markthurston
Posted by markthurston
Oct 28, 2014

Probability

Hi , I'm a newbie. Can anyone advise on what the probability is of the following strategy been successful ( i know in the long run , it wont be successful ) i'm just interested in what the probability equation is and don't know myself how to work it out ....................

1. Playing single zero roulette
2. Wait till 5 x consecutive results occur where they are all alternative even money bets , such as RBRBR , or HLHLH , or OEOEO , etc
3. Once this occurs , and lets say it is RBRBR bet One ( 1 ) unit on the R that it will come next , that is same as the last result ..............
4. If it doesn't win , then Martingale it for a further 6 bets following the last each time , if it comes black then back black next , if it then comes red , then back red next , ensuring that the total number of the Final 12th bet , does not exceed the table limit you are playing with
5. So , the bets would be

6th bet = 1
7th bet = 2
8th bet = 4
9th bet = 8
10th bet = 16
11th bet = 32
12th bet = 64

I know in the long run it will lose , but what i am interested in knowing is by the time we get to the 12th and last bet what is the probability of success according to probability formula and or what is the probability of all bets from the 6th till the 12th ?????

Any advice would be greatly appreciated , thank you

Comments

Zcore13
Zcore13 Oct 28, 2014

You are better off posting this in the forum under roulette. You'll most likely get much better statistical analysis.



The probability every time all the way through is 48.65 to win (or 51.35 to lose), when betting red/black on a single zero game. It does not matter what order the reds or blacks have come up previously.





ZCore13

markthurston
markthurston Oct 28, 2014

Thank you Zcore13 for your comment

mustangsally
mustangsally Oct 28, 2014

so, a 7 step Marty after seeing a 5 trial alternation (without the 0)

could be waiting for some spins (maybe not for all 3)



OK

a 127 unit bankroll is required

how about double or big loss because if you really can not double before a loss (the big one) you do not have enough to keep the 7 step going



the chance to double before the first big loss would be (1 - (19/37)^7)^127

(19/37)^7 = the chance to lose 7 in a row = 0.00941592820025755468171924197108 (copy/paste)

so that leaves the chance to win 1 unit during those max 7 bets

one time at 99.058407179974244531828075802892%



and you need to win 127 times in a row

99.058407179974244531828075802892%^127 = 30.074625857221451517169989060873%



ouch!

of course one might have some bankroll left to try and get back to 127 units but that only adds about 10% to the chance to double win.



Just betting all 127 units one time gives a 48.648648648648648648648648648649%

chance to double



maybe the Marty is more much more fun



here is some other Marty doubles I seen



Why the martingale is always counted till maximum table bet?



do have fun and hope I helped some with the math



Sally

gameterror
gameterror Nov 10, 2014

the probability of success ? i goes this depends on the number of trials you want to run. if the number of trials is infinite your probability to bust goes to 1.



you are looking for the probability of 12 reds in X spins....you need to define your X.



if you want to test your system you can go to https://primedice and set up automatic betting. you can get a new starting stack of 500 satoshi every 3 minutes (just press the B on the lower left and solve the captcha) so you can play around alot. I have witnessed 12 and 13 reds in a row a couple of times while on a marginale...and with primedice there is basically no betting limit but remember sooner or later there is always a sequence that'll wipe you out...