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Thread: Adding AA78mTc side count to High Low

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    Adding AA78mTc side count to High Low

    This is for the High Low player where the HL player adds the AA78mTc (Ace, Ace, Seven, Eight minus Ten count) to the HL and calculates a pseudo Ten count, Tc = HL + AA78mTc, which is used for insurance and standing on hard 12 v 2, 3, 4, 5 or 6. HL is used for betting and for all other strategy decisions.

    Use chips to keep the AA78mTc and keep the HL in your head. If for example 6 chips are to the right of your betting stack, then AA78mTc is +6 and if 9 chips are to the left of your betting stack then AA78mTc is -9.

    AA78mTc which is used with the HL gives the Ace a tag value of +2, seven and eight +1 and Ten-1. There is a lot of cancelling and plenty of time to update the AA78mTc in the shoe game. Update the HL in your head as soon as the cards hit the table. On updating the AA78mTc, when the cards are on the table and before players make any playing strategy decisions, look for cancellations. For example, an Ace will cancel two Tens, a 7 will cancel a Ten and an 8 will cancel a Ten. Calculate AA78mTs (s = seen during the current round) and then add AA78mTs to AA78mTc from the previous round to get an updated AA78mTc. Continue to update AA78mTc as players make their playing strategy decisions and dealer finishes his hand. If you have trouble keeping two integers in your head, chips can be used keep track of the AA78mTc
    as explained above.

    If the Aces, sevens and eights seen on the table during a given round do not conveniently cancel nearby Tens then just calculate AA78s (s = seen during the current round) and then scan for the Ts (Tens seen during the current round) and subtract one from AA78s for each Ts, i.e. you are calculating AA78mTs = (AA78s – Ts) which is then added to AA78mTc from the previous round to get an updated AA78mTc.


    Use Tc = HL + AA89mTc for the following six strategy changes listed below and use HL for all other strategy changes and for betting. dr = decks remaining.

    1.Take insurance if Tc >= 4*dr
    2.Stand on hard 12 v 2 if Tc >= 4*dr
    3.Stand on hard 12 v 3 if Tc >= 2*dr
    4.Stand on hard 12 v 4 if Tc >= 0
    5.Hit hard 12 v 5 if Tc < (-3)*dr
    6. Hit hard 12 v 6if Tc < (-1)*dr
    Last edited by bjanalyst; 12-20-2018 at 09:37 AM.

  2. #2


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    You can do all the side counting you want ! i get pay on 18 vs 18 , that beats any side counting hands down . side counting is rather inefficient . Welcome to the world of practical application.

  3. #3


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    How do you get paid on 18 v 18 which is a push? If you get paid on what should be pushes then no need to count at all - just play every hand!

    If you do not get paid on pushes then you need to count which is the real world. So back to the real world.

    Insurance is the most important playing strategy deviations. HL + AA78mTc >= 4*dr gives almost a perfect insurance decision increasing the HL >= 3*dr CC around 78% to HL + AA78mTc >= 4*dr CC of around 96%. Just the insurance decision alone is enough to justify adding the AA78mTc to the HL which gives a pseudo Ten count. Thorp's original Beat the Dealer used the Ten count and HL + AA78mTc gives a Ten count which you use when you need a Ten count and you still have HL for betting and all other playing strategy changes.

  4. #4


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    You are very detail driven or a driven individual while i am not.

  5. #5


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    Please don't take my comments at "face value".

  6. #6


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    Quote Originally Posted by bjanalyst View Post
    This is for the High Low player where the HL player adds the AA78mTc (Ace, Ace, Seven, Eight minus Ten count) to the HL and calculates a pseudo Ten count, Tc = HL + AA78mTc, which is used for insurance and standing on hard 12 v 2, 3, 4, 5 or 6. HL is used for betting and for all other strategy decisions.

    Use chips to keep the AA78mTc and keep the HL in your head. If for example 6 chips are to the right of your betting stack, then AA78mTc is +6 and if 9 chips are to the left of your betting stack then AA78mTc is -9.

    AA78mTc which is used with the HL gives the Ace a tag value of +2, seven and eight +1 and Ten-1. There is a lot of cancelling and plenty of time to update the AA78mTc in the shoe game. Update the HL in your head as soon as the cards hit the table. On updating the AA78mTc, when the cards are on the table and before players make any playing strategy decisions, look for cancellations. For example, an Ace will cancel two Tens, a 7 will cancel a Ten and an 8 will cancel a Ten. Calculate AA78mTs (s = seen during the current round) and then add AA78mTs to AA78mTc from the previous round to get an updated AA78mTc. Continue to update AA78mTc as players make their playing strategy decisions and dealer finishes his hand. If you have trouble keeping two integers in your head, chips can be used keep track of the AA78mTc
    as explained above.

    If the Aces, sevens and eights seen on the table during a given round do not conveniently cancel nearby Tens then just calculate AA78s (s = seen during the current round) and then scan for the Ts (Tens seen during the current round) and subtract one from AA78s for each Ts, i.e. you are calculating AA78mTs = (AA78s – Ts) which is then added to AA78mTc from the previous round to get an updated AA78mTc.


    Use Tc = HL + AA89mTc for the following six strategy changes listed below and use HL for all other strategy changes and for betting. dr = decks remaining.

    1.Take insurance if Tc >= 4*dr
    2.Stand on hard 12 v 2 if Tc >= 4*dr
    3.Stand on hard 12 v 3 if Tc >= 2*dr
    4.Stand on hard 12 v 4 if Tc >= 0
    5.Hit hard 12 v 5 if Tc < (-3)*dr
    6. Hit hard 12 v 6if Tc < (-1)*dr
    Did you stop using your KO with side counts method then?

  7. #7


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    To answer your question, I do not use the Hi-Low with plus minus side counts. I use the KO. I analyzed the Hi-Low with plus minus side counts because many card counters do not like unbalanced counts and so are reluctant to switch to the KO. For the shoe game, KO is much better. For the double deck or single deck game where the true counts go all over the place and are often outside a table of critical running counts then I would use the High Low. But I only play shoe games so I personally use the KO. The KO indices are very similar the High Low indices - the counts are very similar. But I like the KO since it has a pivot of a true count of 4 and so the closer to the pivot the less the errors in estimating the decks remaining have on calculation of the true count. So I personally use KO with AA89mTc but wrote about HL with AA78mTc because, as I mentioned above, most players like HL which is the count I analyzed for those players and then added side counts to the HL for those players.

    This can be seen in the table below. n = number or decks, dr = decks remaining. The closer to the pivot the less the error in estimating decks remaining has in the calculation of a true count. For the High Low the pivot is zero. At the pivot a HL running count 0 also corresponds to a HL true count of zero. For the KO a KO count of 4*n where n = number of decks corresponds to a true count of 4. So at the pivot the true count calculation is totally independent of the number of decks remaining. The closer to the pivot the less errors in estimating the decks remaining has in the calculation of a true count. So at tc = 5, the HL running count = 5*dr but KO running count = 4*n + dr. So an error in the estimation of dr when tc = 5 using the HL has five times the effect in the calculation of a true count as the same error in estimating dr does with the KO true count calculation.

    KO has a pivot of a true count of 4. So at a true count of 5 you are one true point count away from the pivot of 4 for the KO but with the HL you are 5 true count points away from the HL pivot of a true count of zero. You want true count accuracy at true counts around 4 where your maximum bet is out. Thus KO is better. Also with KO you can use a table of critical running counts to calculate the true count which depends on number of decks, decks played and the KO running count so there is no need to do division. For example, take insurance when KO + AA89mTc > 4*n where n = number of decks. KO + AA89mTc gives a perfect Ten count, i.e. Tc = KO + AA89mTc gives a perfect Ten count and has a CC = 100% for the insurance bet.

    The KO table of critical running counts has many patterns and is very easy to remember since the patterns are there which I will not be covering here. Also all side counts added to the KO are balanced counts. When a balanced count is added to an unbalanced count such as the KO with an unbalance of 4 per deck, the resulting derived count is also unbalanced with the same unbalance as the primary unbalanced count. Thus KO has an unbalance of 4 per deck, AA89mTc is balanced so KO + AA89mTc is unbalanced with the KO unbalance of 4 per deck.


    So to answer your question again. I use KO with AA89mTc. I covered HL with AA78mTc because most players like the HL and want balanced counts. Also many teams insist on using the HL. Thus I analyzed many different side counts to use with the HL. I would recommend the HL for the one and two deck game but as I stated above I would use the KO for the shoe game.

    Below is my analysis of Accuracy of HL vs KO at various true counts.

    err = absolute value of error in estimating decks remaining
    tc HL HL error KO KO error (HL/KO) error
    0 0 0 4*n - 4*dr 4*err 0
    1 dr err 4*n - 3*dr 3*err (1/3)
    2 2*dr 2*err 4*n - 2*dr 2*err 1
    3 3*dr 3*err 4*n - dr err 3
    4 4*dr 4*err 4n 0 infinite
    5 5*dr 5*err 4n + dr err 5
    6 6*dr 6*err 4n + 2*dr 2*err 3
    7 7*dr 7*err 4n + 3*dr 3*err (7/3)
    8 8*dr 8*err 4n + 4*dr 4*err 2
    Last edited by bjanalyst; 12-21-2018 at 08:30 PM.

  8. #8
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    One suggestion. Since you only use the adjustment for insurance and plays for a 12 total you should be side counting AA89mT The insurance play would be unaffected but the play on the 12 totals would be improved.

  9. #9


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    You are correct. Using AA89mTc with the KO, besides giving a prefect insurance decision, it also greatly improves hit/stand hard 12 v 2, 3, 4, 5 and 6 with a CC increase over the KO of 20% to 30%. A huge increase in power. Below crc(t) critical running count of true count of "t". crc(t) = 4*n + (t-4)*dr where n = number of decks and dr = decks remaining. I have created a table of critical running counts for the KO which I)s very easy to memorize because of patterns in the table. 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.

    hard 12 v 2 stand if KO + AA89mTc >= crc(4)
    hard 12 v 3 stand if KO + AA89mTc >= crc(2)
    hard 12 v 4 stand if KO + AA89mTc >= crc(0)
    hard 12 v 5 stand if KO + AA89mTc >= crc(-2)
    hard 12 v 6 stand if KO + AA89mTc >= crc(-1)

    So I hope the above explanation answers your question in more detail.

    Actually I use KO + k*(AA89mTc) for many different playing strategy decisions. Using k = 1 gives a special case of KO + k*(AA89mTc) which is KO + AA89mTc = Ten count (Tc). But you chose the values of k that maximizes the absolute value of the CC between then EoR and the tag values of the derived count and then use LSL to calculated AACpTCp which is then used with full deck house advantage for the given strategy change to calculate the infinite deck index for the given situation. As the CC increases, the indices for any given number of decks converge to the index for the infinite deck case which is what I analyzed. So I only gave you are very, very brief introduction using only k = 1 for insurance and hard 12 v 2, 3, 4, 5 and 6. My preference for the shoe game is KO with AA89mTc but I included HL with AA78mTc because many counters I meet do not want to switch to the unbalanced KO and want to stick with the HL. So attached is one of my spreadsheets for the sample calculations for doubling hard 9 against dealer's up card of 2 using Hi Opt 2 with side count of eights, HL with AA78mTc and KO with AA89mTc. Using plus minus side counts is easier than keeping side count of 8's and is more accurate as it is exact where 8 deficiency or excess is estimated and also both HL and KO are very easy level one count compared to the more complicated Hi Opt 2 level 2 count.

    So I have an entire table of adjustments for both HL + k*(AA78mTc) and KO + k*(AA89mTc). I just used insurance and hit/stand hard 12 v 2, 3, 4, 5 and 6 as those were the most important and easiest to use. Review attached PDF and you will see what I am talking about.

    Attached Images Attached Images
    Last edited by bjanalyst; 12-22-2018 at 10:25 AM.

  10. #10
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    Quote Originally Posted by bjanalyst View Post
    Actually I use KO + k*(AA89mTc) for many different playing strategy decisions. Using k = 1 gives a special case of KO + k*(AA89mTc) which is KO + AA89mTc = Ten count (Tc). But you chose the values of k that maximizes the absolute value of the CC between then EoR and the tag values of the derived count and then use LSL to calculated AACpTCp which is then used with full deck house advantage for the given strategy change to calculate the infinite deck index for the given situation.
    I do the same thing with my two balanced counts. It is pretty powerful for me. Kudos to maximizing the use of extra info gathered. To those that don't understand what he is talking about, he is using a number of different combined playing counts depending on which one is strongest for the matchup at hand. For traditional counting methods .7 is the limit to PE. By using multiple playing counts that ceiling is removed. With having the cards he does in the side count he can make strong index plays where none exist for traditional approaches, or an index exists but is very weak. Many matchups have the 7 and 8 as key cards for the play. When they act similarly, he can weight them properly relative to other cards to make the strongest index play possible for the counts he keeps. He would be altering the weight of the 7, 8, and A simultaneously.

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