Is each spin of a roulette wheel independent? 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?
 A: 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 $0$.
A: @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.
A: 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.    
