Adding AA78mTc side count to High Low

[Norm]

I am amazed by supposedly educated and talented players making such ridiculous, untrue and ludicrous statements.

Count me in as uneducated and ludicrous. This strongly reminds me of discussions about a system many years back that was fine tuned with devices used for back of the envelop estimates and supposedly resulting in more "power" with sims that were not accurate comparisons. ET Fan was also involved with that and gave extremely poor advice on how to compare strategies.

[bjanalyst]

You are not answering my question. You are giving me critical running counts for KO. I am asking about balanced KO not KO. The tag values for KO and balanced KO is different because the KO count is being converted to equivalent balanced count. It is counting the 8s and 9s as -1/3 to make it balanced. You can't just true count KO without getting the running count tag values from balanced KO. You must count the tags in fractions for power.

The 8's and 9's are -(1/13) not -(1/3) in the balanced KO count. But the balanced KO count is equivalaent to the unblanaced KO count where the balanced KO count has integer tag values which humans can use. That the two counts are equivalent can be seen by CC(KO, KO.bal) = 100%.

No human would use the balanced KO count because you cannot count in fractional (1/13)th values. Also the balanced KO has a pivot of a true count of zero because it is balanced and so the advantages of true count accuracy at true counts around the KO pivot of a true count of 4 are lost with KO.bal. So when KO.bal > 0 or equivlaently when KO > 4*dp then there is a very slight excess of 8's and 9's left in the shoe because of the negative tag value of -(1/13)th for the 8's and 9's in the KO.bal count.

You had a question about calculation of the true count for the KO unbalanced count. If you look at the previous exhibits that I gave to you, you will that tc(KO.bal) = KO.bal / dr which you should have no problem with since KO.bal is a balanced count. Also KO.bal = KO - 4*dp also shown in a pervious exhibit. So tc(KO) = KO.bal / dr = (KO - 4*dp) / dr = (KO - 4*(n - dr) / dr = (KO + 4*dr - 4*n) / dr = 4 + (KO - 4*n) / dr where n = number of decks, dp = decks played and dr = decks remaining. That is how the unbalanced counts work.

[seriousplayer]

The 8's and 9's are -(1/13) not -(1/3) in the balanced KO count. But the balanced count is equivalaent to the unblanaced count with integer tag values as can be seen by CC(KO, KO.bal) = 100%. No human would use the balanced KO count because you cannot count in fractional (1/13)th values. Also the balaanced KO with have a pivot of a true count of zero and so the advantages of true count accuracy at true counts around the KO pivot of a true count of 4 are lost with KO.bal. So when KO.bal > 0 or equivlaently when KO > 4*dp then there is a very slight excess of 8's and 9's left in the shoe because of teh negative tag value of -(1/13)th for the 8's and 9's in the KO.bal count.

Than there is a flaw to your "Table of Critical Running Count". You can't figure out the values for KO.bal since no human would use the balanced KO count. This make your work junk. Do you have brain before posting such exhibit? There is a difference to balanced and unbalanced KO.

[bjanalyst]

You will need the running count from the count tag values for KO.bal. Otherwise, you can't apply the formula:

KO.bal + k1*(5m7c) +k2*(AA89mTc) ? (Idr) * dr

Without counting the KO.bal in fractions how do you figure out KO.bal? No, you can't just use the unbalanced KO tag values to plug in for KO.bal because KO and KO.bal count the 8s and 9s with different tag values.

When I originaly talked to Gronbog I had throught of him using the KO.bal in the simluatoin program but KO.bal is not my system. KO.bal is equivalent to the KO count but it is not what humans use. The computer has no problem with fractions but people do. People use the KO count with a unbalance of 4 per deck and so gives very accurate true counts around the pivot of a true count of 4. Using the balanced version of the KO count, KO.bal, your pivot is reset to a true count of zero and you would lose that accuracy of true counts around a true count of 4 that the unbalanced KO system has. You have to simulate the unbalanced KO system since that is what is used.

[bjanalyst]

Than there is a flaw to your "Table of Critical Running Count". You can't figure out the values for KO.bal since no human would use the balanced KO count. This make your work junk. Do you have brain before posting such exhibit? There is a difference to balanced and unbalanced KO.

The unbalanced KO count was used in simulations since that is the count that humans are actually using and that is the count that needs to be simulated.

There is no flaw. If my system was flawed then the simulations using the unbalanced KO count would not have shown that my system is superior to HO2 w ASC for back counting.

I will give you a few pages of proofs in a bit later as I have to leave right now.

[seriousplayer]

The unbalanced KO count was used in simulations since that is the count that humans are actually using and that is the count that needs to be simulated.

There is no flaw. If my system was flawed then the simulations using the unbalanced KO count would not have shown that my system is superior to HO2 w ASC for back counting.

I will give you a few pages of proofs in a bit later as I have to leave right now.

This is a flaw I just pointed it out. I am not talking about a flaw in mathematics. If you can understand English. I am talking about a flaw of being practical. You even admit yourself that the unbalanced KO and balanced KO is not the same because both have different pivots. There is no point simulating the balanced KO since as you said no human would use it. The same with the system you created for yourself no such human would use your system beside yourself.

[Dbs6582]

This is a flaw I just pointed it out. I am not talking about a flaw in mathematics. If you can understand English. I am talking about a flaw of being practical. You even admit yourself that the unbalanced KO and balanced KO is not the same because both have different pivots. There is no point simulating the balanced KO since as you said no human would use it. The same with the system you created for yourself no such human would use your system beside yourself.

And Carla...you’re forgetting Carla. She would use it too.

[Three]

And Carla...you’re forgetting Carla. She would use it too.

I would use any count if I can go to the casino and not risk any money but get to keep all the winnings from when they win. Best AP move there is. 0% chance of losing and will win on many occasions.

[refinery]

HO2 with ASC is VERY, VERY, VERY difficult which is why ho blackjack teams uses it.

You should probably refrain from commenting on what teams do or don't do and why. You are nowhere near qualified for that discussion.

[seriousplayer]

First as far as I am concerned the level 2 HO2 with ASC is VERY, VERY, VERY difficult which is why ho blackjack teams uses it. Level 2 counts are terrible with the HO2 counting the 4's and 5s as +2 and the Tens as -2 and 3, 4, 6 7 and +1 and they all must be done at the same time.

Your KO+5m7c+AA89mTc+b system is even more difficult than Hi-OPT II ASC. If Hi-OPT II ASC is difficult for you why don't you try to use other level 2 counts, like the Zen Count? Most definitely the Zen Count is much easier to count than the Hi-OPT II ASC. You don't know what you are talking about when you said level 2 counts are terrible with countings the 4's and 5s as +2. For stronger count systems you are suppose to give the 4's and 5s higher tag values than other cards because their EOR is higher.

I created a table of critical running counts form the formula tc(KO) = 4 + (KO - 4*n)/dr.

Also, why do you need the formula tc(KO) = 4 + (KO - 4*n)/dr for the balanced KO when it doesn't have an imbalanced of +4. Shouldn't you just divide the running count by the numbers of deck remaining to get the true count for the balanced KO? Maybe you can explain that to me. Your playing indices for balanced KO would be in fractions for: bal.KO = KO - (1/13)*(A23456789Tp) to get bal.KO. Wouldn't it be easier to just count the tag values and divide it the deck remaining?

Mistake number 1 you should use variables consistent with what you have in the legend. You have bal.KO in the legend but in the formula: KO.bal +k1 *(5m7c) +K2*(AA89mTc)>= (ldx)*dr you have KO.bal. bal.KO = (KO – 4*dp), what does KO.bal equal to? No, I am not a mind reader so I am not going to assume you mean bal.KO when you have KO.bal in the formula. Probably your formula should be:
(KO – 4*dp) +k1 *(5m7c) +K2*(AA89mTc)>= (ldx)*dr and not KO.bal +k1 *(5m7c) +K2*(AA89mTc)>= (ldx)*dr or maybe KO - (1/13)*(A23456789Tp) +k1 *(5m7c) +K2*(AA89mTc)>= (ldx)*dr . Variable KO.bal has not been defined in the chart.

You had a question about calculation of the true count for the KO unbalanced count. If you look at the previous exhibits that I gave to you, you will that tc(KO.bal) = KO.bal / dr which you should have no problem with since KO.bal is a balanced count. Also KO.bal = KO - 4*dp also shown in a pervious exhibit. So tc(KO) = KO.bal / dr = (KO - 4*dp) / dr = (KO - 4*(n - dr) / dr = (KO + 4*dr - 4*n) / dr = 4 + (KO - 4*n) / dr where n = number of decks, dp = decks played and dr = decks remaining. That is how the unbalanced counts work.

Again, tc(KO.bal) is not define in the chart. No, I don't know what you mean by KO.bal.

I think what you are trying to say is that the unbalanced KO equals the balanced KO. No, it does not because the balanced KO take into account fractions and the unbalanced KO doesn't. Mathematically, tc(KO) = KO.bal / dr = (KO - 4*dp) / dr = (KO - 4*(n - dr) / dr = (KO + 4*dr - 4*n) / dr = 4 + (KO - 4*n) / dr where n = number of decks, dp = decks played and dr = decks remaining but you fail to take into account that the running count would be in fractions for the balanced KO. The running count (KO - 4*dp) does not equal KO - (1/13)*(A23456789Tp) which gives bal.KO.

(KO - 4*dp) =??? KO - (1/13)*(A23456789Tp) which in terms equal to bal.KO.

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The Noir count counts all non-Tens as +1 and all Tens as -2.

Wrong!! The Noir count counts both the Aces and Tens as -2 all non-Tens as +1. The count that you mention that counts all non-Tens as +1 and all Tens as -2 is call Archer System, also known as, the one-two count.

[Dbs6582]

I would use any count if I can go to the casino and not risk any money but get to keep all the winnings from when they win. Best AP move there is. 0% chance of losing and will win on many occasions.

Good point Three! Sounds like Carla is at the top of the food chain when it comes to APing.

[Gronbog]

I was thinking the wonging criterion may not be equivalent giving one count an advantage over the other. No mention of wonging criterion for both counts. Do they wong in at .5% advantage and wong out when no advantage exists? Plus the software wongs in place rather than table hop so back counting is a misleading term.

The optimal entry count and unit size for the given spread are used in each case. Optimal bets are made at counts at or above the entry count up to the max for the given spread. No bets (i.e. bets of zero) are made at counts below the entry count. Yes, this is equivalent to wonging in place. The methodology and the use of the term back-counting are taken from results of the same nature in BJA3 chapter 10 and conform to the criteria for producing a SCORE in the strict meaning of the term.

[Three]

Thanks Gronbog. I am sure if you considered making entry point equivalent you did that.