I'll just preface this with the fact I know nothing about other peoples roulette strategies and really nothing about gambling or probablities and I would imagine this isnt original but I have no idea.

As far as I can tell, this strategy has, by definition, positive ev assuming you have enough money to stomach the losses.

The basic idea is you start with a small black or red roulette bet. Lets say $10.
If you win you repeat this bet again indefinitely. If you lose you double the bet. Any time you win you go back to your original bet, if you lose you just keep doubling up until you win again.

As far as I can tell, it would be impossible to lose money with this system given enough funds to eat losses. The obvious downside is that you risk putting yourself in a situation where youd need to risk huge amounts of money to avoid losing small gains.


Any thoughts???

Martingale Strategy. Eventually, you'll have a run of luck so bad it wipes you out.

You're about 250 years late:

http://en.wikipedia.org/wiki/Martingale_(betting_system)

It can, and usually does produce small gains for a long time. When it fails, it fails spectacularly.

lol what about betting restrictions, every casino has some max amount that you can put let's on red, for example max bet on colour is 1000$, what to do when you need to put more then that

As long as every bet is -EV, every strategy is -EV since no matter what you do, you're just making a losing bet after another.
The trick is that the only bankroll size where it truly "would be impossible to lose money" is infinity, which is not a real number. Try any other number, be it millions or greater than the amount of atoms in the universe, there's a non-zero chance to go broke.

It's 100% chance to go broke with any size bankroll. Not "non-zero".

100% is non-zero, the probability to go broke in one iteration is closer to 0% than 100%, the probability to go broke with a finite bankroll if playing until either hero or casino is broke is not exactly 100% unless casino has an infinite bankroll (impossible in a real world setting).

With Martingale betting, the casino is always playing with your money so the size of their bankroll is irrelevant. Every time you double up, you are betting to get YOUR money back that they now hold and took from you already. So you are mistaken that the casino needs a larger bankroll than you initially. A Martingale strategy has a 100% probability of going broke within a finite time period.

I don't see what you mean with this, sorry.
I've never said anything like this. Roulette is obviously always -EV for the player with any bankrolls for both sides.
What? I'm pretty sure you're misthinking something.
The probability to bust the bank without ever losing a single bet is (18/37)^X where X is casino bankroll in relation to our first bet. Right?

That's correct. So then subtract that from 100% and see what you have left. Yes it's not exactly 100% if you insist. For all practical purposes in all real worlds, it is.

That's correct. So the subtract that from 100% and see what you have left. Yes it's not exactly 100% if you insist. For all practical purposes in all real worlds, it is. If we just use say, 100,000 bets as the exponent, the denominator wouldn't fit on this web page it would have so many zeroes in it. We're talking thousands of digits to the right of the decimal.

I'm pretty sure all he is saying is this:

p(you bust a casino using martingale) <<<<<<<< p(you bust yourself using martingale)

FWIW, in the first comment that you didn't understand, he is saying that each time you lose a bet and then double your bet, the casino is paying you back with your own money. At least most of what they pay you back is that. Let's say it is a $5 minimum table and you want to bet the minimum to reduce the chances of going broke while martingaling. So you bet $5 and lose. Then you bet $10 and lose. Then you bet $20 and lose. Now $40 and lose. Now you bet $80 and lose. $160 and lose. $320 and lose. $640 and lose. $1280 and lose. $2560 and you finally win. Is the casino upset to lose $2560?

No because they haven't lost that much. Add up your losses to this point and you will get $2555. The casino lost $5. That is all he is saying there.

I remember reading a book by John Scarne where he calculated that the house edge in roulette was the same no matter how the player placed his bets? Something like 5.75%?

5.26% on all bets for a 00 wheel, except for the silly 0-00-1-2-3 bet that has a house edge of 7.89%. All other bets would have no house edge if there were no 0 and 00, so the house edge is just the probability of 0 or 00 which is 2/38 = 5.26%.