Hi KFB!
Unb plan #1 relies upon the probability to get a fair singles vs streaks ratio along with a kind of asymmetrical streaks distribution considered within the doubles/3+ streaks range.
Simplifying, we'll set up a two events vs one strategy, where one event remains as a steady winner (singles) at the same time confiding that specific streaks (doubles or 3+ streaks) will come out clustered at least for 1 point along the way.
So if singles=1, doubles=2 and 3+s=3, we'll bet that the 1-2 or 1-3 two events will last at least one time, thus we're discarding the 2-3 or 3-2 sequences that must be attacked by another point of view.
In a word, before betting we'll wait for two such 1-2 or 1-3 events to come out at least one time, 'hoping' they'll prolong by two different levels: singles and/or the same streak (2 or 3+) happened so far.
When the number of singles is roaming around averages or slightly below averages and unless a long singles streak happened right at the start of the shoe, the number of 1 - (2-3) permutations makes this plan invincible as a possible '2/3+ streaks' hopping needs a kind of 'perfect unlikely' pace to show up.
Naturally whenever the singles/streaks ratio happen to be moderated or strongly shifted toward the left, winning is a joke.
The opposite situations, that is shoes rich of streaks and poor of singles makes this plan less straightforward as now we ought to transfer the distribution problem from horizontal to vertical.
Of course I'm just talking about a simple 'Big Road' result succession, there's another B or P events registration taken separately (not mentioning derived roads, again splitted withing red and blue spots and so on)
Examples.
Aria casino shoe, september 2019.
Big Road (first hand is P)
3-1-1-2-1-1-1-3-3-1-3-1-1-2-3-1-1-3-2-1-2-1-1-2-1-1-3-1-1-3-2-1-3-2-1-2
Unb plan #1 (1-2 and 1-3):
+ - + + + - + + + + + - - + + - + + + + + + - + + + - - - +
Now let's consider my unb plan #1 splitted into two different B and P registrations.
P= 3-1-1-1-3-3-1-3-1-2-2-1-1-3-1-2-3-1
B= 1-2-1-3-1-1-2-1-3-1-1-2-1-1-3-1-2-2
Unb plan #1 applied to those two different P and B registrations (lines):
P= + + + + + + + - + + + - + - -
B= + - + + - + - + + - + + - + - +
Now I'm asking: is that shoe forming whimsical positive spots denying the natural math impact over the outcomes?
No fkng way, eventually every line got all cumulative negative amounts (plus vig).
Nonetheless, I'm certain that most acute bac players will get the best of it by selectively betting some spots.
One more shoe, now a streaks rich shoe.
Aria casino shoe, september 2019
First hand is B.
Big Road
1-2-1-2-1-3-1-2-1-3-1-3-1-3-2-1-1-2-3-3-3-3-3-2-1-3-1-3-1-2-3-...
Unb plan #1:
+ + + - + - + - + + + - + + - - + + + - -
B= + - + + + + + + +
P= - -
Notice that streaky shoes make less likely to happen symmetrical long lines at B and P lines taken from a unb plan #1 point of view.
In this shoe we had just one option to take, that is by folding most of the hands, thus betting very few spots or nothing at all.
It's true that the unlikely 3-3-3-3-3 sequence (or that P side formed all streaks and just one single) would have made many recreational players as winners, yet do not forget the word 'unlikely'.
People making a living about numbers rely upon more probable events to show up and not chasing 'miracles'.
For that matter even this shoe constitutes a wonderful opportunity to win several hands in a row, providing what to look for, now by a 'more likely propensity' to happen, already traced in my thread.
Good news is that whenever the 'horizontal' way seems to fail, the 'vertical' one takes a decisive long term role.
Simplyfing, streaky shoes that tend to deny a steady 'horizontal' unb plan #1 advantage, will distribute by very low variance lines, meaning that some streaks cutoff points won't follow the (unbeatable) expected values.
Back to this 'streaky' shoe.
Let's analyze more deeply the streaks nature at Big Road (number displayed is the streak exact lenght):
2, 2, 3, 2, 6, 5, 3, 2, 2, 5, 4, 3, 3, 3, 2, 4, 3, 4.
If you've read my previous posts you promptly see that no hopping situation will last for long (if any), so only an id.io.t may miss what is more likely to happen at some point of the shoe.
To get a better idea of what I'm talking about, let's consider another streaky real shoe, now considering just exact streak values:
4, 4, 4, 2, 5, 2, 6, 3, 3, 3, 4, 2, 4, 3, 4, 4, 3.
Notice the different cutoff points making 'clustered streak events', anyway even this shoe would be hugely beatable by other tools, so let's provide the complete shoe's texture (now streaks are considered under the 2 or 3 simplyfied form):
First hand is P.
1, 3, 1, 3, 3, 1, 2, 3, 1, 1, 2, 3, 3, 3, 3, 1, 3, 1, 2, 3, 1, 3, 1, 3, 1, 3, 3, (2)
Unb plan #1:
+ + + + - - + - - + + - - + + + + + + (-1)
Unb plan at P side:
- + - + + -
Unb plan at B side:
+ + + + + + + + + + (-1)
In some way we got plenty of room to spot two-step 'all-in' situations without having to wait for 'premium' poker hands.
Just out of curiosity let's see how this shoe perfroms at three derived roads (unb plan à1 dsplayed with + and - signs):
byb:
1, 2, 3, 2, 1, 3, 3, 1, 1, 2, 2, 1, 2, 1, 1, 3, 1, 2, 1, 3, 2, 2, 3, 1, 1, 1, 3, 1, 3, 2, 3, 2, 2, 1, 1.
- - + + + - + + + + + - + - + - - + + + + + - +
sr:
1, 3, 1, 2, 2, 1, 2, 3, 3, 3, 1, 1, 2, 2, 1, 3, 1, 2, 2, 2, 2, 2, 2, 2, 1, 3, 2, 3, 1, 1, 1.
+ - + + + - + - + + - + - + + + + + + + - - + +
cr:
1, 2, 1, 3, 3, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 2, 2, 2, ,2, 3, 3, 3, 2, 3, 2, 1, 2, (2)
+ - + - + + + - + + + + + + - + (-1)
Splitting the derived roads results into blue (B) and red (R) spots:
byb:
B= 1, 3, 1, 3, 1, 2, 2, 1, 1, 1, 2, 3, 1, 3, 3, 3, 2, 1
unb plan #1: + + + - + + + + + - + + + -
R= 2, 2. 3. 1. 2. 1. 1. 3. 2. 3. 2. 1. 1. 1. 2. 2. 1.
unb plan #1: - + + - - + + + + +
sr:
B= 1, 1, 2, 2, 3, 1, 2, 1, 1, 2, 2, 2, 1, 2, 1, 1
unb plan #1: + - - + + + + + + + + +
R= 3, 2, 1, 3, 3, 1, 2, 3, 2, 2, 2, 2, 3, 3, 1
unb plan #1: - + + - -
cr:
B= 1, 1, 1, 2, 1, 2, 2, 1, 2, 2, 3, 3, 3, 1, (2)
unb plan #1: + + + + + + - (-1)
R= 2, 3, 3, 1, 1, 3, 1, 1, 2, 2, 3, 2, 2, 2.
unb plan #1: + + + + - + -
Finally all three derived roads considered by a succession of r and b spots:
byb:
1, 2, 3, 2, 1, 3, 3, 1, 1, 2, 2, 1, 2, 1, 1, 3, 1, 2, 1, 3, 2, 2, 3, 1, 1, 1, 3, 1, 3, 2, 3, 2, 2, 1, 1.
Unb plan #1:
- - + + + - + + + + + - + - + - - + + + + + - +
sr:
1, 3, 1, 2, 2, 1, 2, 3, 3, 3, 1, 1, 2, 2, 1, 3, 1, 2, 2, 2, 2, 2, 2, 2, 1, 3, 2, 3, 1, 1, 1.
unb plan #1:
+ - + + + - + - + + - + - + + + + + + + - - + +
cr:
1, 2, 1, 3, 1, 3, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 2, 3, 2, 1, 2 (2)
unb plan #1:
+ - + + - + + + - + + + + + + - + (-1)
oOoOo
At baccarat there are no 'imperfect informations' to look for, meaning we do not need to overthink a fkng nothing.
We can't bluff of course, but we can't be bluffed either.
Therefore no game theory applies at baccarat, we're just playing a taxed game where at some points some events are more likely to appear than others.
As long as a 312 or 416 cards shoe is ready to be played and cards utilized are burnt from the play, a given number of asymmetrical patterns will take place, either for bac rules and, more importantly, for the actual card distribution.
Ignorant people (math experts first) when talking about baccarat directly fall into the Dunning-Kruger effect category.
The more they think to know about baccarat, greater it's their incompetence on the subject.
Good for us.
as.
Unb plan #1 relies upon the probability to get a fair singles vs streaks ratio along with a kind of asymmetrical streaks distribution considered within the doubles/3+ streaks range.
Simplifying, we'll set up a two events vs one strategy, where one event remains as a steady winner (singles) at the same time confiding that specific streaks (doubles or 3+ streaks) will come out clustered at least for 1 point along the way.
So if singles=1, doubles=2 and 3+s=3, we'll bet that the 1-2 or 1-3 two events will last at least one time, thus we're discarding the 2-3 or 3-2 sequences that must be attacked by another point of view.
In a word, before betting we'll wait for two such 1-2 or 1-3 events to come out at least one time, 'hoping' they'll prolong by two different levels: singles and/or the same streak (2 or 3+) happened so far.
When the number of singles is roaming around averages or slightly below averages and unless a long singles streak happened right at the start of the shoe, the number of 1 - (2-3) permutations makes this plan invincible as a possible '2/3+ streaks' hopping needs a kind of 'perfect unlikely' pace to show up.
Naturally whenever the singles/streaks ratio happen to be moderated or strongly shifted toward the left, winning is a joke.
The opposite situations, that is shoes rich of streaks and poor of singles makes this plan less straightforward as now we ought to transfer the distribution problem from horizontal to vertical.
Of course I'm just talking about a simple 'Big Road' result succession, there's another B or P events registration taken separately (not mentioning derived roads, again splitted withing red and blue spots and so on)
Examples.
Aria casino shoe, september 2019.
Big Road (first hand is P)
3-1-1-2-1-1-1-3-3-1-3-1-1-2-3-1-1-3-2-1-2-1-1-2-1-1-3-1-1-3-2-1-3-2-1-2
Unb plan #1 (1-2 and 1-3):
+ - + + + - + + + + + - - + + - + + + + + + - + + + - - - +
Now let's consider my unb plan #1 splitted into two different B and P registrations.
P= 3-1-1-1-3-3-1-3-1-2-2-1-1-3-1-2-3-1
B= 1-2-1-3-1-1-2-1-3-1-1-2-1-1-3-1-2-2
Unb plan #1 applied to those two different P and B registrations (lines):
P= + + + + + + + - + + + - + - -
B= + - + + - + - + + - + + - + - +
Now I'm asking: is that shoe forming whimsical positive spots denying the natural math impact over the outcomes?
No fkng way, eventually every line got all cumulative negative amounts (plus vig).
Nonetheless, I'm certain that most acute bac players will get the best of it by selectively betting some spots.
One more shoe, now a streaks rich shoe.
Aria casino shoe, september 2019
First hand is B.
Big Road
1-2-1-2-1-3-1-2-1-3-1-3-1-3-2-1-1-2-3-3-3-3-3-2-1-3-1-3-1-2-3-...
Unb plan #1:
+ + + - + - + - + + + - + + - - + + + - -
B= + - + + + + + + +
P= - -
Notice that streaky shoes make less likely to happen symmetrical long lines at B and P lines taken from a unb plan #1 point of view.
In this shoe we had just one option to take, that is by folding most of the hands, thus betting very few spots or nothing at all.
It's true that the unlikely 3-3-3-3-3 sequence (or that P side formed all streaks and just one single) would have made many recreational players as winners, yet do not forget the word 'unlikely'.
People making a living about numbers rely upon more probable events to show up and not chasing 'miracles'.
For that matter even this shoe constitutes a wonderful opportunity to win several hands in a row, providing what to look for, now by a 'more likely propensity' to happen, already traced in my thread.
Good news is that whenever the 'horizontal' way seems to fail, the 'vertical' one takes a decisive long term role.
Simplyfing, streaky shoes that tend to deny a steady 'horizontal' unb plan #1 advantage, will distribute by very low variance lines, meaning that some streaks cutoff points won't follow the (unbeatable) expected values.
Back to this 'streaky' shoe.
Let's analyze more deeply the streaks nature at Big Road (number displayed is the streak exact lenght):
2, 2, 3, 2, 6, 5, 3, 2, 2, 5, 4, 3, 3, 3, 2, 4, 3, 4.
If you've read my previous posts you promptly see that no hopping situation will last for long (if any), so only an id.io.t may miss what is more likely to happen at some point of the shoe.
To get a better idea of what I'm talking about, let's consider another streaky real shoe, now considering just exact streak values:
4, 4, 4, 2, 5, 2, 6, 3, 3, 3, 4, 2, 4, 3, 4, 4, 3.
Notice the different cutoff points making 'clustered streak events', anyway even this shoe would be hugely beatable by other tools, so let's provide the complete shoe's texture (now streaks are considered under the 2 or 3 simplyfied form):
First hand is P.
1, 3, 1, 3, 3, 1, 2, 3, 1, 1, 2, 3, 3, 3, 3, 1, 3, 1, 2, 3, 1, 3, 1, 3, 1, 3, 3, (2)
Unb plan #1:
+ + + + - - + - - + + - - + + + + + + (-1)
Unb plan at P side:
- + - + + -
Unb plan at B side:
+ + + + + + + + + + (-1)
In some way we got plenty of room to spot two-step 'all-in' situations without having to wait for 'premium' poker hands.
Just out of curiosity let's see how this shoe perfroms at three derived roads (unb plan à1 dsplayed with + and - signs):
byb:
1, 2, 3, 2, 1, 3, 3, 1, 1, 2, 2, 1, 2, 1, 1, 3, 1, 2, 1, 3, 2, 2, 3, 1, 1, 1, 3, 1, 3, 2, 3, 2, 2, 1, 1.
- - + + + - + + + + + - + - + - - + + + + + - +
sr:
1, 3, 1, 2, 2, 1, 2, 3, 3, 3, 1, 1, 2, 2, 1, 3, 1, 2, 2, 2, 2, 2, 2, 2, 1, 3, 2, 3, 1, 1, 1.
+ - + + + - + - + + - + - + + + + + + + - - + +
cr:
1, 2, 1, 3, 3, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 2, 2, 2, ,2, 3, 3, 3, 2, 3, 2, 1, 2, (2)
+ - + - + + + - + + + + + + - + (-1)
Splitting the derived roads results into blue (B) and red (R) spots:
byb:
B= 1, 3, 1, 3, 1, 2, 2, 1, 1, 1, 2, 3, 1, 3, 3, 3, 2, 1
unb plan #1: + + + - + + + + + - + + + -
R= 2, 2. 3. 1. 2. 1. 1. 3. 2. 3. 2. 1. 1. 1. 2. 2. 1.
unb plan #1: - + + - - + + + + +
sr:
B= 1, 1, 2, 2, 3, 1, 2, 1, 1, 2, 2, 2, 1, 2, 1, 1
unb plan #1: + - - + + + + + + + + +
R= 3, 2, 1, 3, 3, 1, 2, 3, 2, 2, 2, 2, 3, 3, 1
unb plan #1: - + + - -
cr:
B= 1, 1, 1, 2, 1, 2, 2, 1, 2, 2, 3, 3, 3, 1, (2)
unb plan #1: + + + + + + - (-1)
R= 2, 3, 3, 1, 1, 3, 1, 1, 2, 2, 3, 2, 2, 2.
unb plan #1: + + + + - + -
Finally all three derived roads considered by a succession of r and b spots:
byb:
1, 2, 3, 2, 1, 3, 3, 1, 1, 2, 2, 1, 2, 1, 1, 3, 1, 2, 1, 3, 2, 2, 3, 1, 1, 1, 3, 1, 3, 2, 3, 2, 2, 1, 1.
Unb plan #1:
- - + + + - + + + + + - + - + - - + + + + + - +
sr:
1, 3, 1, 2, 2, 1, 2, 3, 3, 3, 1, 1, 2, 2, 1, 3, 1, 2, 2, 2, 2, 2, 2, 2, 1, 3, 2, 3, 1, 1, 1.
unb plan #1:
+ - + + + - + - + + - + - + + + + + + + - - + +
cr:
1, 2, 1, 3, 1, 3, 2, 1, 1, 1, 2, 3, 2, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 2, 3, 2, 1, 2 (2)
unb plan #1:
+ - + + - + + + - + + + + + + - + (-1)
oOoOo
At baccarat there are no 'imperfect informations' to look for, meaning we do not need to overthink a fkng nothing.
We can't bluff of course, but we can't be bluffed either.
Therefore no game theory applies at baccarat, we're just playing a taxed game where at some points some events are more likely to appear than others.
As long as a 312 or 416 cards shoe is ready to be played and cards utilized are burnt from the play, a given number of asymmetrical patterns will take place, either for bac rules and, more importantly, for the actual card distribution.
Ignorant people (math experts first) when talking about baccarat directly fall into the Dunning-Kruger effect category.
The more they think to know about baccarat, greater it's their incompetence on the subject.
Good for us.
as.