*PATTERN 4*

[Johnlegend]

As long as you don't get past the double loss, that's the killer

P3 is a nasty escalation, and will risk the bank in my opinion especially if playing with more than 1pounds. no difference to standard martingale which can threaten table limit. also the P3 progression is simply to recover ,and gain a net 1 unit. so it makes no sense to risk hundreds of pounds for 1 unit .

e.g.

1,2,4 - not too bad here eh!

8,16,32 >> still okay!

48,96,192 >> yikes...its getting hot in here!

Halba1 you don't try to recover the whole stake in one game that's the classic mistake. You recover 3/7 or there abouts so you would stake something like this

1,2,4

3,6,12

12,24,48

that's what I do but I have the bankroll. Remember the first game of each day I play at level 2 for both Hi lo and od ev so you are pinching extra units on these first games before dropping down to level 1. And when you hit a double figured streak for that first game which you will often. This is where one of your advantages is gained. So when I get a losing game which is nearly always later pulling back the loss is much easier.

And when hit and run works its magic that's when you really go ahead. One step backwards three steps forward is the name of the game. ALWAYS playing with a BR at least 20 times bigger than your level one stakes. Preferably 40 times bigger. I kid you not.

The persom who has a 1,000 unit BR and plays for 5-7 max units per game as I do is set for life. I make no bones about that.

[Johnlegend]

Hi JL,

Since there has been a little 'heat' in this thread I decided to calculate just how likely your results are, according to probability theory. Now I know that you are contemptuous of mathematics, but that's ok, the maths doesn't care; it is what it is.  

Your results are:

TOTAL GAMES PLAYED 510
TOTAL GAMES WON 471
TOTAL GAMES LOST 39

I haven't actually tested your system, but I understand that the chance of a loss is approx 1/8 (ignoring the zero effect).

Now the chance that you will get more than this number of losses is 99.97%, and that's relatively conservative because I haven't taken into account the zero.

I have to say, these results are so impressive that I'm tempted to give your system a trial. What's your secret?

Hi Bayes, there is no secret but an understanding of what this method delivers most of the time played hit and run. That it seldom produces double losses and will be rare to show you a treble loss AGAIN played HIT AND RUN.

Ad to this money management that takes advantage of the impressive winning streaks the first game of the day can deliver. And the results speak for themselves.

Now on paper you are right play this method consecutively with a one level progression and you can't win. Do as the casinos want you to do you are toast.

BUT, we arent playing the way nature intended. We are organising our campaign around the strengths of the method. And its not just working how I play it. Both Strato1985 and Gordonline are outdoing me. Warrior too has taken this to a level I Cannot believe.

It basically works Bayes applied in the right way. So your test will only tell me what I already know. But I and others will still be building BANKROLLS. Maths can't explain everything it really can't. MATRIX VERTICAL 5 totaly defies math.

My arguments have always been simple and successful overall. Random has things it doesn't do very well. Once you identify them you are on the right track. You can't beat random straight off the layout.

I saw someone learn that the hard way the other night playing against six of the same dozen forming. Covering the two other dozens. They had won 120 times in a row. When random decided no only was it going to deliver 6 dozen 1s but throw a zero smack in the middle. He waved 242 units goodbye. Hed have been a lot safer putting that on MV5 but people can't wait to win. that's why casinos never have to worry.

What all the methods I play have in common is they take random away from the layout break up its flow and affect its performance. They ask random to do things over and over again that go against what random is all about.

They set it codes to figure out. And it has a limit to how often its interested in solving the code. The price to the player is time. But the reward is certain. NEVER BE IN A HURRY to beat this game just BEAT IT.

[Halba1]

Halba1 you don't try to recover the whole stake in one game that's the classic mistake. You recover 3/7 or there abouts so you would stake something like this

1,2,4

3,6,12

12,24,48

that's what I do but I have the bankroll. Remember the first game of each day I play at level 2 for both Hi lo and od ev so you are pinching extra units on these first games before dropping down to level 1. And when you hit a double figured streak for that first game which you will often. This is where one of your advantages is gained. So when I get a losing game which is nearly always later pulling back the loss is much easier.

And when hit and run works its magic that's when you really go ahead. One step backwards three steps forward is the name of the game. ALWAYS playing with a BR at least 20 times bigger than your level one stakes. Preferably 40 times bigger. I kid you not.

The persom who has a 1,000 unit BR and plays for 5-7 max units per game as I do is set for life. I make no bones about that.

i confirm the above strategy is fine. P3 even looks safe there. you'll recover 3/7 or so. the rest of the sessions - normal profit will compensate. Just need to play more sessions. maybe a robot can help as it can play more and never gets tired.

bankroll 1000 units is fine.

re: Bayes Now the chance that you will get more than this number of losses is 99.97%, and that's relatively conservative because I haven't taken into account the zero.

where did you get that figure bayes? assuming independent events, it stays at 12.5% chance of loss. but its less than that because john is taking the 4th pattern so it mucks up the random of it.

in my opinion getting 2 losses in a row(i.e. P1>>P2) is prolly around 1% chance of loss I reckon. chance of a pattern repeating is 0125*0.125 but thats on the line above and line below. but taking 4th pattern impossible to calculate odds. all i know it confuses random, thats why it works

[furple]

Thanks Chris, so I take it that your results haven't been as good as JLs. I saw that at least 2 others have done at least as well, any more? don't be shy!

That's a good idea.

I gave it a whirl over the last few weeks off and on playing at lucky live.
Here are my stats.



total games: 42
total games won:31
total games lost:11


double losses:3

[albertojonas]

"Two things have come to my attention of late most losses are caused when betting against three of the same such as H H H. Or by the zero. Most of the success comes from betting against mixed patterns"

There is a solution for it, ignore a triple with the same like HHH/LLL, OOO/EEE and bet the next row if not HHH.
But always bet Ha, Ha, Ha...  ???
Hermes
P.S. albertojonas it is an excellent strategy. I won all sessions on baccarat. Test it and you will accept it because of the results you get. But still the Marty progression is the weakest link on that strategy.

Hermes,
i know we interacted much before in an excellent (by the way) system of yours on streets.
i saw the way you played it bermuda triangle style. very cool. more bets, no sleep.
i also did my twist (we all do) to adequate it to my playing style.
if you notice my posts on twister's thread playing bac with p4, i said precisely that the progression was the weakest link on this one.
[reveal]8 different combinations is true
but...
4 of those put you on the 2nd step
ie
play against
111
you have
222
212
211
221

could take you on the 3rd step
222
221
and finally you only lose your progression on 1/8
222
====

so if you play 1-2-4
and lose 1/8
you get even

you may use hit & run and leave as soon as you hit +7 on a positive Ecart

or

you can play it other ways
i would try at least 4 diff progressions before that one

1st
1-1-1
2-2-2
3-3-3
4-4-4
etc

2nd
regression
7-3-1

3rd
longest fibbo
1-1-1-2-3-5-8-13...

4th
modified D'Alembert (in which you win as soon as W and L balance out and then reset)
ie.
start with 20units bet
increase +3 on a Loss
decrease -4 on a Win
=====================================================[/reveal]

also posted this remark in that thread
[reveal]
1-2-4= 7units

8-16-32 = 56units

16-32-64 = 112units
 

175units if you lose 3 times in a row
LLLLLLLLL scenario
wow that is quite a risk isn't it?
per example:
you would lose 42 in the same scenario if using 7-3-1

it is a bit more than 4 times less
 
[/reveal]

so i do not deny it is a good idea and since the very start of the thread i was saying nothing but that it could be improved.
apart from that, what i can not cope with is the following:

1. the belief that color strikes more than any other even chance.

2. that the odds 1:8 are not given by the progression. odds remain 1:2 apart from the progression.

3. play the first game risking more units is better.

4. hit & run style vs continuous play makes a real measurable difference in long term.

5. results exposed are set in stone (even if hit rate is against all odds) we all must have different results but the norm should be around average.


Cheers,
Al

how is Canada this time of year? Cold?

[chrisbis]

Profit/Loss = +81 Units

Not alot, but I have the patience to slowly build my BR before increasing my unit bets
Gordon

Hi Gordon, nice Gin BTW!.

Could I ask U, since I personally think it affects the "Psych" of how one plays,
how much IS ur base unit bet?

At the moment, my min. unit base bet is 0.10 cents

At this level, even when the progression level lifts somewhat, its nearly always less than 1.00

cheers.

[Bayes]

Bayes could you do a calculation on hitting a treble loss please?

Approximately 1/8³ or 1 in 512.

[Bayes]

By the way bayes how you would calculate zscore of PB4.

add bets,wins form all EC, add it up and then calculate zscore
or
calculate zscore separately for each EC and then take average of them 3.

It shouldn't make any difference. You can use the formula for z-score using p = 1/8 and that will be close enough. 

[Bayes]

where did you get that figure bayes? assuming independent events, it stays at 12.5% chance of loss. but its less than that because john is taking the 4th pattern so it mucks up the random of it.
in my opinion getting 2 losses in a row(i.e. P1>>P2) is prolly around 1% chance of loss I reckon. chance of a pattern repeating is 0125*0.125 but that's on the line above and line below. but taking 4th pattern impossible to calculate odds. all I know it confuses random, that's why it works

No, it's not impossible. Here is the proof:

Since each pattern of the 8 available (ignoring zero for simplicity) is equally likely, and using the 'master formula' for probability:

P = number of 'favourable' outcomes / number of equally likely outcomes

What is the number of favourable outcomes? since you're betting that the 4th pattern will not be a repeat of the first, the first 3 can be any of the 8 available but the 4th can only be one of 7 if your bet is to be successful (ie; all patterns but the first). For the first pattern, there are 8 possibilities, and for each of these, there are 8 for the 2nd pattern, then for each of these 64 ways, there are 8 possibilities for the 3rd, and finally only 7 for the 4th pattern. So the number of favourable outcomes is -

8✕8✕8✕7

Now the number of equally likely outcomes is just

8✕8✕8✕8

So now if you divide the first number by the second, 3 of the 8's cancel out and you're left with a probability of 7/8, which is just what you would intuitively expect without having done all the maths. Also, you can see that it wouldn't matter whether you chose the first pattern to not occur anywhere you like (for example, instead of betting that the 4th pattern won't be a repeat of the 1st, you could require that the 6th, 27th, or 393rd pattern won't be a repeat of the 1st, and you would still get the same result, because the favourable outcomes would consist of a row of 8's multiplied together with one 7 in them, and it doesn't matter where the 7 is in the line, because the equally likely outcomes would be a row of 8's multiplied and there would always be one more 8 in the bottom than the top, so the cancellation would always result in 7/8.  See what I mean?

[Bayes]

Maths can't explain everything it really can't.

No, of course not. If I believed that I wouldn't be here, but it's useful if only to tell you when your results are deviating from expectation. I know the system uses a progression, but the fact is that your losses are significantly less than probability predicts. The numbers always come out the way probability says they will if you're betting randomly (and to my knowledge, no-one has come up with a bet selection which does better than random), so there must be something in the way you're choosing when to bet which explains the results.

[artattack]

Hi, I just thought I would pop in here with a comment on perhaps why H L   O E   seem to perform better than Red and Black.

Could it have something to do with the wheel layout.

0-32-15-19-4-21-2-25-17-34-6-27-13-36-11-30-8-23-10-5-24-16-33-1-20-14-31 -9-22-18-29-7-28-12-35-3-26

If you notice, red and black are alternate on the wheel, but odd/even, high/low are not quite and there are mini groups that could just tip the balance.

For instance 10-5-24-16  you have 3 low numbers out of 4 also 3 even numbers out of the same 4

another example 15-19-4-21  here we have 3 odd numbers in a sector of 4

Hello by the way.

Arthur

[ophis]

It shouldn't make any difference. You can use the formula for z-score using p = 1/8 and that will be close enough.  

Why 1/8 and not 16/37. Should I reread your zscore thread?
this is how zscore is calculated in MST:

Code Select Expand


     //check zscore of the last N amount bets
     n:=MaxBetsAmmount;

     //check how many wins there is in the last N bets
     t_win:=0;
     for i:=1 to n do
       if win_pb4.Items[win_pb4.Items.Count-i]='1' then inc(t_win);

     //calculation
     m:=t_win;
     p:=16/37;
     t:=(m-n*p);
     s:=sqrt(n*p*(1-p));

     pb4_zscore:=RoundD(t/s,2);


is here something wrong?

[esoito]

Welcome, Arthur!

A good first post. The point you make is interesting. 

Hopefully, others will pick up on it.

[Bayes]

Why 1/8 and not 16/37. Should I reread your zscore thread?
this is how zscore is calculated in MST:

Code Select Expand


     //check zscore of the last N amount bets
     n:=MaxBetsAmmount;

     //check how many wins there is in the last N bets
     t_win:=0;
     for i:=1 to n do
       if win_pb4.Items[win_pb4.Items.Count-i]='1' then inc(t_win);

     //calculation
     m:=t_win;
     p:=16/37;
     t:=(m-n*p);
     s:=sqrt(n*p*(1-p));

     pb4_zscore:=RoundD(t/s,2);


is here something wrong?

Not sure why you're using 16/37, but then I don't know how you're counting the successes for the P4 system. All I'm saying is that if I were tracking it, I would take 3 at a time from the stream of R/B (because the P4 is based on 3 spin patterns) and ignore the zero. Then if there was an exact match I would take that as a loss, the probability of which is 1/8 because there are 8 equally likely patterns. It doesn't matter that it's a 'compound' event (ie; made of 3 spins) as long as you're consistent in how you use the z-score formula. Don't forget though that 'n' in this case is only incremented after 3 spins, not 1 (because you've defined an 'event' in this case to consist of 3 spins). That's what I mean about being consistent.

[ophis]

Not sure why you're using 16/37, but then I don't know how you're counting the successes for the P4 system. All I'm saying is that if I were tracking it, I would take 3 at a time from the stream of R/B (because the P4 is based on 3 spin patterns) and ignore the zero. Then if there was an exact match I would take that as a loss, the probability of which is 1/8 because there are 8 equally likely patterns. It doesn't matter that it's a 'compound' event (ie; made of 3 spins) as long as you're consistent in how you use the z-score formula. Don't forget though that 'n' in this case is only incremented after 3 spins, not 1 (because you've defined an 'event' in this case to consist of 3 spins). That's what I mean about being consistent.

oh no i'm calculating it every time it bets.

each row contains 3 bets.

HLH
LLL
HHH
HHL = 3 bets (each of probability of hit 16/37 - betting 16 numbers out of 37) 2 losses 1 win

with progression this would mean we have won because progression would bring us in profit
but with flat bet it would mean you have lost.

that's why we have to count all of them bets so progression will be irrelevant and zscore correct. Is this wrong?

I need to have it sorted so I can release corrected version in case this is wrong.

how is success rate calculated?

well quite strait forward.
you have 2 lists. one with BETS second one with LW

if System have encounter L or W then to the list with BETS is added 1 and to the list with LW is added 1 or 0.

then if I want to calculate zscore of last 12 BETS. then I'm calculating how many "1" there were in last 12 entries of list with LW.

so zscore of those 3 BETS would be:
2,99 .....

well either way I'm doing it like this because I am caluclating zsore of many systems and it need to based on the same amount of bets for each system. if not then comparison would be false.

even if lets say numerology had 35 bets already and pattern breaker 4 had only 6
then MST will show zsore of the last 6BETS for each of this system.

you can setup min and max ammount of bets MST is looking at. if 12 is max then even after 3000 spins Zscore will be shown of the last 12 bets.

This is what this thread was about:
link:://rouletteforum.cc/math-reference/zscore-comparsion/

because at beginning i wanted to compare zscores basing on spins...

zscore of last 100 spins of each system
or
zscore of last 100 bets of each system

mmm is this right?