Indices please - 8v4, 9v7

[k_c]

Before people start getting off track about uncounted cards, The book "Theory of Blackjack" by Peter Griffin (UCLA math professor and the math authority on blackjack) explains the mathematical proof behind cards that are not counted.

Uncounted cards remain normally distributed, no matter what the "count" is.

In a 6 deck shoe, there are 6 x 4 = 24 8s at the top of shoe. Suppose 3 decks are dealt out. Without counting anything, it is safe to assume there are 12 8s left in the half shoe. HiLo counts the 8 as 0, so it is essentially uncounted.

If the HiLo count was a really high +9? It is most likely there are 12 8s left. If the HiLo count was a really low -9? Most likely still 12 8s left.
The "density" of uncounted cards remains normal.

Also note if you are using KO, where the 7 has 1/2 the value of the other positive cards, in a high negative count KO is only half as effective in predicting a 7 as another low card. So in the classic 14 vs T situation, where 7 is the key card, HiLo is completely worthless for a index, and KO isn't much better.

My first efforts analyzing blackjack used infinite shoe composition. I assumed all uncounted cards to be 1/13 of the total cards the same as in a full shoe. So for Hi-Lo the high and low cards could vary from 0-10/13 of the total cards and the percent of 7-8-9 was always assumed to be 1/13 (each) of the total cards. Indices generated using this method are somewhat in accordance to what is published.

More recently I got into computing actual EVs for any shoe composition for a finite number of decks. The hardest part was to be able to do the computing in a reasonable amount of time. Along the way I wrote a program that used a weighted average of all of the possible subsets for a given Hi-Lo or KO count at any given penetration. It turns out that the percent of uncounted cards is equal to 1/13 only when 50% of the shoe is left to be dealt. Below shows a Hi-Lo RC of +4 for a single deck at 3 different penetrations and lists the probability of each rank. I've used this program to refine insurance indices for Hi-Lo and true counted KO to take into account the hand composition. Anyway the percent of uncounted cards isn't always exactly 1/13.

I don't know why I lean on exact calculation analysis, but that's my approach.

Cards in deck=13 (TC=16.0)
    p(2) = 0.04662
    p(3) = 0.04662
    p(4) = 0.04662
    p(5) = 0.04662
    p(6) = 0.04662
    p(7) = 0.07537 (<1/13)
    p(8) = 0.07537 (<1/13)
    p(9) = 0.07537 (<1/13)
    p(10) = 0.43264
    p(1) = 0.10816

Cards in deck=26 (TC=08.0)
    p(2) = 0.06154
    p(3) = 0.06154
    p(4) = 0.06154
    p(5) = 0.06154
    p(6) = 0.06154
    p(7) = 0.07692 (=1/13)
    p(8) = 0.07692 (=1/13)
    p(9) = 0.07692 (=1/13)
    p(10) = 0.36923
    p(1) = 0.09231

Cards in deck=39 (TC=05.3)
    p(2) = 0.06651
    p(3) = 0.06651
    p(4) = 0.06651
    p(5) = 0.06651
    p(6) = 0.06651
    p(7) = 0.07744 (>1/13)
    p(8) = 0.07744 (>1/13)
    p(9) = 0.07744 (>1/13)
    p(10) = 0.34810
    p(1) = 0.08702

 

[cardcounter0]

"Anyway the percent of uncounted cards isn't always exactly 1/13"

Then you are doing something wrong, unless you can explain why a 7, 8, or 9 is more likely or less likely to be dealt out because a lot of Aces came out.

OR try this:

Cut 1/4 of a deck off. Perform a HiLo count on this 1/4 deck that you removed. If the count is +5 or -5 or 0, what assumption can you make about the number of 8s left in the remaining 3/4 of the deck?

According to probability, there should be 3 8s left in the 3/4 deck pack.
Since the HiLo count does not count 8s, you cannot make any other assumption. Because if you removed a bunch of low cards in the 1/4 deck, or removed a bunch of high cards, that event is independent of how many 8s were removed. You have to logically assume that with a deck with four 8s, there is one 8 per 1/4 deck.

If you are trying to do any type of simulations, you need to read "Theory of Blackjack" by Peter Griffin. He came up with all these problems and developed the mathematical formulas to solve them 30 years ago.

 

[cardcounter0]

I think I see the problem:

Cards in deck=13 (TC=16.0)
p(2) = 0.04662
p(3) = 0.04662
p(4) = 0.04662
p(5) = 0.04662
p(6) = 0.04662
p(7) = 0.07537 (<1/13)
p(8) = 0.07537 (<1/13)
p(9) = 0.07537 (<1/13)
p(10) = 0.43264
p(1) = 0.10816

You cannot have .04662 2s in a 13 card subset. You can only have 0, 1, 2, 3, 4, .... 2s. You will never encounter a subset with .04662 2s.

Generate all the possible combinations of 13 card subsets that have a tc of 16. Some of these subsets will have zero 8s, one 8, two 8s, etc. Sum the number of 8s, divide by the total combinations, and if you do not come up with 1 -- you are missing some combinations.

 

[k_c]

"Anyway the percent of uncounted cards isn't always exactly 1/13"

Then you are doing something wrong, unless you can explain why a 7, 8, or 9 is more likely or less likely to be dealt out because a lot of Aces came out.

OR try this:

Cut 1/4 of a deck off. Perform a HiLo count on this 1/4 deck that you removed. If the count is +5 or -5 or 0, what assumption can you make about the number of 8s left in the remaining 3/4 of the deck?

According to probability, there should be 3 8s left in the 3/4 deck pack.
Since the HiLo count does not count 8s, you cannot make any other assumption. Because if you removed a bunch of low cards in the 1/4 deck, or removed a bunch of high cards, that event is independent of how many 8s were removed. You have to logically assume that with a deck with four 8s, there is one 8 per 1/4 deck.

If you are trying to do any type of simulations, you need to read "Theory of Blackjack" by Peter Griffin. He came up with all these problems and developed the mathematical formulas to solve them 30 years ago.

I've read Theory of Blackjack.

I'm not doing anything wrong.

If you consider all of the possible subsets for a given running count/penetration and take a weighted average of them you get the probability of drawing an uncounted card=1/13 only when 50% of the cards in a full shoe are removed or when there is a full shoe. I don't explain things all that well but here is a thread in which I tried to explain - http://www.blackjackinfo.com/bb/showthread.php?t=8440

 

[UK-21]

Hi,

I wasn't going to post anything further on this, but after reading through the thread several times over I decided to throw in a last few lines.

Having thought about the responses, I think I've been confusing the issues of best plays, based on mathematical probability, and risk (the nasty four letter word). We have an established BS table, based on maths, that shows the proven best plays. Nothing equivalent though for risk. Risk is a personal perspective. The level of risk one is prepared to shoulder is reflected in one's own risk aversion profile, comfort zone - call it what you will. Although everyone's will be different, just like a$$holes, everyone has one and some people will have larger ones than others.

Back to my question - doubling 9 v 2,3,4? The index for 9v2 is +1, and 9v3,4 is a BS play. At this point, if playing to a classic spread, the play would be made with only 1 unit in the box - so doubling it risks 2. At +4 or greater, the max will be out so doubling risks 16 or 32 units. Assuming that one plays with a finite number of bets in hand the potential loss increases by a factor of 8 or 16. In my case, where I normally enter the House of Chance with 50 units in hand, a 16 unit loss represents more than 10% of the session role. This would be an acceptable level of risk when playing with, say, a 10 or A v a dealers 5, where the probabilities of making a 20/21 are higher than the dealer making a better hand, but perhaps not so when playing an 8 or 9 v a dealer 2,3,4 where the possibility of losing increases.

As Kewljason has written, it may not be mathematically optimal but would keep someone (me?) within their particular comfort zone. The other issue to consider of course, and which has been highlighted, is that doubling at max bets increases variance (and thereby risk) which needs to be considered when playing with a limited resource. So, to sum up, I think that there is a balance to be struck between the optimal, maths proven, plays and the risk consideration when taking into account the bankroll. Sorry if I've upset any of the purists by saying so.

That's it from me on this one (thread). Points noted and thanks for the contributions, which have been helpful.

 

[k_c]

I think I see the problem:

Cards in deck=13 (TC=16.0)
p(2) = 0.04662
p(3) = 0.04662
p(4) = 0.04662
p(5) = 0.04662
p(6) = 0.04662
p(7) = 0.07537 (<1/13)
p(8) = 0.07537 (<1/13)
p(9) = 0.07537 (<1/13)
p(10) = 0.43264
p(1) = 0.10816

You cannot have .04662 2s in a 13 card subset. You can only have 0, 1, 2, 3, 4, .... 2s. You will never encounter a subset with .04662 2s.

Generate all the possible combinations of 13 card subsets that have a tc of 16. Some of these subsets will have zero 8s, one 8, two 8s, etc. Sum the number of 8s, divide by the total combinations, and if you do not come up with 1 -- you are missing some combinations.

I'm saying the overall probability of drawing a 2 from a +4 RC 13-card slug is .04662 = 4.662%.

If there are 13 cards left and Hi-Lo running count is +4, the possible card distributions are:

(2-6)  (7-9)  (10-A)
4      1      8
3      3      7
2      5      6
1      7      5

Each rank has a probability of being drawn depending on what subset is being referenced. Also each subset has a probability of occurring. When you weight the probabilities of a rank being drawn by the probabilities of the subsets occurring, the probability of a rank being drawn for the condition of running count = +4 with 13 cards remaining can be computed.

Simple arbitrary example:
You have 2 subsets
Subset 1 occurs 4 times as often as subset 2
In subset 1 the probabilty of drawing a 7 is .1=10%
In subset 2 the probability of drawing a 7 is .2=20%
Subset 1 weight = 1
Subset 2 weight = 4
Subset 1 weight + Subset 2 weight = 5
Overall prob of drawing a 7 = .1*1/5+.2*4/5 = .18=18%

 

[cardcounter0]

I mispoke since you are dealing with a finite set of cards.

Consider a full deck, the ratio of 8s is 4/52.
You remove a card and the count is +1 the ratio is 4/51
You remove a card and the count is -1 the ratio is 4/51
You remove a card and the count is 0 the ratio is 3.6666/51
There is a change in weighting, even if the card is not counted.

I think you have captured the correct ratios, I would just do some math to make sure you are capturing all the possible combinations. 52 choose 13 creates a huge amount of combos, even if you only filter those that have a count of +4.

 

[k_c]

I mispoke since you are dealing with a finite set of cards.

Consider a full deck, the ratio of 8s is 4/52.
You remove a card and the count is +1 the ratio is 4/51
You remove a card and the count is -1 the ratio is 4/51
You remove a card and the count is 0 the ratio is 3.6666/51
There is a change in weighting, even if the card is not counted.

I think you have captured the correct ratios, I would just do some math to make sure you are capturing all the possible combinations. 52 choose 13 creates a huge amount of combos, even if you only filter those that have a count of +4.

Well I wrote a program that does the math. That's where my figures come from. It has to handle very large numbers that are well outside the range used in normal computing. It could probably be more efficient but it works.