Roulette Ideas

This topic has come up many times in the forums over the years, and discussions have often generated more heat than light. The differences are subtle and can be confusing, but here's a simulation which may be helpful. I found the link to it on stats.stackexchange, posted by Matt Wonlaw - thanks Matt! A link back to that discussion is also on the site, with links to several other discussions on the topic. These are worth reading.

https://tantaman.com/2021-01-26-regression-mean-vs-gambler/

This is what I have a hard time getting my head around.

Spin a million times and you will have less streaks of 20 reds in a row than you will 10 reds in a row.

Lets say the highest reds in a row are 25. A million spins were only able to reach 25 reds in a row, but by chance 25 reds came up and you get to place your next bet.

Do you bet that a million spins couldn't get 26 reds in a row and bet black.

Or do you assume the table is broken and bet red?

I couldn't edit, but what I'm getting at is that you are more likely to get 4 in a row than 10 in a row. It is not a 50/50 chance of getting either one. The past doesn't predict the future, but ran it a million times and you will never have more 10 in a rows than 4 in rows. The longer the chain is, the less chance of it happening.

I should make one observation that feels kind of like a schrodinger's cat situation. What players will do is wait for 7 reds in a row, then pretend like they played the reds, but not really, then place a bet. But in order for it to be valid, they actually had to play, not pretend they played.

A lot of people have a hard time getting their heads around this concept. They can seem to accept that, for example, there is a roughly 50:50 chance that the next outcome will be red. However, the chance that you'll see 6 reds in a row is very small. So how is that that if you've seen 5 reds in a row, and you bet on red, it's NOT the same as betting 6 times in a row on red?

One way of thinking about it is like this : If you bet for 6 reds BEFORE any outcomes, the chance is (1/2)6 = 1/64, ignoring zero. It's also the case that for ANY series of 6 EC outcomes, at the start the chance of that particular series is 1/64. e.g. the chance of RBBRRB is 1/64, and the same for the other 63 possible patterns.

You now see 5 reds in a row. How many series of 6 are possible at this point? Just two : RRRRRR and RRRRRB. Each of these series started out with a probability of 1/64 but having seen the first 5, the chance is 1/2.

In fact, by similar reasoning, you can show that no bet selection whatsoever, if based on past spins, can change the probability. It doesn't matter whether you're looking for virtual wins or virtual losses (hot numbers or cold numbers).

Let's take another example betting on single numbers. Suppose you favour a "hot number" bet selection, such as that if you see a number repeat in the last 10 spins you will bet on it to repeat in the next 10 spins.

Example : you see 3,15,34,11,25,15,0,9,29,3.  #15 has repeated, signaling a bet on it for up to 10 spins.

You are hoping that having seen the last 10 spins, the 11th spin will be #15. There are 37¹¹ possible patterns consisting of 11 numbers, but having seen the first 10, there remain only 37, namely

3,15,34,11,25,15,0,9,29,3,0
3,15,34,11,25,15,0,9,29,3,1
3,15,34,11,25,15,0,9,29,3,2
3,15,34,11,25,15,0,9,29,3,3

etc,so your preferred outcome of 3,15,34,11,25,15,0,9,29,3,15, where #15 repeats again, has exactly one chance in 37 of occurring.

This logic is the same for each and every spin, and for any possible bet selection based on past spins. Can you see that no matter what it is, the probability of a win hasn't changed?

Quote from: Median Joe on Oct 26, 2022, 11:54 AMfact, by similar reasoning, you can show that no bet selection whatsoever, if based on past spins, can change the probability. It doesn't matter whether you're looking for v

This is the ultimate truth. Everyone should have it in mind.

Your bet selection leads to win but also to loss. In the statistical advantage.

I disagree

When you have a window of events in a larger window of events things are due to happen.
Have no time to prove this for the forum members.
But I have the simulations - excel sheets - that pinpoint pout 3.0 STDV windows (imbalance).

What happens is regression, but they can come in small, medium, or large chunks.
We can get 3.0 and then it continues to grow stronger and sometimes it can stop and hover around zero point and then grow stronger again and then comes regression.
Sometimes you can bet back to back where the window gets a small regression, then grow stronger again, and finally regression.
The combinations are endless, the point is when the state hovers around zero you get your winning events that is part of regression.

Bayes/Jules simulate several million trails and the peak was around 5.54 STDV
I can run different playing models day in and day out and I always get regression after the initial 3.0 window.

My last example is to describe the process.
Let's say you have singles versus series then 14 singles versus 2 series window will not continue the whole 300 trail window.
It will not grow to infinity with singles and small chunks of series.
Series will reach zero point or regression sooner or later at some point.

No one can prove me wrong on that matter.
The tricky part is to develop a march to catch the hovering or regression.
Using indications can be a small regression.
But problem is that you can get several small regressions (indications) and lose several bets.

So even if we know what will happen after a certain window of events.
So is there no way to know when to jump in on a hovering/regression.

Cheers

Quote from: Patrik on Oct 27, 2022, 09:34 AMNo one can prove me wrong on that matter.

Nobody needs to, cause you did it already yourself.

Try and look at this:

If you have a peak of >+- 3StdD then stop your run and start it from that point again.

Then you will have the clear eyewatering way back to mean.

The distance between -5 Dev. and -2 Dev is exactly +3 Dev. did you realise that?
So no need for Roulette to do any back to something and getting up to something to serve both ends of Dev.
In the END there will be exactly as many -5 to -2 than +5 to +2 and any others.
You cant await to see -5 to +5 that´s a very very very ................ rare event.

 @Median Joe

Thank you for the explanation. It does make sense to me now. If I did a 1000 spins, it would create a distributions chart of different lengths. So, if I wait until 10 reds in a row or what ever, I would set the distribution chart right on top of the column and that would be my odds for the next spin. But I had played from the beginning, then the distribution chart would be at the bottom of the column of 10 reds, but then, I would have to deal with all the losses.

We can use enter and stop/out  points , this doesn t mean will always help  but many  times yes.



See the chart(use zoom) ...the M  and W  patterns are allways there. When system start to go up( after one or two virtual wins) start the real  play, and when system  tend to go down ( for example two or 3 Ls)  stop and wait new virtual wins. Sometimes we can avoid short or long bad runs ( LLLL ...s) .

Virtual play until WW  and stop at LLL,

Virt.,
WLLLWW( start )LWLWWLLWWWLWWLLL( stop )LLWLLLLLWW( start )LLWLWWWWWWLWLWWLLL( stop ) LLLLLWW ( start ) ...so on

I've made a thread about using press (as a fundamental inter-session module), related to the ≈3std topic above .. taking advantage off the fact & plowing straight through it.

[1→2→3] .. design & plug aporopriate base system in;
you might even use a bit of margin = a bit lower session (press-step) gosl than the termination threshold.