situation: 6D, S17, 3 decks dealt, Hi-Lo RC = 7, number of 7's dealt = 17?
situation: 6D, S17, 3 decks dealt, Hi-Lo RC = 7, number of 7's dealt = 17?
> situation: 6D, S17, 3 decks dealt, Hi-Lo RC
> = 7, number of 7's dealt = 17?
My question is: how sure are you 17 - 7's were dealt?
You can use the table on p. 76 of Griffin's Theory of Blackjack to figure this problem out, though the math on how to do it is scattered throughout the book. The technical answer is "yes, if you can't surrender" but why are you side counting sevens unless you already have a system in place to answer this?
In real casino play, you should stick with your system, and with hi-lo you always hit 7-7, unless you can surrender and TC >= +4.
ETF
> My question is: how sure are you 17 - 7's
> were dealt?
Very sure. If two counters play at the same table (which I know is a bad thing in general) and one uses Hi-Lo while the other one uses K-O, they have an automatic side count of 7's.
> situation: 6D, S17, 3 decks dealt, Hi-Lo RC
> = 7, number of 7's dealt = 17?
Your question has only theoretical value and frankly I don't know the answer to it.I would be much more intrigued by this question: What to bet when KO and hilo are contradicting each other?
One can easily imagine a situation where KO indicates an advantage while hilo doesn't or vice versa.What do you do?
Francis Salmon
> You can use the table on p. 76 of Griffin's
> Theory of Blackjack to figure this problem
> out,
Soon to be even more accurate and expanded EOR tables in a new Appendix D of BJA3, softcover!
> though the math on how to do it is
> scattered throughout the book.
See pp. 87-90.
Don
> situation: 6D, S17, 3 decks dealt, Hi-Lo RC
> = 7, number of 7's dealt = 17?
According to Griffin (These tables will also be in softcover version of BJA3): If you have a 14 against a 10, you are normally 6.64% better off hitting over standing. Every unseen card will have the following tag values(A to 10) to be TC'd:
.08 .44 .17 -.26 -.77 -1.41 -4.21 .22 .77 1.28
So every deck must be short approx. 1 1/2 7's to make standing the correct play. Tho multi-parameter plays such as this will not add a lot in terms of EV, I think it is always best to use all the information at your disposal, and side counting 7's will definately add a lot of PE, so you will have a little +EV and a little less negative variance and definately more fun.
Hope this helps,
TAO
6 dks, Hi-lo, +7 (RC) and seven remaining 7s, at the 156-card level.
Card ranks 2 3 4 5 6 7 8 9 X A
Quantity 10 11 12 11 10 7 15 14 55 14
This 159-card subset (it will contain 156 cards exactly after removing t,4,t) reflects a +7 (on average we?ll have here 48 tens and 12 aces)
and subsequently 7 low cards less. The 8s and 9s have been properly altered to balance the big deficiency of the others neutralized 7s.
This artificially constructed subset yield the following expectations:
T, 4 vs. T
Hitting = -.586469
Standing = -.553888
Therefore the Hi-lo player despite the fact of having here a 7/3 = 2.33 true count only, is better standing than hitting, for this particular situation.
On the other hand, there is no way a Hi-lo player can play this hand with certain accuracy. The reason being mainly, due to the meagre playing correlation (PE = .41 aprox.) Moral? Skip the index.
Hope this example helps.
Sincerely
Zenfighter
Lovely work, as always. Thanks for jumping in.
Don
> situation: 6D, S17, 3 decks dealt, Hi-Lo RC
> = 7, number of 7's dealt = 17?
I think so. Since Basic Strategy for single deck is stand 7,7 vs T because half the sevens are gone, the key card for this play.
So I would think anytime over half the sevens are gone, 14 vs T would be a STAND.
... pp. 87-90 in ToB or in BJA3? I'm aware of the explanation on p. 86 of ToB, but you also need to know how to set up a "typical" shoe for RC = 7 with 3 decks remaining, and then you need to know how to adjust the 11th column for 6 decks as per Chapter 14. You can come up with an educated guess, as in ToAnyOne's response below, but then right away, someone will want to know whether to stand with 14 or 15 - 7s out.
ETF
> See pp. 87-90.
> Don
> Card ranks 2 3 4 5 6 7 8 9 X A
> Quantity 10 11 12 11 10 7 15 14 55 14
> This 159-card subset (it will contain 156
> cards exactly after removing t,4,t) reflects
> a +7 RC
It seems to me that this subset reflects an RC of +14.
By the way, I made my own calculation with my good old CA-program.I simply replaced the usual probability for a seven 1/13 with 7/156.
The result was that hitting was still about 2% better than standing.
Francis Salmon
> ... pp. 87-90 in ToB or in BJA3? I'm aware
> of the explanation on p. 86 of ToB, but you
> also need to know how to set up a
> "typical" shoe for RC = 7 with 3
> decks remaining, and then you need to know
> how to adjust the 11th column for 6 decks as
> per Chapter 14.
In ToB. And, I think Zenfighter's post addresses some of these issues.
Don
> It seems to me that this subset reflects an
> RC of +14.
Yes, you're right. I think Zen counted the 7 as +1, which isn't the case for Hi-Lo.
Don
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