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If "this", then "that"

Started by Priyanka, Jan 10, 08:21 AM 2017

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Priyanka

Quote from: shazwad on Jan 10, 03:20 PM 2017What win goal / stop loss do you use.
Shazwad - I have been playing this only for the past two days now. Also am not a believer of win goals and stop losses, so unfortunately not able to advice one. Welcome your suggestions for win goal and stop losses and any recovery sessions.
Disclaimer : Roulette systems are subject to laws of probability. If you are not sure about the effects of it, please refer to link:://:.genuinewinner.com/truth. Don't get robbed by scammers.

RouletteGhost

Test it for many many spins

See the average of how low it goes. Create stop loss around that. Make win goal modest nothing extreme
the key to winning with systems : play for a statistically irrelevant number of spins

link:[url="s://m.youtube.com/watch?v=nmJKY59NX8o"]s://m.youtube.com/watch?v=nmJKY59NX8o[/url]

falkor2k15

Quote from: Priyanka on Jan 10, 01:21 PM 2017Unless a bet exists (I am not sure one exists) where stitching of the bets leads to a situation where the accuracy of prediction increases and impacts the odds of an event occurring. Considering none of us are aware of any such bet we have to draw a conclusion that there is no edge.
If true then I guess the event must be based around a repeat, such as Cycle Length 2 or 3? The odds of that event occurring are based on what has already appeared once - preferably halves that are not equally likely rather than the equally likely ones. For example, 6 options based on Dozen Cycles:
CL1o1, CL2o1, CL2o2, CL3o1, CL3o2, CL3o3.

Usually the odds for a repeat on those are as follows:
49%
22%
22%
2%
2%
2%

Let's say we have the following open cycle of unique options 123...

The next repeat event for one of the 6 options is as follows:
16%
39%
41%
1%
1%
1%

In the above situation we expect a repeat on CL2 more than anything else - particularly if CL2 and CL1 have repeated more times than CL3 (=less CLs) so we could stitch our EC > Dozen 5/1 bet so that it finishes on spin 2 to match the anticipated repeat. However, the odds for the 2nd spin happen to be split equally between order 1 and order 2 (the previous defining dozen vs. the last dozen). Therefore, we could look at which dozen has repeated the most (=less dozens) and aim our 2nd stitched bet to land on that dozen; presumably, the stitched bet would then impact further on the 39 or 41% for CL2o1 or CL2o2? I'm just thinking out loud mind you - It could work vice versa - but am I at least thinking along the right lines..?
"Trotity trot, trotity trot, the noughts became overtly hot! Merily, merily, merily, merily, the 2s went gently down the stream..."¸¸.•*¨*•♫♪:

RouletteGhost

I am glad we aren't making so simple into something so complicated.
the key to winning with systems : play for a statistically irrelevant number of spins

link:[url="s://m.youtube.com/watch?v=nmJKY59NX8o"]s://m.youtube.com/watch?v=nmJKY59NX8o[/url]

Priyanka

Quote from: falkor2k15 on Jan 10, 05:12 PM 2017true then I guess the event must be based around a repeat, such as Cycle Length 2 or 3?
Am not sure falkor as am not aware of existence of such a bet.
Disclaimer : Roulette systems are subject to laws of probability. If you are not sure about the effects of it, please refer to link:://:.genuinewinner.com/truth. Don't get robbed by scammers.

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