### Topic: Why bac could be beatable itlr  (Read 44592 times)

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#### AsymBacGuy

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• Posts: 972
##### Re: Why bac could be beatable itlr
« Reply #225 on: November 17, 2020, 01:37:15 am »
• Palms casino, \$25-\$5000.

PPPP
B
P
BBBBBBB
P
BB
P
BBBB
P
B
PPPP
B
P
BBB
P
BBB
PP
B
PP
B
PPP
B
PPPP
BBBB
PPPPPPP
BB
PP
B
P
BBBBBBBB
PPPPP

ub plan #1 both sides: +++-+-+++++++-++-++++-+-+

ub plan #1 B side: -+++++++--

ub plan #1 P side: ++++++-+-+

ub plan #2: ++

actual random walk: --++++++++++--

as.

Next to edge sorting it's me

#### AsymBacGuy

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• Posts: 972
##### Re: Why bac could be beatable itlr
« Reply #226 on: November 23, 2020, 12:11:37 am »
• The plans applied to the above shoes show that in the long run the final total of bets (units) won or lost will be almost always approaching the zero value, burdened by the vig.
We can't disrupt the math action, we can only take advantage of statistical features happening or not at different degrees shoe after shoe.

Of course we do not know when and how long things might come in our favor, yet some card distributions are more likely than others, meaning that certain patterns are more likely than others.
In other words, patterns may be so hugely card endorsed that we could win 8 or more bets for the entire lenght of the shoe without having a single loss and by a probability way greater than math expected.

Notice that getting an aim to win 8-12 bets in a row without a single loss needs to reduce the actual results by a 1:10 ratio. In reality less than that as many hands are formed by "neutral" ties.

Considering for simplicity BP outcomes as mere 50/50 propositions, math will teach us that we'll win 8 bets in a row by a 1/64 probability (1.56%).
If a method applied to a large sample data provides ratios higher than that we are in very good shape. (Of course the same reasoning applies to lower classes of WL probability values).

From a strict probability point of view that means that baccarat must be solved by disproving a perfect random shoe formation acting here and there with all the related falsifications of the hypothesis.

Certain shoes are more well shuffled than others, anyway it's not how deep or how light shoes are shuffled (unless consecutively dealt), what it counts is about how key cards are more concentrated or diluted along a given shoe. Better sayed, the portions where such features will more likely take place.

Biased shoe.

In the 80s some black jack scholars raised the issue that not everytime a positive count will get the player a math edge over the house, thus enlarging a possible "card clumping" problem.
Simply put, not everytime a shoe supposedly rich of high cards will get the player an edge as those favourable high cards might remain silent in the unplayable portion of the shoe.
Recent studies tried to disprove scientifically this suspicion, nonetheless and knowing the actual bj rules adopted by casinos, the original theory seems to take a more sensible impact.

At baccarat this "clumping card" theory is well more interesting for several reasons and by different points of views:

- besides Montecarlo casino where more than two decks are discarded from the play (8-deck shoes), almost every live casino in the world will deal the shoe for the most entirety of it.
Actually whenever the first card is not a picture or zero value card, most part of the shoe is dealt at different degrees by cutting off very few cards.

- at baccarat we can bet any side we wish at any moment we wish and by any amount we wish.

- at baccarat previous patterns belonging to certain random walks are decisive to know whether the future outcomes will get a more or less key card dilution/concentration as some numbers must follow finite sequences having a given probability to show up.

- at baccarat the key card concentration/dilution problem could be assessed by the times (gaps) some favorite precise two-card points will get a real win or a loss, meaning that favorite two-card points distribution must be registered up to given cutoff points. After those cutoff points are surpassed, we ought to consider that cards are shuffled to get too whimsical results to be properly exploited.

Every baccarat shoe dealt is an endless proposition of two-step math oriented results getting certain gaps of appearance.
First step involves the higher two-card point, this is the main step.
Say the third card impact will be just an accident.

If the third card won't intervene, the probability to get a higher point will be symmetrical, but a finite card distribution will put some limits on it. Depending about how much cards were properly shuffled.
I mean that without the third card intervention and baccarat rules, only an idi.o.t couldn't find a way to beat the game.

Especially if we want to disprove a perfect "so called" ndependent random source of outcomes.

as.
Next to edge sorting it's me

#### AsymBacGuy

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• Posts: 972
##### Re: Why bac could be beatable itlr
« Reply #227 on: November 23, 2020, 11:32:32 pm »
• At baccarat we have endless options to consider binomial propositions, Big Road is by far the most commonly used for wagering, then there are the four derived roads (BP, BEB, SR and CR).

No matter how deep we want to dissect outcomes, math experts teach us that after considering the slight asymmetricity, A=B forever and ever.

In reality A=B with all related statistical implications if the propositions are randomly placed at any shoe dealt.
More precisely, if we bet A at any given stage of the shoe, itlr A or B must follow the old 0.5068/0.4932 ratio. An unbeatable ratio, btw.

This kind of thought is failed by several reasons.

- we can't mix results coming from different shoes as the random postulate cannot be working at any shoe dealt. Actually randomness doesn't work for most shoes dealt.

- at baccarat there's no one single hand getting the 0.5068/0.4932 probability to appear (ties considered neutral).

- pc simulated shoes differ from real live shoes.

- players must rely upon successions of short term situations, the long run apply to very large long term data that easily confuse unrandomness with pseudo randomness, that last one more likely approaching the "expected" values.

- at baccarat place selection feature totally denies the perfect randomness of the outcomes.

- at baccarat probability after events feature totally denies the perfect randomness of the outcomes.

- the number and lenght of "runs" (a run is the number of the shifting attitude of changing the winning side) is quite different than what 50/50 or a 0.5068/0.4932 ratios dictate, thus proving the unrandomness of the outcomes.

Actually I can't swear that the partial and unconstant unrandomness will play the decisive role on that, maybe baccarat is vulnerable by its own characteristics, but I'd tend to be very cautious to state this last assumption.

Next time I'll post some ideas about how to build up a winning random walk.

as.
Next to edge sorting it's me

#### AsymBacGuy

• Moderator
• Posts: 972
##### Re: Why bac could be beatable itlr
« Reply #228 on: November 29, 2020, 10:38:59 pm »
• Although baccarat provides innumerable card situations, math advantaged spots falling here and there are surely going to win itlr (now "itlr" is way more widely intended as commonly considered).
By far the largest impact cards will make over the actual results are coming from 7s, 8s and 9s.

There are many card combinations forming final winning hands not involving those cards (or working partially), nevertheless whenever we are going to peak at our cards we better aim to get one of those cards, instead of hoping that our 4 will be followed by another 4 or a 5.
This card class constitutes 23% of total cards dealt and it's more or less concentrated along the shoe with a "memory" as key cards are burnt from the play.

Differently to black jack where the final count of "good" or "bad" cards must be zero (penetration considered), meaning that only few portions of the shoe might be favourable for the player ('good card' concentration after a strong 'good card' dilution), at baccarat there are no good or bad cards for the player, just probabilities to get key cards concentrated or diluted along various portions of the shoe.
Naturally at baccarat we have the advantage to bet any side we wish anytime we want and the disadvantage to not know which side will be kissed by such a possible key cards impact.

Notice that I haven't mentioned "how much we want" as a long term winning plan must win by flat betting.

Anyway if we are here is because we are trying to prove that a key card concentration/dilution approximation acting along any shoe will play a huge role over our winning probability as many times some actual patterns will be more detectable than others.

It's natural to think that conditions not fitting a perfect random world are more likely to produce winning situations (when properly considered), as "perfect" key cards distributions can easily produce too many undetectable patterns.
If the key card distribution would be always close to the expected 23% ratio, well no betting plan could get the best of it.
Fortunately no one single live shoe dealt in the universe will get such constant ratio.

It's interesting to say that some very unlikely BP patterns will get no hint to be attacked, but certain derived AB situation will at some point.

Obviously we'd prefer to place bets when a derived AB situation MUST come out clustered, thus lowering the probability to catch a losing spot.

Math speaking experts say that every bet will make is EV- no matter what.
That's our fortune.

Tomorrow more AB hints.

as.
Next to edge sorting it's me

#### AsymBacGuy

• Moderator
• Posts: 972
##### Re: Why bac could be beatable itlr
« Reply #229 on: November 30, 2020, 11:58:01 pm »
• Best bac players in the world play a A/B game.
More specifically they play baccarat by taking advantage of runs and gaps.
They do not follow anything, they don't like Banker as being more favorite to win.
They bet very few hands.
And, of course, they bet rarely but wagering huge amounts.

Probability world is made upon runs and gaps, I mean the number of BP shifts. Ouch, AB shifts.
Everything depends about how we want to consider opposite results.

Consider this 18-hand shoe portion:

B
PPPP
BB
PPP
B
PPPPP
BB

7 runs, 2 singles, 1 double, 3 3+s  singles/streaks ratio 2:5

A
BB
A
B
A
B
AAA
BBB
A
B

10 runs, 7 singles, 1 double, 2 3+s  singles/streaks ratio 6:3

AA
BB
AA
BB
A
B
AA

7 runs, 2 singles, 4 doubles, zero 3+s. singles/streaks ratio 2:5

A
B
AA
B
A
BB
A
B

8 runs, 5 singles, 2 doubles, zero 3+s. singles/streaks ratio 5.2

In summary the overall singles/streaks ratio is 15:15

The overall doubles/3+s ratio is 11:5

Now this second 18-hand shoe portion:

B
P
BB
P
B
P
B
P
BBB
P
BBBBB

11 runs, 8 singles, 1 double and 2 3+s.  singles/streaks ratio 8:3

A
BBB
AAAA
B
A
BBB
AAA

7 runs,  3 singles, zero doubles, 4 3+s.  singles/streaks ratio 3:4

BB
A
B
AAA
B
A
B
AAA
B
A

10 runs, 6 singles, 1 double, 2 3+s. singles/streaks ratio 6:3

AA
B
AA
B
A
B
A
B
AAA

9 runs, 6 singles, 2 doubles and 1 3+s. singles/streaks ratio 6:3

Overall the singles/streaks ratio is 23:13.

The doubles/3+s ratio is 16:12

Both shoes portions came from a moderate/strong key card concentration, even though they formed quite different BP results.

Of course I've posted the most common derived roads any bac player is familiar of and the fact that some AB patterns are cumulatively superior than counterparts was just a coincidence.
Moreover any B or P result could form opposite patterns on different roads (a single from one part and a streak from another one, etc).

We can put in action more random walks, for example OBL, A=same, B=opposite:

first shoe

A
BB
AAAA
B
A
B
A
BBB
AA

second shoe

B
AA
BBBBB
A
B
A
B
A
BBB

As long as long streaks are not coming in short intervals and as long as "symmetrical patterns" are not coming out consecutively any betting plan has its merit. And odds are that they do not itlr.
Btw, those are the precise situations recreational players are looking for. And I do not know a single recreational player being ahead of the game.

as.
Next to edge sorting it's me