Roulette

Discussion in 'Physics & Math' started by Lakon, Sep 3, 2013.

  1. Tach Banned Banned

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    Let's try again, since you have so much difficulty in understanding:

     
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  3. Fednis48 Registered Senior Member

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    Ok. That's what I thought. The fact that you won while the people you bet against lost is not your whole point; you also think that you had good reason to expect that you would win. That's the part I disagree with, for the following reasons.

    This is the fallacy that I was hoping to dispel by posting actual math. It is mathematically impossible for the predictions of a roulette algorithm to be biased in any way that matters, right or wrong. In post 40, I showed that the average loss per bet is independent of how the bets are chosen. This mean's that its just as impossible to come up with an algorithm that performs worse than chance as it is to come up with one that performs better than chance.

    The difference is that a mechanical bias is possible in principle, while an algorithm bias (without mechanical bias) cannot exist according to basic rules of probability. That's why Lakon and I spent so much time pinning down your agreement that there was no mechanical bias in your case: if there had been, then it would have been technically possible (although very unlikely) for your strategy to be a winning one. Without mechanical bias, your strategy was mathematically guaranteed to be no better than chance.
     
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  5. Tach Banned Banned

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    Nonsense, if the algorithm is biased, the probability of loosing is higher than 50-50.


    Nonsense, there is no difference between a hardware bias and a software bias.
     
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  7. Tach Banned Banned

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    This is precisely where you are making your mistake, in a systematically flawed algorithm \(P_{your color} \ne \frac{1-P_0}{2}\).
    In fact, \(P_{your color} << \frac{1-P_0}{2 \) (depending how bad your algorithm is).
     
  8. Fednis48 Registered Senior Member

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    725
    You can't just make that assertion. You need to back it up. To sum up my argument from post 40:

    -Each spin is an independent, random event that chooses red or black with equal probability.
    -No matter which color you choose for a given spin, you have a 50% chance to be right and a 50% chance to be wrong (if we ignore the probability of spinning zero).
    -Since these odds are the same for either choice you could make, they hold regardless of how you make your choice.
    -Since the spins are independent from one another, these odds hold over any number of consecutive spins.
    -An algorithm is nothing more than a method of choosing bets for some number of consecutive spins, so the 50-50 odds hold for any algorithm.

    What is wrong with my reasoning?

    edit: New response while I was writing. Getting to it now...
     
  9. Fednis48 Registered Senior Member

    Messages:
    725
    By the nature of probabilities, \(P_{red}+P_{black}+P_0=1\).

    Because the wheel is unbiased, \(P_{red}=P_{black}\).

    Putting the above together and solving, \(P_{red}=P_{black}=\frac{1-P_0}{2}\)

    For any given spin, you have to choose red or black, so either \(P_{yourcolor}=P_{red}\) or \(P_{yourcolor}=P_{black}\).

    Plugging line three into either of the two possibilities give the same result: \(P_{yourcolor}=\frac{1-P_0}{2}\).

    Like I've said before, it's just as impossible to devise a worse-than-chance algorithm as it is to devise a better-than-chance one.
     
  10. Tach Banned Banned

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    5,265
    You persist in the mistake of thinking that this is an equiprobable situation, it isn't.
    For the "algorithm developers", the probability of success is NOT \(\frac{1-P_0}{2}\) but rather \(\frac{1-P_0-B}{2}\)

    where \(B\) is their "bias", their systematic error. This is precisely what makes the probability of the person "bidding against the algorithm" to be: \(1-\frac{1-P_0-B}{2}-P_0=\frac{1-P_0+B}{2}\). So, I have \(B\) over the (flawed) algorithm. FACT confirmed experimentally. I suggest that you try to find the people with the laptop and that you bid against their bids, you too, may end up a few hundreds of EUROS richer.



    Bad theories get falsified by experiments. Good theories get confirmed by experiments.
     
  11. Sarkus Hippomonstrosesquippedalo phobe Valued Senior Member

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    10,355
    But there is no "B".
    In roulette, on a wheel that has no mechanical bias, betting on Red or Black (if that is what was being done, or at least what is being exampled) is random. There can be no "B" in such a scenario... only a perceived "B" that actually does not exist.

    Their "algorithm" (if it was a pure Red/Black decision-maker) was no better or worse than Random - and your betting against it would be likewise.
    Any success is, by all the rules of probability, mere chance.


    What the casino undoubtedly saw was an algorithm that simply didn't work and was producing outputs no different than chance. Betting with or against the algorithm would be no different - although with short term gains and losses, as you exprienced.
    Bear in mind that an algorithm that doesn't work is merely an algorithm that produces an output no different from chance: it is NOT an algorithm that produces results worse than chance.
    Afterall, someone who consistently names the wrong colour of unseen playing cards is just as "psychic" as someone who consistently picks the right colour.
    And bad maths gets falsified by mathematicians.
     
  12. Tach Banned Banned

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    5,265
    Incorrect, the \(B\) has nothing to do with any mechanical bias, it is the systematic error of the algorithm proper, you need to pay attention, I must have pointed this out half a dozen times.

    Incorrect. The algorithm does not operate as an equiprobable method, it seeks an "advantage" over the random betting and, in doing so, it introduces an error.



    You keep repeating a fallacy. A fallacy that was put forward earlier by Fednis as well.


    ...provided that they are good.
     
  13. Sarkus Hippomonstrosesquippedalo phobe Valued Senior Member

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    10,355
    Read what I wrote: "In roulette, on a wheel that has no mechanical bias... there is no B."
    There are also no algorithms that can predict randomness - by definition. All algorithms that try will perform to the same level over the long term... they will all tend to the norm. This is not some spurious claim but a mathematical certainty.
    The algorithm is fundamentally irrelevant in this scenario.
    You can not introduce any error into randomness without turning it non-random. Since the randomness is the ball moving around the roulette wheel, the only way to affect the randomness of the result would be to interact physically with the ball. Did their algorithm interact with the ball? No. It did not alter the randomness of the ball, or the result. All it did was predict whether it was Red or Black... and, as has been explained, there is equal probability of either. Betting against a 50:50 chance is also a 50:50 chance.
    Did the algorithm somehow turn the result into a 40:60 chance, or even a 49:51 chance?
    No.
    Every spin (ignoring any zeros), if the wheel is random (i.e. no mechanical bias), results in 50:50 chance.
    It is not a fallacy, and repeating it seems to be warranted.
    The explanations have been given to you. They have been thorough in their detail.
    Your only argument is that "the algorithm does not operate as an equiprobable method..." yet you fail to show how that is even mathematically possible.

    So let's take this one step at a time:
    - Is the roulette wheel free from mechanical bias, such that the spin results in a random result - i.e. (ignoring the zeros) it will, over the long term, land on each number from 1 to 36 with equal probability and with no determined pattern? (and if you think there is a determined pattern, how do you propose this is achieved without some mechanical bias?)
    - Do you agree that this lack of mechanical bias results in a 50:50 chance for each spin to end on a RED or a BLACK?
    - Do you agree that with odds of 50:50 it makes no difference whether you bet on RED or BLACK?
    - Do you agree that this "algorithm" hopes to turn those odds in the user's favour - to say 40:60 (or anything other than 50:50).
    - Do you agree that neither the algorithm, nor the users, had any way to physically affect the result?

    So... one simple question: given the above, how does the algorithm's decision affect the outcome of the result, as you seem to suggest it does?
    The spin remains physically unaffected by the decision of the algorithm, and so, per the above, will land on RED or BLACK with 50:50 chance.
    You (hopefully) agreed above that it therefore makes no difference whether you bet on RED or BLACK.
    So how does it make any difference whether the algorithm tells you to bet on RED or BLACK?


    Hopefully you have followed this... it is a relatively non-mathematical way of trying to show you where the error may lie in your current thinking.
     
  14. Tach Banned Banned

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    5,265
    Wrong. I can see you are not getting it, so there is no point in continuing with you.

    But, but! This is precisely what I have been trying to explain to you that the algorithm authors did! This is not an equiprobable (random) problem.



    Of course that the algorithm doesn't interact with the ball, it only predicts the outcome incorrectly. Therefore, betting against it (not against the house), gives the better an advantage. The same exact way as bidding with dr. Jarecki's correct algorithm gives the better an advantage.

    I am not suggesting such a thing. It is you who , in failing to understand the strategy, constructed this strawman and you are beating it to death.
     
  15. Fednis48 Registered Senior Member

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    725
    This is non-responsive. I know you claim that the algorithm bias lowers its probability of success, but that's just an assertion; assigning the bias to a variable "B" doesn't make it any more rigorous. In post 66, I wrote 5 lines of deductive math that prove there can never be such a "B" for any algorithm. Of those five lines, two of them (lines 3 and 5) just come from applying algebra to previous lines. So which of the other three lines, if any, do you disagree with?

    If you agree with all of these statements, the exact value of \(P_{yourcolor}\) can be found by algebra, so you can't just say that I'm forgetting to include B.
     
  16. Tach Banned Banned

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    5,265
    It is just as much an assertion as your claim of \(\frac{1-P_0}{2}\)


    You wrote a lot of stuff, none of it a valid proof because you insist in shoehorning an approach that doesn't fit the problem. They are assertions just the same.
    Tell you what, experiment confirms my theory (and falsifies yours). So, how about you tried to replicate the experiment, go find the people with the laptop and bid against them?




    The wheel is unbiased but the "algorithm" IS biased. You STILL don't get it (probably never will). This is not your standard problem of probabilities, you are not starting with equal probabilities, you are starting with a biased situation. I know, you can't find this type of exercise in the books but you need to understand that this is a slightly different problem than the ones you find in textbooks.
     
  17. Fednis48 Registered Senior Member

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    725
    Given the exact number and success rate of bets you placed, you might be able to do enough statistical analysis to call your experience a rigorous experiment. As it stands, all we know is that you won a lot in one sitting, which is uncommon but could still be easily attributed to luck. Your empirical results don't decide the debate one way or the other.

    This type of non-response is really frustrating. My claim was: "Given three non-controversial claims about probability and a bit of algebra, we can deduce that no algorithm of any type have odds different from chance." Your response is: "You're ignoring the fact that the algorithm is biased!" You don't attack any of the equations from probability that I use as assumptions. You don't attack the math I use to get from those assumptions to my final result. You just say that the result itself should contain an extra "B" term. That would be a fair counterpoint if I was just stating the probability of success in one step, but I'm not; since the probability of success is the quantity under dispute, I derive it from simpler quantities that we can all agree on. To prove me wrong, you have to address those simpler quantities rather than just skipping to the end.

    I know how much you hate simple yes-or-no questions, but I'm going to try this anyway. My calculation relies on three assumptions:

    Do you disagree with any of these? If so, why? If not, my next post will be to apply algebra to these assumptions and check again for your agreement.
     
  18. Tach Banned Banned

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    5,265
    I bid for one hour. I must have placed about 30 bids and I won all of them. Feel free to try to replicate my experiment.


    But I attacked the fact that you are applying the WRONG approach to the problem. Repeatedly. You are trying to reduce a situation where there is bias to an equiprobable one. What you are doing is patently wrong. It doesn't help writing a bunch of equations that do not address the situation being discussed. It is like insisting to solve a problem where gravitational fields are present by writing a bunch of (correct but irrelevant) equations lifted from SR. Something that you have a lot of practice doing.
     
  19. Dinosaur Rational Skeptic Valued Senior Member

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    4,885
    Fednis48: The following is erroneous:
    You need to check some book or pamphlet describing roulette.

    Zero (Also double Zero on an American wheel) is treated like any other number. You can place a bet on it at 35 to one. True odds are 37 to one for European wheel & 38 to one for an American wheel.
     
  20. Dinosaur Rational Skeptic Valued Senior Member

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    4,885
    Tach:I would never expect to replicate your results at roulette:
    Assuming you are betting on one of the even money payoff options (red/black, high low, odd/even), the probability of your winning 30 bets on a European wheel is:

    (18/37)[sup]30[/sup] = 0.000 000 000 41

    This figures to occur once in circa 2.44 billion attempts.

    I am certain that neither I nor anyone else would replicate your claimed 30 wins.

    An American wheel has a bigger house edge.

    House edge for American Wheel is circa 5% (European wheel edge is about half the American edge).


    A wheel would have to be very biased to allow a bettor to overcome the house edge. A minor bias would not overcome that edge. As mentioned by me in a earlier post (not an exact quote):
    I am sure that wheel manufacturers do their best to assure that their wheels have at most a very slight bias.

    Each bet versus the house has a negative expectation. Claiming that some system can beat that edge is a claim like the following:
    The claim that you can win by betting against a losing bettor is nonsense in a game like roulette which has a fixed house edge & is an independent trials game (unlike Blackjack where the probabilities change as cards are dealt from the shoe).

    Every bet you make at roulette is subject to the same house edge. Anyone you watch for a long series of bets figures to be a loser. Betting against a loser does not guarantee that you will be a winner in a game with a house edge. Most players are losers; Some lose more than others; Few break even or win.

    If you bet against a player who loses more than expected, you will win more than expected & vice versa; Both of you figure to be losers in the long run.

    I posted the following:
    You replied by citing some passages from that book. How about finding some citations other than that book?

    If I claimed that the Blair Mill Witch project was a work of fiction, would you cite passages from it to refute my view? This is like claiming that Alien abduction occur by citing passages from books describing alien abductions.

    Many books have been written about alien abductions & haunted houses. I consider them to be fiction even when presented as true descriptions of actual phenomena. I similarly view the Jarecki story/novel to be fiction. It is only one book. It is less documented than stories about haunted houses & alien abductions.
     
  21. Fednis48 Registered Senior Member

    Messages:
    725
    I know that you can bet on zero like any other number, but my impression was that it was a worse bet than red and black. Getting paid 35:1 on 1/37 odds gives a little under 95% return, while getting paid 2:1 on 18/37 odds gives just over 97% return. So putting some of your chips on zero will indeed perform "worse than chance", as long as by "chance" you mean randomly choosing red or black.

    You're saying the algorithm lost 30 out of 30 times you tested it?! That's not just performing worse than chance - that's anti-predicting the wheel with >90% accuracy. I guess at this point we'll just have to set aside the issue of experiment, because I frankly don't believe your story anymore. We should be able to resolve the math with math by itself, so let's focus on that.

    Ok, but tell me exactly what is wrong with my approach. You say I'm writing a bunch of equations that do not address the situation being discussed, so please tell me which equations do not apply to this problem. Again, here's a handy list of three, from which you're welcome to choose one or more to disagree with:

    If these equations don't apply to a biased algorithm, presumably you can tell me how they need to be modified so that they do apply. Just broadly saying that my approach is wrong is not a response, nor is jumping straight to the end of the calculation and stating \(P_{yourcolor}=\frac{1-P_0-B}{2}\) without somehow deriving it.
     
  22. Tach Banned Banned

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    5,265
    Let's not, since you are clearly unable to understand what I am saying. The "algorithm" lost about 30 bets on red/black and odd/even. It may have won on bets of the type "3 red", I repeatedly pointed out that I stayed on such bets and that I did not follow how they performed.
    This is a physics problem, not a math problem and if you continue to treat it as a math problem, I guarantee that you will never get it. So, I am going to ask you two questions, depending on your answers, I may be able to get through to you. These are not loaded questions, so please answer:

    1. You studied physics, correct? Are you practicing physics or are you doing something different.
    2. If you are practicing physics, are you an experimentalist or a theoretician?
     
  23. Fednis48 Registered Senior Member

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    I'm a practicing theoretician. Respond to that how you will, but while you're at it, answer me this:

    You say this is a physics problem, not a math problem. That really surprises me, because I do not see how physics enters into it. Clearly there is no physics in the algorithm, because it's all software. Earlier I asked you about the possibility of a mechanical bias in the wheel, which would be physics, but that was ruled out. So what in this problem has anything to do with physics?
     

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