Optimum Stopping Theory

[CarpeDiem]

Euler's number

e=2.71828
Taking 37 spins
37/e= 13.6115484791
Int (37/e)= 14 (roundabout)

Wait for 14 spins
Keep in your mind the highest number from the initial 14 spins, lets say name that number A
Keep rolling the spins until you get a number higher than A, say B comes after 5 spins where B>A

You will bet all numbers from 0 to B ( including B) for the reminder of the spins until spin 37

Now change 37 with any number of spins, use the same algorithm as above and any partition you want.

Mathematically, you will get a 33% hit rate.


Enjoy!

[precogmiles]

I think you are on to a winner.
Just mix a little bit of repeaters and a good progression like a martingale and it has the potential to win at least a million spins I think.

Maybe a good hit and run method or maybe the holy grail.

[Taotie]

I think you are on to a winner.
Just mix a little bit of repeaters and a good progression like a martingale and it has the potential to win at least a million spins I think.

Maybe a good hit and run method or maybe the holy grail.

Troll.

....and not a very good one.

[precogmiles]

Troll.

....and not a very good one.

Taotie calling someone else a troll 

[CarpeDiem]

Reference to partitions:
Ignoring 0, we have:
36 1 number partitions
18 2 number partitions
9 4 number partitions
12 3 number partitions
6 6 number partitions
4  9 number partitions
3 12 number partitions
2 18 number partitions

As the strategy involves e, a number > 2, this strategy will work on anything that involves
Int(1/e)>=1, hence there is no EC application to the game of roulette.
The 12 number partitions also do not interest us, as the chance is already 33%

So the only partitions we are interested in are the 9+ numbers(4 partitions of 9 numbers and onward)

There are lots of applications to this mathematical premise in the game of roulette.

That being said, although EC are not per se included, we can play an 1-2-3-4 interpolation of it, by attributing a pair of EC a value

RR value 1
BB value 2
RB value 3
BR value 4

hence making a game within a game
The possibilities are endless.

Ps: i do not care about progressions, repeaters of any kind, if a system works it has to win flat on an infinite permutation basis

[Taotie]

Taotie calling someone else a troll 

deja vu

[Steve]

He's not a troll, just an occasional c*nt.
It's encouraged, as long as you're honest.
Sometimes political correctness gets in the way.

[Taotie]

He's not a troll, just an occasional c*nt.
It's encouraged, as long as you're honest.
Sometimes political correctness gets in the way.

Of course, you are right Steve.

Sorry precogmiles you're not a troll. As Steve said, you're just a bit of a c*nt sometimes.

[cht]

Nobody cares if he's a c*nt or a troll.

He's a sly scammer in disguise, "invest in ME!"

[Drazen]

Mathematically, you will get a 33% hit rate

Hi Carpediem

Interesting theory indeed!

May I ask did you found a way around repeating distribution of roulette numbers which is slightly different than the one presented in the video, where we have only uniques?

If we could come up with a way to have only uniques all the time I am sure we could find a way to beat roulette 

Best

Drazen

[daveylibra]

Hi CarpeDiem

Not sure about the maths here.
Suppose the number B turns out to be 35.
We bet all numbers from 0 to 35 with only a 33% hit rate? 36 would have to come up a lot!!

Maybe the the maths only works for very large numbers, eg the googol in the video.

[Herby]

Int(1/e)>=1

Hi Carpediem,
1/e is the limit of the "law of the turd" as "Firefox" called it

(Rumours say Firefox had to many contact with precox man, so he was ashamed and disappeared)

If e = Eulers number how comes "Int(1/e)>=1" ??

[CarpeDiem]

If e = Eulers number how comes "Int(1/e)>=1" ??

Hi

I made an error in that equation, what i meant was int(k/e)>1, where k equals the max partition space.
I personally do not find any application to EC or dozens in the game of roulette applying this premise, unless we group the partitions in a way that assures a min space of 4 (25% chance).

Hi CarpeDiem

Not sure about the maths here.
Suppose the number B turns out to be 35.
We bet all numbers from 0 to 35 with only a 33% hit rate? 36 would have to come up a lot!!

Maybe the the maths only works for very large numbers, eg the googol in the video.

If in the last 14 number 36 shows up, we are facing a no bet, as we found our MAX value (the Googol in our roulette environment=36) so that game is complete. You will continue the same modality with the first spin after 36-next 14 spins.

The math expectation remains the same,33 % no matter if we choose k=37 , 166 or 1 billion spins. The only constant that changes is the sample size=k/e. In roulette, choose a k 4<k<=37(european wheel) or 4<k<=38 for the American version.

The betting interval will be from int(k/e)+1 until k. Due to the numbers discrete ordinal relation, the 33% math certainty can only be applied to the whole partition (0,B)

You will lose 2/3 of the attempts, and win 1/3.  Every single time.

[daveylibra]

Hi there

I understand what you are saying, but suppose after 14 spins the highest number is, say, 30.
A couple of spins later, 35 comes up.
We then bet 0,1,2,3... to 35 for the next 21 spins.  (37-16=21)
Am I right so far?
And, we will only win 33% of bets? Surely more?

[CarpeDiem]

Hi there

I understand what you are saying, but suppose after 14 spins the highest number is, say, 30.
A couple of spins later, 35 comes up.
We then bet 0,1,2,3... to 35 for the next 21 spins.  (37-16=21)
Am I right so far?
And, we will only win 33% of bets? Surely more?

Hi

Given the same conditions apply in 3 instances, in at least 1 out of those 3 scenarios you will not meet #36 until the end of the k=37 spins. 35 will be the highest number, and not 36.

The 1/3 applies until the k-th spin appeared for the (0,35)  bet in your case.

This deviation you have presented will most likely appear given enough permutations, but once number 36 appears in your example (or any number >B), that session ends as a loss. You only continue betting until spin k if all other numbers are <B.