I heard about a new roulette system the other day that my friend had read about in a book, he’s always finding some new roulette strategy. It was yet another one of those, a guaranteed system and all you needed to do was to visit or logon to an online casino. Ah if only life was that simple, just grab a system from a web page and sign up at your local casino then just enjoy the cash. Obviously not quite how it works but the one thing I do say for many roulette systems is that they don’t actually worsen the odds you are playing against, they just don’t do anything. The roulette system in itself is not usually the problem, it’s more the expectation of the player thinking they can’t lose.

Ok so here’s the latest masterpiece to add too all the other systems that I’ve been though on this blog. It hasn’t got a name but how about the **James Bond Roulette system** although it’s probably not flamboyant enough for 007 tastes.

# The Latest of the Roulette Systems That Work (well perhaps not)

This one does sound quite good and just in case you don’t read until the end, then let me make clear it isn’t exactly a dangerous way to play roulette in fact it’s quite likely to make you last as long as any other system would. The system as much as it is, was developed in the USA and therefore uses an American Roulette Wheel as illustrated here –

Don’t worry about the layout, the major thing to note with the American Roulette wheel is that it has two zeros – ‘0’ and ’00’ which obviously changes the house edge.

So here it is, quite a simple system I think you’ll agree –

**You place $100 on an even money bet – let’s say red or black and stick to that bet each time.****You then put on your other bet – which is a $10 split bet on ‘0’ and ’00’ and that’s basically it.**

The idea behind this system is quite basic, the casino always wins because it has a slight advantage on all bets. That is represented by the zeros on the wheel – for if you bet red or black there’s always the chance it lands on one of the zeros and the house wins.

So we can see the plan here is to bet against that house advantage, and reclaim that edge by betting on the zeros as well. Sounds like a plan but does this roulette system actually work? Well here’s a summary of the maths behind it, presuming that you bet on black.

- If the ball lands on black (probability=18/38) then you win $90 ($100 win – $10 for the split bet)
- If the ball lands on one of the zeros (probability=2/38) then you win $70($170 win – $100 for the bet on black)
- If the ball lands on red (probability=18/38) you lose the full $110 on both bets.
**E = [18/38x(+90)]+[2/38x(+70)]+[18/38x(-110)]**

which works out as -220/38 or -5.7895**E per dollar bet =-5.7895/110=-00526**

**The final equation is as follows (where E=Expected Outcome)**

Which means the House Edge against this particular system is 5.26% which is actually exactly the same as the normal house edge of 5.26% that everyone faces without using this system. It’s not a bad way of gambling as long as you don’t presume it gives you some real edge.

It’s just losing in a slightly different way, but there is an important point to made here especially if you use any sort of system or technique for playing roulette. The odds in roulette are very reasonable compared to many gambling games and therefore you genuinely have a good chance of winning in any particular session with any sensible method of picking numbers. A roulette system cannot be proved to work in one particular session or even many of them – you have to keep playing it to prove it works. The unfortunate fact is that if you play any roulette over the long term you will definitely lose.

Thanks I enjoyed this, these roulette systems often look so convincing but it’s good to see one explained using maths I can understand.

Yep although the theory sounds promising, when you get to the maths there’s no advantage at all. Sometimes that doesn’t matter though – there are many roulette systems like this, the problem is that it alters expectations so people can end up losing more.

So you’re saying that this method is exactly the same as random chance? Is that right?

@arindam (sorry lost your post in the thousands of spam comments I get here!)

“This is same as what you have calculated. So, I don’t understood the necessity of putting 10 extra dollar for the split bet. Please let me know what I am missing here.”

You’re not missing anything, that’s exactly the point. The odds are identical in this system and covering the 0s with the split bet does nothing to your chances!

Ooops you’re right – Thanks Moshe!

Updated post.

Just a little remark: you meant to write RED instead of Black when calculating the lose of -110$ 🙂 and also the sign = should be replaced by + in the middle of the equation.

meanwhile im really enjoying reading your post!

Hmm perhaps I should check my maths – perhaps it does work !

Anyway glad you guys won 🙂

WORKS FOR ME, TOO…

Well I didn’t follow the math but actually I tried this last night and won. Thanks !

I don’t think I understood why I would put the $10 split bet on 0 and 00. Follwoing your logic, if I only put $100 even bet on black (say), the expected outcome is as follows.

E = 18/38 * (+100) + 20/38 * (-100)

= -200/38

= -5.26315

So, E per dollar is -5.26315/100 = -0.0526315 = -5.26%

This is same as what you have calculated. So, I don’t understood the necessity of putting 10 extra dollar for the split bet. Please let me know what I am missing here.

Thanks for your post though.