Royal Panda roulette

Author Topic: Street Repeat  (Read 1764 times)

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ignatus

Street Repeat
« on: January 04, 2017, 09:23:55 AM »
I was inspired by the system i found on loothog, where they track x numbers then start to bet them, and continue to bet every number until hit, and looking at the profitchart it only went higher and higher, although that must be a crazy progression for straightups. So, i though well? i try the same thing, but with STREETS...and it seems to work pretty good, extreme cases of 9 uniqe streets happened three times during my first 10 sessions, but i got the progression to cover upto 9 uniqe streets. no more than that. IF the 10th uniqe street hits, that would be game over.

Procedure: BET every uniqe street that hits, in a progressive bet, until hit. That would be 1+1+1+1+1+1+1+1+1 =9 STREETS (stop & restart at any hit) IF the 10th uniqe street hit, that is game over. IF zero hit play the same progression as usual but with a x2 unit bets.

Progression: 1 1 1 1 2 4 10 30 120 STOP

Bankroll: I haven't calculated on this yet, but upto covering all 9 steps in the progressionline.

First tests was played with 1u bets,

10/10 Games won

8)
« Last Edit: January 04, 2017, 09:29:52 AM by ignatus »
 
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ignatus

Re: Street Repeat
« Reply #1 on: January 04, 2017, 09:25:13 AM »
more spins...
 

ignatus

Re: Street Repeat
« Reply #2 on: January 04, 2017, 09:25:57 AM »
more spins...
 

ignatus

Re: Street Repeat
« Reply #3 on: January 04, 2017, 02:01:24 PM »
This system just failed. I ask for this thread to be deleted...sry  :(
« Last Edit: January 04, 2017, 03:20:54 PM by ignatus »
 
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ignatus

Probability calculations for 10-12 uniqe streets
« Reply #4 on: January 04, 2017, 02:17:33 PM »
I have two questions, related to my "Street Repeat"-system, to those who know maths and probability calculations:

*What is the probability for 10 uniqe Streets to hit (no repeats?)

*What is the probability for 12 uniqe Streets to hit (no repeats?)

Thx
 

kav

Re: Probability calculations for 10-12 uniqe streets
« Reply #5 on: January 04, 2017, 02:30:09 PM »
Hello ignatus,
interesting questions.
I know this is NOT what you ask, but I though it may help.
These are the probabilities for streets (3 numbers) in general.
3-number bet

Spins	Probability % to...	...hit at least once	...hit exactly once	...hit more than once	...sleep all spins1	8.108108108108098	8.108108108108098	0	91.89189189189192	15.558802045288516	14.90138787436085	0.657414170927666	84.441197954711483	22.405385663238093	20.53975085384874	1.865634809389352	77.594614336761914	28.696840879732292	25.165820865976837	3.5310200137554553	71.303159120267715	34.47817810569995	28.906686129838263	5.57149197586169	65.521821894300056	39.790758259291835	31.87548092155138	7.915277337740456	60.2092417407081657	44.672588670700605	34.17281287986139	10.499775790839216	55.3274113292993958	49.15859499469784	35.888050591977986	13.270544402719857	50.841405005302169	53.28087107620883	37.100484733598876	16.180386342609957	46.7191289237911710	57.06890855651622	37.88037480307392	19.1885337534423	42.9310914434837811	60.54980786274463	38.28989236851256	22.259915494232068	39.4501921372553712	63.748472090089656	38.38397072814036	25.364501361949294	36.25152790991034413	66.68778516386618	38.21106995909469	28.476715204771494	33.31221483613382614	69.38877555598512	37.813865489665424	31.574910066319696	30.61122444401487315	71.87076672712146	37.22986756704512	34.640899160076344	28.12923327287853316	74.15151537086837	36.49197829995053	37.659537070917835	25.8484846291316317	76.24733844890606	35.62899232664089	40.61834612226517	23.75266155109393318	78.17322992602179	34.66604658808304	43.50718333793875	21.82677007397820619	79.94296804012814	33.62502416802048	46.317943872107655	20.05703195987186820	81.56921387471233	32.52491669168411	49.04429718302822	18.43078612528766421	83.06360193892485	31.382149348462782	51.681452590462065	16.9363980610751522	84.43682340333635	30.21087221705297	54.22595118628338	15.56317659666365623	85.69870258684962	29.023221220805198	56.67548136604442	14.30129741315038524	86.85826724196991	27.82955172288724	59.028715519082674	13.14173275803008625	87.92381314126965	26.638647482493422	61.28516565877622	12.07618685873034926	88.90296342711265	25.457907431918034	63.44505599519462	11.0970365728873527	89.80272314923866	24.293512497402034	65.50921065183663	10.19727685076134828	90.62952938038146	23.150574471998738	67.47895490838272	9.37047061961853729	91.38929726845865	22.033268754238183	69.35602851422047	8.61070273154135730	92.08746235479983	20.944952590235733	71.1425097645641	7.91253764520016831	92.72901946116741	19.888270297395017	72.84074916377239	7.27098053883258732	93.31855842377546	18.865246803457524	74.45331162031793	6.681441576224540533	93.86029692995582	17.877370703952153	75.98292622600367	6.139703070044173
 
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kav

Re: Probability calculations for 10-12 uniqe streets
« Reply #6 on: January 04, 2017, 02:37:43 PM »
I'm not sure this is right, but if we multiply the 0,38 chance of a street appearing exactly once in 10 spins in the 10th degree, then we get 0,04% probabilities (or 4 in 10000 sequences of 10 spins)  for 10 streets appearing exactly once in 10 spins. The chance for no repeats in 12 spins is even lower.
« Last Edit: January 04, 2017, 02:59:57 PM by kav »
 
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ignatus

Re: Probability calculations for 10-12 uniqe streets
« Reply #7 on: January 04, 2017, 02:51:34 PM »
we get 0,04% probabilities (or 4 in 10000 sequences of 10 spins)  for 10 streets appearing exactly once in 10 spins. The chance for no repeats in 12 spins is ev en lower.

I hope so, what you're saying is that it's a probability for 10 uniqe streets to hit is 4 times in 10 000 spins? or did i misunderstood? 0.04% chance ! :D nice! well, i reached 9 uniqe streets hit 3 times during my 10 game session (that's about 10* 100= 1000 spins) :S

thx
« Last Edit: January 04, 2017, 02:57:25 PM by ignatus »
 

kav

Re: Probability calculations for 10-12 uniqe streets
« Reply #8 on: January 04, 2017, 03:02:10 PM »
The statistics are like this:
You take 10.000 of 10-spin sequences, and of these 4 have different streets in each spin.
You need 10 spins to perform each test, right?
 
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ignatus

Re: Probability calculations for 10-12 uniqe streets
« Reply #9 on: January 04, 2017, 03:12:27 PM »
ok Yes.

Well, this "unlikely" event just happened. 10 uniqe streets hit. in just ~1100 spins. I don't know if it's worthwhile to test this system further.. and a 1400u bankroll needed, that is i have to win more than 14 sessions before profit with my wingoal of 100u  :-\

Anyway thanks for the help Kav
« Last Edit: January 04, 2017, 03:19:10 PM by ignatus »
 

kav

Re: Street Repeat
« Reply #10 on: January 04, 2017, 03:25:43 PM »
Quote
Progression: 1 1 1 1 2 4 10 30 120 STOP
This progression is a disaster waiting to happen. Maybe with a milder prog...
 
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Bayes

Re: Street Repeat
« Reply #11 on: January 04, 2017, 03:27:40 PM »
I have two questions, related to my "Street Repeat"-system, to those who know maths and probability calculations:

*What is the probability for 10 uniqe Streets to hit (no repeats?)

*What is the probability for 12 uniqe Streets to hit (no repeats?)

Thx

Hi ignatus,

I worked out a formula for this kind of problem:

N! / NR (N - R)!   (note that 0! = 1 by definition).

Where N = number of equally likely outcomes (in this case N =12 because there are 12 streets).
R = no. of spins.

So for 10 unique streets in 10 spins, R = 10 and the probability is 12! / 2 x 1210 = 0.00387

I'll leave the second one to you.  :)
« Last Edit: January 04, 2017, 03:33:07 PM by Bayes »
 
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ignatus

Re: Street Repeat
« Reply #12 on: January 04, 2017, 03:29:55 PM »
Quote
Progression: 1 1 1 1 2 4 10 30 120 STOP
This progression is a disaster waiting to happen. Maybe with a milder prog...

Yes, if i had a mild progression working i would have the grail along time ago, but that's what everyone wants, and that's what everyone cannot figure, except the guy who claimed to have the grail, and he say it's all about the progression, not the betselection(?)  :(
 
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ignatus

Re: Street Repeat
« Reply #13 on: January 04, 2017, 03:43:33 PM »

So for 10 unique streets in 10 spins, R = 10 and the probability is 12! / 2 x 1210 = 0.00387


You got the same answer to this riddle as Kav did! :) 4 times in 10 000 spins, Thanks Bayes!

Guess i was very unlucky then that it happend once in just about 1100 spins?? :S

Is it worthwhile to test this one further, with another progression? You decide??