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Bet method based on mini-games

Started by GLC, Mar 27, 01:27 AM 2011

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mr.ore

QuoteYes,the problem is that the 995 losing players on straight numbers,lose a huge amount of money,much more than each EC players.

The nearly optimal strategy you mention is what I play in Montecarlo Casino.
My win goal is very low(no greed),15% at the most of my bankroll.
Progression:1,2,4,8,10,12,14,16,18,20:105 units(Marty 4 terms and D'Alembert 6 terms).
Risk of ruin/bankruptcy 1,2% or 1 every 83 attacks
(you are expert in Computers,can you check these % as risk of ruin?Thanks)

Risk of ruin of your progression is computed there:

link:://rouletteforum.cc/math-reference/how-to-compute-probability-of-a-progression-winning-or-losing/

Enjoy :thumbsup:

beretta28

Hi and many thanks.
It's very kind of you to send me the computed risk of ruin of my progression.

1)I'm a bit surprised by the high probability of losing.
My calculation was close to 1,20%.
Unluckily i'm not as good as you in software and computer ,even I have already download Octave on my MAC.
I have calculated 1,20% of losing all bkr, with a software of a friend of mine, based on Krigman formula.
Is in your simulation the partage option taken into account?
I have also checked with the informations of the following book: Roulette odds and profits by Catalin Barboianu (available on the web) and my result is what I said above.
But I'm afraid that your calculation is much more accurate,so I tend to trust your figures.
In both cases,my or your risk of ruin,is a progression that I must give up asap.

2)I have given a look to your OSBD 128MM table.
Very interesting,useful,but dangerous.
All my conclusions about roulette,that I have studied for 25 years are a bit disappointing.It's normal.
If you play even chance with partage option,you must play the lowest possible number of spin.
The best way of betting,from a mathematical point of view, is to bet the maximum allowed on a straight number.But you need a lot of money and the risk of beeing ruined is still there.
No roulette players enter a Casino and like the two examples above.

That's why I appreciate your yesterday table even if they have the goal to last more and to lose less,but not to win,it's impossible IN THE LONG TER:
Because the bet selection doesn't exist,now I'm am studying a way of playng that of course will lose but the risk of ruin is very far away,maybe behind my life of player or may tomorrow,but very very rare.
Of course the Murphy law is still there.
If you are interested I'll give you  a snapshot of my studies.

mr.ore

Sad truth about roulette is, that with a math help you can get as near coin flipping as possible. You are coin flipping for your bankroll. If your target is to double your bankroll, then best probability you can get is a very near 1/2*36/37, but never 1/2 or higher. Your chance is always somewhat smaller than fair coin. With bad systems, your chance is singificantly lower. With good systems, your chance of winning should be very near (36/37)*(bankroll/(bankroll+target)).

As for bet selection, I am not absolutely sure nothing can be done, but it is most probably true. There are ways how to control chaos a little, but roulette is random and not chaotic. But random system should sometimes act as chaotic system because the chance is there. Then some advanced method could be used there.

I am interested in this idea - if you use singles vs. series on even chances, then if you observe them, then you moved to another dimension, because while probability of seeing 10 series in a row is same as probability of seeing ten reds, you will play more spins on average before this happens, because series last always longer than one spin. That's why I will look into bet selections even if there is no reason for them to work. There must be some way to do something...

beretta28

1)Even if you succeed in reaching  exactly 1/2 probability,you will lose in any case,because the money available  of a Casino is ,for sure, higher than  your bankroll.
I'm sure you know the formula that confirms that!

2)If sometimes roulette behaves cahotic and not random,because there is the chance,is the confirmation that to win is a matter of chance only!

3)Here in Europe all serious players have read Marigny de Grilleau book,600 pages,500$ on the web if available, the biggest book ever written on roulette.
The topic about singles vs series on even chances has been explored in all the ways in this book and the conclusion of Marigny is to wait for a "difference" of 3 between singles and series,difference calculated like that:4/5*square root of total number of spins.
When you met this situation,play FLAT BET at the most 5 spins for reducing this "difference"(ecart in French).
It seems that it doesn't work,computer tested.
The behaviour of singles and series is the same of Red and Black!

4)I'm testing with quite good result a bet selection based on "Law of periodicity" of streets and lines.
Please look at :.win-maxx.com for more informations


beretta28

Mr ore

Very simple question:I play Red. European roulette,en prison/option.
My probability of losing is 51,35% or 50,675%?
Thanks

mr.ore


beretta28

Thank you very much.
I have appreciated all your previous post that I've read yesterday
The best,in my opinion, is the bold policy( 40 units ,wingoal 55 units),that I used today in a real casino with success.
I'm wondering if with the en prison option,your table must include some bets on even chances or the difference is negligeble?

albertojonas


3)Here in Europe all serious players have read Marigny de Grilleau book,600 pages,500$ on the web if available, the biggest book ever written on roulette.
The topic about singles vs series on even chances has been explored in all the ways in this book and the conclusion of Marigny is to wait for a "difference" of 3 between singles and series,difference calculated like that:4/5*square root of total number of spins.
When you met this situation,play FLAT BET at the most 5 spins for reducing this "difference"(ecart in French).




THIS IS WRONG. BAD INTERPRETATION. 1st is not the total number of spins and maybe that's why the computer tests failed... if they did...

outsider

Quote from: mr.ore on Mar 28, 05:50 AM 2011
Yes, I know what a bold policy is. Those are simple bold policy. Simple because there is always bet only on one location.

Policy to win 11 units:


bankroll    bet    W    L    Pwin 
  no bet    -    -    0.0000% 
1           1 unit on 8:1    9    0    8.7203% 
2           1 unit on 8:1    10    1    17.5148% 
3           1 unit on 8:1    11    2    26.4321% 
4           1 unit on 5:1    9    3    35.2262% 
5           3 unit(s) on 2:1    11    2    44.2668% 
6           1 unit on 5:1    11    5    53.3046% 
7           2 unit(s) on 2:1    11    5    62.3424% 
8           3 unit(s) on 1:1    11    5    71.3802% 
9           1 unit on 2:1    11    8    80.6623% 
10           1 unit on 1:1    11    9    90.0698% 
11    no bet    -    -    100.0000% 

For comparison, even chance only version - anti+Martingale:


bankroll    bet    W    L    Pwin 
  no bet    -    -    0.0000% 
1           1 unit on 1:1    2    0    8.1882% 
2           2 unit(s) on 1:1    4    0    16.8313% 
3           3 unit(s) on 1:1    6    0    25.7125% 
4           4 unit(s) on 1:1    8    0    34.5976% 
5           5 unit(s) on 1:1    10    0    43.7549% 
6           5 unit(s) on 1:1    11    1    52.8534% 
7           4 unit(s) on 1:1    11    3    61.8523% 
8           3 unit(s) on 1:1    11    5    71.1174% 
9           2 unit(s) on 1:1    11    7    80.4107% 
10           1 unit on 1:1    11    9    89.9406% 
11    no bet    -    -    100.0000% 


You start on line with same number as your starting bankroll and according to W,L columns you continue. If you win, go on line in W column, if you lose, go to line with L column. At any point, you know your probability of winning. In this case, start on line 10 to win 11 units.

Number of spins is on average low enough, so house dos not eat so much, that's what I think. While my policy improvement algorithm is not always that good and needs some tuning, computed probabilities are correct, it is just a matrix multiplied many times. So if algorithm finds a policy, it might not be absolutely optimal, but if I see that computed probabilities are higher than Marty, then I know it is very near.

Also remember that those are just fancy versions of parachute...
i'm not american/English, so sorry for my idioma, however it's possible to increase the probability of this matrix
one chance can is split in dozzina (12 num) and 3 terzine (9 num)
so the probability of canche is 48,64
12 num + 9 num have 48.86


ozon

Quote from: mr.ore on Mar 27, 05:42 PM 2011
I am sure I am not mistaken. It works because on average you bet less than five units to reach your target, so they are not exposed to house edge. It is same idea as instead of betting two dozens, bet 1 unit on ec and if it lose, bet 1 unit on dozen. If you win, you are 1 unit up, if you lose, you are two units down. Probability of winning is 18/37 + 19/37*12/37 = 0.65303140978816654492.

If you bet on two dozens, each of them 1 unit, probability of winning is 24/37 = 0.64864864864864864865.

0.65303140978816654492-0.64864864864864864865=0.00438276113951789627

That is a difference of 0.4%, not much, but it is there.

We have two units, want to have 3 units. If we avoid betting one unit sometimes, probability of winning is increased. You do not bet on dozen, if you win on even chance. If you bet two dozens, you ALWAYS lose one unit while winning two.

Another example:


Let's have two methods to win one unit with five units bankroll, both are based on probability:

method 1: bet 1 unit on five lines, we cover 5*6 numbers.
Probability to win is p(method 1 wins) = 5*6/37=0.810810.

method 2: bet 1 unit on even chance, if it loses bet 1 unit on dozen, if that also loses bet last 3 units on even chance. Anytime it wins, profit will be 1 unit, while still risking 5 units bankroll like with method 1. Probability is different:
p(method 2 wins) = 18/37+19/37*12/37+19/37*25/37*18/37 = 0.821826

method 2 is better than method 1

Difference is 0.821826 - 0.810810 = 0.011016. That is 1.1% difference in probability. There are those little nasty tricks in gambling. Similar trick can be used in craps, for example. You will still lose in the end, and the difference is actually negligible. But you will play more spins this way for same price :)




Sometimes it happens that I will discover something very interesting in this forum, this is just the post.
Mr.Ore after the simulation, shows that the use of 3 step progresion gives a 1.1% advantage over the normal flat placement.
If his calculations are correct, it gives a very strong reduction of HE, using roulette with the la partage principle and HE only 1.35%, we are already close.
Now you need to determine all the methods that even theoretically cause a delicate relaxation of the HE by the bet selection.

ozon

Fast, I get two options, the first option that I tested a long time ago is to play EC and after two lost bets waiting for virtual wins, it gives about 0.50% edge in longrun, to confirm that I had 2 simulations after 100k spins and let two confirm it .
Unfortunately, we are playing the first EC bet the second bet is a dozen, so you will have to combine it a little differently.

The second option is to play, trends with dozens, using High / Low
I have long simulations, where I can see on the wheel without zero, that playing the last two dozen has an advantage. You can also combine to create EC bets and play them.


If someone has an idea and in a simulation on the wheel without a zero some strategy for the EC bets showed him better results than the random selection, add it on this topic, or maybe something will come up with an interesting

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