Can you beat Roulette using maths?

Hurkyl

There's an easier way to win most hands: simply place a single-number bet on every number from 1 through 34. 34 out of every 37 spins you come out ahead.

Exercise: explain why the house still has the edge when you adopt this strategy.

(P.S. in the U.S., roulette wheels have 38 numbers)

AlephZero

The key to winning at roulette is to own the casino. When was the last time you saw a casino that had to close down because it ran out of money?

scepticus

Thanks Guys
But the Heading was" can you beat roulette with maths" and I took it at face value.
I was hoping someone would show me where my maths was wrong and not just post soundbites.

Hurkyl

Try my exercise. Really, the best way to understand is to do the computation for yourself, and compute your expected winnings. Do you know how to do that?

Fortunately, the computation isn't that hard.



It's important to actually consider all of the cases. One of the classic traps is the following strategy:

• Bet 1 dollar.
• If you win, stop. Otherwise bet 2 dollars.
• If you win, stop. Otherwise bet 4 dollars.
• If you win, stop. Otherwise bet 8 dollars.
• Et cetera

At 1:1 payoff it looks like this strategy guarantees that you'll come out a dollar ahead, right? The problem is that there is a tiny problems

There's an upper limit on how much money you have available to spend gambling

Sometimes, there's also a limit on how much the casino will allow you to bet.

Your odds of coming out ahead are around 85724 in 85725. That's very likely, right? Surely this is a good strategy to beat the house!

It's easy to lull yourself into that false sense of security -- the problem is that one time in 85725, you lose all of your money. You are risking your entire million dollars for a chance at coming out a single dollar ahead. The house expects to come out over 11 dollars ahead each time a millionaire tries this strategy.

scepticus

Thanks Hurkyl
What you describe is known as the Martingale system
I am talking about level stakes.
You don't really answer my question - tell me where my maths is wrong.

PlayingMonk

I came up with a system that I believe allows one to win at roulette in the long run but I'm not 100% sure that it's valid (I haven't tested it in practice) and I want to share the basic idea here and see if any of you guys can tell me why it is wrong or if there is anything to it.

To increase the odds we'll assume we're using a European roulette wheel. This is basically a variation on the famous martingale system. The algorithm I would use would be something like this:

Step 1 - Bet $1 on black
Step 2 - If you win go back to step 1, if you lose bet $2 on black
Step 3 - If you win go back to step 1, if you lose bet $4 on black
Step 4 - If you win go back to step 1, if you lose bet $8 on black
Step 5 - If you win go back to step 1, if you lose bet $16 on black
Step 6 - If you win go back to step 1, if you lose bet $32 on black
Step 7 - If you win go back to step 1, if you lose bet $64 on black
Step 8 - Go back to step 1

So, the main difference between this and the martingale system is that here you set a cap for yourself that you will not double and will just accept the loss at the point instead of continuing on until you go broke. In this example you will lose $64 every time you lose money.

Now, to do the math. There are 37 spaces on a European roulette wheel and the chances of getting black and winning on any of these spins is 18/37. The chances of losing on a given spin are 19/37. The chances of losing seven times in a row are (19/37)^7 which is roughly .0094159282. The odds of going through the algorithm and not losing seven times in a row should then be .9905840718. Therefore, on any given run through of the algorithm there is about a 99.06% chance of winning $1 and about a .94% chance of losing $64.

So, (1)(.9905840718) + (-64)(.0094159282) = .9905840718 - .6026194048 = .387964667

Because we get a positive number we should win money in the long run.

(I did the math for an american wheel and got a value of about .27, so it should still work on an american table but not quite as much as quickly).

So, would this work or am I messing something up?

Hurkyl

You lose all 7 bets, not just the last one. That totals to $127.

Hurkyl

I can't tell you where your math is wrong unless you actually do the math and show your work.

scepticus

Hi Hurkyl
Possible scenario
We choose the 3rd Dozen and the 1st Column
So, to begin with ,we choose the numbers
1,,4,7,10,
13,16,19,22
25,26 27 28, 29 30 31 32,33,34,35,36
a total of 20 numbers an advantage of 20/37
We also consider all the Red / Odd and Black / Even numbers
These are ;
1,2,3,4,5,6,7,8,9,10
19,20,21,22,23,24,25,26,27,28 - again 20 numbers an advantage
of 20 / 37
The chances of both winning on the same spin are 20 / 37 multiplied by 20 / 37 which is 400 / 1369.
Choosing only those numbers which they have in common we now have 1,4,7,10,19,22,25,26,27,28 a total of 10 numbers
Each bet costs 10 units and for each win we receive 36 units
So over an average 1369 spins we spend 13690 units and receive 400 times 36 which is 14400 a surplus of 710 units
So we can use maths to win at roulette unless my maths is wrong which I am asking people to show.
BUT this can only be theoretical as each spin of the wheel gives a random number so no one can claim, with certainty ,that we MUST lose or win at roulette, Where uncertainty exists certainty can be claimed by clairvoyants but not by mathematicians.

turbo

Unless the wheel is rigged and you find out why and how it is rigged, you will always lose at roulette. There is no mathematical program that will beat that wheel. You will lose.

Millennial

Martingal can beat the wheel in theory, but not really in practice (they have rules to prevent it, and it consumes your money fast.)

turbo

"Betting strategies" cannot overcome the house advantage in roulette. It is not possible, if the wheel is honest. The house wins. Why is this thread still alive?

Number Nine

No, not with a finite amount of money. "In theory", with a finite amount of money, you will lose.

Hurkyl

Why would that be the chances? Presumably, you are applying the theorem that, for independent events A and B (where A and B are each of your 20/37 events):

P(A and B) = P(A) P(B)

but why would you think A and B are independent? They are clearly related in a non-trivial fashion depending on the layout of the roulette wheel, so it should be at least somewhat surprising if they turned out to be independent.

But we don't need to speculate: we can directly compute the odds of one of those 10 out of 37 numbers coming up in a spin of the wheel and see that the odds aren't 400/1369.

daveb

Assuming the wheel is fair, you can minimize your chance of losing by starting a bet at 1 dollar on some number. If you lose, increase the bet by 1 dollar but bet the same number. Up to the 70th bet, assuming you win, you come out ahead. The 71st bet means you come out even (at 2556 dollars), and beyond that you are behind. Once you win, go back to 1 dollar with a different number (if you can remember all numbers rolled up to that point your loss is minimized further by picking a number that still hasn't been rolled).

Now, the chance of a specific number not coming up in 71 consecutive tries is (37/38)⁷¹, or about 15%, so 15% of the time, this strategy doesn't work

skeptic2

How much money have you won with this strategy?

scepticus

Thanks hurkyl for taking the trouble to reply
I am not a mathematician but I understood them to be independent since they do not depend on each other.
so if we ASSUME they ARE independent does my maths stack up ?
There is no real need for a real roulette wheel as we could use a RNG
Thanks