Why do you constantly obsess about variance, and incorrectly, to boot? If you bet more money, you increase variance. Period. You seem to think that it's wonderful to be able to bet more at lower counts, with tighter bell curves, and all this nonsense, while never once acknowledging or understanding that calculating variance has NOTHING to do with the edge you have at any given TC!
In Table 2.1 of page 20 of BJA3, do you see any advantages anywhere?
Don
I did make that point in one of my posts previously today. Maybe it was in another forum. As Norm doesn't think we should copy posts from another forum I won't do so here. My point here is that you decrease the decisions variance. A decrease in overall variance can be taken as reduced ROR using suboptimal bets or be spent to make higher optimal bets. Obviously if you spend the the reduced RoR and keep bets optimal for the same RoR, variance isn't reduced since you spent the variance reduction by increasing bets which increases variance.Originally Posted by DSchles
1,0,0,0,0,1,2,2,-2.5,4
He wants to use the addition of two counts to cancel every non-face card. If you really want to try this, another way is to use -4 tag on Q and -2 tag on 10, J, K to avoid 2.5 tag. Basically you hope to cancel Q and A on the second count. Q is chosen to help Lucky Lady bet.
Last edited by BJGenius007; 12-24-2018 at 12:36 PM.
Personally I would switch to either the HL with AA78mTc or better still, if you do not mind unbalanced counts, the KO with AA89mTc. But if you must keep this more complicated level 2 count then I have done my analysis below.
FELT count
Let Fc =FELT count
Fc = 2p +2*(3456p) + 7p – 2*(TAp)
KO = 234567p– TAp
So 2*(KO) =2*(234567p) – 2*(TAp)
2*(KO) - 27p= 2p + 2*(3456p) + 7p – 2*(TAp) = Fc = balanced version of KO count.
In my 4th book to be published soon, I have analyzed what I called the HL level two count = HL2= (1/2)*(2p) + 3456p + (1/2)*(7p) – TAp. I chose to use halves for 2 and 7 so that the HL indices and the KO indices and the HL2 indices are all very close to each other and so you do not need to memorize different indices for the HL2.
Your Fc is EXACTLY double my HL2 count so you just need to double my HL2 indices and you will get your Fc indices.
So Fc is a balanced version of the KO count with most tag values doubled so that the indices of the Fc are APPROXIMATLY twice the indices of the KO but are EXACTLY twice my HL2 indices.
Since Fc includes the 7 in its count then I would suggest that the AA89mTc be used with the Fc just as the AA89mTc is used with the KO.
So I have calculated the indices for Insurance and hard 12 v 2, 3, 4, 5 and 6 directly so that there are no estimations.
I will list KO with AA89mTc and Fc directly beneath KO so that they can be compared. I will list Fc with AA89mTc for these strategy changes. For all other strategy changes and for betting just stick with your Fc indices.
I will also list the calculations for insurance and hit/stand hard 12 v 2 using Fc andAA89mTc so you can see how these strategy changes were calculated.
crc(t) = critical running count corresponding to KO true count of t = 4*n + (t-4)*dr where n = number of decks and dr = decks remaining or easier just "memorize" my table of critical running count that has patters so it is easy to remember.
Insure if KO+ AA89mTc >= crc(4) CC = 100%
Insure if Fc+ 2*(AA89mTc) >= 8*dr CC = 99%
Stand on hard12 v 2 if KO + AA89mTc >= crc(4) CC = 92%
Stand on hard 12 v 2 if Fc + 2*(AA89mTc) >= 7*dr CC = 92%
Stand on hard12 v 3 if KO + AA89mTc >= crc(2) CC = 95%
Stand on hard 12 v 3 if Fc + 2*(AA89mTc) >= 3*dr CC = 95%
Stand on hard12 v 4 if KO + AA89mTc >= crc(0) CC = 97%
Stand on hard 12 v 4 if Fc + 1.5*(AA89mTc) >= 0 CC = 98%
Stand on hard12 v 5 if KO + AA89mTc >= crc(-2) CC= 99%
Stand on hard 12 v 5 if Fc + 1.5*(AA89mTc) >= (-4)*dr CC = 98%
Stand on hard12 v 6 if KO + 1.5*(AA89mTc) >= crc(-1) CC = 99%
Stand on hard 12 v 6 if Fc + 3*(AA89mTc) >= (-3)*dr CC= 99%
F count insurance.jpg
F count hard 12 v 2.jpg
Last edited by bjanalyst; 12-24-2018 at 04:05 PM.
Typo in one of my previous posts about KO critical running counts that I would like to correct here.
IT SHOULD READ:
The table of critical running counts depends on number of decks, decks played and KO true count and give the corresponding KO running count. the formulas used to make the table of critical running counts is t = (KO - 4*dp) / dr = 4 + (KO - 4*n)/dr where t = KO true count, dp = decks played, dr = decks remaining and n = number of decks.*
The typo was t = 4 * (KO - 4*n) / dr. I hit * instead of + key. I noticed the error when I was rereading my posts. If I am writing quickly I sometimes make mistakes. If you see any mistakes please let me know.
So the correct formula is
t = (KO - 4*dp) / dr = 4 + (KO - 4*n)/dr
I have a question for you. How much improvement in BC, PE, and IC is gain using your AA78mTc with Hi-lo and KO ? KO without side count already have a BC of .98. KO's PE is .55. How much gain will there be for KO's playing efficiency along with the insurance correlation?
Last edited by seriousplayer; 12-25-2018 at 09:33 AM.
Please read all the threads in these postings and your questions should be answered. But I will give short answer here.
1. KO + AA89mTc is the Noir count with 100% insurance efficiency
2. AA89mTc helps with playing strategy only. It increases hard 12 v 2, 3, 4, 5, 6 decisions by 20% to 30% . Not sure how this translates into PE
3. AA89mTc does not help with BE. Also both KO and HL are 96.5% for S17, DAS, LS game, not 98%
I checked the BC with the Blackjack Card Counting Efficiency Calculator for KO. The BC is 97.7% if you round it you it becomes 98%. Playing Efficiency of 55% and Insurance Correlation of 78%. I think you mean KO + AA89mTc becomes the Unbalanced Ten Count (also known as the original Jacques Noir ten count from "Casino Holiday") and not the Noir Count. The Unbalanced Ten Count have an IC of 100% and not the Noir Count which have an IC of 84%. With the Unbalanced Ten Count you can't just use one index for insurance. You have to take insurance when the RC >= 2*deck for prefect insurance. Why are you taking insurance with KO+ AA89mTc at TC = 4 when the Unbalanced Ten Count say to take insurance when the RC >= 2*deck? The reason AA89mTc helps with hard 12 vs 2, 3, 4, 5 and 6 decision is because you are counting the 8s and 9s.
Last edited by seriousplayer; 12-25-2018 at 05:03 PM.
I would like to answer your questions one at a time.
1. Betting Efficiency
Betting efficiency of KO is the CC between betting EoR and the tag values of the KO. Attached below is the calculation of KObetting efficiency using betting EoR for S17, DAS, LS game from Don Schlesinger’sBlackjack Attack which is what I quoted in my previous post. If you have a different game and differentrules then you will probably have different EoR. Not sure what game your Blackjack Card Counting Efficiency Calculator used. Take a look at CC in the row X1 = KO in the attached file and you will see CC = 96.51% 01 Betting.jpg
2. Insurance
Please look at attached file. You will see insurance efficiency of KO is 78.32% (which agrees with what you mentioned in your post) and insurance efficiency of KO + AA89mTc is 100%. Also note that the last column shows KO +AA89mTc which shows that all non-Tens are counted as +1 and Tens as -2 and has an unbalance of 4 per deck. This is the unbalance Ten count which I mentioned in earlier posts and which you mentioned in your post above. I called this in prior posts the Noir count in honor of Jacques Noir ten count from "CasinoHoliday" that you also mentioned in your post above. I used that term Noir count for the unbalanced Ten count = KO + AA89mTc. Maybe Noir count actually referred to a differ count that I was not aware of. What I called the Noir count is KO + AA89mTc.
And this give perfect insurance but you take insurance when KO + AA89mTc >= 4*n where n = number of decks. So you take insurance when KO + AA89mTc >= four times the number of decks, NOT two times the number of decks which you mentioned in your post.
01 Insurance.jpg
3. Hard 12 v 2
You state that the reason AA89mTc helps with hard 12 vs 2, 3, 4, 5 and 6 decision is because you are counting the 8s and 9s. That is correct but it Is not the full story. AA89mTc also helps because it counts the Ace as +2. KO counts the Aces as -1 so KO + AA89mTc counts the Ace as +1 which is what is needed for hit/stand hard 12 v 2, 3, 4, 5, 6 decisions. Attached are calculations of standing on hard 12 v 2 using KO and AA89mTc which increases KO CC for this play of 70% to 92%. Stand on hard 12 v 2 if KO +AA89mTc >= crc(4) = 4*n.
01 hard 12 v 2.jpg
Last edited by bjanalyst; 12-25-2018 at 06:27 PM.
Posting Permissions
-*-
Bookmarks