I'm not sure if it is useful, but while studying the behaviour of repeats, I found something interesting:
If I wait for a N repeats of a number and start betting on all numbers that repeated N times, the higher N, the higher the less unique numbers I need to bet on.
So for example N=2: I wait till a number hit 2 times, than I start betting on that number till a number hits 3 times. I attached pictures of the distribution for N=1, N=2 and N=20 (~500 sessions, so the tail would of course be longer, but it's about the general behaviour of the distribution).
The way this could be utilized is via a standard progression (maybe there are other methods): the power of the reduces when N increases. Of course this is not a viable method: the earnings per hour are minimal, but maybe I can trigger some thought.
grts reddwarf
OK, I created a script that bets on repeats (N=20), max table limit=500, max unit per number=50, european roulette, restart session when in +:
win probability: 99. 74%, but because it is a progression, a loss will hammer you. So if a different approach would be needed here.
Anyway, it was worth a try.
hi reddwarf, I have been messing with repeats for a while now, wait until a number hits twice then bet on that number (and any others with the same amount of hits) what do you mean by N=20?
Hi Superman:
With N=20 I mean waiting till a number occured 20 times. I know, this is not feasible in a B&M casino, but I still think that the effect shown is interesting.
does this help?
grts reddwarf
Hi reddwarf,
What you detected is the normal deviation.
If you wait until a number has hit x times and then bet on it, it may hit at once or will be missing 500 times. x might be 0 or 500 it makes no difference.
This selection is described by "Haller" in his Roulette-Lexikon. He is waiting for a number that hit 8 times.
br
winkel
Hi Winkel,
Thanks for your reply. I do agree that the previous spins do not bear any information about future spins, so independant from X (or N) they might or might not sleep.
What I found is, in my humble opinion, different from what you describe (of course I will think hard and long about your answer!): the more repeats we use as a start, the less unique numbers we will need for an additional repeat (or in other words, if we start spinning, we must wait up to say 24 spins for a repeat, if we have however a number that has been hit 20 times (just for arguments sake), we will have to bet on maximum (more or less) 11 unique numbers.
This is indeed a result of normal random behaviour. Can this be exploited? I don't know. Not if we use the normal progressions.
But please, correct me if I'm wrong, I'm here to learn (and of course to find the holy grail ;))
grts reddwarf
By the way, I encountered this phenomenon while studying repeating behaviour, for me this is just a side result.
Quote from: reddwarf on Dec 07, 09:41 AM 2010
Hi Winkel,
... I do agree that the previous spins do not bear any information about future spins, so independant from X (or N) they might or might not sleep.
Upps, I didnÃ,´t say that!
It is long tested: betting only on favorites or on sleepers makes the known results.
The only way to get informations is to detect which of them is on a hitting streak at the moment.
br
winkel
Hi Winkel,
OK I think I now understand what you are meaning: i was confused because I do not think in terms of sleepers and non-sleepers: i do not believe in them. You are right, what i did not mention is how many spins are needed for a next repeat, for N=20 this might well be a couple of 100!!.
What I find interesting (but again, I'm not walking that avenue, for me it is only a helping knowledge while building a beting system based on 3 very different principles): the progression, if used, is less steep when N increases.
grts reddwarf