A trial with only two possible outcomes (one "success" and one "failure") is often called a Bernoulli trial. It looks like that's what you have here. Let's call a success "lands on 00" and a failure "lands everything else".
An obvious extension of this idea is a binomial experiment, where we run a fixed number of Bernoulli trials ($n=500$ from your description), each of which has a the same probability of success ($p=\frac{1}{38}$ for a standard American roulette wheel). These define a Binomial experiment, or binomially distributed random variable. We keep track of the number of successes too, and call that $k$. Then, for a Binomial experiment, the probability of seeing exactly $k$ successes is:
$$
P(k) = \binom{n}{k} p^{k} (1-p)^{n-k}.
$$
Plugging in your numbers, we get
$$\begin{alignat}{3}
P(k) &= \binom{500}{1} &\cdot \bigg(\frac{1}{38}\bigg)^1 &\cdot \bigg(\frac{37}{38}\bigg)^{500-1}\\
&= 500 &\cdot \bigg(\frac{1}{38}\bigg) &\cdot \bigg(\frac{37}{38}\bigg)^{499}\\
&\approx 2.19 \times 10^{-5}\end{alignat}$$
If you take that equation apart, it is easy to see where each of the three pieces come from. The first, the binomial coefficient $\binom{n}{k}$, tells you how many ways you could have observed $k$ successes during $n$ trials. In this case, it's 500 because it could have been the first, second, ..., or last trial. The next component tells you the probability of observing those $k$ successes, while the last one tells you the probability of observing the remaining failures. As you may, the recall that the probability of failure must be $1-p$ since a) success and failure are our only options b) they're mutually exclusive and c) mutually exclusive and exhaustive probabilities sum to one.
Going through the binomial distribution is nice because we can adapt this formalism to calculate other things, like the probability of seeing $k$ or more (or $k$ or fewer) successes instead, which might be more relevant.
Note that these values are for a single, pre-specified spot on the wheel. Perhaps you always play your mother's birthdate, heard a rumor that the casino was cheating players and want to check. The $p$-value you get from the above calculation will be correct in this case.
Now, consider a slightly different scenario. Suppose you kept notes on the outcome of each spin (it's a weird casino) and reviewed them when you got home. You notice that "00" was the least likely outcome and start to wonder if something fishy is happening. In this case, the $p$-value you calculate is going to be misleading, because you have used the same data to generate your hypothesis and test it. This effect is surprisingly strong, as you can see in a simulation. I started by generating 500 integers between 1-38 and tabulating how often each number occurred. This mimics the outcome of a single session. I then repeated this procedure 10,000 times. On average, each number comes up 1/38th of the time (13.15/500 trials), as you'd expect from a a fair roulette wheel. However, the rarest number in each session only comes up 6.1/500 times. The probability of seeing exactly six successes is quite small (~0.01), as is the probability of seeing six or fewer successes (~0.02), even though our simulation is, by construction, fair. One way to combat this problem is to split the data. You could use one part of your data to generate a hypothesis, then test it on the remaining data.
Alternately, you might use a $\chi^2$, or chi-squared, test. This test compares your observed counts to the expected number of counts (1/38 * 500). This directly tests your underlying hypothesis--the roulette wheel is fair--without dredging through the data to find a specific space to test.
