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Calling All Math Guys!!!!

Started by Drogan, Feb 09, 01:00 AM 2012

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0 Members and 1 Guest are viewing this topic.

vundarosa

Cover the 0/00, set up a stop win and pray for good luck
vundarosa

Bayes

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.
"The trouble isn't what we don't know, it's what we think we know that just ain't so!" - Mark Twain

reddwarf

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

reddwarf

Sorry, I can not undo the "bold"

reddwarf

Drogan

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......

reddwarf

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...

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