# Odds of a number appearing only one time in 500 spins on roulette

I recently observed a roulette wheel where one number appeared only once in five hundred spins. This is an American roulette wheel with 0 and 00, so the odds of any number hitting should be 1/38.

I am wondering if this suggests that there could be some type of bias on this roulette wheel that is causing one number to not appear as often as expected, or if this would be considered a normal result.

I seem to recall that most expected results fall within three standard deviations of the expected result, but I'm not confident in my ability to do the math.

• Welcome to our site! There's no need to sign your post or give thanks at the end of a question - to keep things to the point, we generally prefer you don't - even though it might make the site look a bit impolite at times! – Silverfish Apr 23 '16 at 19:36
• Since you have written the question in terms of "odds" rather than "probabilities", you may be interested in the question odds made simple, which talks about the relationship between odds and probabilities – Silverfish Apr 23 '16 at 20:14

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.

• Nice answer! I wondered whether it's worth contrasting P(there is some number which occurs once in 500 spins) with P(a particular, pre-selected number occurs once in 500 spins). I also toyed with the idea of "as or more extreme" (i.e. probability of one or fewer) in the manner of a p-value. – Silverfish Apr 23 '16 at 20:08
• Thank you so much for your answer Matt. That is so much more than I expected! I really appreciate the time you took to give me such a detailed answer. I decided to run the chi square test using the calculator at graphpad.com and this is what it said:Chi-square test results P value and statistical significance: Chi squared equals 53.814 with 37 degrees of freedom. The two-tailed P value equals 0.0364 By conventional criteria, this difference is considered to be statistically significant. I was unable to post all the data as it was too many characters for one post. – user113266 Apr 23 '16 at 20:28
• Instead of calculating whether one specific number turns up much too rare, you could also check whether any of the 38 numbers turns up much too rare. The probability is roughly 38 times higher. – gnasher729 Apr 23 '16 at 22:16
• @Silverfish, good point--I added a bit about that. – Matt Krause Apr 29 '16 at 1:35

I intended this to be a comment but it got way too long.

One thing that doesn't appear to have been mentioned in the answers so far (or if it was, I missed it) is the fact that it sounds as if -- rather than specifying the event of interest beforehand (i.e. before any spins, saying "let's spin 500 times and see if a particular number -- say 36 -- comes up 1 or fewer times"), instead simply observing 500 spins and - about any sufficiently bizarre-seeming outcome asking post hoc: "Wow, what are the chances of that??" -- that is, there the data you observe themselves lead you to ask the question you wish to then test on the same data (such as "does 36 come up too rarely?" after you see it come up only once) - that is, a data generated hypothesis.

The answers you presently have attempt to address a question of the first kind but are not really sufficient to answer a question of the second kind, because it's not clear what other outcomes would have led us to ask a question of the form "Wow, what are the chances of that??".

For some explanation of the issues, see this wikipedia article Testing hypotheses suggested by the data, or any of a fair number of posts on site (such as The Risk game dice problem$\!$).

## Definitely not a normal result.

If the roulette is fair then in 500 spins is expected to get about 13 appearances of a certain number(500/38$\approx$13).
If you use the Bernulli for $k=13$ you'll find 11.13%, which means there is 11.13% chance the certain number appearing exactly 13 times in 500 spins. In other words, if you repeat the 500-spins trial many times, it is expected in 11.13% of those trials, to have exactly 13 appearances of the number. Not less, not more, but exactly 13 appearances. Also if you try the Bernulli for "$k$" other than 13, you will always find less than 11.13%. Meaning, 13 is the most possible frequency of apperances of a certain number in 500 spins.

For k=1, as shown in another answer, $P(k)\approx0.00219\%$. Apparently is very unlikely to have only one success in 500 spins. It is very likely that the appearances of a number will be around 13. Let's try for all $k\in[13\pm7]$. This will be the probability of having 6 up to 20 appearances of a number in a 500-spins trial.
$$\sum_{k=6}^{20}\binom{500}{k}(\frac{1}{38})^{k}(1-\frac{1}{38})^{500-k}\approx96.48\%$$
I hope this makes clear what the answer is. Either the roulette is biased or...

As already pointed out by @MattKrause at the end of his answer, if you actually want to know whether the fact of observing only one appearance of a given number when spinning the roulette 500 times confirms that there is a bias on the roulette wheel, then you should perform an appropriate statistical test.

More precisely, a test on proportions would apply, its null and alternative hypothesis being:

$$\left\{ \begin{array}{llc} H_0:&p = 1/38 \\ H_1:&p \neq 1/38 \end{array} \right. \, \text{.}$$

The alternative hypothesis $H_1$ could be also $p < 1/38$, if you actually suspect that the number object of interest appears fewer times than expected (as it seems to be the case).

To solve this hypothesis test basing on your observed data, a chi-square test or an exact binomial test could be used, for instance.

NOTE: If you have recorded how many times all the numbers on the roulette appeared in 500 spins, then you could also check for wheel bias by performing a multinomial test.

• The important part of Matt Krause's answer is the second half, where he points out the problems of formally testing post hoc hypotheses. Your answer seems to ignore that critical issue by failing to clearly describe what $H_0$ actually states: is it the chance of a specific number or is it the chance of all the numbers? It's also unclear how either a $\chi^2$ or binomial test could validly be applied until somehow you have modeled what really was observed--and that's possible only by assuming the hypothesis was formulated before the observations were made. – whuber Apr 29 '16 at 15:01