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Messages - AsymBacGuy

#1261
AsymBacGuy / Roulette: a sure long term finding
May 18, 2016, 10:31:33 PM
Even though roulette is a perfect independent results' game, there are some interesting long term features that could be easily tested by everyone.
The strategy  was conducted over 1.500.000 real spins (single zero).

I mean some events are more likely than others. Unfortunately zero tax and some other practical features will lower a lot the value of such aknowledge.

The trigger we are looking for is really simple: we take note of the last number produced then we bet all the 3 EC belonging to this number. And this procedure is made per every last number sorted out.

For example number 33 sorted out, next spin we want to bet black, odd and high.

Obviously such betting will get 4 different outcomes (zero ignored):

- winning all 3 EC (sorting of 29,31,33,35): +3

- winning just one unit (sorting of 11,13,15,17,19,21,22,23,24,25,26,27,28): +1

- losing just one unit (sorting of 1,2,3,4,5,6,7,8,9,10,20,30,32,34,36): -1

- losing all 3 EC (sorting of 12,14,16,18): -3

Well, in the long run the number of spots winning all 3 bets are greater than the number of spots losing all 3 bets and it will increase the more the hands are played.

Of course to try getting the best of it from this finding needs also to take into account the spots where we'll either win or lose 1 unit.

Despite of what many may think, there are no better numbers or worse numbers to spot as triggers.

True, red-odd-low numbers or black-even-high numbers should have the theorical best probability to match the same EC on the next spin but this wasn't the case, at least on our quite long sample.

In a word and transferring the plan on the statistical field, we'll expect to have more single total different EC outcomes than streaks of different EC outcomes, more double different EC outcomes than 2+ streaks of different EC outcomes and so on. And the reverse is also true regarding the same EC situations (more streaks than single, more 2+ than doubles, etc).

The variance and the weight of zero could be quite high, yet the final result will be sure.

as.   

 

   














 









#1262
AsymBacGuy / Re: A progression that can't lose
May 16, 2016, 10:59:18 PM
Baccarat world moves from deviations and equilibrium points and we know that some events tend to be deviated infinitely whereas others won't, those last touching many many times the zero (equilibrium) cutoff.
Nonetheless we know that such features may easily disappoint this assumption, let's think about B/P gap favoring the right side, B singles overcoming B streaks and many other related issues.

Anyway we know that thanks to the bac features, the ranges of intervention of certain events will be more restrained than others by 100% accuracy.

Which movements should help us most in order to control the outcomes?

Of course those events touching more likely the zero point several times so that the progression presented can "cover" the slight expected deviations going toward a side or another.

We see that in such effort we don't want to classify a given event as good or bad, we want just to pick up the situations where some deviations are more likely to go back to an ideal zero point.

Let's call it a kind of "equilibrium" goal, but it really isn't as every event slightly going back and forward to a point different to zero will allow us to get a profit by a very slow MM.

So we know that even aiming to points different to zero we could find possible valid spots to start our betting.

In reality, most part of authoritative gambling experts of the past had stated that searching for deviations is the best tool to try to get the best of it.
A perfect opposite thought of what just discussed.

Since I'm not in the position of confuting such claimings, I'll take for grant this thought.
Better going toward deviations than hoping for a return to equilibrium.
Still the problem is posed about how much to look for deviations. In a word, when to stop the betting or, better, when to start the betting?

Good questions.

Yet we get answers.

Without a doubt, the probability to get deviations toward a given side and per any event considered will be increased after many equilibrium points will be crossed, and of course we know that we don't necessarily need to reach zero eq. points to win.

On the other hand, a lot of equilibrium points (widely considered) crossed must be considered as a deviation itself.

Then our strategic plan should be focused about deviations. Period.

Easy to say it, but what about the practice?

Actually there are no positive or negative deviations. There are deviations. Each deviation will get its peaks in a way or another. It's just a matter of time. 
Some deviations are more likely than others, nevertheless even an unexpected deviation (multiple clusters of B singles vs B streaks, for example) will reach high points. We can't do anything to prevent such occurence if we have chosen to bet toward B streaks. 
But we know that this unlikely occurence cannot last for long.

Imo, the probability to get simultaneous high deviations on multiple events (deviations also intended as many equilibrium spots reached) is so small to allow us to set up a profitable plan.

as.

   
#1263
AsymBacGuy / Re: A progression that can't lose
May 12, 2016, 08:39:29 PM
Of course the goal on the W side progression will be to get a deviation instead of an equilibrium. Hence utilizing an opposite way of thinking (positive progression).


as.


#1264
AsymBacGuy / Re: A progression that can't lose
May 12, 2016, 08:23:41 PM
Quote from: TheLaw on May 12, 2016, 07:25:29 PM
Hey AsymBacGuy,

What if you looked at how often this method would lose.......and then created a progression around that W/L record?

Example :

If you lose all 66 units let's call it 1 total bank loss. Now, the next series is the same..........trying to win the 66 units back.

Then if that is a loss, then we multiply the total bank x 2. Basically we are using a Labouchere with the total bank instead of each bet.

You would need to lose many banks (66 units each) to lose your entire stake. In this example your total stake might be 10 banks or more.

I call this a nested progression.......and you can use as many as you like to create a method. :thumbsup:

Yep. I called it as a multilayered plan but the concept is the same.

And your idea is appliable even before starting the original progression so further diluting the already very low risk of busting.

Even better is to form separate progression banks per any distinct event played.

At baccarat whenever certain events are cold, generally some derived and indirect opposite situations are hot or at least not cold simultaneously.
Therefore we might set up this progression working on both cold (L) and hot (W) different situations, so reducing the stress to forcely reach an equilibrium point.

Now we don't necessarily need to reach the equilibrium in a way or another PER EACH BANK (EVENT) PLAYED as we can even let it go the bank(s) performing bad, as we have gained the profit elsewhere.


as. 











 



#1265
AsymBacGuy / Re: A progression that can't lose
May 12, 2016, 04:56:54 PM
Quote from: alrelax on May 12, 2016, 04:12:56 PM
Just don't let me get too side tracked looking out for the black panthers, SoS Flags, dogs and cats on skateboards, Vanilla Ice, Black Horses running through the casino, Wizards, Gnomes, etc., etc.   ;)

What about dragons?  ^-^

as.
#1266
AsymBacGuy / Re: A progression that can't lose
May 12, 2016, 04:08:51 PM
Quote from: alrelax on May 12, 2016, 03:46:11 PM
Absolutely each of us has considerations, which IMO weigh more than the percentages of what can or cannot prevail according to 10,000 or 100,000 or 1 million shoes or in fact, past shoes, etc.

I can't wait to play with you (as well as with some others), I'm confident we might improve our strategies reciprocally.


as.
#1267
AsymBacGuy / Re: A progression that can't lose
May 12, 2016, 03:20:15 PM
Al, I understand your points and in your fairness nobody demonstrated a possible value of statistics applied at baccarat.

But I'm firmly convinced that besides certain considerations, what it counts is the W/L ratio and its related issues.

Are there some features helping me to occasionally guess which direction will take place some events? Imo, yes.


as. 



#1268
AsymBacGuy / Re: A progression that can't lose
May 12, 2016, 02:52:21 PM
Quote from: Albalaha on May 12, 2016, 02:59:33 AM
Asym,
             You started with a wrong direction. There is no equilibrium in a game of house edge as every bet is subjected to that and in long run, all bets will go far from equilibrium in terms of "extra losses". Variance can take them even more far. Even in a game without any house edge, a bet might not get equilibrium even after a billion trials.
               D'alembert is a classic comedy of errors and based on ideas that do not work in real life. It has no mathematical basis to make it a winner.

True, but equilibrium should be intended in several ways, let's say a sort of RTM effect or something like that.

D'Alambert is the worse progression ever invented, sure.

as.   
#1269
AsymBacGuy / Re: A progression that can't lose
May 12, 2016, 02:43:04 PM
I understand your skepticism but look, if we can't win hoping for equilibrium or a kind of it, how can we think to win hoping for positive deviations always shifted on our side?

Yes, ideally we can if we are able to spot more likely events restricted in their variance.

So, B/P results on Zumma books cannot teach us nothing because B is more likely than P but the variance is astronomical.

Naturally the word equilibrium must be intended in a wide sense.

This progression might not lose as Alrelax correctly pointed out but still has some merits.

And its value is enhanced by finding the proper situations where it might work better, especially if we're using it after some fictional play.

Slow progression and a slow proper betting selection are the best tools, imo, to try to control the random world.

The rule is to lose, first let's focus about NOT to lose then we will think about winning.

 
as.





#1270
AsymBacGuy / Re: A progression that can't lose
May 12, 2016, 02:29:28 AM
Quote from: 21 Aces on May 12, 2016, 01:59:55 AM
Ok I now understand you are going for two consecutive wins after the losses.  Then I don't understand the last column 'gain after the equilibrium will be reached'.  you are just trying to find two wins after a possibly huge series of losses?

Nope, but I understand as I haven't clarified well the topic.

The first column is the most important. It tells us how many losses we got.
That is HOW MANY BETS I'M BEHIND TO GET THE EQUILIBRIUM (W=L)

If I'm losing 7 bets (consecutively or not, what it counts is the overall total) I'll have to stay at this level for 7 bets as I'll get the equilibrium after those 7 bets. Giving me a profit of $280.

In a word, we have to take into account the L number and bet what that level dictates up to the equilibrium point.

If we're at level #7 and we win the first hand we keep playing this level up to the equilibrium. If after a win at a given level we lose, we don't go to #8 level as the number of losses hasn't change (always 7). Only if we'll have a first loss at a given level we'll go to #8 and so on.

Obviously final levels can get a profit even though we won't reach the equilibrium point.

But under normal circumstances, it would be a mistake to stop the searching of the eq. point as we want to get a profit PROPORTIONALLY placed to the risk involved.


as. 
 

 


#1271
AsymBacGuy / Re: A progression that can't lose
May 12, 2016, 02:11:43 AM
As many times mentioned, a progression must hold up to the most likely variance features any game will provide. And talking about 50/50 games, variance cannot act other than on W/L or on/off outcomes.

It's reasonable to put the variance limit into a deviation range of -25 and +25 (5 sr).
Here we got an even higher amount, a 5.56 level.

Never forgetting that the winning or losing sequences must be always put in relationship with the ideal zero point.
Hence a -18 overall losing sequence following a +12 situation isn't a sort of statistics disaster: here we're just 6 step far from the zero point.

In this example, starting the L progression at the zero point would be a far better idea than starting it at the +12 level (6 losses vs 18 losses).
That's why we have to adjust the progression even on the W side, this time lowering the bets.

as.     

 

   
#1272
AsymBacGuy / Re: A progression that can't lose
May 12, 2016, 01:38:04 AM
Quote from: 21 Aces on May 12, 2016, 01:06:04 AM
The gain after the equilibrium will be reached looks incorrect.  If you bet $110 the 2nd bet and your total value it risk from the start with that bet is $210 and you win then you are at -$100 P&L (-$100 1st Bet + $110 2nd Bet = -$100).

??

210 is the loss amount if you lose both 1st and 2nd hand.
If you are at a -210 level, you need to win two hands at level #3 ($200 + $30) getting a gross profit of $30. 

No way the progression can be wrong, it was carefully studied more than one century ago.
But unfortunately it was devised for roulette where it cannot work by obvious reasons.   

as.



#1273
AsymBacGuy / Re: A progression that can't lose
May 12, 2016, 01:32:53 AM
Quote from: soxfan on May 12, 2016, 12:57:55 AM
There is an Armenian cat on a member only dice forum that has come up with a progression style that is, imho, nearly unbreakable. That's cuz it covers the mathematical expectation of a certain event popping within so many toss of the cubes. So, if the action at the dices table always play out according to expectation then he would never bust a progression. But there are practical reasons it's hard to play in that you need a minimum 10 thousands unit bankroll, and they stones to make the max bet of slightly more than 800 unit. And ya gotta have access to a joint that gives you a nice fat spread between min-mx bets, hey hey.

Yep, limiting the bankroll should be of paramount importance at the cost to play a very slow game.

as.
#1274
AsymBacGuy / Re: A progression that can't lose
May 12, 2016, 01:31:00 AM
Quote from: Tomla on May 12, 2016, 12:20:33 AM
are you saying bet 1 unit 10x then 2 units 20x?

Nope. You have to start the progression whenever you get some losses and the first column (# of losses) dictates how mush to bet.
You keep staying at the same level if you win up to the point where you get the zero equilibrium point, in that case you get a profit.
If you lose, you keep track of the losses (again column #1) and act accordingly.

Of course a perfect progression/regression should start at a higher standard bet working counterwise, let's say $200. Whenever you win you go down the same you would do with the L progression.

When we win the house owns something from us as well as we expect something from it after some losses. 

as.

 
   
#1275
AsymBacGuy / A progression that can't lose
May 11, 2016, 11:19:31 PM
We know that any progression will get the best of it whenever a zero equilibrium point will be reached within a fair amount of trials. Of course some progressions could do even better, that is getting the player a profit even when the W/L ratio is shifted toward the right.
Notice that the well known D'Alambert progression will win 1 unit after the equilibrium is reached but not everytime as everything depends about the DISTRIBUTION of W and L.

Here I'm talking about the almost absolute impossibility to lose our entire bankroll and this is a total different thing than stating that we will win easily. Nonetheless knowing that we won't lose in the longest possible runs isn't a vulgar accomplishment.

I have to forcely consider a $100 standard unit bet and the total bankroll is $6600 (66 units).
For simplicity we won't take into account the commission when applied.
Remember that our goal is to reach at a given point a zero equilibrium point, meaning we want to get the W/L ratio = zero.
Later more on that.

Columns are: L deviations, betting amount in $, financial exposure, gain after the equilibrium will be reached

0  $100           100   -
1  $100 + $10 210 10
2  $100 + $20 330 30
3  $100 + $30 460 60
4  $100 + $40 600 100
5  $100 + $50 750 150
6  $100 + $60 910 210
7  $100 + $70 1080 280
8  $100 + $80 1260 360
9  $100 + $90 1450 450
10 $100 + $100 1650 550
11 $200 + $10 1860 660
12 $200 + $20 2080 780
13 $200 + $30 2300 780
14 $200 + $30 2430 910
15 $200 + $30 2760 910
16 $200 + $30 2990 910
17 $200 + $40 3130 1050
18 $200 + $40 3370 1050
19 $200 + $40 3610 1050
20 $200 + $40 3850 1050
21 $200 + $50 4100 1200
22 $200 + $50 4350 1200
23 $200 + $50 4600 1200
24 $200 + $50 4850 1200
25 $200 + $50 5100 1200
26 $200 + $60 5360 1360
27 $200 + $60 5620 1360
28 $200 + $60 5880 1360
29 $200 + $60 6140 1360
30 $200 + $60 6400 1360
31 $200 + $60 6600 1360

We see that to lose our entire bankroll we need either a 5.56 sr negative deviation (like looking at 31 negative hands in a row, a 31 streak) or, most likely, a W/L gap of 31.

Every roulette player knows that a gap between even chances could easily reach and surpass the W/L amount (btw a 31 streak is a very very very rare finding also at this game) but at baccarat we have a lot of ploys to find two opposite events that cannot reach the 31 negative (or less likely positive)value by any fkn means.
Especially if we want to prolong the progression by another 10 or so steps. 

So we know that adopting this slow progression we can't lose or, better sayed, that the probability to lose is really very very low, let's say almost impossible.

And, wonder of wonders, with proper adjustments we may use it betting only the Player side, hence knowing that we won't pay a bit of commission.

In a word, we can even regularly bet the unfavourable side knowing that we can't lose itlr.

A further example why we have to play slowly and with a lot of patience.

as.