Can you beat Roulette using maths?

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Roulette is fundamentally a game where the house maintains an edge, making it impossible to beat in the long run. Progression betting systems merely shuffle risk without providing a true advantage, often leading to larger losses. Players may believe they can predict outcomes based on dealer behavior, but the randomness of the game undermines such strategies. Historical attempts to use mathematical models or devices to gain an edge have met with limited success and significant challenges. Ultimately, while some may experience short-term wins, the statistical reality ensures that consistent profit is unattainable in roulette.
  • #61
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.
 
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  • #62
Martingal can beat the wheel in theory, but not really in practice (they have rules to prevent it, and it consumes your money fast.)
 
  • #63
Millennial said:
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?
 
  • #64
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.
 
  • #65
scepticus said:
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? :confused: 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.
 
  • #66
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
 
  • #67
How much money have you won with this strategy?
 
  • #68
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
 
  • #69
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.
 
  • #70
scepticus said:
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.
 
  • #71
---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
 
  • #72
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 !
 
  • #73
scepticus said:
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)
 
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  • #74
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.
 
  • #75
scepticus said:
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. :wink:
 
  • #76
"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.
 
  • #77
scepticus said:
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
 
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  • #78
scepticus said:
"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.
 
  • #79
This thread has gone on for long enough.
 

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