Step In To My Game

[luckyfella]

Please keep in mind.  If you can prove any of the current assumptions wrong. Then that means you can use that proof to create a way to play. (except the house edge one).

Spot on MoneyT101👍

That's the reason why I wrote all those math facts down in all my posts.
(Read my posts, the answer lies in there)

Challenge any one of those to be not always true.

Still calls it off-topic.   

[luckyfella]

Readers think I don't teach other than spam what they think they already know.

MoneyT101 gave you the list.

Read it a miilion times.

[falkor2k15]

rrbb once said that there are unlimited number of dependent situations for groups of numbers. It's very simple to find one, using cycles. The cycle start/end H/L positions was just one example. Here's an other one.
10000 random spins have 5265 dozen cycles
now if I check if the double street that hit at the end of a dozen cycle is the same same as the previous double street that hit at the end of the previous dozen cycle, I get less than 50%, but with 5 to 1 payout the result is +4948 (+5 if same, -1 if different)
And since I love charts, it will look like this 



Lines are inside dozens, so they are dependent. This of course won't make anyone a winner, because we don't know when to bet for this to happen, and all the losing bets will cancel out the winnings.

How about if we switch to lines when a CL2 opportunity arises on the dozen cycles? But the switch depends on how many times the same line has hit on previous Dozen CL1s?

[Still]

rrbb once said that there are unlimited number of dependent situations for groups of numbers. It's very simple to find one, using cycles. The cycle start/end H/L positions was just one example. Here's an other one.
10000 random spins have 5265 dozen cycles
now if I check if the double street that hit at the end of a dozen cycle is the same same as the previous double street that hit at the end of the previous dozen cycle, I get less than 50%, but with 5 to 1 payout the result is +4948 (+5 if same, -1 if different)
And since I love charts, it will look like this 



Lines are inside dozens, so they are dependent. This of course won't make anyone a winner, because we don't know when to bet for this to happen, and all the losing bets will cancel out the winnings.

Wait. Is this a fact?  We're still working on the assumptions list.  I'm assuming this doesn't work.  If it does work, please switch over to private communications.  If it doesn't work, i'm assuming we can switch over to cycle 2 (CL2) after the third line switch (TLS3), followed by the switchover to CL4 simultaneously when CL3 starts and ends?  ??

Anybody want to add to my assumptions?

[MoneyT101]

Thank you lucky, I appreciate your posts

[MoneyT101]

How about if we switch to lines when a CL2 opportunity arises on the dozen cycles? But the switch depends on how many times the same line has hit on previous Dozen CL1s?

Why does the same line hitting before matter?  If your in a new cycle and it was not the defining line. 

Why would that line hit more then any other line?

[falkor2k15]

Why does the same line hitting before matter?  If your in a new cycle and it was not the defining line. 

Why would that line hit more then any other line?

Well, I've not tested it yet, and it certainly doesn't make sense - but I'm just trying to understand ati's reasoning. There isn't many variables in just a single cycle consisting of only dozens and lines.

[MoneyT101]

Well, I've not tested it yet, and it certainly doesn't make sense - but I'm just trying to understand ati's reasoning. There isn't many variables in just a single cycle consisting of only dozens and lines.

Ati combined the dozen cycle ending with the ‘same’ defining element of the double street

Your question spoke about same defining on CL1.

Same defining takes into consideration that the one that started it also ends.

Has no bias towards cl1 or cl2 etc

[ati]

Don't get too excited about the chart. What I posted is not some miracle event, there are two lines in each dozens so they are dependent. You cannot hit a dozen without hitting a line.
We know that on average 63% of the time the next defining dozen will be the same as the previous one, and with 2 to 1 pay out, I would probably get a similar chart looking at only the cycle ends. But it will not help us win. You should not try to predict what comes next or wait and play for a winning event.

[ati]

Assumptions List

-Assumption: Roulette is random

-Assumption:  Roulette is not beatable

-Assumption: Roulette is not beatable because of the house edge

-Assumption: No dependency in roulette

-Assumption: no mechanical system can beat the game of roulette

-Assumption: Roulette has no limit, Infinite combinations

Interesting, because some of these have also been stated as facts.
Everyone's favorite facts: the spins are independent, spins are random, the odds of next spin never changes
These are I think all provable facts.

[ati]

See attached for example...

Hi Falkor, I'm not allowed to post in the random thoughts thread for some reason so I'll reply here. Apologies for the off topic.
What you did in that test is I think also qualifies as a waiting game. Even though every spins are played, you are waiting for a winning event. (the defining position to repeat)
Also, the 63% is only true, because of cycle length 1's are included. As soon as you have two different dozens in the cycle, the probability of a repeat happening on the starting position is reduced to 45%, and after three unique dozens it becomes 33%. But I'm sure you know this after many years of testing.

[MoneyT101]

Interesting, because some of these have also been stated as facts.
Everyone's favorite facts: the spins are independent, spins are random, the odds of next spin never changes
These are I think all provable facts.

Ok we are going to question these ‘Facts’ one by one and see what happens

[MoneyT101]

Hi Falkor, I'm not allowed to post in the random thoughts thread for some reason so I'll reply here. Apologies for the off topic.
What you did in that test is I think also qualifies as a waiting game. Even though every spins are played, you are waiting for a winning event. (the defining position to repeat)
Also, the 63% is only true, because of cycle length 1's are included. As soon as you have two different dozens in the cycle, the probability of a repeat happening on the starting position is reduced to 45%, and after three unique dozens it becomes 33%. But I'm sure you know this after many years of testing.

No apologies, still on topic

Ok so if waiting cause that number?  What happens when you don’t wait?

I know it sounds silly

Read this a few times....

[falkor2k15]

Hi ati, Priyanka mentioned before about playing a game within a game, i.e. first to X repeats, as well as increasing the span of a biased game. She also talked about increasing predictability - but since then the Non-Random game seems to have become severely handicapped in terms of recommended play.

[MoneyT101]

Assumptions List

-Assumption: Roulette is random

-Assumption:  Roulette is not beatable

-Assumption: Roulette is not beatable because of the house edge

-Assumption: No dependency in roulette

-Assumption: no mechanical system can beat the game of roulette

-Assumption: Roulette has no limit, Infinite combinations


FACTS List

Let’s tackle this one

Roulette has no limit...infinite combinations

Alright when something has no limit it has no end to it.  So it can go to infinity but this game can’t say that.  It only has 37/38 numbers.  So to even think of the possible combinations when you combine spins is ridiculous.

So let’s take 2 spins and combined them.

Your results can only be 38 numbers first spin and 38 numbers second spin

First spin you can’t get 39, 147, 1820, etc

Roulette has a limit to the amount of numbers it can give you in that spin.  Limited to 38 numbers.

Now with combined spins roulette again can only give you numbers based on the amount of number it has to offer. So when it runs out of numbers to give you.  YOU HAVE TO GET A NUMBER TO REPEAT.

Something that has no limit can go on forever without repeating!

After 38 spins on spin 39 you can only have a repeat of previous numbers.  So once again the game hit a limit!

So can we continue with the assumption that it has no limit? 

-The limit is 38 possible numbers in a spin!

-The limit is after it runs out of consecutive new numbers it needs to repeat numbers already used.

Fact - Roulette has a limit!

Anyone care to debate on this specific assumption before I make it official on the facts list