QuoteTalk to any programmer like Reyth and he will confirm this.

Reyth has played only once on actual roulette wheel and this was online studio, not real casino, so I don't think him as the proper person to ask.

When you see a summary of millions results you are looking at the ''destination'' not how you've arrived there...

If you really believe that roulette is unpredictable then the only two things which remaining are progressions and luck!
Sorry if I don't share the same point of view.

Quote from: Albalaha on March 09, 2016, 11:30:52 AM
I have simulated over 50 million spins of RNG and over 5-7 millions real spins. I do not see anything like that. It is an illusion merely.

Perhaps you were not paying attention in details...

Quote from: Albalaha on March 09, 2016, 10:22:34 AM
Really?
What made you form such opinion? Observation, experience or simulations?

All these together

Quote from: sqzbox on March 09, 2016, 04:57:39 AM
Matt - I'm not sure if this answers your question or not, because TBH your question is not clearly stated. You might be surprised by the answer in which case I would surmise that you haven't made your point clearly.

Approx. 12.33

We use the Geometric distribution for finding the number of trials until the first success. The formula for the Geometric distribution is just the reciprocal of the probability of the event. The probability of the event, in this case, is 3/37 and so the reciprocal is 37/3 and so the answer to "For a street bet, what is the expected number of spins until the first hit?" is 12.33. In this case the phrase "For a street bet" means that it has been specified before the trials begin exactly WHICH street we are waiting for.

Since you state "in cycles of 12 spins" I am guessing that the question I state above is not the question you are asking. The thing is, you say "average spin number that a street will hit on" but which street? One nominated ahead? Then the above applies. If not nominated ahead then which street? Any street? OK - then the answer is 1 because whatever street hits on the first spin qualifies as "any street" - yeah?

No, he didn't question about this but about in which spin the first repeat for a street is going to happen.
For example:
1st spin, number 13, no bet
2nd spin, number 23, bet 5th street, -1 unit
3rd spin, number 32, bet 5th and 8th street, -3 units
4th spin, number 4, bet 5th,8th and 11th streets, -6 units
5th spin, number 18, bet 5th,8th,11th and 2nd streets, -10 units
6th spin, number 17, bet 5th,6th,8th,11th and 2nd streets, -3 units (first repeat on 8th street)
7th spin, number 23, bet 5th,6th,8th,11th and 2nd streets, +4 units
8th spin, number 26, bet 5th,6th,8th,11th and 2nd streets, -1 unit
9th spin, number 5, bet 5th,6th,8th,9th,11th and 2nd streets, +5 units

In the following 3 spins till the 12th spin there is remaining 1 more repeat because the average expectation is 4 and so far we got 3 repeaters.
In such case it's not in our best interest to bet the remaining 3 spins because we expect 2 loses and 1 win while we win and lose equal units as we are betting 6 streets.
I hope was clear.

Bet 1$ straight up on 00,8,29

Bet 1st & 3rd column 10$ each

You are betting this way 23 continuous numbers plus a few on the other side.

Also you can bet from 5$ on columns instead of 10$ each.here is the 23 number group.

[attachimg=1]

The progression suggestion:

The point is to keep an account for each and every bet separately AND as 1 group simultaneously.
By applying a positive progression on 2 columns we're expecting the streaks of 2 columns to be longer than the streaks of 1 column.

What kind of positive progression would accommodate successfully such expectation?
My answer is a conservative one which doesn't add unnecessary risks into the equation.

As long as our 2 columns hit and win we keep on flat betting 1 unit on each of them, when the ''losing'' column appears we write down minus 2 and keep adding the negative outcomes as they occur, when the losing streak eventually stops we check the negative balance by deducting the last win.

Let's say the negative balance is -5 (3 losses/1 win) now we're going to bet 2 units on each of our 2 columns, thus a possibility to win 2 units net, if successful the next bet would be for 3 units on each column providing the chance to win additionally 3 units net and reducing the negative balance to 0.

In such case we have neutralize the negative balance by offsetting 3 loses with 3 wins.

Whenever we are reducing a negative balance to 0 and/or changing it into positive, the next bet on 2 columns is always 1 unit each.

If we had lost after 2 consecutive wins the negative balance would have been -9 units and the next bet would be 1 unit on each column.

Let's say another losing streak of 2 occur before a win breaks it, the negative balance would have been -9 + -4 = -13 - 1 = -12 units, the next bet would be where you left off from the previous winning streak, 3 units on each column.

Can you calculate how many more wins we need in order to neutralize the negative balance?

If your answer is 3 you are right, 3+4+5=12
In other words a positive progression which increases by 1 unit ONLY when there is a negative balance till it becomes neutralized or positive.

We are also betting 3 straight up numbers, the method is the counting in multiples of 12 spins...

As group of 3 we need 1 win out of 12 results, if that win would not occur within the 12 first spins keep on flat betting 1 unit on each of 00, 8, 29 but now we are looking for 2 wins in the next 12 spins.

Let's say that after 24 spins one or none of our three numbers appeared, now we need 2 or 3 wins in the next 12 spins, if this wouldn't happen, we are checking the TOTAL balance that far (including our 2 columns), if the total is in the positive after 36 successive outcomes we wouldn't raise bets on numbers 00,8,29.

But if the total is negative, we would increase on 2 units on each number for the next 36 successive spins.

The betting schedule for the three numbers must be reseted to 1 unit for each number as soon as you reach a new high on your bankroll, regardless if this has been achieved by the 3 numbers or 2 columns betting.

There can be no proof of such matter.

I've an opinion though which you might find false but I'm going to say it anyway;

RNG's have a tendency to repeat more than the actual roulette wheels.

Of course what I just said is only my subjective opinion, you don't have to agree.

I remember a time when I was testing a theory.

My theory was that streets is the most suitable bet section to expose the law of thirds...
Why? Because each time a street hits includes 1 shown number to 2 sleepers, so if we take this ratio to the 36 spins scale it would be interpreted like 12 repeaters to 24 shown numbers.

But the repeaters are coming from the 24 shown numbers, so actually it's quite the opposite...

In order to reflect the average expectation of "LOT" with a bet selection, that section should contain 2 shown to 1 non appeared and expect that 2 of the already shown to be repeated.
Also the spins range or cycle if you prefer should be 36 spins instead of 12.

So here it is; bet every street which includes 2 hits in 2 DIFFERENT numbers for just 1 win, no progression, just 36 spins bet limit (including the tracking).

You might think, does every street with 2 hits in 2 DIFFERENT numbers hits for third time within 36 spins?
The answer is no but doesn't have to in order to be an overall winner.
The principle is 2 shown to 1 sleeper, this proportion reflects "LOT", there isn't more precise section than the streets (except numbers) to practice it.

Let's say a street hits for 3rd time but it's not a repeat from the previous 2 shown number, but the 3rd number, in such case my personal preference is to continue bet the street till one from the three numbers repeats (within 36 spins).

Of course you could decide that you got your win and stop betting the specific street, you can choose the option which you think works better.

Overall it's not bad, but I've a better game now for straight up numbers...

QuoteThe earliest 24 nbrs could be a nice bet but it would seek about 40 spins of tracking

Have you heard Ion Saliu, his "super roulette strategy" was about this, first get 24 unique but after didn't bet every spin, he was using triggers like after 1 hit for the 24 numbers bet them that are going to win 1 more time or, after 2 hits for the 13 numbers bet that one of the 24 numbers will break the losing streak.

Trigger number 1: after W bet for 1 or 2 more W in a row

Trigger number 2: after LL bet for 1 win within 2 or maximum 3 spins

Quote from: BEAT-THE-WHEEL on March 07, 2016, 05:49:09 AM
Instead of waiting for 18nos, why not wait for 24,
and bet the 24nos???
Since the law of third, says, always 24???
isn't this better?
Anyone???

IMO 24 numbers are not better than betting 18, yes law says that on average are appearing 24 numbers in 37 spins but you are betting them for the next cycle and only 66% of them are going to hit.

Quote from: Albalaha on March 07, 2016, 05:03:11 AM
Dear Angelo,
   It would be far better if you do things upside down here. Instead of betting the unseen/sleeper 18 nbrs, you bet the 18 nbrs that came first. You get the first 18 nbrs almost always before first 30 spins. If you bet them further, they are most likely to do far better than the sleeper 18s.

  It seems you hand test most ideas and that makes you get confused regarding what will work in long term and what not.
Remember, the unhit numbers will have about 2-3 numbers at least that you should not expect to win even once in 100 bets. That makes a huge difference.

Rest is only money management.

I agree with this one, the least repeats I've seen in 37 spins are 7, by the time there are 18 unique numbers to bet it could have already from 0 up to 6 repeats, thus there will be at least 1 more in the following spins.

Quote from: Albalaha on March 07, 2016, 04:43:43 AM
So, we bet all 18 unseen numbers, so far and remove those which hits after we start to bet yet we do not get a net win? Is it roughly you prescribe?

Can you tell me how will you win, this way? What is your suggested progression?

The overall win could happen from the first bet but since we are not sure about when we continue betting all 18 (nothing being removed) with the expectation of 1 win out of 2 spins as ratio, if you don't get 1 win in 2 spins you are expanding the event's horizon proportionally: 2 in 4, 3 in 6, 4 in 8, 5 in 10, 6 in 12 and so on till the expectation meets the average probability.
For 18 numbers there shouldn't be great deviations because we are not talking about 1,2,3 or 4 numbers which could disappear for too much time!

QuoteI don't care who you are, a nobody in my world.

Glad we agree because you are the same for mine.

QuoteAnyone here knows, how many spins, the virtual limit for two-numbers to sleeps??

666 / 2 = 333

QuoteYou forgot 3 numbers and I don't use that G word. Only rookies use those terms.

What's 3? Something between 2 and 4, that makes sense only for you.
The success and value don't depend on terminology.

Quote@BA,
         I read your opening post in this topic, yet failed to understand what are you going to bet, why and when.

I got it that it is something to do with law of the third. You assume there will be at least 18 to 30 unique numbers in the 37 spins cycle.

Now, how do you propose to extract profit here? How and when you will start to bet and what exactly you propose to bet?

I have many ways to approach the game but because you spoke about law of thirds I'm going to analyze this:

Think it this way, in how many spins will appear 19 numbers?
22? 23? 24? more?
So those 18 numbers are like an EC which has slept for 22,23,24...etc spins!
By the time there have left 18 unseen numbers the reversed countdown begins, I've created a progression specifically for such case.
In first 2 spins aiming for 1 hit, in 4 spins 2 hits, in 6 spins 3 hits and so on...

Quote from: Mr J on March 07, 2016, 03:52:44 AM
I think it went over your head, no biggie.

Ken

That was a good one! LOL!

I'm Kavouras disguised as BlueAngel! LOL!

OMG, laughing so hard!

Quote from: Mr J on March 07, 2016, 03:41:18 AM
"Therefore is impractical for someone to wait so long before bets" >> ?? SAY WHAT??

I thought with "you guys", if it works on paper, it COUNTS as a winner?

(Laying out 163 dollar chips in a couple seconds for example).
Hey, if it works on paper, it's a winner! (lmao)

Ken

What are you blathering about buddy??

Your theory is interesting but I think the virtual limit is not 481 spins (13 x 37) it must be 666 (18 x 37).

Therefore is impractical for someone to wait so long before bets.

Personally I'm using the hottest number per 37 spins cycle, if there are 2 with 4 hits each I bet both but I expect 2 hits regardless from which of these 2 numbers, if I have 3 numbers with 4 hits each I bet all 3 of them but I expect 3 hits in the next 37 spins, regardless from which of these 3 numbers will hit.

If I have 6 numbers with 3 hits each I'm betting all these 6 numbers for the next 37 spins and I'm expecting 6 hits, regardless from which of these 6 numbers will hit.