Winning More towards the end. You really do.

[H Bomb]

Being a KO user, I might be wrong on this, but I thought the TC was 0 off the top of a new shoe.

To have an apples to apples comparison of the last deck of a shoe to the first deck, the TC = 0 condition is BEFORE the last deck but AFTER the first deck.

Well, after pondering this some more, I’m going to have to backtrack some of what I said. Does the last deck of a shoe given TC = 0 play like a SD? NO. Obviously it doesn’t play JUST LIKE a regular deck because you could have 5 aces left. And I’m not even sure if it makes sense to talk about whether, “on average”, it play like a SD. Does it play like 8D? Actually it’s probably better to phrase the question as: Does it play like a randomly selected 1D slug from 8D? The answer is also no because the TC = 0 condition. I’ve heard that TC = 0 occurs 37% or so of the time. Unless I’m missing something this means that the universe of possible final deck compositions given TC = 0 is only roughly 37% of the universe of possible 1D slug compositions from 8D. To say one should use SD BS at this point is too strong of a statement. On the other hand I’m not sure if it’s appropriate to use the 8D BS. Theoretically, there is an optimal BS for our scenario and I’d be shocked if it’s exactly the 8D BS or SD BS. Even if sims can spit out the optimal BS, I don’t think it’d be practical to remember additional BSs for XD game, YD left, TC = 0 (or Z). But if sims can show if another “off the rack” BS has a higher EV than 8D (if not SD maybe 2D, 4D, 6D) for this and some other TC = 0 scenarios, then we can increase our EV using info a lot of us already have at our disposal.

I’m ready to surrender too. :laugh: This was not the way I envisioned jumping in on this board. I think I showed some idiocy from some of the things I said but hey, it happens! :laugh:

 

[k_c]

Probabilities for card #261 of 6 deck shoe given Hi-Lo RC=0

The Hi-Lo player is admitting to a bit of ignorance. All he knows in this case is that an equal number of high (T-A) and low (2-6) cards have been dealt in the first 260 cards of a 6 deck shoe. Could Have been 120 high, 120 low, and 20 (7-9). Could have been 94 high, 94 low, and 72 (7-9). Could have been any Hi-Lo subset in between. Each subset has its own probability of occurrence.

At the start of a 6 deck shoe Hi-Lo RC=0
There is only 1 Hi-Lo subset possible; the full shoe subset
The probabilities of drawing each rank are:
(ace through 9) 1/13 = .076923
(ten) = 4/13 = .307692

At the start of the last 52 cards of a 6 deck shoe given Hi-Lo RC=0
There are 27 Hi-Lo subsets possible;
As a result of a weighted average of all possible subsets, the probabilities of drawing each rank are (from a program I wrote to enumerate this type of thing for Hi-Lo and KO):
Cards in deck=52 (TC=00.0)
p(2) = 0.07677
p(3) = 0.07677
p(4) = 0.07677
p(5) = 0.07677
p(6) = 0.07677
p(7) = 0.07742
p(8) = 0.07742
p(9) = 0.07742
p(10) = 0.30710
p(1) = 0.07677

 

[sagefr0g]

The Hi-Lo player is admitting to a bit of ignorance. All he knows in this case is that an equal number of high (T-A) and low (2-6) cards have been dealt in the first 260 cards of a 6 deck shoe. Could Have been 120 high, 120 low, and 20 (7-9). Could have been 94 high, 94 low, and 72 (7-9). Could have been any Hi-Lo subset in between. Each subset has its own probability of occurrence.

At the start of a 6 deck shoe Hi-Lo RC=0
There is only 1 Hi-Lo subset possible; the full shoe subset
The probabilities of drawing each rank are:
(ace through 9) 1/13 = .076923
(ten) = 4/13 = .307692

At the start of the last 52 cards of a 6 deck shoe given Hi-Lo RC=0
There are 27 Hi-Lo subsets possible;
As a result of a weighted average of all possible subsets, the probabilities of drawing each rank are (from a program I wrote to enumerate this type of thing for Hi-Lo and KO):
Cards in deck=52 (TC=00.0)
p(2) = 0.07677
p(3) = 0.07677
p(4) = 0.07677
p(5) = 0.07677
p(6) = 0.07677
p(7) = 0.07742
p(8) = 0.07742
p(9) = 0.07742
p(10) = 0.30710
p(1) = 0.07677

well that's neat.
lol, not sure how to assimilate the meaning of it all, but i'm curious at first glance about the higher probability for the 7's, 8's & 9's compared to Ace's through 6's. Ace's through 6's having the same probability seems intuitively sensible, and the ten's having a higher probability than all seems intuitively sensible as well. but i'm wondering why the 7's, 8's and 9's have a bit higher probability than the ace's through 6's?
maybe something to do with how the number of cards needs to be structured to end up with a tc=0 for fifty two cards? or maybe something to do with that weighted average of the subsets stuff you refered to? i don't really understand that weighted average concept in this sense.:whip:

 

[k_c]

well that's neat.
lol, not sure how to assimilate the meaning of it all, but i'm curious at first glance about the higher probability for the 7's, 8's & 9's compared to Ace's through 6's. Ace's through 6's having the same probability seems intuitively sensible, and the ten's having a higher probability than all seems intuitively sensible as well. but i'm wondering why the 7's, 8's and 9's have a bit higher probability than the ace's through 6's?
maybe something to do with how the number of cards needs to be structured to end up with a tc=0 for fifty two cards? or maybe something to do with that weighted average of the subsets stuff you refered to? i don't really understand that weighted average concept in this sense.:whip:

Best way to get some sort of understanding is to start simple.

Single deck, 1 card remaining, RC=-1
1 subset possible: 1 (2-6), 0 (7-9), 0 (T-A)
Cards in deck=1 (TC=-52.0)
p(2) = 0.20000
p(3) = 0.20000
p(4) = 0.20000
p(5) = 0.20000
p(6) = 0.20000
p(7) = 0.00000
p(8) = 0.00000
p(9) = 0.00000
p(10) = 0.00000
p(1) = 0.00000

Single deck, 1 card remaining, RC=0
1 subset possible: 0 (2-6), 1 (7-9), 0 (T-A)
Cards in deck=1 (TC=00.0)
p(2) = 0.00000
p(3) = 0.00000
p(4) = 0.00000
p(5) = 0.00000
p(6) = 0.00000
p(7) = 0.33333
p(8) = 0.33333
p(9) = 0.33333
p(10) = 0.00000
p(1) = 0.00000

Single deck, 1 card remaining, RC=+1
1 subset possible: 0 (2-6), 0 (7-9), 1 (T-A)
Cards in deck=1 (TC=+52.0)
p(2) = 0.00000
p(3) = 0.00000
p(4) = 0.00000
p(5) = 0.00000
p(6) = 0.00000
p(7) = 0.00000
p(8) = 0.00000
p(9) = 0.00000
p(10) = 0.80000
p(1) = 0.20000

This has more than 1 subset, so subsets are weighted according to their probability of occurring.
Single deck, 2 cards remaining, RC=0
2 subsets possible: 1 (2-6), 0 (7-9), 1 (T-A); 0 (2-6), 2 (7-9), 0 (T-A)
Cards in deck=2 (TC=00.0)
p(2) = 0.08584
p(3) = 0.08584
p(4) = 0.08584
p(5) = 0.08584
p(6) = 0.08584
p(7) = 0.04721
p(8) = 0.04721
p(9) = 0.04721
p(10) = 0.34335
p(1) = 0.08584

This is 1 way to do the calculation:

Prob of subset A 1 (2-6), 0 (7-9), 1 (T-A) = 20/52*20/51*2
Prob of subset B (2-6), 2 (7-9), 0 (T-A) = 12/52*11/51
(Prob subset A)/(Prob subset B) = 800/132
Subset A is 800/132 times more likely than subset B
In other words subset A will occur 800/132 times for every single occurrence of subset B
Subset A relative weight = 800/132/(1+800/132) = 800/932
Subset B relative weight = 1/(1+800/132) = 132/932

Prob of (2-6) each in subset A = .10; prob of (2-6) each in subset B = 0
Prob of (7-9) each in subset A = 0; prob of (7-9) each in subset B = 1/3
Prob of (T) in subset A = .40; prob of (T) in subset B = 0
Prob of (A) in subset A = .10; prob of (A) in subset B = 0

Overall prob of (2-6) each = 800/932*.10 + 132/932*0 = ~.08584
Overall prob of (7-9) each = 800/932*0 + 132/932*1/3 = ~.04721
Overall prob of (T) = 800/932*.40 + 132/932*0 = ~.34335
Overall prob of (A) = 800/932*.10 + 132/932*0 = ~.08584

That's the basic idea. Each subset contributes to the overall probability according to its weight relative to the other subsets. A subset with a greater probability of occurring has more influence on the overall probability than one with a lesser probability of occurring.

 

[H Bomb]

The Hi-Lo player is admitting to a bit of ignorance. All he knows in this case is that an equal number of high (T-A) and low (2-6) cards have been dealt in the first 260 cards of a 6 deck shoe. Could Have been 120 high, 120 low, and 20 (7-9). Could have been 94 high, 94 low, and 72 (7-9). Could have been any Hi-Lo subset in between. Each subset has its own probability of occurrence.

At the start of a 6 deck shoe Hi-Lo RC=0
There is only 1 Hi-Lo subset possible; the full shoe subset
The probabilities of drawing each rank are:
(ace through 9) 1/13 = .076923
(ten) = 4/13 = .307692

At the start of the last 52 cards of a 6 deck shoe given Hi-Lo RC=0
There are 27 Hi-Lo subsets possible;
As a result of a weighted average of all possible subsets, the probabilities of drawing each rank are (from a program I wrote to enumerate this type of thing for Hi-Lo and KO):
Cards in deck=52 (TC=00.0)
p(2) = 0.07677
p(3) = 0.07677
p(4) = 0.07677
p(5) = 0.07677
p(6) = 0.07677
p(7) = 0.07742
p(8) = 0.07742
p(9) = 0.07742
p(10) = 0.30710
p(1) = 0.07677

Now I'm REALLY confused. Is your point that because the probability for each card is essentially the same as the start of 6D, we should still use the 6D BS with 1 deck left, TC = 0? Or is your point to address what I said below?

Unless I’m missing something this means that the universe of possible final deck compositions given TC = 0 is only roughly 37% of the universe of possible 1D slug compositions from 8D.

 

[k_c]

Now I'm REALLY confused. Is your point that because the probability for each card is essentially the same as the start of 6D, we should still use the 6D BS with 1 deck left, TC = 0? Or is your point to address what I said below?

My point is only to show the probabilities.

The probabilities are similar to a full shoe but not exactly equal. Also when you begin to remove specific cards from the 52 card slug, you see some interesting effects.

I started doing this to investigate the possiblity of doing a combinatorial analyzer based upon a count. I think theoretically it would be possible but it would be way too slow for my tastes.

I can give you my opinion on what I've seen and tried.

I have tried all sorts of counting systems with the goal of finding one that can define indices that yield the mathematically correct play from the start of a shoe to the end for any pen. I have concluded that no such count exists even if you have perfect knowledge, which I wouldn't, but I assumed this for the sake of argument.

This kind of stuff is not easy to explain but I'll make a feeble attempt at an example. Let's say you're counting (2-5)=low, (T,J,Q,K)=high, and (6,7,8,9,A)=neutral. Let's also say you have perfect knowledge of exactly how many neutral cards remain. Let's say you know that zero neutral cards remain and RC=0 so that you know that there are an equal number of high and low cards remaining to be dealt. Let's also assume that cards will be dealt down to the last card.

If there are lot of cards left at RC=0 with zero neutral cards, it would be right to hit hard 16 versus ten. Ballpark EV for hitting = ~-35% and EV for standing = ~-40% depending on exact situation. What if there were exactly 2 cards left? You have a 50% chance of making a hand if you hit. Your hand will be 18,19,20, or 21 each 25% of that 50%, but every time you make a hand dealer will have a ten in the hole so you win when you have 21 and push when you have 20 but otherwise lose. The other 50% you will bust and lose. Your EV for hitting=1/2*1/4(-1-1+0+1)-1/2*1=-62.5%.

If you stand, dealer will have 20 50% of the time and 50% of the time he will have a stiff and will always bust. Your EV for standing=1/2*1+1/2*(-1)=0.

So at the end of a shoe the same situation of TC=0 with no neutral cards present, standing is clearly the better play.

What happened? When you get near the end of a shoe you get what you could call the principle of restricted choice. If you make your hand, dealer will be restricted to having a pat 20. If you don't make your hand you will bust and lose.

Just how far into a shoe does this kick in? I have no idea but I think this is why there is no "perfect" counting system as far having an index that always works.

On one hand a counter is wishing for as much pen as possible. On the other hand very deep pen may cause him to misplay his cards using fixed indices.

These are only observations and it boils down to how often misplays are made. Probably on balance indices increase EV as is shown by sims.

Back to the original question. In my opinion the last 52 cards are similar on average to a single deck but are not the equivalent of a single deck. The fact that the starting point is 6 full decks makes it different. A full single deck has one defined composition whereas the last 52 cards of a 6 deck shoe does not. One thing that would be true for a full single deck is that there is no composition like my feeble example composition whereas that composition would be theoretically possible for the last 52 cards of a 6 deck shoe.