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Andre Chass

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System 4
Mar 01, 04:06 PM 2019
The way of playing it is betting ten splits: 1-2, 4-7, 11-12, 13-14, 15-18, 16-19, 20-21, 28-29, 27-30, 33-36. These numbers will be called 'Positive Numbers', and the remaining ones will be called 'Negative Numbers'.
If you take a closer look at the wheel and the numbers to bet, you can see how the sectors are accommodated.

Wait some consecutive Negative Numbers and then start. If you want to play it safely, wait for at least five Negative Numbers.

First ball: 1 chip to each split

Second Ball: 1 chip to each split. If you win here you will end up losing just 2 chips.

Third Ball: 4 chips to each split.

Fourth ball: 10 chips to each split.

Note: You will rarely have to bet 10 chips if you wait for at least 5 or 6 Negative Numbers.
Nothing ventured, nothing gained...

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sugtips

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Re: System 4
Mar 05, 11:33 AM 2019
Thanks God and Good Morning All.
Thank you Andre Chass.

Sorry Andre I don't like it but in fact I LOVE IT. Great system. Thanks for sharing.

Love and Light
SugTips
If you think you can, You can. If you think you can't, you are right.

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Andre Chass

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Re: System 4
Mar 05, 06:45 PM 2019
Thanks God and Good Morning All.
Thank you Andre Chass.

Sorry Andre I don't like it but in fact I LOVE IT. Great system. Thanks for sharing.

Love and Light
SugTips

It's a pleasure my friend!
Nothing ventured, nothing gained...

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Firefox

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Re: System 4
Mar 05, 10:33 PM 2019
Here's an example of how a compact system like this can be tested as a discrete probability distribution without simulation on a computer.

Formula:

Mean  = Expected value of w = sum of w x probability(w)

Standard Deviation = SQRT(sumw^2xpr(w) - (sumw x pr (w))^2)

w is the amount won or loss (-) after each of 5 events. Winning on the first, second, third, or fourth balls. Or losing on the fourth ball.

Pr (Win) = 20/37 = 0.541
Pr (loss) = 17/37 = 0.459

The Prob of an event is obtained by multiplying the probabilities for wins and losses in that event together.


Event  Prob   w      wpr(w)   w^2pr(w)


W        0.541 +8      4.328     34.634
LW      0.249  -2     -0.498     0.996
LLW    0.114  +12   1.368     16.416
LLLW  0.052  +20   1.040     20.800
LLLL    0.044 -160  -7.040    1126.4


Sum    1.0               -0.802    1199.24


Expected value of the win is -0.802 or a loss of 0.802 units.

Standard deviation
is

SQRT (1199.24 - (0.802)^2)

= (1199.24-0.64)^1/2 = 34.6 units

The standard deviation is quite high due to the swingy event of losing 160 units after 4 tries which will happen 4.4% of the time or one every 22.7 attempts.

If anyone can spot an error, please advise as I didn't do a huge amount of checking.

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sugtips

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Re: System 4
Mar 05, 11:08 PM 2019
The standard deviation is quite high due to the swingy event of losing 160 units after 4 tries which will happen 4.4% of the time or one every 22.7 attempts.

Many thanks for your reply firefox.

Ok now it's very hard for me to understand math part, but what I understand from the quoted paragraph above is that after 4 tries the winning chance is 4.4%.

But am playing only after 3 virtual losses, for 4 spins. So what will be the winning chance at 7th try?

(sorry for my bad English, it's not my first language)

Thanks
SugTips
If you think you can, You can. If you think you can't, you are right.

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Firefox

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Re: System 4
Mar 05, 11:51 PM 2019
Many thanks for your reply firefox.

Ok now it's very hard for me to understand math part, but what I understand from the quoted paragraph above is that after 4 tries the winning chance is 4.4%.

But am playing only after 3 virtual losses, for 4 spins. So what will be the winning chance at 7th try?

(sorry for my bad English, it's not my first language)

Thanks
SugTips

After 4 (or less) tries you will either have won some number of units (about 6.525 on average) with probability 96.6% . Expectation 6.238 units

Or after 4 tries you will have lost them all (-160) with probability 4.4%. Expectation 0.04x-160 = -7.04 units.

The difference 6.238-7.04 is the total expected change to the bankroll -0.802 units.

Unfortunately the hypothetical losses make no difference to the calculation. The wheel has no memory. After you start betting, those losses have aready occurred with a probability of 1.0 .

You can multiply everything by a probability of 1.0 if you want, but it will make no difference to the results. The fact those losses are helping you are just an illusion.

You may want to consider if we toss a fair coin and it comes up heads 10 times in a row....

What is the chance it will now come up tails? It's still 50%.... and still 50% for heads..... assuming it is a fair coin. The coin, like the wheel, has no memory. We start from scratch every play.

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Firefox

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Re: System 4
Mar 06, 12:09 AM 2019
With this system, most of the time you will post a steady win.

That's great but one time in 22.7 tries you lose 160. Those big losses will be just enough to make it a steady loser in the long term.

I'm not knocking the system, you can be the judge of it!  1 in 22.7 is quite small. You may go a whole evening and rack up 220 units without a loss hitting.

The next evening you may get two quick losses setting you back -320. That's the way it can work. You're losing only -0.802 units every trial of the system so over 100 trials that's 80 units loss on average.

I'm just giving the mathematics of the long term and how to calculate them without computer simulation. It's a pretty swingy system with Standard deviation 34 units and it could be profitable short term if you dodge the 4 losses in a row with a bit of luck!

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sugtips

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Re: System 4
Mar 06, 12:46 AM 2019
it could be profitable short term if you dodge the 4 losses in a row with a bit of luck!

Sir, what are the chances of 7 losses in row for this system?

Thanks
If you think you can, You can. If you think you can't, you are right.

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Andre Chass

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Re: System 4
Mar 06, 01:07 AM 2019
Okay, I understand all the math here. But as everyone already knows, it is impossible to have any chance of winning without using some kind of trigger and a progression. If so, please tell me otherwise.

No hg here
Nothing ventured, nothing gained...

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Firefox

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Re: System 4
Mar 06, 01:33 AM 2019
Sir, what are the chances of 7 losses in row for this system?

Thanks

The chances of 7 losses in a row are (17/37)^7 = 0.004 or 1 chance in 232.5.

But that's not the same thing as waiting for three losses and then start betting!

The chances of 4 more losses now making 7 in total are (1.0) x (17/37)^4 = 0.044 or 1 chance in 22.7. 1.0 at the front because the three losses have already occured with probability 1.0

Thinking they are the same thing is what is known as gambler's illusion or gambler's fallacy. It's one of the things which keeps casinos in business and pays for the plush carpets, the hotels, and the suits.

Don't take my word for it though. Give it a go.  Wait for three losses and then start betting. It should soon become clear after a few hundred trials which is the truth for the chance of 4 more losses ... 1 in 22 chance or 1 in 232 chance  :smile:

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Steve

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Re: System 4
Mar 06, 01:35 AM 2019
But as everyone already knows, it is impossible to have any chance of winning without using some kind of trigger and a progression. If so, please tell me otherwise.

Why would a trigger change odds?

Why would changing bet size change odds?

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sugtips

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Re: System 4
Mar 06, 01:39 AM 2019
Don't take my word for it though. Give it a go.  Wait for three losses and then start betting. It should soon become clear after a few hundred trials which is the truth for the chance of 4 more losses ... 1 in 22 chance or 1 in 232 chance 

Many thanks for your reply sir.

I will take that chance and play it after virtual losses. I understand what is the gambler's fallacy, but that what is making money for me continously, so I can't give up easily. Or may be I am so lucky, so let it be.

Thanks and regards,
SugTips
If you think you can, You can. If you think you can't, you are right.

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Andre Chass

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Re: System 4
Mar 06, 01:45 AM 2019
Sir, what are the chances of 7 losses in row for this system?

Thanks

Wait for at least 5 losses in a row.

I usually wait for 6 or 7 losses.

My two cents
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sugtips

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Re: System 4
Mar 06, 01:48 AM 2019
Wait for at least 5 losses in a row.

I usually wait for 6 or 7 losses.

My two cents

Thanks for your reply sir.

I understand waiting for 6 or 7 virtual losses make this system very concrete but I mostly play at B&M casinos and for 6-7 virtual losses I have to wait a long, so I play with very small amount to get trigger and then play big once it passes more losses.

regards
If you think you can, You can. If you think you can't, you are right.

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Firefox

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Re: System 4
Mar 06, 02:01 AM 2019
Okay, I understand all the math here. But as everyone already knows, it is impossible to have any chance of winning without using some kind of trigger and a progression. If so, please tell me otherwise.

No hg here

I don't know about that. I could flat bet 1 unit on black for 100 spins and get 60 blacks, 2 zeros, and 38 reds.

I've won 60 lost 38 and lost 1. A tidy profit of 21 units with not a trigger or progression in sight!

On average though I will post a very small steady loss flat betting.

Using a negative progression such as Marty, Labby,  or any of the others my win/loss pattern will be different. Most of the time I'll post a small steady win until I get a bad run. Then I'll have a thumping loss. Triggers do not affect the a priori odds in any way. That's mis application of a Baysian model to a random game.

It depends how you want to play. Steady wins and a big crash, or steady losses with slightly less steady wins.

In the end though the progression players will lose more because they are exposing bigger bets/action to the house edge.

I'm not knocking your dark and light system by the way.  There's always a lot of talk of systems testing or paying for software or paying someone to code.

I just wanted to show people how they could save money/time and easily get a handle on a compact system like this with a back of envelope calculation. This system is a good one to apply discrete probability theory to.


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