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I've seen quoted in a number of places that the standard deviation for a 1-number bet on a 38 number Roulette wheel (0, 00, 1, 2, .., 36) is 5.76. I can't seem to find a single source that shows how this calculation is made.

I am a student, this is for further understanding of a homework problem, so please assume I'm clueless and be as detailed as possible.

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  • $\begingroup$ What are your possible outcomes if you bet $1 on a single number, meaning how much do you win if that number comes up and how much do you lose if that number does not come up? What are the probabilities of these events? $\endgroup$
    – jsk
    Commented Jun 10, 2014 at 3:03
  • $\begingroup$ I didn't notice the mention of homework before. Since it's in connection to a homework problem, please add the self-study tag, and check the tag wiki info. When you say 'further understanding' I assume you're not actually doing a homework problem, just trying to understand one... is that right? $\endgroup$
    – Glen_b
    Commented Jun 10, 2014 at 3:11
  • $\begingroup$ One of the first hits on a Google search is the Wikipedia article which shows the calculation and relates it to the theory of the Binomial distribution. $\endgroup$
    – whuber ♦
    Commented Jun 10, 2014 at 5:45

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Your question should make explicit what quantity you want the standard deviation of.

You say "a one number bet" but you don't clarify what the outcome is that you're considering.

I will assume you mean the following:

Let there be a bet of one unit (say \$1), with a payout of 35-1 -- that is you stake \$1 and you end up with an outcome of either \$0 or \$36 (equivalently, a profit of either \$-1 or \$35). The probability of the outcome \$36 is 1/38. What is the standard deviation of the outcome?

Let's start first with the variance. (The variance is the square of the standard deviation.)

See the definition of variance here, and specifically for a discrete random variable here.

The outcome $0$ has probability 37/38 and the outcome $36$ has probability 1/38.

So the variance is $E(X^2)-E(X)^2 = \frac{36^2\times 37}{38}-(\frac{36\times 37}{38})^2 = 33.20776...\quad$

and the standard deviation is the square root of that.

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  • $\begingroup$ PERFECT! Thanks so much, this makes a lot of sense. $\endgroup$
    – Johnny Bones
    Commented Jun 10, 2014 at 3:18
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sorry, I don't have enough reputation for a comment, this is a comment to Glen_b's answer. When $X$ is the outcome, isn't $E(X) = 0\cdot 37/38 + 36 \cdot 1/38$ ?

It leads to the same variance though.

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    $\begingroup$ So it's boiled down to (36^2/38) - (36/38)^2? $\endgroup$
    – Johnny Bones
    Commented Jun 10, 2014 at 12:56
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    $\begingroup$ exactly. It gives 33.207 like @Glen_b said. $\endgroup$
    – spore234
    Commented Jun 10, 2014 at 14:22
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The standard deviation is the square root of the sum of squares of deviations from the mean. So I suggest you finding the mean first, which is in this case the Expected Value. You have two X values, the gain amount and the loss amount.

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  • $\begingroup$ The question asks about theoretical SD rather then empirical SD. Theoretical SD is (as stated in other answers) $\sqrt{E(X^2) - E(X)^2}$. Also: mean is not the same as expected value. Meas is a estimator of expected value. $\endgroup$
    – Tim
    Commented Jul 22, 2017 at 17:38
  • $\begingroup$ If you have 0 and 00 along with the integers from 1 to 36 you may have a uniform distribution for thirty eight outcomes. But the set is not entirely numeric. So I don't see how it is meaningful to calculate even a theoretical standard deviation. It seems to make more sense to test for uniformity based on a large sample. $\endgroup$
    – Michael R. Chernick
    Commented Jul 22, 2017 at 21:52

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