Is this possible? ???
Drag - " However, in roulette you only have a 94.7% chance on winning every spin because of the 0 and 00."
Are you really that inept?
I didn't read le thing but its only true if you cover 36 numbers out of 38. It gives you 94.7% chances of winning.
That doesnt sound very new or practical to me.
Jismtron- Yes. You have me pegged. I am inept. As a matter of fact, I am so inept that everytime I walk up to a Roulette table I walk away a winner.
Now....would anyone else like to make a rude comment, or do you think we can actually have some intelligent conversation about the topic?
No. This system is flawed. You need simply to look at the initial rule of the method. Each player bets one unit on pass and dont pass respectively.
The author states this bet will break even. It will not break even. As a matter of fact this 'combination' bet can never net a profit but can and will net a loss. The 12 is your enemy here.
The same holds true for every other combination bet. They are all net losers.
If you want to play this type of method ... which is fine ... but dont expect to win long term ... You need to understand the idea of differential betting. You do not need a partner and truth be told your results will be better if you never make any type of 'combination' bet as described in nthis method.
Good luck.
Hi Drogan,
Why ask us if you can test? Or, think about it for a minute and than do not ask. Of course this is a looser, and a fast one at that. Why do you think that it could be a winner at all?
Do not spend a second on "systems" or strategies that:
1. require you to wait ("wait for 20 spins and write blah blah")
2. use progressions: a lousy system will not become a winner because of a progression: on the contrary you will loose more
3. uses logic like: "this is so unlikely that overall you will win"
4. use some magic or "secret loophole" in either maths or statistics or programming or wheel design or table layout
reddwarf
Gads, you don't need the second player for one thing:
1, 2, 3, 5, 8,13 (yours)
1, 2, 4, 7, 12 (mine)
All you need is to find the first loser and then bet the progression against it. On each win you have the loser to restart on for the next bet. The second player's results always reduce the progression steps by one unit as a net result. So why not leave the partner to find something better to do?
Quote from: Gizmotron on Feb 09, 01:36 AM 2012
Drag - " However, in roulette you only have a 94.7% chance on winning every spin because of the 0 and 00."
Are you really that inept?
--Expect that some of us are playing enprison rule only...and that makes difference.
Reply: Reddwarf:
"4. use some magic or "secret loophole" in either maths or statistics or programming or wheel design or table layout"
Like that last one Reddwarf :)
Wow...I am not exactly sure what I have done to invoke all of the hostility. I certainly didn't start this thread in order to insult anyone.
Perhaps I wasn't clear on what I was asking.
The reason I named the thread Calling All Math Guys is because I was looking for some sort of mathematical explanation as to why this bet would or would not work.
I have no interest in testing this system. I am not looking for opinions from people who say, "don't forget about the zero", "don't play systems that require progressions", "don't trust systems that require pen and paper", "no ec bets work", "don't forget that each spin in an independent event of......" blah, blah, blah, f@ing blah....
I know all that. I too am of the opinion that it won't work. But what I am looking for is mathematical evidence as to WHY it won't work. That is why I am asking for input from the math guys.
GLC, Bayes, Hermes, AMK, Mr. J, and anyone else who wants to weigh in with any sort of mathematical explanation. And please remember that the book talk about Craps and Baccarat as well. No zero to consider.
I am not interested in any opinions. I don't care about your experience or how many years you have been playing. I don't want to hear stories about whosiewhatsits in Nevada in 1964 and he tried this and yada yada yada.......none of that is helpful.
Just the math.
Shall we begin again?
I didn't read everything but I see similarities with diferential betting, something that is being discussed in another topic in this forum.
Drogan,
How did you find this method? Would be an interesting story............
Part of getting to the bottom of this mathematically is in identifying the real bet's true value for each step of the progression and how the progression is triggered.
1) The progression is triggered by any loss.
2) The true values of the progression are: 1, 2, 4, 7, and 12.
If you can see it. The second player cancels 1 unit from each bet of the progression.
So all you have is a weakened Marti.
The second player is an unnecessary redundancy.
@Giz
Thank you. Now we are getting somewhere.
@AMK
A simple pdf search of google. i.e. roulette filetype:pdf
@GameBreaker
Your post did not show on my computer until just now. Weird.
Thank you for taking the topic seriously. Your post was very helpful.
@vladir
Thank you for bringing that to my attention. I will make a search to find the topic you are referring to.
Cover the 0/00, set up a stop win and pray for good luck
vundarosa
Hi Drogan,
I haven't read the whole pdf, but I see the method is based on differential betting. There is some advantage to this, as demonstrated on Mr Oops' site here (link:://xerxx.se/oops/main/reading/rr000011.html).
Hi Drogan,
I actually didn't mean to be nasty; If you want a mathematical proof: here it goes:
Preliminairy:
a. probability of win= outcomes that make you win/ all possible outcomes (of course the same holds for probability of loss)
b. a static bet: both betting amount and betted number do NOT change (so actually it is a flatbet)
c. a numberset: any combination of numbers that I can bet on on the roulette table
d. expectation value: win probability X return - lose probability X loss (so this is what I "expect" in the long term)
e. we assume that the spins are independant, unpredicatable and unbiased (and yes, if I have all starting condition so fthe ball at the wheel, I have a slight edge over the wheel, but this has nothing to do with the maths of "guessing game" most people play, it is a different game all together!!)
The proof consists of a few steps"
1. step1: we proof that any static bet is a long term loser
2. step2: we proof that a progression is actually a combination of static bets
3. step3: draw your own conclusions
Step1: here we "proof" that any static bet is a long term loser
Let F be any static bet. We bet the amount B on N numbers. Please note: if I put 1 unit on a line, this can also be seen as putting 1/6 unit on 6 numbers!.
EV=(N/37)*(36-N)*B - (1-N/37)*N*B
Do the math yourself, the result is:
EV=-bN/37
This means that on the long run a static bet will lose you money. how much? Your action/37 (Action total amount bet)
Step1: ANY static bet combination
Now we are ready to proof the next thing: ANY combination of static bets is a losing bet.
We only have to consider the following: lets assume that I combine 2 static bets, say, I bet on a line AND a dozen. I can now find 2 static bets that have the SAME amount of money bet on the number: there are 4 numbers with 2/12 units on it, and 16 numbers with 1/12 units on it.
In other words: every combination of bets will lead to a set of static bets where each static bet consits of numbers with the same betting amount. lets assume that we have m of such sets.
Now we calculate the EV again:
EV= EV1 + EV2 + ... + EVm.
because we showed that the EV of 1 set is negative, the sum of all negative bets is also negative!
This means a lot: imbalances on the table are NOT exploitable using flatbets (at least, we will show however that prograssions are also flatbets of some sort).
This by the way also means that "waiting on certain patterns" will not work either: I'm not going to proof this, but it is ralated the the above
step2: we proof that a progression is actually a combination of static bets
OK Drogan, your system is using a progression. Actually you make it really easy for me, because it is a limited progression:
A limited progression is a progression where the static bets are limited. Do the counting yourself: it is limitited. Actually: the tabel limit ensures that all progressions are limited!!
Anyway: so we have a limited progression. Now we do the following, lets assume that there are P static bets in the progression: for example
static bet1: 1 unit on red
static bet2: 2 units on red etc
static betP:...
So lets now calculate the EV:EV=EV1+EV2+EV3+...+EVp
per static bet we have shown that the expectation value is negative. This means that the expectation value of the progression is negative.
step3: draw your own conclusions
So Drogan, draw your own conclusions! Did I proofr that roulette is unbeatable? No, not at all, I only showed that the "guessing game" is a losing proposition. Are ther other games within roulette? Sure the are: the afore mentioned calculation game, the "look for a biased wheel game", and just maybe there are more...
reddwarf
Sorry, I can not undo the "bold"
reddwarf
So Bayes' guy Mr. Oops says it WILL work, as long as you bet different than last.
And then we have RedDwarf's formula showing that it WON'T work.
Interesting......
Hi Drogan,
read the mr.Oops page again: nowhere he claims that it is a winning method. he only states that it reduces the amount of money betted...