# Casino (Roulette) system.

by kyleball
Tags: casino, roulette
 P: 6 Excuse my appalling maths; and also my audacity to even propose a Roulette 'system'. I understand the concept that each roulette spin (if we disregard 0, and bet on a 2 to 1 option) is 50/50. It has no memory of previous spins. However, can probability not be determined by the amount of spins? Since it usually even out (or somewhat) to a 50/50 ratio. With that in mind, and adapting the Marlingale system, I wanted to see what the thoughts of this system is: Online betting (where there is no minimum), and document the cases of Red/Black, Evens/Odds. Bet 10p on both Red AND Black; this allows you a move to 'forecast'. If you lose, it is only 20p (by getting 0). Then, once you have 4 reds come up, for example, then you could bet black and then utilise the Marlingale system of doubling up. In my mind; probably due to my limited maths, this works. I tried this with £50, and within 2 hours I had £2000. I know that some will not believe that, and I only use it as an example of the system in use, not to brag (since I do not know anyone here). I admit it could be luck; but probability wise, its not likely to get 5 reds in a row rather than a black? Thanks for any input here; and please note that I set myself a £50 limit purposely to avoid life altering losses or the such.
 P: 772 No. Flatly, no. Absolutely not. No. Each spin is independent; the probability of landing red (or black) does not in any way, shape, or form depend on the outcome of previous spins. This is the definition if independence. Example: What is the probability of obtaining heads on a coin toss? The answer is 0.5 (50%) What is the probability of obtaining heads given that the last 100 coin tosses have been tails? The answer is 0.5 (50%). Each toss is statistically independent; the probability of the outcome is not influenced by past outcomes. Probability is not a force that influences events. It is not going to force the ball to land on black just to "even the odds". Don't even get me started on the martingale strategy, which is provably flawed and guaranteed to result in bankruptcy.
P: 4,573
 Quote by Number Nine Probability is not a force that influences events. It is not going to force the ball to land on black just to "even the odds". Don't even get me started on the martingale strategy, which is provably flawed and guaranteed to result in bankruptcy.
The scary thing about the martingale strategy is that when you have access to unlimited funds then it's not seen as a crazy thing to do.

But yeah I agree with you in the way that it is not only completely irresponsible, but also completely insane.

P: 800
Casino (Roulette) system.

 Quote by Number Nine No. Flatly, no. Absolutely not. No. Each spin is independent; the probability of landing red (or black) does not in any way, shape, or form depend on the outcome of previous spins. This is the definition if independence.
Interestingly, when we turn our attention to actual roulette wheels rather than theoretical ones, the opposite is true. Since a roulette wheel is a physical system, it exhibits subtle biases due to mechanical imperfections. These biases can be observed and quantified; and accurate predictions of future wheel behavior can be made.

There have been practical, successful roulette computers built using these principles.

http://en.wikipedia.org/wiki/Eudaemons

kyleball, roulette can be beaten. But not by gambling schemes; rather, by analyzing the physics of the wheel and ball.
P: 6
 Quote by Number Nine No. Flatly, no. Absolutely not. No. Each spin is independent; the probability of landing red (or black) does not in any way, shape, or form depend on the outcome of previous spins. This is the definition if independence. Example: What is the probability of obtaining heads on a coin toss? The answer is 0.5 (50%) What is the probability of obtaining heads given that the last 100 coin tosses have been tails? The answer is 0.5 (50%). Each toss is statistically independent; the probability of the outcome is not influenced by past outcomes. Probability is not a force that influences events. It is not going to force the ball to land on black just to "even the odds". Don't even get me started on the martingale strategy, which is provably flawed and guaranteed to result in bankruptcy.
So do they not have any kind of relationship?

Can you not say, since I start betting after 4 in a row have shown up, that we're looking at the relationship of 5 spins?

Is there no math term for this? Or is it simply just 'luck' that I was able to make £2000 from £50?

I've noted down my spins each time, the maximum I got of one kind in a row was 9 (that was from around 900-1000 spins). I researched, the maximum (documented) is 27. I admit - at that point I'd lose my money (but I start with such small amounts its inconsequential).

Can you not use that information to say; If I do small bets (£0.1 on both red and black) and then wait until four appear to bet £1, or £5, etc. It's likely (although of course no where near certain), that there will not be more than 8 in a row of the same kind.

I apologise if my questioning is tiring; I am just interested in whether this is credible or just luck.

Thank you :)
 P: 497 Making £2,000 was (probably) not luck, if your initial bet is £4 that is what an unrestricted Martingale 'strategy' makes most of the time from 1,000 turns. You were reasonably lucky to only get a losing run of 9 though, there is a roughly 60% chance of a longer losing run in 1,000 turns. And there is a 1 in 10 chance of a losing run of 13 or more, by which time you would be £16,000 in the hole and need to place a bet of that much again to keep up with the strategy. If you have recorded the results, why don't you run through them again, using the opposite stratgy that a run of 4 reds likely to be continued with another red 'because the wheel is in a groove'? I bet your winnings will be more or less the same.
P: 772
 It's likely (although of course no where near certain), that there will not be more than 8 in a row of the same kind.
The probability is low, but you don't seem to fully understand what that means.
If I flip a coin 10 times, the probability of landing heads ten times in 0.510, which is very low. If, however, I have flipped a coin nine times and received heads nine times, the probability of my tenth throw also being heads is 0.5, which is exactly the same as the probability of the coin flip coming up tails. There is no mystical force that compels the coin to come up tails just to "even the odds".

The probabilities of any specific sequences of outcomes are exactly the same: Again, the probability of receiving heads ten times in a row is 0.510 (0.5 for the first flip times 0.5 for the second flip...). The probability of receiving nine heads and then tails is [i]still 0.510 (0.5 for the first flip, times 0.5 for the second flip...). Google the Gambler's fallacy if you want to know why your system doesn't work.
 P: 6 Can you not talk about the 'unlikelihood' of getting 6 in a row, or 7 in a row, or 8 in a row, and bet on the basis of that? Rather than betting on the individual spin; you're betting on the unlikelihood (that's evident) that it would be rare (albeit not impossible) to get 6, 7, 8, 9 blacks in a row, etc. There must be some kind of logic to this 'system' and I am sure it is undeniably better than just going into a casino and placing £50 on red. It would be better, I would think, to see a string of 7 blacks, and bet on red; which is betting on the unlikelihood of 8 blacks coming up (because you know that is rare). There must be some way to explain it; rather than only focussing on the independence of the spins; causing the gamblers fallacy - since you know that long strings is rare - that is what you can focus your bet on.
P: 6
 Quote by MrAnchovy Making £2,000 was (probably) not luck, if your initial bet is £4 that is what an unrestricted Martingale 'strategy' makes most of the time from 1,000 turns. You were reasonably lucky to only get a losing run of 9 though, there is a roughly 60% chance of a longer losing run in 1,000 turns. And there is a 1 in 10 chance of a losing run of 13 or more, by which time you would be £16,000 in the hole and need to place a bet of that much again to keep up with the strategy. If you have recorded the results, why don't you run through them again, using the opposite stratgy that a run of 4 reds likely to be continued with another red 'because the wheel is in a groove'? I bet your winnings will be more or less the same.

The way that this system works is that you would stop after three failed bets, and 'absorb' the loss into your profits.

After analysing 2000 spins - this has been shown to make a profit.

So it would work like this:

You place 10p on both red and black (since losing 20p on 0 is meaningless), and do this numerous times until you get 4 of the same in a row. Then bet your set figure, e.g. £4, and if you lose double this up to a maximum of 3 spins that you're properly bet on.

If you lose a third time; absorb that loss and then either start from £4 again on the same one you've been betting or absorb that £28 loss. The £28 loss will be made back in 7 wins. From analysing 2000 spins - this achieves quite a substantial profit. The most important thing is starting after identifying 4 in a row and stopping after 3 bets.

Of course - you could get poor 'luck' (for lack of a better word) and get 20 losses in a row. The likelihood, however, from what I've analysed, shows that you will turn a profit.
 P: 199 You should heed the advice of Number Nine. You are committing a common mistake with probability. The probability of any sequence of blacks and reds is equally likely. There is nothing special about 7 black in a row. The probability of brbrbrb is equal to the probability of bbbbbbb or rbrbrbr or rrrrrrr in a finite number of seven spins and tells you nothing about the next spin assuming the wheel is fair, i.e. black and red occur with equal probability of 1/2. Of course, there is some theoretical bases for claiming that if you had an infinite amount of money, could play the game for an infinite amount of time, and could count an infinite number of spins, then maybe you could have some winning streaks. Not likely though.
P: 6
 Quote by alan2 You should heed the advice of Number Nine. You are committing a common mistake with probability. The probability of any sequence of blacks and reds is equally likely. There is nothing special about 7 black in a row. The probability of brbrbrb is equal to the probability of bbbbbbb or rbrbrbr or rrrrrrr in a finite number of seven spins and tells you nothing about the next spin assuming the wheel is fair, i.e. black and red occur with equal probability of 1/2. Of course, there is some theoretical bases for claiming that if you had an infinite amount of money, could play the game for an infinite amount of time, and could count an infinite number of spins, then maybe you could have some winning streaks. Not likely though.
Then that completely disregards the proven unlikelihood of 7 in a row occurring compared to 3 in a row, for example.

You would bet for three turns, on the understanding that you could lose all three; then absorb that cost. However, you will win doing that more times than lose.

I'm not sure how that can even be disregarded - and say that 3 in a row of black is the same probability of 7 in a row of black.

I apologise to keep questioning it, but it's still been proven by any analysis of 10,000 spins, for example, that you will get less 7 blacks in a row than 3 blacks in a row. What I am saying is that you would bet on that basis.
 P: 199 I didn't say that 3 in a row has the same probability as 7 in a row. I said that any sequence of 7 spins has the same probability. Many people have run many simulations. What you find with any martingale strategy is a normal curve about some value of earnings greater than zero. But you also find a low, broad curve over a wide range of negative values. The expected earnings from the higher frequency of wins are exactly cancelled by by lower frequency but much larger losses. Contrast this with the flat bettor who has a normal distribution with mean zero. The traditional martingale strategy, of which yours is just a variation, is to bet 1, then double if you lose and bet 2, etc, doubling each time you lose. When you do win you will receive all of your losses back plus 1. You win much more often than you lose. But you will lose because you have a finite amount of money. When you do lose, the average losses are much larger than the average winnings and the net result is zero. Of course, we are ignoring the house edge from the two extra spots. In that case your expected value is negative, not zero. 10,000 spins really doesn't prove anything. This is really kind of like a random walk in 1 dimension. You can show that every sequence of steps is certain to occur if you can sit around for an infinite number of steps. But in any finite number of steps your expected position is zero. Consider this. Casinos have been around for hundreds of years and thousands, if not millions, of very smart people have devised hundreds of betting strategies yet casinos are still in business. If there existed a successful betting strategy then the casinos would be gone.
 P: 199 Here's a very good simulator from the math dept at University of Illinois that allows you to set a lot of parameters. What you will notice, as I mentioned above, is that you actually do win more often than you lose with your strategy but the losses are larger and you always go broke.
 P: 199 Sorry, I forgot the link: http://mste.illinois.edu/pavel/java/martingale/
P: 688
 Quote by kyleball Then that completely disregards the proven unlikelihood of 7 in a row occurring compared to 3 in a row, for example.
Sure; but 7 reds in a row have exactly the same chance of occurring than a black, then a red, then 2 blacks, and *then* your 3 reds in a row.

Just the same, 5 reds in a row have exactly the same chance as 4 reds in a row, followed by a black. The 5 reds are more striking to our human brains, but that's all.
P: 6
 Quote by alan2 Here's a very good simulator from the math dept at University of Illinois that allows you to set a lot of parameters. What you will notice, as I mentioned above, is that you actually do win more often than you lose with your strategy but the losses are larger and you always go broke.
I've looked at it - but can't establish how to set the parameters of my system; of 4 in a row of the same, and then betting a maximum of three doubling up bets after that (then starting again at a 10p bet on both red and black).

I can't see how to set ALL of that on that applet?
 P: 199 Well, your 4 in a row on both to begin is irrelevant because the wheel has no memory. I think you're getting stuck on thinking that when you start doubling depends on what has already occurred. That's not true. It's like the horse player who figures a long shot is "due". Why not try a simulation by flipping a coin? Heads=black, tails=red. Start with 100 and apply your strategy by starting with an initial bet of 1. You will win more often than you lose but you will eventually go broke. No fair adding more money to your bankroll. This is a common way for gamblers to lose track. They lose it all today and start over tomorrow, forgetting how much they've previously lost. They are essentially playing with an infinite bankroll but in they end they come up even or lose the house edge.
 P: 199 I pose a question. Suppose you sat down at the wheel and got 4 blacks in a row on the first 4 spins. You decide that it's time to start doubling on red but the guy next to you says "you should have been here earlier, there were 20 reds in row just before you sat down". How does that affect your strategy?

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