I'm writing this as much as a note to myself as anything else. An electronic post-it, if you will...........
What if we we started a new trot and waited for twelve numbers to come in a row with no repeaters? What then if we only bet the six oldest to repeat?
Thoughts?
Sam
From my statistics:
in 32000 rotations
there were after 13 spins no repeaters: 2988
your chance to hit one of the first 13 numbers
in spin %
14 36,44578313
15 35,24096386
16 35,74297189
17 36,0107095
18 35,10709505
19 34,00267738
20 34,57161981
21 35,9103079
22 35,24096386
23 35,07362784
24 34,63855422
25 35,37483266
average 13/37 = 35,13513514
you need to kill the house-edge 13/36 = 36,11111111
to beat american roulette 13/35 = 37,14285714
br
winkel
in 32000 rotations
there were after 13 spins only one repeater 9011
11 numbers appeared
your chance to hit a number which appeared:
14 29,11898678
15 30,05221642
16 29,36340407
17 30,14109543
18 29,61893123
19 29,16342629
20 30,17442506
21 29,81890901
22 29,29674481
23 29,63004111
24 29,33007444
25 30,01888679
average 29,72972973
to beat the house 30,55555556
to beat AR 31,42857143
your chance to hit the repeater again
14 2,97744695
15 2,621930897
16 2,84412843
17 2,666370403
18 2,555271636
19 2,988556827
20 2,733029663
21 3,099655594
22 2,599711143
23 2,866348184
24 2,77746917
25 2,755249417
average 2,702702703
to beat the house 2,777777778
to beat AR 2,857142857
br
winkel
Thank you, winkle.
I would delete the whole thread, but your statistics may be helpful to others.
I see the folly of my idea.
Sam
How long do we bet for? Are they the same numbers? Win goals and stop loss?
geoff
I wouldn't even try it; it was just a fleeting idea. Winkel proved mathematically that it won't work.
Sam