Length of Plus Count: 6-deck vs. 8-deck

[aslan]

I have a question. I know that 8-deck has a higher house edge than 6-deck, same rules. But what about the length of plus counts? I have noticed in experience, which nowhere near approaches a scientific test, that in 8-deck with 1 deck cut off (7 deck penetration) I get longer plus counts than I do with 6-deck with 1 deck cut off (5 deck penetration).

When I play at a crowded table, with the 6-deck game (83.3% pen), I get one or two rounds of plus counts if I get them at all. With the 8-deck game (87.5% pen), I seem to get 4, 5, or 6 rounds of plus counts about every three shoes. Is this consistent with what should occur mathematically, or am I just lucky to get the longer plus counts with the 8-deck games?

My results make me favor an 8-deck game when I must play at a crowded table. Of course, if I can find a 6-deck game that is not crowded, I prefer it both because of the lower house edge, and the fact that I will see longer plus counts and see them more often than with 8-deck game. Also, I can spread to more hands in plus counts if the table is not crowded. Also, I don't need the extra ploppies to eat up the larger number of negative counts of the 8-deck game.

 

[k_c]

I have a question. I know that 8-deck has a higher house edge than 6-deck, same rules. But what about the length of plus counts? I have noticed in experience, which nowhere near approaches a scientific test, that in 8-deck with 1 deck cut off (7 deck penetration) I get longer plus counts than I do with 6-deck with 1 deck cut off (5 deck penetration).

When I play at a crowded table, with the 6-deck game (83.3% pen), I get one or two rounds of plus counts if I get them at all. With the 8-deck game (87.5% pen), I seem to get 4, 5, or 6 rounds of plus counts about every three shoes. Is this consistent with what should occur mathematically, or am I just lucky to get the longer plus counts with the 8-deck games?

My results make me favor an 8-deck game when I must play at a crowded table. Of course, if I can find a 6-deck game that is not crowded, I prefer it both because of the lower house edge, and the fact that I will see longer plus counts and see them more often than with 8-deck game. Also, I can spread to more hands in plus counts if the table is not crowded. Also, I don't need the extra ploppies to eat up the larger number of negative counts of the 8-deck game.

Hopefully the following data can shed some light on your question.

I know you use KO so I input KO tags. Tags are relative to what remains in shoe, so just reverse sign of tags if what you are used to is to sign the tags relative to what has been removed. It makes no difference in the end result.

Program records all possible subsets relative to a given count and number of decks. Data shown assumes half of each of a 6 and 8 deck shoe have been seen and counted and next round will be dealt from the remaining half.

The point chosen is KO's pivot point. If initial running count is something different than -4*decks then running count will not be equal to zero but will equal 4*decks + IRC but this still will correlate to what is below. (If IRC = -4*decks + 4 then pivot = +4 instead of 0).

Probability of being exactly at pivot for 3/6 decks = 0.0164866
Probability of being exactly at pivot for 4/8 decks = 0.00990101

In general probability of a favorable count is less for 4/8 decks than 3/6 decks. However if you do reach pivot then you are in a better position for 4/8 decks rather than 3/6 decks.

Count tags {1,-1,-1,-1,-1,-1,-1,0,0,1}
Decks: 6
Cards remaining: 156
Initial running count (full shoe): -24
Running count: 0
Specific removals
A: 0
2: 0
3: 0
4: 0
5: 0
6: 0
7: 0
8: 0
9: 0
T: 0

Number of subsets for 156 cards: 5773
Prob of running count 0 from 6 decks: 0.0164866

p[1] 0.0845073 p[2] 0.0704228 p[3] 0.0704228 p[4] 0.0704228 p[5] 0.0704228
p[6] 0.0704228 p[7] 0.0704228 p[8] 0.0774634 p[9] 0.0774634 p[10] 0.338029

Press x or X to exit program (it may take some time to close,)
any other key to enter more data for same count tags/decks:

Count tags {1,-1,-1,-1,-1,-1,-1,0,0,1}
Decks: 8
Cards remaining: 208
Initial running count (full shoe): -32
Running count: 0
Specific removals
A: 0
2: 0
3: 0
4: 0
5: 0
6: 0
7: 0
8: 0
9: 0
T: 0

Number of subsets for 208 cards: 10193
Prob of running count 0 from 8 decks: 0.00990101

p[1] 0.0845071 p[2] 0.0704226 p[3] 0.0704226 p[4] 0.0704226 p[5] 0.0704226
p[6] 0.0704226 p[7] 0.0704226 p[8] 0.0774647 p[9] 0.0774647 p[10] 0.338028

Press x or X to exit program (it may take some time to close,)
any other key to enter more data for same count tags/decks:

 

[aslan]

Hopefully the following data can shed some light on your question.

I know you use KO so I input KO tags. Tags are relative to what remains in shoe, so just reverse sign of tags if what you are used to is to sign the tags relative to what has been removed. It makes no difference in the end result.

Program records all possible subsets relative to a given count and number of decks. Data shown assumes half of each of a 6 and 8 deck shoe have been seen and counted and next round will be dealt from the remaining half.

The point chosen is KO's pivot point. If initial running count is something different than -4*decks then running count will not be equal to zero but will equal -4*decks + IRC but this still will correlate to what is below.

Probability of being exactly at pivot for 3/6 decks = 0.0164866
Probability of being exactly at pivot for 4/8 decks = 0.00990101

In general probability of a favorable count is less for 4/8 decks than 3/6 decks. However if you do reach pivot then you are in a better position for 4/8 decks rather than 3/6 decks.

Count tags {1,-1,-1,-1,-1,-1,-1,0,0,1}
Decks: 6
Cards remaining: 156
Initial running count (full shoe): -24
Running count: 0
Specific removals
A: 0
2: 0
3: 0
4: 0
5: 0
6: 0
7: 0
8: 0
9: 0
T: 0

Number of subsets for 156 cards: 5773
Prob of running count 0 from 6 decks: 0.0164866

p[1] 0.0845073 p[2] 0.0704228 p[3] 0.0704228 p[4] 0.0704228 p[5] 0.0704228
p[6] 0.0704228 p[7] 0.0704228 p[8] 0.0774634 p[9] 0.0774634 p[10] 0.338029

Press x or X to exit program (it may take some time to close,)
any other key to enter more data for same count tags/decks:

Count tags {1,-1,-1,-1,-1,-1,-1,0,0,1}
Decks: 8
Cards remaining: 208
Initial running count (full shoe): -32
Running count: 0
Specific removals
A: 0
2: 0
3: 0
4: 0
5: 0
6: 0
7: 0
8: 0
9: 0
T: 0

Number of subsets for 208 cards: 10193
Prob of running count 0 from 8 decks: 0.00990101

p[1] 0.0845071 p[2] 0.0704226 p[3] 0.0704226 p[4] 0.0704226 p[5] 0.0704226
p[6] 0.0704226 p[7] 0.0704226 p[8] 0.0774647 p[9] 0.0774647 p[10] 0.338028

Press x or X to exit program (it may take some time to close,)
any other key to enter more data for same count tags/decks:

Right. I use IRCs of -20 and -28 for 6 and 8 decks respectively, but I don't think that will change your analysis. As I suspected, your probability of reaching pivot would be better for 6 than 8 deck, since the overall house edge is better for 6 deck than for 8 deck. That seems somewhat intuitive.

But you also say that once you reach pivot it fares better for 8 over 6 deck. Is this because once reached there are more cards remaining which could constitute a run of plus counts in 8 deck? This has been my contention. Playing 6 deck I must wait an entire shoe for the usual one or two rounds at a near plus count or plus count. At 8 deck, yes, I must wait longer for a shoe to deplete, but once a plus count is achieved, I usually find several rounds, not one or two, to take advantage of the plus count. (All of this assumes a full table.) This is equivalent to two or three runs through a 6-deck shoe in which the count turns positive at the end of each, and some will not turn positive.

This is why I feel I have a better game with 8 decks and one deck cut than 6 decks and one deck cut. The added pen with extra cards is king. But did your analysis support what I just said, or did I interpret it to my own ends? lol

 

[k_c]

Right. I use IRCs of -20 and -28 for 6 and 8 decks respectively, but I don't think that will change your analysis. As I suspected, your probability of reaching pivot would be better for 6 than 8 deck, since the overall house edge is better for 6 deck than for 8 deck. That seems somewhat intuitive.

But you also say that once you reach pivot it fares better for 8 over 6 deck. Is this because once reached there are more cards remaining which could constitute a run of plus counts in 8 deck? This has been my contention. Playing 6 deck I must wait an entire shoe for the usual one or two rounds at a near plus count or plus count. At 8 deck, yes, I must wait longer for a shoe to deplete, but once a plus count is achieved, I usually find several rounds, not one or two, to take advantage of the plus count. (All of this assumes a full table.) This is equivalent to two or three runs through a 6-deck shoe in which the count turns positive at the end of each, and some will not turn positive.

This is why I feel I have a better game with 8 decks and one deck cut than 6 decks and one deck cut. The added pen with extra cards is king. But did your analysis support what I just said, or did I interpret it to my own ends? lol

I think it's intuitive that you're in a better position at a +EV count such as pivot with more cards remaining. After all wouldn't you just as soon dump enough low cards into the discard tray at the start of a shoe to get to pivot? :grin: That would mean you would have the edge on average throughout the entire shoe with no waiting.:laugh: With more cards you get more hands at an advantage that is on average positive.

 

[caramel6]

Right. I use IRCs of -20 and -28 for 6 and 8 decks respectively, but I don't think that will change your analysis. As I suspected, your probability of reaching pivot would be better for 6 than 8 deck, since the overall house edge is better for 6 deck than for 8 deck. That seems somewhat intuitive.

But you also say that once you reach pivot it fares better for 8 over 6 deck. Is this because once reached there are more cards remaining which could constitute a run of plus counts in 8 deck? This has been my contention. Playing 6 deck I must wait an entire shoe for the usual one or two rounds at a near plus count or plus count. At 8 deck, yes, I must wait longer for a shoe to deplete, but once a plus count is achieved, I usually find several rounds, not one or two, to take advantage of the plus count. (All of this assumes a full table.) This is equivalent to two or three runs through a 6-deck shoe in which the count turns positive at the end of each, and some will not turn positive.

This is why I feel I have a better game with 8 decks and one deck cut than 6 decks and one deck cut. The added pen with extra cards is king. But did your analysis support what I just said, or did I interpret it to my own ends? lol

you are right, Aslan, I am not as scientific as you and oher AP, but for me it is obvious that 8 decks have more sustainable count, (I use hi low ), if it comes , then it is there for a while, however, I notice that advantage is higher when max 2 boxes are opened.I would not spread for more boxes with an advantage.

 

[aslan]

I think it's intuitive that you're in a better position at a +EV count such as pivot with more cards remaining. After all wouldn't you just as soon dump enough low cards into the discard tray at the start of a shoe to get to pivot? :grin: That would mean you would have the edge on average throughout the entire shoe with no waiting.:laugh: With more cards you get more hands at an advantage that is on average positive.

Are you agreeing with me? :laugh::whip:

 

[aslan]

you are right, Aslan, I am not as scientific as you and oher AP, but for me it is obvious that 8 decks have more sustainable count, (I use hi low ), if it comes , then it is there for a while, however, I notice that advantage is higher when max 2 boxes are opened.I would not spread for more boxes with an advantage.

The trouble with playing AC and other east coast stores is that during normal business hours, the tables tend to be crowded. At times, I am able to be at the table during slow times (late, late night), but at other times I have no choice. Yes, I too like to spread to two hands, even three at times, but more likely at 75% or 80% of max for the extra box(es) during plus counts. If I am playing two boxes "play all," then I would play max bet on both (and not play a third), if that is what you mean. Thanks for the response.

 

[k_c]

Are you agreeing with me? :laugh::whip:

Long story short: I think so. :devil:

 

[aslan]

Long story short: I think so. :devil:

:laugh: Now that that's settled.

 

[Blue Efficacy]

Longer high counts isn't necessarily a good thing. You don't make money from high counts, you make money from the count dropping.

 

[aslan]

Longer high counts isn't necessarily a good thing. You don't make money from high counts, you make money from the count dropping.

And the further it drops when you are in a plus count situation, the better for you. So, if it goes up to 140, as someone recently told me they experienced, the downhill trip to zero must be one h$ll of a sleigh ride! But don't forget, too, that the odds of getting good cards is higher in any plus count. You don't know whether the count will continue to go up or begin to come down, so it is a moot point, though accurate, that you make money when the count is dropping.

I will give you all my short plus counts, if you will give me all your long plus counts. Deal? We'll just have to leave it to chance whether they begin to drop before reaching the cut card.