> I guess in practice you would round up 2.7,
> 2.8, and 2.9 to 3. So practically there is
> no difference.
The point is that a system seller would likely simply tell you to use +3 for all of the above values.
Don
> I guess in practice you would round up 2.7,
> 2.8, and 2.9 to 3. So practically there is
> no difference.
The point is that a system seller would likely simply tell you to use +3 for all of the above values.
Don
"Doubling 11 against a dealers 10 or Ace can be dangerous if there are an excess number of aces remaining in the unplayed cards; because for this play the aces are counted in the wrong direction as a minus card when actually they react as a plus card for this double, therefore I subtract 2 points for each excess remaining ace from your running count prior to converting to a true count and then apply the index to make the playing decision."
This is a departure from Wong's technique.
"For Betting Decisions: This is a really tough one in a shoe game because extra ace information is of limited value for betting purposes unless you know where they are located within the shoe or you are really deep into the shoe have excess aces remaining and have a high count after your ace adjustment. In that situation I'll double my normal bet for the count and increase my betting ramp making it steeper because of the potential of catching the bj's."
This, also is a departure from Wong's technique. There is currently a post concerning this on "Theory and Math."
Mathprof has posted that aces are actually less advantageous than 10's in H17 for betting purposes. He also asserts that their value is only approximately 1/4 more than 10's for betting purposes in S17.
-- JL
Don
You have accurately described Wong's method. Do you feel this is indeed the optimum method for calculating insurance using hi lo with the ace side count?
A paragraph from my post higher in the thread:
"It is curious. also, that for insurance calculation Wong does indeed follow Griffin's guideline, and counts aces as +1, requiring a 2 index point adjustment to the RC for each ace imbalance. It is not obvious to me why the ace would be handled differently in this situation than in any other playing decision, given that we are merely trying to perform a 10 count."
Would you please explain the rationale for Wong's departure from his primary ace ajustment technique for the insurance decision?
Thank you,
JL
Short simulations:
6dks, das, spl3 and spa1. 5/6 dealt out.
Bets: 1 unit at TC = 1 or > otherwise null.
Indices: Basic strategy to play the hands except Insurance Hilo index (+3)
The software looks to see how many aces are out of the deck. If there are more than the proportional share, than this excess out is multiplied by the multiplier as a negative and adjusted conveniently with the main count before the final calculation. The opposite case means the number will be positive instead. Same procedure.
1) Traditional Hilo count without adjustments:
wr = .89 se = .02
2) Using an ace multiplier = 1 with the proportional share of aces
wr = .92 se = .02
3) Using an ace multiplier = 2 with the proportional share of aces
wr = .84 se = .02
At least here, you can see that there is no base to support the notion to deduct 2 for every ace in excess still remaining.
Let?s keep this open to further double checks and more accurately sims.
Heading West? Hmmn?.. :-)
Best regards
Zenfighter
Zen
Thank you for the data.
In calculating this data, did you (or Cacarulo, or whoever did the work) use Wong's method presented in earlier PBJ editions? Or another technique?
Also, am I correct in assuming that this SCORE data includes gain from improved insurance correlation? If so, was this calculated using Wong's method (discussed by Schlesinger and others further down the thread)?
As we discussed 2 years ago on a different board, I strongly suspect that standard ace-adjusted hi lo counts are deficient.
Wong did indeed discuss the ace side count to hi lo in earlier editions of PBJ (I am relying on my 1977 edition for these comments). His technique was to value imbalanced aces as 0 for all playing decisions.
Examining Wong's SD ace side count index tables, note that these tables appear to be essentially identical to his standard SD index tables. (These are somewhat abbreviated tables in my edition, however; no values are given for counts below -2 in the SD tables. Such a limited index range is inadequate in SD.) Thus Wong does not appear make any special accomodations for ace adjusted playing strategy other than the simple one described above and stated in his text.
It is curious that for insurance calculation Wong alters his ace adjustment technique and counts aces as +1, requiring a 2 index point adjustment to the RC for each ace imbalance. It is not obvious to me why the ace would be handled differently in this situation than in any other playing decision, given that we are merely trying to perform a 10 count when considering insurance. I hope someone will explain this.
The ace almost certainly counts as a high card in certain situations: double 8 through 10, and split 88,99, and 10,10. These hands should not be ace-adjusted for playing purposes. (This, also, is alluded to in other posts in this thread.)
Also, perhaps imbalanced aces should actually be counted as small cards for certain plays, and given a value of -1 vs 0. For example, double 11 (as per Cardkounter's post) and splitting aces.
Thus traditional SCORE and playing efficacy calculations, based on a uniform treatment of the ace for playing strategy, would presumably underestimate gain from the ace side count if the above premises are correct.
I hope someone with the requisite skills will examine this interesting hi lo question. If a greater yield from ace side counting could be demonstrated the technique might be an attractive one for some hi lo players.
Thanks
John
Exactly the response I was looking for.
>> I guess in practice you would round up 2.7,
>> 2.8, and 2.9 to 3. So practically there is
>> no difference.
> The point is that a system seller would likely simply tell you to use +3 for all of the above values.
And they would be doing you a favor.
I'll again say that 95% of those playing BJ could not come up with a quotient of 2.7, 2.8, or 2.9 in the solitutde of a math class let alone under the bright lights of the casino.
+3 is a good answer.
"Example for SD:
26 cards already gone, RC = 2 and 3 Aces played
Hand = 12 vs. 2 (Rules= sd, h17, das, spl3 and spa1)
Here the pack is one Ace poor, therefore our Ai = -1
RC = 2-1 = 1
TC = 1/(1/2) = 2
12 vs. 2 ? Hit or stand. Well you should hit.
Without adjustments:
RC = 2
TC = 2/(1/2) = 4
12 vs. 2? Hit or stand. Well you should stand."
Zen
Your example gives an instance of 12 v 2 at 1/2 deck SD with an RC of +2, one extra ace dealt (3 vs 2 at 1/2 deck). One more 10 remains in the undealt deck than expected. Thus the RC should be adjusted upwards by one for this play (adjusted TC of +6), rather than downwards, making the stand decision even more desirable (adjusted TC of +6).
Am I correct?
Thanks,
JL
> Your example gives an instance of 12 v 2 at
> 1/2 deck SD with an RC of +2, one extra ace
> dealt (3 vs 2 at 1/2 deck). One more 10
> remains in the undealt deck than expected.
> Thus the RC should be adjusted upwards by
> one for this play (adjusted TC of +6),
> rather than downwards, making the stand
> decision even more desirable (adjusted TC of
> +6).
> Am I correct?
Yes, you are. To make the point Zen wanted to make, we should assume that only one ace had been seen, rather than three.
Don
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