Can you beat Roulette using maths?

[Mentallic]

Sorry scepticus, but you're way off base. The contents of this thread was mainly about discussing if the Roulette wheel has any uneven bias that would cause the randomness of a number to appear to not be completely random afterall - thus allowing someone to beat Roulette.

I'm feeling optimistic that the users on this Math forum can understand the definition randomness and its relationship to gambling.

 

[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 betting is Bet Selection.

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]

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.

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]

In this example you will lose $64 every time you lose money.

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

 

[Hurkyl]

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.

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]

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

"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]

Martingal can beat the wheel in theory

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

 

[Hurkyl]

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

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

 

[scepticus]

Hi Skeptic 2
I do not use this method I use another but using maths.
All I am doing here is trying to prove that ,yes,we CAN beat roulette by using maths
but can does not mean certainty.

 

[Hurkyl]

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.

They do depend on each other, because they are both interdependent upon what numbers are available and how they're laid out and what colors are assigned to them.

so if we ASSUME they ARE independent does my maths stack up ?

If they were independent, then they would intersect in about 10.8 numbers. Obviously in the real world you can't have a fractional number of numbers, but it's no problem to consider it in the hypothetical. So, each bet would cost about 10.8 chips. Now, repeat your calculation on how much you spend and receive.

 

[scepticus]

---They do depend on each other, because they are both interdependent upon what numbers are available and how they're laid out and what colors are assigned to them.----

But this is inevitable as there are only 37 numbers available

Should we choose the 3rd dozen and 3rd Column there are 9 numbers they have in common so, according to your own observations this is clearly a viable method.

Thanks Hurkyl

 

[scepticus]

Keyed into a closed thread here in which Sylas claimed that the Caro system bucked the odds

"Finally, you need to be very disciplined in excluding the number 30 and the group of consecutive numbers that begins with 11 and continues clockwise through and including 14."
This fails to take into account ALL the Probabilities available.And , if we choose only Red/ Even and Black / Odd numbers why exclude 30, 11 etc.? No explanation. I would guess that over his trials he has found these numbers interfere with his theory and has abandoned them.
Gamblers should get the message. Because the wheel gives a Random result no one can predict with certainty the outcome .But for the same reason, no one can say with certainty that we must lose. All we can do is guess and an educated guess is better than a random guess.
Gambling is gambling is gambling. As the financial whizkids who gave us the banking crisis did not understand !

 

[Hurkyl]

Keyed into a closed thread here in which Sylas claimed that the Caro system bucked the odds ... No explanation.

If you try to use the strategy, the reason becomes quite clear: the Caro system is just a long winded way to say "don't place a bet".

But for the same reason, no one can say with certainty that we must lose.

But we can say with certainty that, if the wheel is fair, then your statistically expected result is to lose 2 chips out of every 38 you bet no matter what strategy you try -- unless you make a 5-number bet, in which case your expected losses will be even greater. Actually, that number is for an American roulette wheel, with 38 numbers. The 37-number wheel has different payoffs, and I never learned the figures for it.

And we can be nearly certain that if you play long enough, your actual losses will well approximate the expected value. (I'm probably assuming a cap on bet sizes in that statement)

 

[scepticus]

I am glad to see, Hurkyl, that you have abandoned claims of certainty and opted for
" nearly certain ".
Probability Theory is about uncertainty not certainty from which we can infer EXPECTED results but not certainty.Anyway, unless we will live till Infinity THE LONG RUN is irrelevant
As is the claim that the wheel has "no memory" for we can indeed use previous winning numbers as a guide.They are statistics and stats are used in other forms of gambling.
I think that bettors would be much better off using maths in their betting but we are all entitled to our opinion.There would be no gambling without these different opinions.

 

[Hurkyl]

I am glad to see, Hurkyl, that you have abandoned claims of certainty and opted for
" nearly certain ".

Eh? I'm pretty sure I've changed nothing about my claims. I think, maybe, you are vastly overestimating the difference between "nearly certain" and "certain".

e.g. I think your odds of winning the grand prize in a (6 from 49) lottery in one try are slightly better than coming out ahead after placing 10,000 18-number bets on a (fair) 38-number roulette wheel.

Note the lottery also requires a much smaller bet and gives a much greater payoff.

Probability Theory is about uncertainty not certainty from which we can infer EXPECTED results but not certainty.

We can infer lots of things, not just average outcomes.

Anyway, unless we will live till Infinity THE LONG RUN is irrelevant

The long run comes sooner than you think. A lot sooner than you think, apparently.

As is the claim that the wheel has "no memory" for we can indeed use previous winning numbers as a guide.

Not if the wheel is fair.

If a wheel is unfair, discovering that fact and making use of it is going to require mathematics, not mysticism, especially if you're going to find a useful bias before the casino does.

I think that bettors would be much better off using maths in their betting

Then stop trying to go against the math.

 

[scepticus]

"Then stop trying to go against the math. "
i have already given a method using math.so don't tell me i am wrong, tell me WHERE it is wrong.You haven't so far which is surprising.Note that .I did not. and do not claim that it would be profitable only that it gives the bettor an "Edge " And if the casinos' Edge of 2.6% means that they will win in the long run then so must a bettor's Edge.

" the long run comes sooner than you think "

If that's logic then I'm a banana and I ain't a banana

Your answers reveal desperation to justify your fundamentalism.

 

[Diffy]

Your answers reveal desperation to justify your fundamentalism.

No, his answers are factual and based on actual math and research.

Your questions are horribly written, and you demonstrate no mathematical understanding.

You want to know WHERE your method is wrong. How about everywhere. You use phrases like "unknowable until" .

Furtheremore you CLAIM things like:
"If we choose to bet only RED/ODD numbers plus BLACK/EVEN numbers we bet 20 numbers which means that we should win 20 times in every 37 spins.

You do not justify this claim. There is no math to back it up. If you want to know where your math is wrong, then SHOW US SOME DAMN MATH FIRST!

Great you bet 20 numbers out of a total 37 numbers.

That means your chances of winning one spin are 20/37 or ~54%.

So say you bet 1 dollar on each number each spin. That means you bet 20 bucks each time with a 54% chance of winning 37 dollars. 20 dollars each time for 37s times mean you will bet $740

I ran a simulation of doing this 100,000 times That is betting 20 dollars, having a 54% chance of winning 37 dollars for 37 spins. The average amount won in 37 spins came to be ~$618

That means on average you lose 122.

I fail to see how that is winning.

You are wrong and your math wrong because you haven't done any.

Edit: I made one mistake in my simulation the average should be ~$629.39 Still the point stands you lose $110.61

 

[pwsnafu]

"Then stop trying to go against the math. "
i have already given a method using math.so don't tell me i am wrong, tell me WHERE it is wrong.You haven't so far which is surprising.

That is not how we do things here. Hurkly has already did so in post #65. You then change your stance rather than addressing the criticism. Hurkyl then asks you to do redo the calculation post #70. You have not done so. You need to do the math.

Note that .I did not. and do not claim that it would be profitable only that it gives the bettor an "Edge " And if the casinos' Edge of 2.6% means that they will win in the long run then so must a bettor's Edge.

Goal post shifting. To "beat" roulette you must create a strategy and prove said strategy gives a positive expectation. That is what the phrase means mathematically. You wrote in post #50

The answer to the question as put is YES.

You seem to have misunderstood the question. So be it.

" the long run comes sooner than you think "

If that's logic then I'm a banana and I ain't a banana

Your answers reveal desperation to justify your fundamentalism.

Statements like this do not help.

 

[micromass]

This thread has gone on for long enough.