Cac