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Originally Posted by
Freightman
Don’t thank me so quickly. Your posts are repetitive, monotonous and absolutely maddening. I as simply saying that, buried within your lengthy novellas, are some advanced concepts. You need to simplify your absurdly complex posts into short simple easy to understand comments. You fail miserably at that task.
I’m loathe to say this, fearing an avalanche of verbosity, however, it seems that your research does not touch on an important area of EV maximization, coupled with variance reductions. Perhaps the unnamed poster from post 7 can spout some additional nonsense in his/her quest Of pr9mot8ng his mediocrity.
i always try to increase expectation and reduce variance.
Doubling increases variance so you better absolutely have the advantage when you double. That is why whenever I double or split I want to make sure that I am one or two true count points over the expected value index for doubling or splitting. For example, doubling hard 8 v 6 at KO = crc(2). But at exactly KO = crc(2), the expected value of doubling and just hitting are equal. That is, if you double at exactly the expected value index you do not increase your expectation but you do increase risk. That is why I add one or two true count point \to the expected value doubling or splitting indices. Thus for hard 8 v 6 double , instead of doubling when KO = crc(2), I wait unit KO = crc(3) or even KO = crc(4). But you already know all of this as that is what is called Risk Averse indices which have been widely talked about.
Now let's talk about the situation that happened to me a few days ago and also risk averse indices, which I sort of touched on, for hard 11 v 9 and hard 11 v T doubling.
Actually i use brc = betting running count = KO + (1/2)*(5m9c) which has a BCC = 98.6% as compared to the KO which has a BCC of 96.5% which also happens to be the BCC of the HL for the S17, DAS, LS game. Remember, I keep the KO with AA89mTc and 5m9c where the 5m9c helps with betting and some playing strategy variations as well. But for simplicity I will ignore 5m9c for betting in the discussion below so I can concentrate on what happened.
In this situation I came across I had my maximum bet out of $100 on two hands because KO = crc(4) = 4*n = 24 for n = 6 decks.
I then had a hard 11 v 9 which you would normally just double without even thinking. Now i keep AA89mTc and 5m9c side counts with chips. The situation was n = 6 decks, dp = 4, KO = 24, 5m9c = 2 and AA89mTc = (-24). (And just a note here, brc = KO + (1/2)*(5m9c) = 24 + (1/2)*(2) = 25 so I had a bit more advantage than just KO = crc(4)). I glanced at my AA89mTc stack of chips and I use one green chip to replace 5 red chips in my stack of chips when the AA89mTc because large. So my AA89mTc stack of chips was to the left of my betting chips which indicates a negative AA89mTc. I noticed five green chips and nine red chips in my AA89mTc stack of chips. So that made the AA89mTc = (-24) since each green chip was (-5) since the AA89mTc was on the left.
So I immediately knew that I was in trouble here. With such a large negative AA89mTc I knew there was a great deficiency of Tens and excess of Ace which would make doubling hard 11 v 9 very risky. But because I never thought there would be a situation where you would not double hard 11 v 9 I had never analyzed it before. Since I had not analyzed it I really did not know what to do so I doubled my hard 11 v 9 and surprise, surprise I picked up one of the extra Aces and lost the doubled $200 bet. I should have known that this was a risky double and just hit but I didn't.
So when I got home I analyzed doubling hard 11 v 9 and hard 11 v T and what I found was that you should hit instead of doubling whenever (AA89mTc + 5m9c) < (-6 - tc(KO)*dr.
I already explained that the HL has a CC of only around 60% for these doubles. The KO w AA89mTc and 5m9c has a CC of 89% for hard 11 v 9 double and 96% for hard 11 v T double so the KO with AA89mTc and 5m9c will catch these situations which you would never catch with the HL that has an index of (-6) for hitting hard 11 v 9 or hard 11 v T. Obviously you would never be playing at such negative counts and so if you were just using the HL you would always double hard 11 v 9 and hard 11 v T.
But I am using KO w AA89mTc and 5m9c so I have these side counts anyhow so I might as well use them for this rare situation. This situation is very rare but can happen as it did to me and I misplayed this hand by doubling my maximum bet of $100 and lost $200 because I wasn't paying close enough attention and I never derived the formula for not doubling hard 11 v 9 and hard 11 v T since I thought it would never happen.
Here is an explanation of why this can occur and did occur to me a few days ago;
In the 6 deck game with 5 decks dealt, maximum ABS(HL) = 30. (AA89mTc + 5m9c) = (AA58mTc). The SD(AA89mTc) = SD(AA58mTc) = SD(HL). Thus in the 6 deck game with 5 decks dealt, maximum ABS(AA89mTc) = maximum ABS(AA58mTc) = maximum ABS(HL) = 30. Also CORREL(AA89mTc, KO) = 19.1% and CORREL(AA59mTc, KO) = 28.7%. Therefore AA58mTc, which is used in psrc = playing strategy running count = KO + AA89mTc + 5m9c = KO + AA58mTc for doubling hard 11 v 9, T and A, is relatively independent of KO and so KO.balanced = KO - 4*dp can be a large positive number while AA58mTc can be a large negative number. This is what happened when for n = 6 decks and dp = 4, AA89mTc = (-24) and 5m9c = (+2) so that (AA89mTc + 5m9c) = AA58mTc = (-22) with dr = 2 while tc(KO) = crc(4) = 24 and so KO.balanced = KO - 4*dp = 24 - 4*4 = 8 so that tc(KO) = (KO - 4*dp) / dr = (24 - 4*4) / 2 = 8 / 2 = 4. AA58mTc was very negative at (-22) while KO.balanced was positive at (+8) which is possible because these two counts are very loosely correlated with a CC of only 28.7%.
So for the risk averse player that I alluded to in a previous post here, if (AA89mTc + 5m9c) was very negative then I would avoid doubling hard 11 v 9 and hard 11 v T as I do not want to lose twice my maximum bet by losing this double.
So the exact formula when to hit instead of doubling is hit if (AA89mTc + 5m9c) < (-6 - tc(KO))*dr. So if KO = crc(2), for example, you would hit instead of doubling when (AA89mTc + 5m9c) < (-6 - 2)*dr = (-8)*dr and if I wanted to be risk averse in this situation I would hit instead of doubling when (AA89mTc + 5m9c) < (-6)*dr.
So there you have it. Risk Averse indices for hitting hard 11 v 9 and hard 11 v T when using KO with AA89mTc and 5m9c. And that is what you want which is to maximize expectation and reduce risk.
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