How to play and win with VdW (from "Random Thoughts")

[winkel]

Friend, it was meant as a joke! Forget it, it´s useless.

Not counting with the zero, in the long run the LW registry always ends like this:
If you play ECs, your Ls will equal your Ws.
If you play single dozens, your Ls will be twice as much as your Ws.
If you play double dozens, your Ls will be half as much as your Ws.
And so F on.

In nine spins, playing ECs, both L and W have equal opportunities to show up in the mutual bets.
In nine spins, playing single dozens, the Ls have double the opportunities to show up in the mutual bets.
In nine spins, playing double dozens, the Ws have double the opportinities to show up in the mutual bets.

None of it works.

Thats a common problem. People start thinking and stop before they put all informations and possibilities together.
You can use LW as information with this and any other strategy. you just have to sort it out correctly.

[psimoes]

Analysing the LW with the vdW theorem won´t produce any signs of profit, trust me.

Even if you look at it like this:

VdW says two different colours will form one or more APs every 9  spins.
People immediately think of ECs. But they can be any two opposing chances. Two dozens against one, three streets against nine, five double streets against one, for example.
When playing ECs you refrain from betting the mutual bets because any outcome has 50%, of hitting. Playing two dozens against one, during the mutual bets stages, the double dozens will hit twice as much as the single dozen.
You´d think you got an edge, that the application of the vdW will let you increase the accuracy of your predictions, but the truth is the single dozen will still hit when you´re betting the opposites, and the double dozens will still hit when you don´t bet trying to avoid the single dozens. Each time that happens you still lose twice as much as what you win, and you miss valuable  opportunities of winning.
One L equals two Ws in value.
You can pick any bet selection, use any mechanical method. The odds still don´t change.

[winkel]

So what is the reason that you still are interested in Roulette?
"It always can hit the other chance!" Thats a very depressive view.

The odds will not change
That is something that is for sure. And it is something you can rely on. So you can do something with it.

[psimoes]

Just coming in for a chat.
And checking out how things develop.

There is always a spark.

[winkel]

I had to go back to page #7 to find my last post of handling rules. It is redicoulos how much rubbish is written in between. I really don´t know where I stopped explaining anymore.

Are there some questions to the strategy? Pls no comments to random / non-random or other rubbish.

[stringbeanpc]

lets take my cluster of 6:

1 1 1 + 1 1 1  = 4 wins
1 1 1 + 1 1 2  = 3 wins 1 loss
1 1 1 + 1 2 1  = 2 wins 1 loss
1 1 1 + 1 2 2  = 2 wins 1 loss  total 11 wins + 3 losses + some no AP as trigger
---
1 1 2 + 1 1 1 = 1 loss 1 win
1 1 2 + 1 1 2 = 2 loss
1 1 2 + 1 2 1 = 1 loss
1 1 2 + 1 2 2 = 1 loss           total 5 losses 1 win
---
1 2 1 + 1 1 1 = 2 win
1 2 1 + 1 1 2 = 1 win 1 loss
1 2 1 + 1 2 1 = 1 loss
1 2 1 + 1 2 2 = 1 loss           total 3 win 3 losses
---
1 2 2 + 1 1 1 = 1 loss 1 win
1 2 2 + 1 1 2 = 1 loss 1 win
1 2 2 + 1 2 1 = 1 loss
1 2 2 + 1 2 2 = 1 loss             total 4 losses 2 win

17 wins + 15 losses!!!! and some no AP as Trigger

One bad habit of mine, is browsing over the a forums posts and NOT reading/studying them in full.
Then believing that I have a full understanding of the concept.
There are probably many other members on here who do the same.
It is always good to go back and take a 2nd look.

Regarding VdW, as I understand from spins 1 to 6 there are only two possible AP lengths, being AP1 & AP2

AP1
123
234
345
456

AP2
135
246

In winkel's example above, I believe he is using ONLY the AP length of 1, therefore a win would be three consecutive spins of the same color (if we are using Red/Black)

With respect, I think there is a mistake in the following line.

1 2 2 + 1 1 2 = 1 loss 1 win

I cannot see a winning AP in this line, instead I see two losses

221 - Lose (spins 234)
112 - Lose (spins 456)

This would change the final tally from

17 wins + 15 losses!!!! and some no AP as Trigger

to

16 wins + 16 losses!!!! and some no AP as Trigger

Leaving us again a 50/50 chance, with no edge.

Refer to my attached spreadsheet

Winkel, hope that I am wrong and we can still find an edge. Maybe I am missing something, Please Help.

Thank You

[winkel]

Hi stringbeanpc,

you are right. This is a severe mistake.
I´m sorry for that.

But now I know what I wanted to explain next. Thx.

[stringbeanpc]

No apology needed. I'm Looking forward to your next explanation.

[winkel]

Hi stringbeanpc,

I found another win:
1 1 2 + 1 2 1 = 1 loss
1 1 2 is loss
   1 2 1 2 1 is a win!

[winkel]

Even if we had just a 50:50 Situation there is one thing we can rely on:
In about 68% we meet an AP int the first 6 spins.

But lets stay at the 50:50 Barrier ( dont´t forget: the Zero is included)
We cannot be sure not to start with one or two losses or some losses in a row.

If we play flatbet every loss needs two winning spins. and that could last a very long time, still we are back to +/- 0.

So our Job is to find a suitable progression.

I prefer the PowerMartingale: (a*2)+1 means 1 3 7 15 .....
The advantage is that for every bet we win 1 unit.

Martingale basic: a*2 means 1 2 4 8 16 ...
We have 1 step more but have the same risc-amount. I think its not worth it.
But if you will bet with big units this could be an alternative.

D'Alembert: a+1 means 1 if loss +1 if loss +1 ..... if win -1 if win -1 til 1
Can take a lot of Time to come back, but offers 1/2 unit per bet.

Fibonacci: 1 1 2 3 5 8 13 (last und last before last)
it needs to wins either WW or WLW to win, but then it wins quite good.
Version A:  loss loss loss 13win next 21win
Version B: loss loss loss 13win next 13 win
Version C: loss loss loss 13 win next 8 and win
you should compare these 3 Versions and take some which is suiting your risc-management.

Fibonacci Light (an invention of a friend) 1 1 1 2 3 4 6 9
It starts as a Labouchere 1 1 1 if these 3 are lost then add like Labby first and 3rd
1 .. 1 = 2
.. 1 .. 2 = 3
.. .. 1 .. 3 = 4
.. .. .. 2 .. 4 = 6
and so on
needs three wins

[psimoes]

lets take my cluster of 6:

1 1 1 + 1 1 1  = 4 wins
1 1 1 + 1 1 2  = 3 wins 1 loss
1 1 1 + 1 2 1  = 2 wins 1 loss
1 1 1 + 1 2 2  = 2 wins 1 loss  total 11 wins + 3 losses + some no AP as trigger
---
1 1 2 + 1 1 1 = 1 loss 1 win
1 1 2 + 1 1 2 = 2 loss
1 1 2 + 1 2 1 = 1 loss
1 1 2 + 1 2 2 = 1 loss           total 5 losses 1 win
---
1 2 1 + 1 1 1 = 2 win
1 2 1 + 1 1 2 = 1 win 1 loss
1 2 1 + 1 2 1 = 1 loss
1 2 1 + 1 2 2 = 1 loss           total 3 win 3 losses
---
1 2 2 + 1 1 1 = 1 loss 1 win
1 2 2 + 1 1 2 = 1 loss 1 win
1 2 2 + 1 2 1 = 1 loss
1 2 2 + 1 2 2 = 1 loss             total 4 losses 2 win

I don´t understand one thing: the clusters of 6 posted above seem more like two chained clusters of 3, because any second cluster of 3 never begins by a 2. In a true series of 6 spins one should find something like this 1 2 1 2 1 2, but in your examples the spin #4 always begins by a 1.

[psimoes]

Never mind, I think I got it.

Hi stringbeanpc,

I found another win:
1 1 2 + 1 2 1 = 1 loss
1 1 2 is loss
   1 2 1 2 1 is a win!

B B R + B R B \
                          Both of these outcomes are winners, if the vdW is focused on the order of appearance of the colours.
B B R + R B R /

Cheers!

[psimoes]

Now, unless spin#4 must always be the same colour as spin#1, otherwise no bet, the AP 2-3-4  becomes a mutual bet since you don't know the colour of the fourth spin before it hits. Ex. spin#4 is never a 2, so a bettable 2-3-4 AP must be 1 1 1 + 1.  The 1-3-5 or 2-4-6 APs are bettable because by then we know which colours are 1 and/or 2 in the second cluster of three to bet.

If we avoid such bet when spin#4 is unknown, we have 13 wins and 13 losses.

111 111 - 3W
112 111 - 1W 1L
122 111 - 1W
121 111 - 2W

111 112 - 2W 1L
111 122 - 1W 1L
111 121 - 1W 1L

112 112 - 2L
112 122 - 2L
112 121 - 1W 1L

122 112 - 1L
122 122 - NB
122 121 - NB

121 112 - 1W 1L
121 122  - 1L
121 121 - 1L

13W vs. 13L

[psimoes]

I probably forgot one or several combinations, but true clusters of six seem to provide 30 wins against 10 losses!

111 111 - 4W
111 112 - 3W 1L
111 122 - 2W 1L
111 222 - 2W 1L
112 222 - 2W 1L
122 222 - 3W
122 221 - 2W 1L
122 211 - 1W 1L
122 111 - 1W 1L
121 111 - 2W
122 212 - 2W 1L
122 122 - 1L
121 222 - 1W 1L
112 221 - 1W 1L
112 211 - 2L
112 212 - 2L
112 112 - 2L
121 112 - 1W 1L
121 122 - 1L
121 212 - 2W
121 211 - 1W 1L

30W vs. 20L

[psimoes]

121 221 - 2L. Any others I missed?
For now, 30W vs. 22L...