Random Thoughts

[Big EZ]

Priyanka,

What happens in situations like these?  Is it always a no bet? Is there any tricks to handling this?

Falkor,
How did you handle these in your coding

2
2
1 x
1
2 x 
1   
1
1 nb

2
1
1
2 x 
1   
2
1 nb

2
1
1
2 x
1
1
2 nb

2
1
2
1
1 x
1 w   

1
1
1 w

1
2
1
1
2 x
2
2 nb

Can you go into some more detail so we can learn what you are talking about with the EDGE betting R/B

[falkor2k15]

I just started looking at this.

These are the combinations where the arithmetic progression can "win"

123
135
147
159   = all these can only "win" on an ODD spot (3-5-7)

234
246
258
269  = all these can only "win" on an EVEN spot (4-6-8)     the 9 spot would be a no bet because either could win

345
357
369  =  only "win" on ODD spot

456
468  =  only "win" on EVEN spot

567
579  =  =  only "win" on ODD spot

678

Falkor...in your data is it possible to look at what combinations win more or less if that's possible?


Just thinking out loud here..I could be totally off base and just stating the obvious ..dont chastise me if wrong or have wrong thought process Priyanka

Not sure if you have the right understanding? Every 9 spins 3 of the same colour will form an arithmetic sequence.
1 2 3 4 5 6 7 8 9

So we can guarantee that at least one of these 3 will be of the same colour (all red or all black), but we don't know which one:
1,2,3
2,3,4
3,4,5
4,5,6
5,6,7
6,7,8
7,8,9
1,3,5
2,4,6
3,5,7
4,6,8
5,7,9
1,4,7
2,5,8
3,6,9
1,5,9

So we start looking for 2 matching colours: if spin 1 and 2 is red then spin 3 could be red; if spin 1 is black and spin 5 is black then spin 9 could be black, etc. So we would bet the 3rd spin to be red or the 9th spin to be black. But actually we would never bet on the 9th and final spin - instead forfeit the set - since red and black would both have an arithmetic progression. Priyanka is better at explaining it than me, so maybe re-read first 2 pages.

[falkor2k15]

Priyanka,

What happens in situations like these?  Is it always a no bet? Is there any tricks to handling this?

Falkor,
How did you handle these in your coding

2
2
1 x
1
2 x 
1   
1
1 nb

2
1
1
2 x 
1   
2
1 nb

2
1
1
2 x
1
1
2 nb

2
1
2
1
1 x
1 w   

1
1
1 w

1
2
1
1
2 x
2
2 nb

Can you go into some more detail so we can learn what you are talking about with the EDGE betting R/B

1 = red; 2 = black? Or are these dozens?

There's no trick to handling them. If there's multiple sequences available for both red/black then we can't say which one is more likely, but if there's only a possible sequence for red then it can only be red (or a different sequence will match further down the line before the 9 spins are over). Again, Priyanka explains it better than me.

The strategy to gaining edge is secondary to the basics: certainty that 3 of the same colour will match in an orderly fashion within 9 spins.

[Big EZ]

Not sure if you have the right understanding? Every 9 spins 3 of the same colour will form an arithmetic sequence.
1 2 3 4 5 6 7 8 9

So we can guarantee that at least one of these 3 will be of the same colour (all red or all black), but we don't know which one:
1,2,3
2,3,4
3,4,5
4,5,6
5,6,7
6,7,8
7,8,9
1,3,5
2,4,6
3,5,7
4,6,8
5,7,9
1,4,7
2,5,8
3,6,9
1,5,9

So we start looking for 2 matching colours: if spin 1 and 2 is red then spin 3 could be red; if spin 1 is black and spin 5 is black then spin 9 could be black, etc. So we would bet the 3rd spin to be red or the 9th spin to be black. But actually we would never bet on the 9th and final spin - instead forfeit the set - since red and black would both have an arithmetic progression. Priyanka is better at explaining it than me, so maybe re-read first 2 pages.

Falkor, I understand perfectly whats going on with the simple explanation, thanks. In fact my numbers match yours except I forgot 7-8-9 in mine 

When I say has to win on ODD I mean the last place holder in the arithmetic sequence is ODD
123
135
147
= all these can only "win" on an ODD spot (3-5-7)

I hope this explains what I was trying to say a bit more clear

When there is multiple sequences for red and black, do you cut the 9 spin window at the no bet spin and start over? Is that how it is done in your programming?

Oh, and 1 and 2 are just representing any even chance (red/black,player/banker,etc)

[falkor2k15]

Falkor, I understand perfectly whats going on with the simple explanation, thanks. In fact my numbers match yours except I forgot 7-8-9 in mine 

When I say has to win on ODD I mean the last place holder in the arithmetic sequence is ODD
123
135
147
= all these can only "win" on an ODD spot (3-5-7)

I hope this explains what I was trying to say a bit more clear

When there is multiple sequences for red and black, do you cut the 9 spin window at the no bet spin and start over? Is that how it is done in your programming?

Oh, and 1 and 2 are just representing any even chance (red/black,player/banker,etc)

OK, but if you understand then why are you concerned with novelty attributes like whether numbers are odd/even during an arithmetic sequence of numbers?

Yes, if there are multiple sequences (including both red and black) then the law governing the sequence is guaranteed to be satisfied on the following spin, but we can no longer say which colour is more likely, so we don't bet.

[falkor2k15]

OK, I think I understand what Big EZ means: yes, we aim to win on the 3rd number within an arithmetic sequence. I guess one could aim for the 2nd number as well, but it's already quite complicated as it is.

[falkor2k15]

Playing both Normal AND the suggested Mod together:


Average of both:

[falkor2k15]

First data set:

[Priyanka]

Priyanka,
What happens in situations like these?  Is it always a no bet? Is there any tricks to handling this?

Yes, always no bet. There is no magic to speak against the mathematical truth.

Can you go into some more detail so we can learn what you are talking about with the EDGE betting R/B

I will refer back to the paradox that i had highlighted earlier. Just by looking R and B as R and B, will not help the cause. You could play, four or five games here. If you refer to my earlier posts, i was pointing to play multiple games before a session is complete. As an example, you could play every alternating spin to be part of the 9 spins and hence 18 spin as one game instead of just 9 spins. You could play single and series formations to complete an AP. You could play three sets of alternating spins with one set for completing AP and two sets for not forming an AP. The possibilities are endless, but the key is finding that set of games where 1+1 <> 2

Let me take another example of a game, to illustrate a different game you can play. You can play the fastest colour to reach 3 to complete an AP. Take a set of spins that we saw earlier.

Spin	R/B	Fastest to 3
18	R
19	R
19	R	Red wins. Outcome 1
9	R
31	B
21	R
17	B
25	R	Red wins. outcome 1. Now play for AP to complete on red to become fastest to achieve 3.
26	B
27	R
36	R	Bet red.
31	B	Black appears. Loss. Bet red.
17	B	Loss. Outcome 2. Our outcomes read 112
34	R
13	B
0	0
12	R
26	B
12	R	Red is fastest. Our outcomes read 1121
12	R
10	B
36	R
12	R	Red fastest. Outcomes read 11211. Outcome 1 on next set will complete the AP
18	R
23	R
0	0	Loss.
1	R	Win. End of set. It read LLLW for this set.
10	B
6   B
30   R

Am not suggesting you play the ways i have played here as is. All am trying to get at is look at non-random possibilities that has a limit, as opposed to random variables (AP is just one example), create multiple games that can be played at a frequency can lead to a potential edge you could look at.

[Priyanka]

Hopefully that short break has helped in pondering over what is written a couple of times. Before I move my thought process into an interesting concept of Parallel universes, I would like to explore another aspect of non-randomness.

People always say Roulette is a random game. But they do forget that it has its limits. They do forget that non-randomness is part and parcel of this game and embedded in it. There are numerous situations which are really finite in roulette.

To understand this further, lets take directly jump into an example. As usual, we will ignore the zeroes throughout until we get to a place where we have managed to explore an edge. Let us consider that we are playing dozens. Can you predict the next dozen? If I bet on the negative, the odds will be better than what i will get from playing roulette. However, what we can say for sure is there will be at least 1 repeat of a dozen in 4 spins. Hmm! Is that random? Or is it a finite characteristic and hence non-random?

See the following spins. Construct them into sets of 4.

21 - Dozen 2
17 - Dozen 2. At least one repeat of a dozen
24
12

36 - Dozen 3
18 - Dozen 2
29 - Dozen 3. At least one repeat of a dozen
2

17 - Dozen 2.
17 - Dozen 2. At least one repeat of a dozen
19
10

16 - Dozen 2
7 - Dozen 1
11 - Dozen 1. At least one repeat of a dozen
20

How can we take advantage of this non-randomness. Now here is where Probabilistic and non-probabilistic approach has to go hand in hand.

How many of us have wondered why a few systems always work well at the start and then the graphs grow towards the south? If you are not one of those who has experienced this, then you have not played enough roulette. The law of large numbers always catches up. This is why when some one tests thousands of spins, you always get a southward graph. So what is the issue? Your playing sessions are not short enough to stay ahead of the curve for forseeable future.

Unfortunately, playing the game as is will always lead to the session being long enough to catch up on the game edge. For some it could happen in a minutes. For some it could happen after building a solid bankroll over a year or two. However, if you see roulette as a game made up of a number of finite non-random events, it can help you constructing your sessions short. Short not in its literal sense of minutes or seconds or few spins, but short enough to avoid the game edge catching you forever.

[Nickmsi]

Hello . . .

I have coded a system that uses a 4 spin cycle.  Excel Tracker attached.

1st Spin you bet the last Dozen(FTL).  If it wins, then No Bet the next 3 spins.
2nd Spin you bet the last 2 Dozens, win or lose No Bet the next 2 spins.

Also attached is a resulting graph for 20,000 spins.

The progression used was a mild to aggressive one.  There is a separate progression for the Single Dozen Bet (spin #1) and another progression for the Double Dozen Bet (spin #2).

You can adjust the Progression Divisors. For a more conservative progression, set them for  1,000 which would be Flat Betting.

Enjoy

Nick

[Priyanka]

I have coded a system that uses a 4 spin cycle.  Excel Tracker attached.

Cheers Nick. Always useful.

Creating your own playing positions

Roulette is a game that can give you endless possibilities to play. And there lies the beauty of the game and the beast. It is easy to get oneself lost into the complexities of the game. But if you are able to breakdown those complexities into simple principles, then you will be able to effectively play it with a better understanding of what to expect at the end of every session. Sure one or two odd session may turn out to be exceptional, but you will figure out that a 98-99% of the games will fall within your expectation (win or loss!).

Too many numbers! Too easy that a child can play! â€" Two differing views on playing ECs. But stringing together ECs we can create an odd placement that we like like quads, dozens, so on and so forth. We don’t even have to look at the numbers or wheels. How is this possible. See this example below on Red and Black.

Instead of playing one position of just R and B, what if we play RR, RB, BR and BB. Instead of giving odds of 1/1 we have converted ECs to give odds of 3/1. An example play is below. For simplicity, what we will be looking to play is for getting the outcome RB.

25  - 1 unit on red. Win.
27 â€" Place both units on blck. Loss.

7 â€" 1 unit on red. Win.
29 â€" 2 units on black. Win. We got the win at odds of 3/1

4 â€" 1 unit on red. Loss
18 

27 â€" 1 unit on red. win
10 â€" 2units on blck. Win. We got 3/1 odds

14
28 â€" Won this sequence

34
27 â€" lost this one

6
16 - lost

12
20  - won

This is not a progression. This is not letting it ride. This is an example of stitching together simple EC components to create an odd that is better than even return. Now the possibilities are endless and everyone can create opportunities based on their comfort and style of play. You can create dozens, quads, splits, all possible odds through stitching together these components.

[Priyanka]

Now when it comes to the topic of stitching together bets, it is also important to understand which combinations are profitable and which ones are not. The combinations which might seemingly give better odds at first sight may not be the ones that will be profitable and vice versa. Taking a simple example.

Red and Odd. If we need to stitch together these two, will you place one bet on red and one on odd or one bet on red and 8 bets on the black odd numbers? Any creative ideas and view points?

[Drazen]

Hello Pri

I am very carefully watching this thread on daily basis and trying to figure out what you are trying to show here.

Also downloaded all your videos of play in hope it will be useful in better understanding. Your bank demand and DD is impressively low and I think it is not coincidence.

Although my favorites are EC-s here I am showing something about dozens and hoping to get your thought and input.

It is about number of combinations and It is what I call betting against perfect state.

Okay we would have 6 perfect states:

123
132
213
231
312
321

And twelve other states:

122
121
232
233
311
313
211
212
322
323
131
133

So if we betting two different dozens in a row not to become 3, we are winning on 12 patterns and losing on 6. Or we can reverse and have odds of 3/1 but winning on 6 possible combinations.

Cheers

Drazen

[Priyanka]

I am very carefully watching this thread on daily basis and trying to figure out what you are trying to show here.
Also downloaded all your videos of play in hope it will be useful in better understanding. Your bank demand and DD is impressively low and I think it is not coincidence.

Thanks for the interest shown Drazen. The more interest that you are showing, the more you read these topics again and again, the more iterations you do differently, you will be able to get the answers to the questions you have asked yourselves. There will come a day when you will be able to see past whats happening on the surface and free from the wheel, felt and the statistics that the pit bosses want you to keep your attention to.

Although my favorites are EC-s here I am showing something about dozens and hoping to get your thought and input.

My favourites are double street/lines/6 numbers. But I agree with you, that the more you push your boundaries and come out of your comfort zone, you will be able to take the learnings back to your favourite playing position and play a completely different game.

So if we betting two different dozens in a row not to become 3, we are winning on 12 patterns and losing on 6.

Lets not do simulations. Lets not do 1 millions spins. Lets not apply this in practice. Lets break down what you are trying to say.

6 combinations against 12 combinations. Making it simple once against twice. On the outset it looks as if 12 combinations are better than 6 combinations hence this looks like a very good selection. But what is the return on those 12 combinations. You will have to play 2 dozens and hence the return is 1/2 (Apologies for people who don’t understand English odds, it is 1$ returned on a win for every 2$ staked). If you see it that way, you are essentially going back to 6 against 6 (1/2 multiplied by 12). Now you are left to the mercy of deviations, variations and statistic reality to either fail or win.

This is the reason I was pointing back to find out finite, non-random methods within the bet selection process. You have taken the step in the right direction, but I would really encourage you to look deeper.

• What is that you are finding common in those 12 combinations that you have selected?
• Why have you not taken into consideration that other combinations where the dozens have repeated in the first two positions?
• Is there a way you can stitch these dozens together?
• Are you able to find out a common theme between the first and second spins or first and third spins?
• Is three spins sufficient for you to derive that commonality?

I know, a lot of questions . But all I am trying to do is setting your thought process in a direction that can help you think beyond what you see in the surface. I like your avatar. Someone someday said, "Even Einstein cannot find a way to beat the roulette, unless you can cheat the dealer". But world doesnt always stay flat. It became a sphere one day. You have given Einstein some young muscles. Probably he would look at it differently with those fresh muscles added.