1. Did you find this post helpful? Yes | No
Originally Posted by Cacarulo
I see a problem in your shoe composition: for Hi-Lo: 7, 8 and 9 should have the same value. Same for 2, 3, 4, 5 and 6. The ten should be four times the value of the Ace.
Notice that your seven is different from the eight and the nine.

This is my shoe composition for a RC = -13 and 52 cards remaining:
Shoe comp (A-T): {2.7, 5.3, 5.3, 5.3, 5.3, 5.3, 4.0, 4.0, 4.0, 10.8}

Sincerely,
Cac
I am allowing for the specific removal of one ten and one seven since that is the composition of the input hand. It would be better to allow for the specific removal of each up card individually also but I don't presently do that except for insurance where it is a given that an ace is the up card.

I then weight each possible subset of 52 cards that has a running count of -13 and compute the prob of each rank.

Code:
Count tags {1,-1,-1,-1,-1,-1,0,0,0,1}
Decks: 2
Cards remaining: 52
Initial running count (full shoe): 0
Running count: -13
Specific removals (1 - 10): {0,0,0,0,0,0,1,0,0,1}

Subgroup removals: None

Number of subsets for above conditions: 12
Prob of running count -13 with above removals from 2 decks: 0.0017529

p[1] 0.053838  p[2] 0.10249  p[3] 0.10249  p[4] 0.10249  p[5] 0.10249
p[6] 0.10249  p[7] 0.068502  p[8] 0.078288  p[9] 0.078288  p[10] 0.20862

Press any key to continue:
Multiplying each of above probs by 52 should result in the shoe comp before up card.

k_c

2. Did you find this post helpful? Yes | No
Originally Posted by k_c
I am allowing for the specific removal of one ten and one seven since that is the composition of the input hand. It would be better to allow for the specific removal of each up card individually also but I don't presently do that except for insurance where it is a given that an ace is the up card.

I then weight each possible subset of 52 cards that has a running count of -13 and compute the prob of each rank.

Code:
Count tags {1,-1,-1,-1,-1,-1,0,0,0,1}
Decks: 2
Cards remaining: 52
Initial running count (full shoe): 0
Running count: -13
Specific removals (1 - 10): {0,0,0,0,0,0,1,0,0,1}

Subgroup removals: None

Number of subsets for above conditions: 12
Prob of running count -13 with above removals from 2 decks: 0.0017529

p[1] 0.053838  p[2] 0.10249  p[3] 0.10249  p[4] 0.10249  p[5] 0.10249
p[6] 0.10249  p[7] 0.068502  p[8] 0.078288  p[9] 0.078288  p[10] 0.20862

Press any key to continue:
Multiplying each of above probs by 52 should result in the shoe comp before up card.

k_c
This topic is really very interesting and I must admit that it caught my attention. It is always enriching to see how other researchers arrive at solutions through different paths and what I try to understand is why sometimes we do not agree. In no way do I want this to be understood as a criticism. It is not.
Having said this, I have a few questions.
Apparently you would be calculating an index for T7vA that is not the same as for a generic 17 (which would also involve a 98vA and even more than 2 cards). For T7vA we would need to remove 3 cards (T, 7 and A) but in the case of a 17vA only one Ace would need to be removed. But what bothers me is that for the calculation of the index you would be removing only the T and the 7 but not the Ace.
Trying to reproduce the composition of your shoes, I don't agree with them either. These are my shoe compositions:

a) RC = -13 and 52 cards remaining without removing any cards.

2.700000 5.300000 5.300000 5.300000 5.300000 5.300000 4.000000 4.000000 4.000000 10.800000

b) RC = -13 and 52 cards remaining removing only one ace.

2.118712 5.283702 5.283702 5.283702 5.283702 5.283702 4.054326 4.054326 4.054326 11.299799

c) RC = -13 and 52 cards remaining removing an ace, a ten and a seven.

2.287944 5.345533 5.345533 5.345533 5.345533 5.345533 3.148547 4.198062 4.198062 11.439720

d) RC = -13 and 52 cards remaining, removing only a ten and a seven which, from what I understand, would be your case.

2.908158 5.362750 5.362750 5.362750 5.362750 5.362750 3.101591 4.135454 4.135454 10.905593

BTW, this does not mean that I am correct.

Sincerely,
Cac

3. Did you find this post helpful? Yes | No
Originally Posted by Cacarulo
c) RC = -13 and 52 cards remaining removing an ace, a ten and a seven.

2.287944 5.345533 5.345533 5.345533 5.345533 5.345533 3.148547 4.198062 4.198062 11.439720

Cac
This is very thoughtful! You considered all possible scenarios: T-7vA, 9-8vA.
I am learning about the index generating process. At RC=-13 and when there are 52 cards remaining, there are millions of possible integer deck-compositions, but you listed a certain decimal deck-composition. Does this mean you averaged by some means of these millions of possible deck-compositions?

4. Did you find this post helpful? Yes | No
Originally Posted by Cacarulo
This topic is really very interesting and I must admit that it caught my attention. It is always enriching to see how other researchers arrive at solutions through different paths and what I try to understand is why sometimes we do not agree. In no way do I want this to be understood as a criticism. It is not.
Having said this, I have a few questions.
Apparently you would be calculating an index for T7vA that is not the same as for a generic 17 (which would also involve a 98vA and even more than 2 cards). For T7vA we would need to remove 3 cards (T, 7 and A) but in the case of a 17vA only one Ace would need to be removed. But what bothers me is that for the calculation of the index you would be removing only the T and the 7 but not the Ace.
Trying to reproduce the composition of your shoes, I don't agree with them either. These are my shoe compositions:

a) RC = -13 and 52 cards remaining without removing any cards.

2.700000 5.300000 5.300000 5.300000 5.300000 5.300000 4.000000 4.000000 4.000000 10.800000

b) RC = -13 and 52 cards remaining removing only one ace.

2.118712 5.283702 5.283702 5.283702 5.283702 5.283702 4.054326 4.054326 4.054326 11.299799

c) RC = -13 and 52 cards remaining removing an ace, a ten and a seven.

2.287944 5.345533 5.345533 5.345533 5.345533 5.345533 3.148547 4.198062 4.198062 11.439720

d) RC = -13 and 52 cards remaining, removing only a ten and a seven which, from what I understand, would be your case.

2.908158 5.362750 5.362750 5.362750 5.362750 5.362750 3.101591 4.135454 4.135454 10.905593

BTW, this does not mean that I am correct.

Sincerely,
Cac

I don't have a lot of time right now but maybe I can slowly get started.

First we agree on the composition with nothing specifically removed. My method shows that exactly at midshoe with nothing removed the probability of a zero tagged card = 1/13. At any other point other than full shoe this is not true by my method.

a couple of things to note:
-there are 12 52-card HiLo count subsets that compute to a running count of -13
21(2-6) 23(7-9) 8(T-A)
22(2-6) 21(7-9) 9(T-A)
23(2-6) 19(7-9) 10(T-A)
24(2-6) 17(7-9) 11(T-A)
25(2-6) 15(7-9) 12(T-A)
26(2-6) 13(7-9) 13(T-A)
27(2-6) 11(7-9) 14(T-A)
28(2-6) 9(7-9) 15(T-A)
29(2-6) 7(7-9) 16(T-A)
30(2-6) 5(7-9) 17(T-A)
31(2-6) 3(7-9) 18(T-A)
32(2-6) 1(7-9) 19(T-A)
-specific removal of one or more of a rank means that the max number of that rank that can be in any subset is reduced from the number of that rank present in a full shoe

I would be happy to find some other way. My question to you is how did you come to the value of 4.054326 for number of (7-9) for RC of -13 with 52 cards remaining given 1 ace has been specifically removed? My value is 4.028648.

Thanks,

k_c

5. Did you find this post helpful? Yes | No
Originally Posted by k_c
I don't have a lot of time right now but maybe I can slowly get started.

First we agree on the composition with nothing specifically removed. My method shows that exactly at midshoe with nothing removed the probability of a zero tagged card = 1/13. At any other point other than full shoe this is not true by my method.

a couple of things to note:
-there are 12 52-card HiLo count subsets that compute to a running count of -13
21(2-6) 23(7-9) 8(T-A)
22(2-6) 21(7-9) 9(T-A)
23(2-6) 19(7-9) 10(T-A)
24(2-6) 17(7-9) 11(T-A)
25(2-6) 15(7-9) 12(T-A)
26(2-6) 13(7-9) 13(T-A)
27(2-6) 11(7-9) 14(T-A)
28(2-6) 9(7-9) 15(T-A)
29(2-6) 7(7-9) 16(T-A)
30(2-6) 5(7-9) 17(T-A)
31(2-6) 3(7-9) 18(T-A)
32(2-6) 1(7-9) 19(T-A)
-specific removal of one or more of a rank means that the max number of that rank that can be in any subset is reduced from the number of that rank present in a full shoe

I would be happy to find some other way. My question to you is how did you come to the value of 4.054326 for number of (7-9) for RC of -13 with 52 cards remaining given 1 ace has been specifically removed? My value is 4.028648.

Thanks,

k_c
Take your time, I'm not in a hurry.
We both agree that there are 12 subsets that make up -13. Regarding the difference between 4.054326 and 4.028648, I would have to check my code since I haven't touched it for more than 20 years.
As I told you before, I am not 100% sure that my number is correct. The algorithm was never compared to other researchers as no one was interested in subset generation at the time.
What my algorithm does is to generate a subset given the following data: RC, remaining cards, and a card counting system. Before generating it I have the possibility to remove the cards that I want. Although the algorithm works perfectly, it does not mean that what it generates is correct.

For checking purposes and whenever you have time, could you pass me your generated subset with an Ace removed for an RC of +1 and 38 cards remaining? This would be SD.

Thank you.

Sincerely,
Cac

6. Did you find this post helpful? Yes | No
Originally Posted by k_c
-there are 12 52-card HiLo count subsets that compute to a running count of -13
21(2-6) 23(7-9) 8(T-A)
22(2-6) 21(7-9) 9(T-A)
23(2-6) 19(7-9) 10(T-A)
24(2-6) 17(7-9) 11(T-A)
25(2-6) 15(7-9) 12(T-A)
26(2-6) 13(7-9) 13(T-A)
27(2-6) 11(7-9) 14(T-A)
28(2-6) 9(7-9) 15(T-A)
29(2-6) 7(7-9) 16(T-A)
30(2-6) 5(7-9) 17(T-A)
31(2-6) 3(7-9) 18(T-A)
32(2-6) 1(7-9) 19(T-A)

k_c
Let me try to calculate the frequency of the first subgroup of the above 12 that you listed.
For this subgroup 21(2-6) 23(7-9) 8(T-A), we have this many possible card arrangements:
[C(40, 21) xC(24, 23) xC(40,8)] xP(52,52)
= [1.3x10^11 x24 x7.7x10^7] x8.1x10^67
=2.4x10^20 x8.1x10^67.

I start to get what you are talking about. Is my above calculation correct?

The most likely subgroups are these two:
26(2-6) 13(7-9) 13(T-A)
27(2-6) 11(7-9) 14(T-A)

If we average only these two, we get the most likely deck composition as follows,

26.5(2-6) 12(7-9) 13.5(10-A)

7. Did you find this post helpful? Yes | No
Originally Posted by Cacarulo
Take your time, I'm not in a hurry.
We both agree that there are 12 subsets that make up -13. Regarding the difference between 4.054326 and 4.028648, I would have to check my code since I haven't touched it for more than 20 years.
As I told you before, I am not 100% sure that my number is correct. The algorithm was never compared to other researchers as no one was interested in subset generation at the time.
What my algorithm does is to generate a subset given the following data: RC, remaining cards, and a card counting system. Before generating it I have the possibility to remove the cards that I want. Although the algorithm works perfectly, it does not mean that what it generates is correct.

For checking purposes and whenever you have time, could you pass me your generated subset with an Ace removed for an RC of +1 and 38 cards remaining? This would be SD.

Thank you.

Sincerely,
Cac

Here are the rank probs I get for single deck, RC=+1, 38 cards remaining, 1 ace specifically removed
If you want number of rank just multiply each rank prob by 38.

Code:
Count tags {1,-1,-1,-1,-1,-1,0,0,0,1}
Decks: 1
Cards remaining: 38
Initial running count (full shoe): 0
Running count: 1
Specific removals (1 - 10): {1,0,0,0,0,0,0,0,0,0}

Subgroup removals: None

Number of subsets for above conditions: 6
Prob of running count 1 with above removals from 1 deck: 0.11848

p[1] 0.062504  p[2] 0.073909  p[3] 0.073909  p[4] 0.073909  p[5] 0.073909
p[6] 0.073909  p[7] 0.078199  p[8] 0.078199  p[9] 0.078199  p[10] 0.33336

Press any key to continue:
It helps me to solve a simple problem and then work to apply to other problems. Below is data for the simplest subset (1 card) dealt from 2 decks with various specific removals:

Code:
2 decks, HiLo

Cards           Specific
Remaining   RC  Removals  Rank Probs                                                       Prob of Subset

1           -1  none      p[1] 0  p[2] 0.2  p[3] 0.2  p[4] 0.2  p[5] 0.2                   0.38462
p[6] 0.2  p[7] 0  p[8] 0  p[9] 0  p[10] 0

1           -1  A         p[1] 0  p[2] 0.2  p[3] 0.2  p[4] 0.2  p[5] 0.2                   0.38835
p[6] 0.2  p[7] 0  p[8] 0  p[9] 0  p[10] 0

1           -1  A,7       p[1] 0  p[2] 0.2  p[3] 0.2  p[4] 0.2  p[5] 0.2                   0.39216
p[6] 0.2  p[7] 0  p[8] 0  p[9] 0  p[10] 0

1           -1  A,7,T     p[1] 0  p[2] 0.2  p[3] 0.2  p[4] 0.2  p[5] 0.2                   0.39604
p[6] 0.2  p[7] 0  p[8] 0  p[9] 0  p[10] 0

1           -1  7,T       p[1] 0  p[2] 0.2  p[3] 0.2  p[4] 0.2  p[5] 0.2                   0.39216
p[6] 0.2  p[7] 0  p[8] 0  p[9] 0  p[10] 0

1           -1  2         p[1] 0  p[2] 0.17949  p[3] 0.20513  p[4] 0.20513  p[5] 0.20513   0.37864
p[6] 0.20513  p[7] 0  p[8] 0  p[9] 0  p[10] 0
______________________________________________________________________________________________________

1           0   none      p[1] 0  p[2] 0  p[3] 0  p[4] 0  p[5] 0                           0.23077
p[6] 0  p[7] 0.33333  p[8] 0.33333  p[9] 0.33333  p[10] 0

1           0   A         p[1] 0  p[2] 0  p[3] 0  p[4] 0  p[5] 0                           0.23301
p[6] 0  p[7] 0.33333  p[8] 0.33333  p[9] 0.33333  p[10] 0

1           0   A,7       p[1] 0  p[2] 0  p[3] 0  p[4] 0  p[5] 0                           0.22549
p[6] 0  p[7] 0.30435  p[8] 0.34783  p[9] 0.34783  p[10] 0

1           0   A,7,T     p[1] 0  p[2] 0  p[3] 0  p[4] 0  p[5] 0                           0.22772
p[6] 0  p[7] 0.30435  p[8] 0.34783  p[9] 0.34783  p[10] 0

1           0   7,T       p[1] 0  p[2] 0  p[3] 0  p[4] 0  p[5] 0                           0.22549
p[6] 0  p[7] 0.30435  p[8] 0.34783  p[9] 0.34783  p[10] 0

1           0   2         p[1] 0  p[2] 0  p[3] 0  p[4] 0  p[5] 0                           0.23301
p[6] 0  p[7] 0.33333  p[8] 0.33333  p[9] 0.33333  p[10] 0
_______________________________________________________________________________________________________

1           1   none     p[1] 0.2  p[2] 0  p[3] 0  p[4] 0  p[5] 0                          0.38462
p[6] 0  p[7] 0  p[8] 0  p[9] 0  p[10] 0.8

1           1   A        p[1] 0.17949  p[2] 0  p[3] 0  p[4] 0  p[5] 0                      0.37864
p[6] 0  p[7] 0  p[8] 0  p[9] 0  p[10] 0.82051

1           1   A,7      p[1] 0.17949  p[2] 0  p[3] 0  p[4] 0  p[5] 0                      0.38235
p[6] 0  p[7] 0  p[8] 0  p[9] 0  p[10] 0.82051

1           1   A,7,T    p[1] 0.18421  p[2] 0  p[3] 0  p[4] 0  p[5] 0                      0.37624
p[6] 0  p[7] 0  p[8] 0  p[9] 0  p[10] 0.81579

1           1   7,T      p[1] 0.2  p[2] 0  p[3] 0  p[4] 0  p[5] 0                          0.38235
p[6] 0  p[7] 0  p[8] 0  p[9] 0  p[10] 0.8

1           1   2        p[1] 0.2  p[2] 0  p[3] 0  p[4] 0  p[5] 0                          0.38835
p[6] 0  p[7] 0  p[8] 0  p[9] 0  p[10] 0.8
Finally you asked why I do not include up card as a specific removal when generating indexes. I agree that would be better but my CA computes EVs for an input hand comp for all up cards. When I went into generating indexes I adapted to my CA as is. I do include an ace up card as a specific removal as well as hand comp when generating insurance indexes, however.

Basically what I do is to weight each of the possible count subsets by their probabilities. There was a previous thread that asked about the probability of a running count at a given pen. I added this page to my website and it includes how I compute subset probs. I use these to compute rank probs. http://www.bjstrat.net/RC_prob.html

It would be nice to somehow simplify this method somehow though. Maybe yours could.

k_c

8. Did you find this post helpful? Yes | No
Originally Posted by k_c
Basically what I do is to weight each of the possible count subsets by their probabilities. There was a previous thread that asked about the probability of a running count at a given pen. I added this page to my website and it includes how I compute subset probs. I use these to compute rank probs. http://www.bjstrat.net/RC_prob.html

It would be nice to somehow simplify this method somehow though. Maybe yours could.

k_c
Wonderful work! I’ve read the details you posted. Let me just do a little rough math to estimate the whole thing.

The probability of the subgroup 26(2-6) 13(7-9) 13(T-A) is

[C(40, 26) xC(24,13) xC(40,13)] /C(104,52)
= [(2.3x10^10) x(2.5x10^6) x(1.2x10^10)] /(1.6x10^30)
=4.3x10^(-4).

The probability of the RC=-13 can be approximated as
2 x4.3x10^(-4)= 8.6x10^(-4)
These numbers sound good for me. Thank you!

9. Did you find this post helpful? Yes | No
Based on all these considerations, I propose these two HiLo surrender/stand indices for the hand 17vsA:
For the hand T-7vsA, TC=+1;
For the hand 9-8vsA, TC=+2.

10. Did you find this post helpful? Yes | No
After reviewing my code I realized that the method I was using was a Gaussian approximation. Back in the days when computers weren't that fast, doing it under strict combinatorial analysis took a lot of processing time, especially in shoes. In fact, to obtain the probability of a certain RC with N cards remaining, it was much faster to use a Normal approximation (there is an explanation of how to do it in TOB). My idea is to try to improve the results using the Gaussian approach.
Anyway, I can also do it using combinatorial analysis. Here are my numbers to 12 digits of precision so we can see if our programs match:

Code:
1) RC =  +1 | CR = 38 | A removed (1D)

|   0.118480844761 |   0.062504209740   0.073908841110   0.073908841110   0.073908841110   0.073908841110   0.073908841110   0.078198599809   0.078198599809   0.078198599809    0.333355785282 |
|   0.118480844761 |   2.375159970136   2.808535962172   2.808535962172   2.808535962172   2.808535962172   2.808535962172   2.971546792759   2.971546792759   2.971546792759  12.667519840725 |

2) RC = -13 | CR = 52 | No cards removed (2D)

|   0.001319524919 |   0.051923076923   0.101923076923   0.101923076923   0.101923076923   0.101923076923   0.101923076923   0.076923076923   0.076923076923   0.076923076923    0.207692307692 |
|   0.001319524919 |   2.700000000000   5.300000000000   5.300000000000   5.300000000000   5.300000000000   5.300000000000   4.000000000000   4.000000000000   4.000000000000  10.800000000000 |

3) RC = -13 | CR = 52 | A removed (2D)

|   0.001748370518 |   0.046449210406   0.101757691595   0.101757691595   0.101757691595   0.101757691595   0.101757691595   0.077474361349   0.077474361349   0.077474361349    0.212339247570 |
|   0.001748370518 |   2.415358941111   5.291399962953   5.291399962953   5.291399962953   5.291399962953   5.291399962953   4.028666790158   4.028666790158   4.028666790158  11.041640873651 |

4) RC = -13 | CR = 52 | A,T,7 removed (2D)

|   0.002324475807 |   0.048194174755   0.102325104020   0.102325104020   0.102325104020   0.102325104020   0.102325104020   0.069010552983   0.078869203409   0.078869203409    0.213431345344 |
|   0.002324475807 |   2.506097087263   5.320905409029   5.320905409029   5.320905409029   5.320905409029   5.320905409029   3.588548755129   4.101198577291   4.101198577291  11.098429957881 |

5) RC = -13 | CR = 52 | T,7 removed (2D)

|   0.001752858519 |   0.053838176772   0.102492222353   0.102492222353   0.102492222353   0.102492222353   0.102492222353   0.068501931969   0.078287922250   0.078287922250    0.208622934993 |
|   0.001752858519 |   2.799585192160   5.329595562356   5.329595562356   5.329595562356   5.329595562356   5.329595562356   3.562100462394   4.070971957022   4.070971957022  10.848392619621 |

Sincerely,
Cac

PS: These days I am going to publish the exact insurance indices for Hi-Lo taking into account different penetrations.

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