Blackjack and Card Counting Forums - BlackjackInfo.com

  #1  
Old November 24th, 2007, 03:47 PM
k_c k_c is offline
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Default Probs: Hi-Lo TC=0, Single Deck

I'm not sure I should post this, but I thought some might find it interesting.

One of the programs I have written computes the probabilities of drawing any rank for either Hi-Lo or KO given any running count at a given pen starting with from 1-8 decks with optionally including specific removals in the calculation. It uses a weighted average of all of the possible subsets for a given running count/pen combination. The optional removals are good for getting the insurance decision to a greater degree of accuracy. For example, single deck A-A v A insure at Hi-Lo TC >= -2.36! (yes minus.) A hand such as 7-9 v A would be insured single deck at a Hi-Lo TC >= 0 wheras T-T v A insured at HiLo TC >=+ 3.71. The greater number of decks, the less variation in the comp dependent index.

Insurance isn't why I started this post, though. It just came out that way. It turns out that the only times the probabilities of drawing a non-ten=1/13 and drawing a ten=4/13 exactly at a Hi-Lo TC=0 for a single deck is when there are exactly 52 or 26 cards remaining to be dealt. For any number of decks, the starting probabilites of 1/13 and 4/13 only occur exactly when no cards or 1/2 the cards are removed, assuming no specific removals.

What does this mean? In general, it is a little more likely than you might think that you will draw either a low card or ten/ace and less likely to draw a 7-9 at a Hi-Lo TC=0 when more than half the deck remains and just the opposite when less than half the deck remains at a Hi-Lo TC=0. When very few cards remain, the probabilities fluctuate greatly. The probabilities also fluctuate when the deck is nearly full.

I'm not too good with spreadsheets, but I created one with a chart. I'll attempt to attach it to this post.

The effect of the differing probs at TC= 0 probably doesn't amount to much, but I thought it may be of interest.

k_c
Attached Files
File Type: zip SingleDeck_789.zip (3.6 KB, 146 views)
  #2  
Old November 27th, 2007, 08:52 PM
Kasi Kasi is offline
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Originally Posted by k_c View Post
I' It turns out that the only times the probabilities of drawing a non-ten=1/13 and drawing a ten=4/13 exactly at a Hi-Lo TC=0 for a single deck is when there are exactly 52 or 26 cards remaining to be dealt.
Not sure what a non-ten=1/13 means but anyway it seems to me the probabilities could be exactlly the same at 13 and 39 cards too.

Anyway interesting stuff.
  #3  
Old November 28th, 2007, 01:29 AM
k_c k_c is offline
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Originally Posted by Kasi View Post
Not sure what a non-ten=1/13 means but anyway it seems to me the probabilities could be exactlly the same at 13 and 39 cards too.

Anyway interesting stuff.
Well, the "=1/13" line is just there for reference and has nothing to do with non-tens. The points on the chart represent the probability of drawing a single zero tag card from the corresponding number of cards remaining. You would think the probability of drawing a 7 at a Hi-Lo TC=0 would always=1/13 (assuming no other cards have been specifically removed,) but it only =1/13 exactly when there are 52 or 26 cards remaining. The same is true for the 8 and 9 since they are part of the same group of ranks in Hi-Lo.

Take 2 simple cases: 51 cards remaining and 1 card remaining dealt from a single deck where RC=TC=0. When there are 51 cards, the card removed has to be a 7,8, or 9 so the probability of drawing a 7,8, or 9 < 1/13. When there is 1 card remaining, that card has to be a 7,8, or 9 so the probability of drawing a 7 = prob of drawing 8 = prob of drawing 9 = 1/3 and obviously 1/3 > 1/13.

To get the other points on the chart, I have a program that can compute the probabilitiies of drawing each rank given a running count and a given number of cards remaining for either Hi-Lo or KO (I used Hi-Lo) and can optionally account for specific removals. It's based on a weighted average of all of the possible subsets for the given count/pen combo. So far the only reasonable use I have found for it is to more accurately get insurance indices for Hi-Lo and KO. You can get count comp dependent indices for hands of L-L, L-M, L-T, L-A, M-M, M-T, M-A, T-T, T-A, and A-A where L means low, M means medium (cards with 0 tag,) T means ten and A means ace. The extra accuracy would be the most worthwhile for a single deck game, declining in value as more decks are added.

Anyway, I probably didn't explain it well, but hopefully the general idea can be seen.

k_c
  #4  
Old November 28th, 2007, 02:30 AM
Kasi Kasi is offline
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Originally Posted by k_c View Post
Anyway, I probably didn't explain it well, but hopefully the general idea can be seen.k_c
No - you explained it well - I think I see what you are saying but I don't see how any probabilities change if the first 13 cards dealt from a single deck are all spades, for example.

I mean at that point, with 39 cards left, RC=TC=0, all ratios are the same, so what's the diff from 26 cards left?

Anyway, it did make me wonder what count I should I use if I ever encounter a 2-1 BJ game.

I mean we all know that assigning +1 to all cards except 10, and -2 to all 10's, that when this RC (no TC necessary) gets above 4 times the number of decks used, an insurance bet becomes profitable no matter how many decks have yet to be dealt. So it's a perfect count for insurance decisions.

Of course such a count is not very useful for estimating advantage for other betting decisions.

Great as a second count if you can do it lol.
  #5  
Old November 28th, 2007, 12:40 PM
k_c k_c is offline
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Quote:
Originally Posted by Kasi View Post
No - you explained it well - I think I see what you are saying but I don't see how any probabilities change if the first 13 cards dealt from a single deck are all spades, for example.

I mean at that point, with 39 cards left, RC=TC=0, all ratios are the same, so what's the diff from 26 cards left?
When the running count = 0, it simply means that the number of (2-6) present in the deck equals the number of (T-A). If there are 13 cards remaining, the possible Hi-Lo deck compositions are:
Code:
6 (2-6), 1 (7-9), 6(T-A)
5 (2-6), 3 (7-9), 5 (T-A)
4 (2-6), 5 (7-9), 4 (T-A)
3 (2-6), 7 (7-9), 3 (T-A)
2 (2-6), 9 (7-9), 2 (T-A)
1 (2-6), 11 (7-9), 1 (T-A)
When you calculate the probability of each of the above subsets and get a weighted average of them, it turns out that the probabilities of drawing each rank are:
Code:
Cards in deck = 13
    p(2) = 7.64734978774833E-02
    p(3) = 7.64734978774833E-02
    p(4) = 7.64734978774833E-02
    p(5) = 7.64734978774833E-02
    p(6) = 7.64734978774833E-02
    p(7) = 7.84216737417225E-02
    p(8) = 7.84216737417225E-02
    p(9) = 7.84216737417225E-02
    p(10) = 0.305893991509933
    p(1) = 7.64734978774833E-02
Compare this to full deck composition probabilities and you can see that the probabilities aren't the same:
Code:
Cards in deck = 52
    p(2) = 7.69230769230769E-02
    p(3) = 7.69230769230769E-02
    p(4) = 7.69230769230769E-02
    p(5) = 7.69230769230769E-02
    p(6) = 7.69230769230769E-02
    p(7) = 7.69230769230769E-02
    p(8) = 7.69230769230769E-02
    p(9) = 7.69230769230769E-02
    p(10) = 0.307692307692308
    p(1) = 7.69230769230769E-02
It turns out that the probabilities are the same only when 26 cards remain in a single deck and in general when half of the cards remain in any number of decks.
I'm sorry for all the decimal places and scientific format of the numbers. I just pasted them from a previous output file.

Quote:
Anyway, it did make me wonder what count I should I use if I ever encounter a 2-1 BJ game.

I mean we all know that assigning +1 to all cards except 10, and -2 to all 10's, that when this RC (no TC necessary) gets above 4 times the number of decks used, an insurance bet becomes profitable no matter how many decks have yet to be dealt. So it's a perfect count for insurance decisions.

Of course such a count is not very useful for estimating advantage for other betting decisions.

Great as a second count if you can do it lol.
Right on. You can make perfect insurance decisions with the insurance count. I wish I could maintain 2 counts at a time in the casino, but no way.

k_c
  #6  
Old November 28th, 2007, 09:25 PM
Kasi Kasi is offline
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Originally Posted by k_c View Post
but no way. k_c
Me either

Well maybe I only meant that it's possible that the original deck probabilities could exist at 39 and 13 cards lol. Not that they must, if that was your point.

What does your program say if the 1st 13 cards are all spades?

What does your program say if the 1st 12 cards are all the 7,8,9's followed by 4 2's, 4 10's, 3 3's and 3 J's? 26 cards left, RC=TC=0 but no 2's or 10's left? Must be that weighted thing.

Anyway no big deal. I have no idea what I'm talking about anyway. No surprise there.
  #7  
Old November 28th, 2007, 11:02 PM
k_c k_c is offline
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Originally Posted by Kasi View Post
Me either
Well maybe I only meant that it's possible that the original deck probabilities could exist at 39 and 13 cards lol. Not that they must, if that was your point.
What I'm saying is that if you say that if you set the condition that Hi-Lo TC=0, single deck, then the probabilities of drawing each rank are different from the full deck probabilities except when exactly 26 cards remain. If there is no condition, which means you aren't counting, it's always like drawing from a full deck, that's all.
Quote:
What does your program say if the 1st 13 cards are all spades?
Suit means nothing. It just deals with rank.

Quote:
What does your program say if the 1st 12 cards are all the 7,8,9's followed by 4 2's, 4 10's, 3 3's and 3 J's? 26 cards left, RC=TC=0 but no 2's or 10's left? Must be that weighted thing.
This is the output w/ Hi-Lo insurance EV if next hand is dealt from this deck. Insurance EV is something I recently added.
Code:
Cards in deck=26 (TC=00.0, ins EV=1.04%)
    p(2) = 0.00000
    p(3) = 0.03846
    p(4) = 0.15385
    p(5) = 0.15385
    p(6) = 0.15385
    p(7) = 0.00000
    p(8) = 0.00000
    p(9) = 0.00000
    p(10) = 0.34615
    p(1) = 0.15385
Quote:
Anyway no big deal. I have no idea what I'm talking about anyway. No surprise there.
I don't necessarily always know what I'm talking about either, but I try Sometimes I think I have my ducks in a row, but it turns out I'm not quite there yet .

k_c
  #8  
Old November 29th, 2007, 02:30 AM
k_c k_c is offline
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Default Remove all spades

Quote:
Originally Posted by Kasi View Post
Me either
What does your program say if the 1st 13 cards are all spades?
I see now what you mean. Removing all 13 spades is the same as removing 1 each ace through 9 and 4 tens. In that case my program outputs the probability of ace through nine = 1/13 and the probability of ten = 4/13. This is because with those specific removals and 39 cards remaining, there is only 1 subset that satisfies RC=0. If any other RC is input the program simply outputs that no subsets are possible.

Good question .

k_c
  #9  
Old November 29th, 2007, 11:59 PM
Kasi Kasi is offline
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I see now what you mean.k_c
That's all I meant lol. But I wish I knew what it meant
  #10  
Old November 30th, 2007, 01:31 AM
k_c k_c is offline
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Originally Posted by Kasi View Post
That's all I meant lol. But I wish I knew what it meant
At the risk of confusing you, I'll try to reference what happens when 13 spades are removed, which you seem to understand.

When 13 spades are removed, there is 1 and only 1 Hi-Lo composition possible:
15 (2-6), 9 (7-9), 15 (T-A)

When 13 cards are removed and you stipulate that running count = 0, there are 6 possible Hi-Lo compositions possible:
14 (2-6), 11 (7-9), 14 (T-A)
15 (2-6), 9 (7-9), 15 (T-A)
16 (2-6), 7 (7-9), 16 (T-A)
17 (2-6), 5 (7-9), 17 (T-A)
18 (2-6), 3 (7-9), 18 (T-A)
19 (2-6), 1 (7-9), 19 (T-A)

Without going into any calculations, there is more to figure out in the second case. Each composition has it's own probability of occurring and also it's own probability of drawing each rank. To get the overall probability of each rank, you give more numerical weight based on their probability to the compositions more likely to occur and less weight to the ones less likely and compute an average.

In the first case you are stipulating that a specific number of each rank be removed (specific removals.) In the second case you are only stipulating that 13 cards be removed and that the running count = 0 (no specific removals.)

Can you see now why it is possible to have different probabilities of drawing each rank at a Hi-Lo RC = 0 with different pen levels?

That's about as clearly as I can say it. I guess I'm crazy to go this far lol.

k_c
 

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