08﻿ Why bac could be beatable itlr

### Topic: Why bac could be beatable itlr  (Read 20552 times)

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#### AsymBacGuy

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##### Re: Why bac could be beatable itlr
« Reply #135 on: May 12, 2020, 12:01:17 am »
• More P2/P1-P3 results randomly taken:

1-1-1

3-1-1-2-1-*

1-2-1-1-1

1-2-1-1

1-1

1-1

1-2-1

1-1-1-1

1-1-1

1

3

1-1-1

2-1-3

2-1-2-1

1-1-1-*

1-1-2-2

1-1-1-3

1-1-1-2-1-2

2-2

1-1-1

3-2-2

1

1-1-2-1

1-1-1-2

1-1-1

1-2-2-1

1-1

2-1-1-1

1-2-1-1

1-1-1

1-2

2-1

1-1-1

2-4

1-1-1-1

2-1-1-1

1-1-1

2-1-1-1

1-2-1

1-1-1-1

1-1-1-1

Now only a real id.iot could lose at those different B/P situations that MUST happen along each shoe.
Especially at 8-deck shoes.

as.

Next to edge sorting it's me

#### AsymBacGuy

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##### Re: Why bac could be beatable itlr
« Reply #136 on: May 13, 2020, 09:53:49 pm »
• Is B plan better than P or vice versa?
What about a plan considering both strategies simultaneously as a whole?

Let's start with the both sides plan, that is always wagering toward getting a B1/B3 or P1/P3 after a B2 or P2 trigger up to some levels.

Obviously we'll get many losses when many BB or PP doubles are coming consecutively, a kind of costant symmetrical situation but acting asymmetrically after one single hand is dealt, for each single hand considered itlr has a Bp=0.5068 and Pp=0.4932.
We shouldn't give a fk whether a given BBPPBBPPBBPP pattern (or when many other B/P doubles patterns provide more consecutive doubles) will be only formed by symmetrical situations, itlr and on average per every 12 resolved hands one asymmetrical hand favoring B side must happen (for simplicity here I omit the asym hand apparition producing a tie). And we know that many B favored hands can easily make the Player side winning.

Moreover unless a third card is exactly a zero value card, asym hands involve various degrees of B advantage, sometimes even unfavorite math situations as when Banker gets an initial 4 point and the third card is an Ace (slight negative EV as B should draw and not standing).

Baccarat is a game governed by asymmetrical states for rules and card distribution and when certain asymmetrical situations tend to produce symmetrical second-level (or higher) states we might endure some harsh times.

If by various causes, the asymmetricity will be so balanced along the vast or even the entire portion of the shoe, we're not going anywhere, thus imo not every shoe is playable.

A strong predominance of one side could be a kind of an extreme asymmetrical state being so simple to be detected. Unfortunately vast majority of shoes dealt do not belong to such category and moderate/light predominances are assessed after such state happened.
In addition, a simple B or P predominance is just a back to back unidirectional issue, mostly taken without considering the actual conditions that favored one side for long.

Thus we shouldn't bet on how long the asymmetricity works but about when it's more likely to produce given results on the side chosen.

as.
Next to edge sorting it's me

#### AsymBacGuy

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##### Re: Why bac could be beatable itlr
« Reply #137 on: May 13, 2020, 10:54:34 pm »
• Let's see what happens on those 20 live shoes taken randomly:

B plan: 1-1  P plan: 2-1-2

B plan: 2-1  P plan: 1-1-1

B plan: 1-1-2  P plan: 2-1

B plan: 1-1-1  P plan: 1-1-1-2-2

B plan: 1-1-1-1-1  P plan 2-1

B plan: 1-1-1  P plan 2

B plan: 1-1-2  P plan: 1

B plan: 1-1-2-1-1  P plan: 2-1

B plan: 1-2  P plan: 2-2-1-1

B plan: 1-2  P plan: 1-1-1

B plan: 2-2-2  P plan: no triggers

B plan: 1-1  P plan: 1-2-2

B plan: 2-2  P plan: 1-1-1-1

B plan: 3-1  P plan: 1-1-1

B plan: 1-2-1  P plan: 2-1-2-1

B plan: 1-1-1-3  P plan: 2-2-2

B plan: 1-1  P plan: 1-2

B plan: 1-1-1  P plan: 1-1-2

B plan: 3-1-1  P plan: 1-1

B plan: 1-1-1-1-2  P plan: 1-2-2-1-1

Fortunate shoes?
Probably not, as  1=76; 2=34 and 3=3

Since any 2 or 3 (or higher) occurence causes a -3 unit deficit providing a 1-2 mini progression made toward the 1 appearance and 1 just means +1, we'll get (before tax) 76 unit wins and 103 unit losses for a net loss of -27 units.
More interesting is that in this sample betting not to get 3 after 2 means 34 units of profits vs a 9 (3x3) unit loss. That is (before tax) a 25 units profit.

Does this ridiculously small sample suggesting that betting 2 after 2 vs 3+ will provide an advantage whereas the 1 vs 2+ proposition is a long term losing bet?
No way, naturally.
Those short term frequencies just suggest that the asymmetricity overall acted lightly at 1-level degree and very well at 2-level degree.
Indeed we could face shoes getting very different values of asymmetricity, anyway we are pretty sure that smaller classes will overwhelm superior values, all depending upon how good or bad are shuffled the cards.

Of course and regardless of the asymmetricity value acting on the actual shoe, by both place selection and probability after events tools use, many random walks can be built getting ridiculously (now on the positive side) low dispersion values.

as.
Next to edge sorting it's me

#### AsymBacGuy

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##### Re: Why bac could be beatable itlr
« Reply #138 on: May 14, 2020, 12:59:14 am »
• More shoes:

B: 1  P: 1-1-1-1

B: 1-1-2-3  P: 1-1-1-1

B: 1-4-1  P: 1-3

B: 1-1-1-1  P: 3-1-1

B: 1-1-3-1-2  P: 1-1-1

B: 1-1-1-1  P: 1-1-1

B: 1-1-2-2-2*  P: 1-2*

B: 1-1-1-1-2  P: 1-1-3-1-1

B: no triggers  P: 1-2*

B: 1-1  P: 3*

B: 1-1  P: 1-3-1-1

B: 1  P: 1-3

B: 2-1-2  P: 1-4-1-1

B: 1-2*  P: 1-2*

B: 1-1  P: 1-4-1

B: 2-1-1-1  P: 1-1-1

B: 1-1-1  P: 1-1-1-1-1

B: 1-1-1  P: no triggers

B: 1-2-1  P: 1-1

B: 1-1-1-1-1  P: 1-1-1-2

Total 1=91  2=15 3+=11

Now betting 1 vs 2+= +13 before tax; 2 vs 3+= -18

Now asymmetricity considered the way discussed so far went right on the first level.

as.

Next to edge sorting it's me

#### AsymBacGuy

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##### Re: Why bac could be beatable itlr
« Reply #139 on: May 17, 2020, 10:31:38 pm »
• In reality the above P plan doesn't get the same variance features happening on the same B plan.
Even if shoes presented above were randomly taken, P side formed too many 1 or 2 situations than expected as itlr a lot of 3, 4 or higher numbers will be produced, especially whether consecutively considered. That's why my ub #2 didn't consider P side.

Let's try to give a formal answer to this.

If in order to set up our future betting plan we take BB and PP as symmetrical triggers we are making a mistake at the start.

Itlr every BB pattern is already a natural asymmetrical situation as math tend to shift the probability to B side after any given value that now we set after a single fresh B apparition.
On the other hand, itlr PP is already an artificial asymmetrical pattern as in some sense was slightly fighting against the math.
Therefore BB and PP patterns cannot be considered triggers springing up from the same probability. Actually most of the times are, but not itlr.

No matter what happens in between (just to simplify the things here), any new fresh B situation must fight with a new probability after any previous BB pattern had formed.
If we decide to always wager toward a B streak after any BB pattern previous production, we are simply implying that the asym value must act again just on this limited section of the shoe or, that whether the asym didn't act on the previous BB pattern, now it's more likely to work.
In addition, itlr the BB trigger involves a certain degree of "exhaustion" of asym force as the next hand is P.
That's why we could infer that itlr any fresh B appearance next to another exact BB pattern will be somewhat restricted to produce another B streak, thus orienting us to bet one time P side.
Whether this bet went wrong, we are challenging the actual card distribution to give another precise BB pattern, that is missing our plan two times (or more) in a row.

The PP counterpart is easier to be considered as an actual "artificial" asym strenght already worked. Thus after PP itlr the more likely outcome will be a P single and not a P 3+ streak whether the first bet failed (meaning a PP occurence). Thus lowering a lot the winning probability of our second attempt.

Overall and itlr the B2/B1 strenght will be more powerful than the P2/P3 strenght of course considering that both bets are mathematically facing the same 1:1 payment.

Player side must be attacked by other weapons.

as.
Next to edge sorting it's me

#### AsymBacGuy

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##### Re: Why bac could be beatable itlr
« Reply #140 on: May 19, 2020, 11:20:11 pm »
• Player side is more difficult to be assessed despite of its slight lesser probability to appear.
When betting P side we are simply wagering that key cards must be shifted toward this side at various degrees and in the meantime that no asymmetrical B favoring situation will arise.

Since we know that almost every shoe isn't immune to such asym probability, we could infer that  is virtually impossible to wager Player getting a steady 0.5 winning probability fairly payed (1:1).

In a sense when betting Player we are hoping about two orders of things:

a- no asym hand will take place at the time of our betting

b- key cards are shifted toward Player side

Oppositely thinking, we could reckon that B side is really advantaged only when an asym hand will come out within a restricted range of hands as the key cards shift is anyway burdened by a 5% vigorish.

Now let's think about the probability where our plan will get all positive Player betting situations upon a given shoe. Say this is our gold standard.

1- wagering toward getting all P singles.

2- wagering toward getting all P doubles

3- wagering toward getting all P 3+ streaks

4- any mix of the above situations

No need to test many shoes, almost no one single situation belonging to #1, #2 and #3 category will provide all winnings.
Then in order to increase such probability even at the risk of losing more money, we try to couple two different scenarios.

1-2: well, this situation may happen, mostly when many P doubles are formed or when P singles are interpolated by long B streaks.

1-3: situation less likely than the previous one, yet it could happen.

2-3: no way an 8-deck shoe is likely to show all P streaks, of course here the winning/losing probability remains confined at 0.5 at best.

If we aim to get all wins on our bets obviously we must rely upon the probability that things are going right just at the start.
Therefore plans 1-2 and 1-3 are more likely to provide this kind of jackpot, either as they involve a 0.75% or so probability to win and as 2-3 plan isn't going to form winnings at the whole played shoe.

Naturally such jackpot is just an ideal situation thus forcing us to build our betting plan upon lower degree probabilities. Yet some quality factors endorse the probability to get or not the expected long winning streak we should aim for.

Moreover those 1-2 and 1-3 plans are just considered by a mere B/P pattern random walk point of view.
That is not properly considering the actual conditions where those results were formed.

A thing discussed next

as.
Next to edge sorting it's me

#### AsymBacGuy

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##### Re: Why bac could be beatable itlr
« Reply #141 on: May 24, 2020, 10:18:43 pm »
• The decisive tool to test any B/P system is by considering the limiting values of relative frequency of EVERY possible shoe's pattern, thus covering how it fares through every possible card distribution.

The ploy to restrict the outcomes into three classes will help us a lot for two reasons.

First, baccarat features the very slight propensity to produce the opposite result already happened;

Secondly,  after the 3 level is reached we may consider all 3+ superior classes the same as 3s.

Since it may appear so easy to simply bet toward shorter patterns as singles and doubles, we should focus our interest about those 3s distribution.

3s and 3+s are by definition asymmetrical situations even if a given 3+ is composed by a BBBBBB sequence or PPPP pattern as they get or not a given probability of taking advantage (B side) or shifting (P side) the asym force determined by the rules.

Of course pure 3s (streaks of just three B/P hands) are more likely to be the product of sym situations as the overall asym probability is confined to 8.6% over the total hands dealt. The longer any streak is forming higher will be the probability to cross an asym situation as virtually (and practically) no shoe is producing all symmetrical events.
And we know that not all asym hands will form a B decision, of course.

It could easily happen that asym hands may come out within shorter BP patterns, for example after a single B result or after a single P hand or after a couple of the same situations.
Thus, for example, betting itlr toward P singles and P doubles just mean to hope that the asym force will happen right on those spots as the mere symmetrical force cannot be of any help other than for short term variance issues.

Itlr, our profit can only and only come out just when the sum of our Player bets were placed on sym hands payed 1:1 and when our Banker bets were getting a quite higher than 8.6/91.4 ratio.
Naturally those P bets must involve more than a strenght of sym value, mostly in form of more likely card distribution, whereas B bets generally rely upon a selected endorsed math probability.

Back to the "everything is possible" shoe production.
We could think the bac shoe situation as a continuous 1-2-3 succession, knowing that homogeneous 1 or 2 or 3 situations aren't going to happen. But two situations out of three are more likely to happen along the entire lenght of the shoe and we know we had to discard 2-3 situations unless happening at B side (with the additional help of B2/B1 apparition).
We are so sure about that that a multilayered progression made on B doubles consecutiveness will cross very soon a certain "jackpot" situation, the same but at a lower degree when considering two or more consecutive wins when applied to the 1/2 and 1/3 method.

as.
Next to edge sorting it's me

#### AsymBacGuy

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##### Re: Why bac could be beatable itlr
« Reply #142 on: May 24, 2020, 10:51:52 pm »
• Examples taken from Wynn and Gold Coast live shoes data.

1-2 and 1-3 plans joined with B2/B1-B3 attacks made on the entire shoe regardless of asym/sym quality assessment.

+ + - - - + + + + + + + - + + + - + + - + + - - + + + + + + + -

B2/B1-B3:  - - - + - -

+ + + + + + + - + - + + + - -  + + -

B2/B1-B3: + - + +

+ + + - + - + - + + + + + - + - + + + + - + + - + + - + -

B2/B1-B3: + - + + + +

- + + - + - - - + + + + + + + + + - + + - + + + + + -

B2/B1-B3: - + - + + +

- + + + + + + + + + + - - + - + + + + + + - - + + + - + - *

B2/B1-B3: - - + + +

+ + - + + + - + - + + + + + + + - + + + + + - + + + + + -

B2/B1-B3: + + +

+ + + - + + + + - + + + + + + + + + - - + - - + + + + + + + +

B2/B1-B3: + + - - - + + - +

+ + + + + + + - + + - + + + + + + + + + + + + + + - + + - + + + + + - + + -

B2/B1-B3: - + - + +

Not surprisingly in the first shoe presented most asym hands went "wrong" for B side despite of the math advantage.

as.

Next to edge sorting it's me

#### AsymBacGuy

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##### Re: Why bac could be beatable itlr
« Reply #143 on: May 26, 2020, 10:09:42 pm »
• Let's consider our old three different states where every pattern in the universe will belong to.

Generally speaking, the less will be the number of states occurring at a given shoe, better will be the probability to get long winning streaks as a single state or, more likely, a couple of states may be present for long without the "intrusive" effect of the unwelcome third one.

On the other end, we've seen that another strategy relies just upon the opposite thought, that is that certain spots must change their shape in a way or another.

Let's start to examine the theorically "perfect" situations capable to get the highest number of states change happening along any shoe.

When three different states are involved, only six possibilities getting ALL change states come around :

An "endless" succession of 1-2-3-1-2-3-1-2-3.... or 1-3-2-1-3-2-1-3-2... or 2-1-3-2-1-3-2-1-3... or
2-3-1-2-3-1-2-3-1... or 3-1-2-3-1-2-3-1-2... or 3-2-1-3-2-1-3-2-1....

Everything in between gets at least one "winning" situation, that is the third state must be silent for more than the 3-step steady pace featured on the above six patterns.

Notice that all six patterns came out by a 1/3 singles/streaks ratio instead of the more natural 1/1 ratio, meaning that those patterns are "biased" at the start.
Yet we are not interested about the numbers but about the pace.

In a sense we're trying to put in relationship those 6 different "biased" (hence asymmetrical) patterns with the actual natural asymmetrical production, not assigning a precise value to any side or value (as in no way itlr B1=P1, B2=P2 and B3=P3, not mentioning that in the overwhelming majority of times the "3" category inglobes very different patterns).

Even though many "natural" big road or derived roads registrations may offer some profitable opportunities, we need to set up more intricated random walks applied to the actual results' production.

as.
Next to edge sorting it's me

#### AsymBacGuy

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##### Re: Why bac could be beatable itlr
« Reply #144 on: May 26, 2020, 11:23:30 pm »
• A couple of examples taken randomly.

Original shoe results: 2-1-2-1-2-2-1-1-2-1-3-2-1-3-3-1-3-1-3-1-1-1-1-1-1-2-2-3-1-2-2-2-1-1-1

123) +,+,+,-,-,+,-,+,+,-,+,+,-,-,+,+,+,+,+,+,+,-,-,+,+,+,+,+,-,+,-,+

132) +,+,+,-,-,+,-,+,+,-,+,+,-,+,-,-,+,+,+,+,+,-,+,+,-,-,+,+,+,+,+,-,+,-,+

213) -,-,+,+,+,+,+,-,+,+,+,+,+,+,-,+,+,+,+,-,+,+,-,+,+,+,+,+,-,+,-,+,+,+,-

231) -,+,+,+,+,+,+,+,+,+,-,+,+,-,+,+,-,+,+,-,-,+,+,-,+,+,-,+,+,+,+,+,+,+,+,-,+,+

312) +,-,-,+,+,-,+,-,-,+,+,-,+,+,+,+,+,+,-,-,+,+,-,+,+,+,-,-,-,-,+,+,+,+,-

321) +,+,+,+,-,+,+,+,+,+,+,+,+,+,+,+,+,-,-,+,-,+,+,-,+,-,+,-,+,+,+,-,-,+,+

Second shoe: 1-1-1-3-1-2-3-3-2-1-1-2-1-3-2-3-3-1-2-1-1-3-1-3-1-3-1-1-2-3-3-1-1-1-3-1-2-1

123) -,+,+,+,+,+,+,+,+,-,+,+,+,+,+,+,+,+,+,+,+,+,+,-,-,+,+,-,-,-,+,+,+,-,+,+,+,+

132) -,+,+,+,+,-,+,-,-,+,+,-,-,-,-,+,-,+,+,+,+,+,+,+,-,-,+,-,+,++,+,+,-,-,+,+,+

213) +,-,+,+,-,+,+,+,+,+,-,+,+,+,+,+,+,+,-,-,+,+,-,-,+,+,+,+,+,-,+,-,+,+,+,+,-,-

231) +,+,-,+,+,+,+,-,+,+,+,+,+,-,+,+,-,-,-,+,-,+,-,-,-,+,+,+,+,+,+,+,-,-,+,+,+,+,+,-

312) +,-,+,-,-,-,-,+,-,+,-,-,+,+,-,+,+,+,+,-,+,-,-,+,+,+,+,+,+,+,-,-,+,+,+,+,+,-

321) +,+,-,-,+,+,-,+,+,+,+,+,+,+,+,-,+,-,+,+,-,-,+,+,+,+,-,+,-,+,-,+,-,+,+,-,+,+

Even though original shoes were presented by the stupi.dest way of registration (big road) and that many - signs are getting us a -3 unit loss and nearly half of + signs are getting us an inferior +1 payment, some +/- situations are more "due" than others.

Notice that unb plan #1 worked wonderfully on first shoe but quite tremendously bad on the second one.
First shoe presented 21 states change and second shoe 27 states change.

1-step level unb plan #2 results got respectively a LWWW and WW events.

as.
Next to edge sorting it's me

#### AsymBacGuy

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##### Re: Why bac could be beatable itlr
« Reply #145 on: May 27, 2020, 12:06:25 am »
• Another live shoe taken from the now defunct Lucky Dragon casino:

1-2-1-3-2-1-2-1-2-1-3-2-2-2-2-2-1-1-3-1-1-1-1-1-1-1-2-3-2-1-1-1-1-1-1-1-2-2-3-1-1-2-3

123) -,-,+,+,-,+,+,+,+,+,+,+,+,-,+,+,+,+,+,+,+,-,+,+,-,+,+,+,-,+,-,+,+,-,+,+,+,-,-,-,+,+,+

132) -,+,+,+,+,+,+,+,-,-,-,-,+,+,-,+,+,+,+,+,+,-,+,+,-,+,-,+,+,+,-,+,+,-,+,+,+,+,+,-,+,+,+

213) +,+,+,+,+,+,-,-,+,+,+,+,-,+,+,-,-,+,+,-,+,+,-,+,+,+,+,+,+,+,+,-,+,+,-,+,-,+,-,+,-,+,+

231) +,+,-,+,+,-,+,+,+,+,+,+,-,+,+,-,+,-,+,+,-,+,+,-,+,+,+,+,+,-,+,+,-,+,+,-,-,+,+,+,+,+,+

312) +,+,+,+,+,+,+,-,-,+,+,-,+,+,-,+,-,+,-,-,+,+,-,+,+,-,-,-,+,+,+,-,+,+,-,+,+,+,+,+,-,-,-

321) +,-,-,-,-,-,+,+,+,+,+,+,+,-,+,+,+,-,-,+,-,+,+,-,+,+,+,-,-,-,+,+,-,+,+,-,+,-,+,+,+,+,-
Next to edge sorting it's me

#### AsymBacGuy

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##### Re: Why bac could be beatable itlr
« Reply #146 on: May 27, 2020, 01:14:09 am »
• In the last shoe notice how would fare a cumulative strategy applied simultaneously to every 6 possible "highest state" number pattern:

1. - -  + + + +

2. - + + + + -

3. + + + - + -

4. + + + + + -

5. - + + + + -

6. + + + - + -

7. + + - + + +

8. + + - + - +

9. + - + + - +

10. + - + + + +

11. + - + + + +

12. + - + + - +

13. + + - - + +

14. - + + + + -

15. + - + + - +

16. + + - - + +

17. + + - + - +

18. + + + - + -

19. + + + + - -

20. + + - + - +

21. + + + - +

22. - - + + + +

23. + - - + - +

24. + + + + - -

25. - - + + + -

26. + + + + - -

27. + - + + - -

28. + + + + - -

29. - + + + + -

30. + + + - + -

31. - - + + + +

32. + + - + + +

33. + + + - - -

34. - - + + + +

35. + + - + + +

36. + + + - - -

37. + + - - + +

38. - + - + + -

39. - + + + + +

40. - - - + + +
Next to edge sorting it's me

#### RickK

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##### Re: Why bac could be beatable itlr
« Reply #147 on: Yesterday at 09:08:44 pm »
• as...really lost here on what you are recording when you are charting shoes...any chance you could drop back a few steps and maybe take the first shoe you charted and explain a little more about what the numbers represent, i.e. numbers of a particular event, where they developed in the shoe, or anything that might explain it a little more ?...sorry if it's obvious..just not getting it here...Rick

#### RickK

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• Posts: 5
##### Re: Why bac could be beatable itlr
« Reply #148 on: Today at 12:02:02 pm »
• At the top of page 10, you show P2/P1-P3 results. Does each number represent a P2-P1 (2 events) and/or  a P2-P3 ?
A 1 would be the 2 event combination occurred a single time ? And a 2 would be it occurred two times in a row ?

#### AsymBacGuy

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##### Re: Why bac could be beatable itlr
« Reply #149 on: Today at 08:02:19 pm »
• Hi Rickk.

Numbers register how many P2 doubles come out after an initial P2 "trigger": if P2 is limited by an immediate P1 or P3 the number registered will be 1.
If a couple of P2 patterns come around consecutively, we'll write 2. If three P2 patterns show up we'll write 3 and so on.

Example.

BPPBBBPBPPBBPPBBBBBPBPBBPPBPPBBPPBPPBPPPBPPBBBPPPPB according to the P2/P1-P3 r.w is:

1-2-4-2

In the same sequence the B2/B1-B3 r.w. is read as:

1-1-1

as.

Next to edge sorting it's me