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I Beat the roulette wheel with this?

  1. Mar 23, 2017 #1
    I am perplexed because it seems that I have come up with a system for beating the roulette which has a positive mathematical expectancy/hope... but only in the calculations, which I think that I've made wrong.
    So I want to know in what I have failed doing the calculus.
    Well basically the method consists in betting on the three dozens... at the same time!. We start with 0'5 on each. Then we multiply by 1,5 the dozens that didn't turn out.
    (1'5 is the minimum by which we have to multiply the initial bet so that it keeps giving the same benefit as the first one, while covering the losses.) Well. It is a type of martingale as you can see.
    What we do next, is to stop the sequence, the progression of a determined dozen, once it reaches 3 consecutive spins without turning out, so that it goes back to the initial bet (0'5), and yeah, we accept the losses of that dozen.
    The thing is that if we calculate the mathematical expectancy/hope, we see that we're supposed to have benefits, that is, that it is positive. At least it is so in the way I calculated it by, which I suspect that is mistaken
    We will lose money with this system in 2 cases:
    -a determined dozen is not turning out (at least during 3 spins), whilst the other two including the zero are. Chances of this are: (25/37)^3=0.3085. (30.85%).
    -a determined dozen plus the zero turn out 3 consecutive times. Chances of this are (13/37)^3=0.043 (4'3%).

    With this results we have that, each 100 spins, in average, 30'85 of them we will have a 3 consecutive dozen without turning out, so we would have the following losses: 30.85*2,375=73.27€. (2,375 is the acumulated amount from the 1st spin to the last in which we accept losses and reinitiate) (0'5+0'75+1,125= 2'375).
    The same happens in the second case, but now, as only one dozen is turning out, we would lose that amount in the other 2 remaining dozens! So we would lose:
    4'3*(2*2'375) = 20.4€.
    There is a third case that I have not taken into account as its losses are really unsignificant. That case is when the 0 turns out 3 consecutive times.
    Well, On the other hand, we are wining 1€ each spin, whatever the outcome (well.. we don't when the 0 is the outcome though). So in 100 spins, we will win roughly 97€. (As the 0 is expected to turn out 2'7 times).
    If we add the losses and subtract it to the earnings, we clearly see that the expectancy is positive.. so we should win with this system in the long term.
    However I computed it with a sample of 5000 roulette spins and...... the result is that I lose.
    What did I calculate wrong or I didn't take into account?

    Thanks beforehand
  2. jcsd
  3. Mar 23, 2017 #2


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    Staff: Mentor

    As far as I can see, your strategy is the same as the doubling on red/black outcomes. To the best of my knowledge, Casinos forbid this betting strategy. In addition, there is always the question what to do with the green zero, which isn't - again as far as I remember - in the first dozen either, so your assumption on probabilities isn't correct. Furthermore this strategy usually requires really high numbers of rounds and available money to bet.
  4. Mar 23, 2017 #3
    I have taken into account the zero, in the 2 calculations of the probabilities (13/37, and 25/37). I am not so interested in winning at the casinos, I just want to know if the calculations are mistaken and how it should be calculated.
    Thanks for replying
  5. Mar 23, 2017 #4
    I can't follow your computation of the case where you get a payoff for two of the dozens.
    There are 2 subcases here: when you get one zero, and without any zero's.

    If the zero is in the first spin there are 36 possibilities for the second spin and 24 for the third,
    so the probability of this happening is 36*24/(37^3). The zero can be in any position, so the total probability for the first case is 3*36*24/(37^3)

    The second subcase is where there are no zero's. Now 2 of the spins will be in the same dozen, and the 3rd will be different. There are 3*(36*12*24) ways of doing that. (any pair of 2 spins can be in the same dozen, there are 3 different pairs) so you have a probability of 3*(36*12*24)/(37^3)
    The sum of those 2 probabilities is 0.6652...
  6. Mar 23, 2017 #5

    We "run" the three dozens at the same time, (performing the type of martingale already described above), so that for each spin we earn at least 1€ (because the initial bet is 0'5, and the payoff 0'5*3= 1'5, minus the bet on that dozen 1'5 -0'5= 1.) (We regard each dozen isolatedly). Of course we get 1€, while losing X amount of money in the other 2 dozens, by this reason we use the martingale on that dozens (which have not turned out).
    We increase the bet on them so that we cover all the bets carried out on that dozen isolatedly, plus we get 1$ of return. (when that dozen is the outcome).

    I don't understand some stuff

    "If the zero is in the first spin there are 36 possibilities for the second spin and 24 for the third"
    I intuit that here you're considering that in the two remaining spins we can get either one dozen or the other. (Having obtained a 0 in the first one).
    Wouldn't it be the same as (25/37)^3?
    24 would be the two dozens, +1 because of the zero.
    You're calculating ( I suppose) the probabilities of not having a 0 again, and having only as outcome 2 dozens, but we don't mind if there's a zero again or not, we focus instead, on the probabilities of not getting a determined dozen during 3 spins.
    And I divided it in two cases because in both we lose different amounts of money.
    In 25/37 we lose the amount of just one dozen.
    But if there are two dozens that are not got as outcome during 3 spins, we lose twice more than before. Chances of so are 13/37.
    (Which means, being the outcome just one dozen or the zero)

    I don't understand why did you choose that two cases?
    One in which the 0 is got just once, and anotherone in which the 0 is never an outcome. Could you explain it better please?
    Thanks for replying :)
  7. Mar 24, 2017 #6
    To compute the total expected winnings, I need to sum up these cases
    1 none of the dozens ever pays off (3 zeros)
    2 one of the dozens pays off (2 zeros, or 1 zero and 2 spins on the same dozen, or no zeros and 3 spins on the same dozen,
    3 two of the dozens pay off, (1 zero + two diffferent dozens hit, or no zeros and 2 spins of one dozen and 1 of another)
    4 all of the dozens pay off (no zeros and 3 different dozens hit)

    What I computed was the probability for case 3, split in two subcases, with or without a zero.
    I thought this should be the same as:
    What you are computing however is the probability that one specific dozen is not spun, not the probability that exactly one of the dozens is not spun.

    Of course a much easier way to compute the expected winnings is the compute what happens to the bets on one dozen. (25/37)^3 probability of losing 19/8 and 1-(25/37)^3 of winning 1
  8. Mar 24, 2017 #7

    Mmm I think I still don't get it xD.

    "What you are computing however is the probability that one specific dozen is not spun, not the probability that exactly one of the dozens is not spun."

    Is not it the same? I mean, what do u mean?.

    I see that I have to calculate the probability of the 4 cases. But, why is it not the same as what I did? (25/37 and 13/37). That is, what's the difference between computing the 4 cases separatedely and computing just the two that I've mentioned? (Which "include" and stand for the 4 cases).. I mean, why are mine wrong? If in the end, they stand for the same! 25/37 are the chances of a dozen not being spun, and 13/37 of two dozens not being spun. That's what I dont understand haha. Excuse my ignorance.

    "(25/37)^3 probability of losing 19/8 and 1-(25/37)^3 of winning 1"

    What is/ what does 19/8 stand for?
    And why do you set the another calculus as the probability of winning 1? If it is supposed that we win one each spin, no matter the dozen that is spun. (Because we are betting on the three at the same time, so regarding just that dozen itself, once it is spun, we win 1). It is so unless the 0 is spun. So wouldn't the chances of getting 1 be 36/37?

    Thank you for your help ^^
  9. Mar 25, 2017 #8
    I'm not following your logic. Why don't you simplify the problem? First, get rid of the zero. If you don't make a profit without the zero or the house edge you certainly won't make a profit when you put those back in. Stop worrying about "dozens". All you are doing is flipping a three sided fair coin and playing 3 games simultaneously. Each bet has a probability of 1/3 of a payoff of 3 for each 1 bet and a probability of 2/3 of a payoff of zero for each 1 bet for an expected payoff of 1. Each of the three individual games has an expected payoff of one and so all three games played simultaneously have an expected payoff of one because one is a winner and two are losers. It's completely irrelevant how much you bet on each game on each spin because over time you will get back exactly what you bet. Betting schemes that increase bets on the basis of previous losses are entirely ridiculous because each spin is an independent event. The "gambler's fallacy" is well known.
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