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My friend loves roulette. She has a simple strategy: if a number hasn't come up in a while, then that number is more likely to come up, and therefore you should start betting on that number.

I tried to tell her that each spin is an independent event and that any given number has the same probability on each spin. Then, after thinking about it more, I came up with the following idea:

If we choose a number, say, $0$, on the (American) roulette wheel the probability that any other number would come up would be $37/38$. So, if we continue to spin the wheel $x$ amount of times the probability that a number that is not $0$ would appear would be $(37/38)^x$.

Therefore, as $x$ increases the probability that a number that is not $0$ gets smaller and the probability that $0$ comes up gets higher.

Is this correct?

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  • $\begingroup$ Related: math.stackexchange.com/questions/2755438/… $\endgroup$
    – Ethan Bolker
    May 23, 2018 at 2:00
  • $\begingroup$ Your reasoning about $(37/38)^x$ applies to every number: it shows that as the number of spins increases, it gets more and more unlikely that you'll never see any given result. $\endgroup$
    – Théophile
    May 23, 2018 at 2:07

3 Answers 3

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No. You were right the first time. If the wheel is fair the the probability of any particular number is the same $1/38$ every time. Successive spins are independent. Thinking otherwise is the well-known gambler's fallacy.

That said, if the casino's records showed that $0$ never occurred in many thousands of spins, I would suspect that the wheel wasn't fair. Successive spins would still be independent, but the probability of seeing $0$ would be very small. Perhaps the wheel was rigged so that it never showed $0$.

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@Ethan Bolker has already provided a very good answer, but I would like to add that while it is true that the probability of getting no zeros after $x$ spins is $(37/38)^x$, which goes down with time, the probability of getting $n$ zeros after $x$ spins is $(1/38)^n\cdot (37/38)^{x-n}$, which will always be lower. It will always be more likely to get no zeros than getting exactly $n$ zeros, as there is only one zero and 37 numbers that are not zeros.

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I cannot comment so I'm offering some answer. The probability of any result is 1/N, where N is the number of possible outcomes. Your formula (1/N)^x does not express the likelihood of any single outcome. It is the probability of a specific sequence of outcomes, where each is independent (my wording might be off). It is the probability of NOT seeing 0 in x spins. That should in fact be low. That means (as I interpret it) that you are not very likely to never see 0. This is somewhat consistent with your original expectations. Again, you are calculating the probability of a sequence of outcomes, not the likelihood of a specific outcome after many spins.

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