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