Holy Grail By Winkel.

[Azim]

Winkel,

I think you know by now that i don't mind taking no as an answer.



By the way: I got this 3 crossing from a file i tried to look for crossings. Tried to run for spin 25 only.

Let's look at this 3 crossings at spin 25:

Question is do we bet or we don't?

Crossing 1: 17 16  3  1  0.
Crossing 2: 17 17  1  2  0.
Crossing 3: 17 17  2  0  1.


I am looking at your 25 spin page. Working my answer from that. I will give you the actual result. once we done discussing this. At the time of writing this I haven't checked, for reason is don't want to discuss something knowing what the out come
is.
This is based on the fact that number's on the right are the count that hit after facing the left result/crossing on page 25.
If the above is wrong. My observation below is useless. However, the concept is the same.

I look at Crossing 1 from your 25 page and get, 29728 times 0 hit, 27833 times 1 hit, 5184 times 2 hit and 1747 times 3 hit.
Crossing 2: 3491 times 0 hit, 3507 times 1 hit, 204 times 2 hit and 400 times 3 hit.

Crossing 3: 2719 times 0 hit, 2633 times 1 hit , 347 times 2 hit and 148 times 3 hit.

I AM IGNORING THE FACT OF 2.7% HOUSE ADVANTAGE.


I am going to rule crossing 2 out. Why, the 1's hit more than the 0's.

Crossing 3 I would not bet again. Even though the numbers show that 0's hit way often. Why  look at 29000 compared to 2719..  The Ratio against spins observed is way way below.

if I look at crossing 1,  0's have hit more often than not.  Also this is the first observation. at spin 25. We also have to analyze and see if we did miss on spin 26 did we hit on spin 27.  If the hit difference on spin 27 is way ahead of spin 26 or vice versa.  If we have to wait it out. We can't bet.

Considering every thing is right I would bet this crossing. WHY? because the hit rate of 29000 is way to high of a hit rate based on total observed spins. which is about 55%.(even though 55 at this stage is low, we will get better opportunity next spin maybe where the hit rate is way higher. Or 55% might be the highest. It all needs to be taken into account. I am not factoring in a lot for this discussion. trying to minimize to get to the point.


Winkel,

Don't take this the wrong way. But I think, who ever attempts to accomplish this this way, will have to use his / her own data  and thinking.

Yes, I asked and you gave it to us. But after looking at this. Looks like one will have to work their way from spin 0 to 25 and note the path that was taken to reach the crossing observed at spin 25. Easier said than done.

I don't know how many more spins pages you have and are willing to share. I totally understand if you say no.
On the other hand, one will still have to run their own data to understand whats happening before making any conclusions.

[winkel]

Hi Azim,

lets do it step by step:

The first possible crossing over all is 19 - 18

The earliest spin after starting the trot is after spin 18.
That means there were 18 unhit in a row! no repeaters.

So therefore we call this trot fast! What do you expect after this rush? another unhit or seems it more possible that there will be 1´s to hit?

17 - 17 - 3 first appears at spin 23 so this trot is also fast.

therefore to control the "speed" I gave the steps for spin(s) 13 - 25 - 37 - and 50
If trots are fast, don´t bet the early crossings
if trots are slow don´t bet the late crossings

What I wanted you to see is:
at spin 25 we have
26   2   4   5                  
26   2   5   3   1               
26   3   2   6                  
26   3   3   4   1               
26   3   4   2   2               
26   4   1   5   1               
26   4   4   1      2            
26   5   1   4         1         
26   5   2   1   2   1            
26   5   2   2   1      1         
26   5   2   3            1      
26   6      4            1      

at spin 26 we have
26   2   4   4   1   0   0   0
26   3   2   5   1   0   0   0
26   4   3   2   0   2   0   0
26   5   1   3   1   0   1   0
26   5   1   4   0   0   0   1
26   5   2   1   1   2   0   0

at spin 27 we have
26   3   1   6   1   0   0

[Azim]

I think I understand your explanation.

I will see if I can analyze it that way.

[Azim]

What I wanted you to see is:
at spin 25 we have
26   2   4   5                  
26   2   5   3   1               
26   3   2   6                  
26   3   3   4   1               
26   3   4   2   2               
26   4   1   5   1               
26   4   4   1      2            
26   5   1   4         1         
26   5   2   1   2   1            
26   5   2   2   1      1         
26   5   2   3            1      
26   6      4            1      

at spin 26 we have
26   2   4   4   1   0   0   0
26   3   2   5   1   0   0   0
26   4   3   2   0   2   0   0
26   5   1   3   1   0   1   0
26   5   1   4   0   0   0   1
26   5   2   1   1   2   0   0

at spin 27 we have
26   3   1   6   1   0   0

Winkel,

I tried looking at this. Spent 2 days. This makes no sense.

You can have at spin 25 -> 12 different combination of result set.

At spin 26-> this should change to become more than 12-> it can't go below 12 or stay at 12.(i am saying stay at 12 because you have =2 to =8 or =9) this has to go up can't go down or stay even. Your spin   26 has got only 6 from 12...Something wrong.

At spin 27 -> Even worse, it has dropped to 1.  It's not even possible.


The number of result set is supposed to increase with every spin not decrease or be even.

The only way it can actually stay even is if you minimize your elements( meaning =1, =2, =3 , =4 , = 5, =6 ,= 7) if i drop that to =1,=2 i can understand same number but it shouldn't drop.

Are you sure we comparing the data from the same analysis of spin 25?

[winkel]

The number of result set is supposed to increase with every spin not decrease or be even.

No!

look at spin 1:
you have 37 0 0 0 0 0  at the beginning
after the first spin 36 1 0 0 0 0
after the second spin  35 2 0 0 0 0 or 36 0 1 0 0 0
after the 3. spin: 34 3 0 0 0 0 or 35 1 1 0 0 0 or 36 0 0 1 0 0

at spin 25 we have only 26 unhit left at spin 27 they disappeared at all.
same at the end and the beginning of the stats.

And same in the middle of any statistical range:

14 unhit earliest appear after spin 23.
from then on they will disappear by 14/37 and stand by 23/37
and renew themselfs by 15/37of the amount of 15 .. ..

imagine you have 1000 times 14 unhit at spin 23
you know exactly that there will be 378 hits of that 14 numbers in the next spin.
you know exactly that there will be 235 hits of that 14 numbers in the next spin out of the 1000

in total:

Code Select Expand

24 1000 378,3783784
25 621,6216216 235,2081812
26 386,4134405 146,210491
27 240,2029495 90,88760251
28 149,315347 56,49769885
29 92,81764812 35,12019118
30 57,69745694 21,83147019
31 35,86598675 13,5709139
32 22,29507284 8,435973508
33 13,85909933 5,243983532
34 8,615115802 3,259773547
35 5,355342256 2,026345718
36 3,328996537 1,259620311
37 2,069376226 0,783007221
38 1,286369005 0,486734218
39 0,799634787 0,302564514
40 0,497070273 0,188080644
41 0,308989629 0,116914995
42 0,192074634 0,072676889
43 0,119397746 0,045177525
44 0,07422022 0,028083327
45 0,046136894 0,017457203
46 0,028679691 0,010851775
47 0,017827916 0,006745698
48 0,011082218 0,004193272
49 0,006888946 0,002606628
50 0,004282318 0,001620337
51 0,002661981 0,001007236
52 0,001654745 0,00062612
53 0,001028625 0,00038921
54 0,000639416 0,000241941
55 0,000397475 0,000150396


now you can see (calculate) at what spin all these 1000 groups of 14unhit after spin 23 will have disappeard.

[Azim]

Thank you. Now I think I probably was doing it wrong.

Let me ask you this, If we have 37 spins.
we do statistics on 37 spins we get:

spin 23: 20 '0' and 17 '>0'  (there is also other possibility, but we get this in average)

spin 25: 18 '0' and 19 '>0'  (there is also other possibility, but we get this in average)

Now we are playing.. if at spin 24: we get this: 20 14 1 1 1

I know we don't bet any thing over 17.

Just for this example. Since we know we are behind by 2 '0' from average, can we not play the 0 to hit for next 2 spins? Or will that be a bad decision?

[winkel]

Now we are playing.. if at spin 24: we get this: 20 14 1 1 1

Just for this example. Since we know we are behind by 2 '0' from average, can we not play the 0 to hit for next 2 spins? Or will that be a bad decision?

we can´t do a decision only on these informations!

When did the 1 1 1 appear?
at the very beginning on that trot or somewhere in between or just before spin 24?

we could bet that 20 only if the repeaters all/mainly appeared in the last 6-8 spins
otherwise I would bet on that 14 1s for 2 spins

br
winkel

[KoolKat]

Hi Azim,

I think what Mr Winkel is saying is what got us to 1 1 1 the hidden formula,


Mr winkel I am still struggling how to dictate what is classed for a fast trot or slow?

Koolkat

[Azim]

Hi Azim,

I think what Mr Winkel is saying is what got us to 1 1 1 the hidden formula,


Mr winkel I am still struggling how to dictate what is classed for a fast trot or slow?

Koolkat

I realize that.  What I am after is to try and create basic statistics as to what is reliable.

Example: if I was going from A(20) to B(18). I don't want to worry about taking route from A(20) to C(20) to B(19), or taking route A(20) to D(19) to B(18).

What I am saying is from spin 1 to spin 37. There has to be a favorable crossing that can be hit on a spin for a chance of 2 spins.

In my example: i was waiting to be behind by 2 '0's and than bet from 2 spins.

I know Winkel, is saying can't do that.

I am saying, if out of 100 identical chances If I can hit 90-95% of the time without worrying about the 1 1 1 thats a good crossing to bet over a long period of time.  You have to wait for that crossing. It becomes a long grind. If you now pick 5-6 good spins ant there stats it becomes a worthy grind.

All I am after is if i am going from A to B. As long as I can reach B within 2 stops I am good. 

[KoolKat]

Azim, Sorry if I was stating the obvious, I have looked at many of spins trying looking at the same way but it is to Systematic. I was trying to illustrate that spin before last:

if R is doing its job
if N is doing its job
if >N is doing its job

Koolkat

[winkel]

.., if out of 100 identical chances If I can hit 90-95% of the time without worrying about the 1 1 1

....

All I am after is if i am going from A to B. As long as I can reach B within 2 stops I am good.

you have to worry about that 1 1 1, otherwise you can´t reach the 95% hitrate.

don´t try to make it shorter as I explained. Otherwise you will add yourself to the people who claim that GUT doesn´t work.

[winkel]

again:

KonFuSed proofed that every possible crossing ends up at -2.7%. That is correct and it has to be that way otherwise GUT wouldn´t work if we couldn´t trust in that.
but he checked every crossing despite where/when it appeared.

If you look at my statistics you will see, that the rare situations have a better hitrate than the "normal".

What we have to do is to separate the losing crossings from the winning ones.
If we can define a crossing it might hit. We bet on it.
If we can´t do so, we are able to bet the other positions even if they ar not a crossing, or we better bet on crossings like 5 vs 6.

With my above explanation we know, that e.g. the situation 14 23 could last from spin 24 to spin 39.
to do that it has to have hits on 1s and then also on 2s, 3s etc. in the most spins

br
winkel

[Azim]

you have to worry about that 1 1 1, otherwise you can´t reach the 95% hitrate.

don´t try to make it shorter as I explained. Otherwise you will add yourself to the people who claim that GUT doesn´t work.

Winkel,

Be assured, I will never say that. I know for a fact it does work. I am willing to put my money on it.

What I am trying to find is an easier way out. I guess there isn't. 

I have a question in respect to your data.  I know you have mentioned you have from spin 25  to spin 50.

Was your data analysed based on 1-50 and 51-100 or 1-50, 2-51, 3-52.....49-98, 50-99 and 51-100.

I wonder if it would make any difference?  I think it  shouldn't make any difference, I could be wrong ...Am I?

[Azim]

again:

KonFuSed proofed that every possible crossing ends up at -2.7%. That is correct and it has to be that way otherwise GUT wouldn´t work if we couldn´t trust in that.
but he checked every crossing despite where/when it appeared.
I think you have said this. if the trot is fast we don't bet. so you can't compare that. The only way to over come the -2.7% is to grind down to a perfect betting opportunity. With GUT it is doable but will become a very long grind. But the end result is worth it for the grind.
If you look at my statistics you will see, that the rare situations have a better hitrate than the "normal".
This is what I am looking for, how often do the come is a trot of 50?
What we have to do is to separate the losing crossings from the winning ones.
If we can define a crossing it might hit. We bet on it.
If we can´t do so, we are able to bet the other positions even if they ar not a crossing, or we better bet on crossings like 5 vs 6.

With my above explanation we know, that e.g. the situation 14 23 could last from spin 24 to spin 39.
to do that it has to have hits on 1s and then also on 2s, 3s etc. in the most spins

br
winkel

[winkel]

What I am trying to find is an easier way out. I guess there isn't.  absolutely not

Was your data analysed based on 1-50 and 51-100 or 1-50, 2-51, 3-52.....49-98, 50-99 and 51-100.

I wonder if it would make any difference?  I think it  shouldn't make any difference, I could be wrong ...Am I?

In Roulette anything can happen and will never know at which point you are getting into the game. It could be spin 56 or 119.
But there is always a winning streak inbetween 37 spins. If you come into a losing one, then just try to jump into a winning streak.

And please don´t bet down to spin 50 or even more spins. Stop in a and restart. Or if you are unsure, just jump!
Don´t bet before spin 25 (as long as you are not able to read every possible sequence.