Is there an easy way to calculate standard deviation for dice rolls?

So I've got all these polyhedral dice for role playing games. I can use classical statistics to calculate quite easily the assumed uniform distribution for a single die roll, and what the mean is.

Now if I roll the same die several times, and add the results, the probability for any particular number starts to form a bell shaped curve.

For example, if I roll a six-sided die three times, there would be a sort of bell shaped curve, with 10.5 being the mean, with 3 and 18 being the extremes.

It seems to me that I should be able to calculate the standard deviation for these sort of bell shaped curves without too much trouble, but our statistics teacher thinks the only way to do it is to roll the dice about a billion times, and go through these complicated equations for each individual roll. Isn't there an easier way?


What a mess! I might as well make a table for each one and estimate it off of that!

4 Answers

  • gp4rts
    Lv 7
    1 decade ago
    Favorite Answer

    EDIT: This answer has been corrected. See note below.

    In the case of the dice, you have discrete values in the probability distribution. It is not a "normal" (bell-curve) distribution. But you can calculate the std dev. as follows:

    For every possible value v of a throw i, compute P(vi). In a large number of throws N, the expected number of times a value vi occurs, nvi = N*P(vi). The mean value would then be m = (1/N)∑nv*v = (1/N)∑N*v*P(v) = ∑v*P(v). The standard deviation would then be

    d^2 = [1/(N-1)]∑(xi - m)^2 = [1/(N-1)]*{∑[N*vi*P(vi)]^2 - 2*N*m*∑v*P(vi) +N*m^2}

    d = √ [1/(N-1)]*{∑[N*vi*P(vi)]^2 - 2*N*m*∑vi*P(vi) +N*m^2}

    You can't eliminate N (the total no of triails) but you can compute the result for a large N without actually rolling the dice.

    EDIT: There is an error in the formula: the terms N and P(vi) should not be squared in the first term of the sum. The formula should be

    d = √ [1/(N-1)]*{∑N*P(vi)*vi^2 - 2*N*m*∑vi*P(vi) +N*m^2}

    Then, for large N, N-1≈1 and the N's cancel out to give

    d = √ [∑P(vi)*vi^2 - 2*m*∑vi*P(vi) +m^2]

    So for all possible values of the dice (vi), compute the probabiity of that value P(vi), the mean value m = ∑vi*P(vi), and sum over all value according to the above formula.

    Plugging the mean value into the formula gives the simplification:

    d = √ [∑P(vi)*vi^2 - 2*m^2 +m^2] = √ [∑P(vi)*vi^2 - m^2]

    FINAL: Made final correction. Above formula is now correct; the m^2 is not under the summation sign.

    Calculated for an ordinary pair of dice: m = 7, d = 2.415

    You can see the calculation here

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  • 1 decade ago

    There are two points here.

    For calculating standard deviation, you use a firmula or you estimate it.

    When you calcualte it , what you do is

    1. Compute the mean for the data set.

    2. Compute the deviation by subtracting the mean from each value.

    3. Square each individual deviation.

    4. Add up the squared deviations.

    5. Divide by one less than the sample size.

    6. Take the square root.

    When you estimate, you use some distribtions.

    Now, you must have heard of cenral limit theorem.

    In the case of taking samples and finding means, the central limit theorem tells us that sample means are normally distributed, when n(number of expereiment) tend to infinity.

    So, the standard deviation value becomes more and more precise as you increase the number of experiements. Thats the reason why your teacher ask you perorm the experiment many times.

    Its sometimes beyond hand calcualtion capabilities and at these points making experiment for atleast 30 times would give you a correct approximation(beyond this it just precises by a very small fraction). So, you got to do it atleast 30 times.

    I hope, its clear now.

    All the best and regards to your teacher.

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  • 3 years ago

    you may no longer get a probablility over a million. If order isn't a attention then on the 1st roll all six numbers are available 6/6 on the 2d roll you have 5 selections out of the 6 5/6 on the 0.33 roll you have 4 selections that would take place 4/6 on the fourth roll, you're decreased to 3 numbers that isn't reason duplication 3/6 on the 5th roll in basic terms 2 numbers are left 2/6 ultimately on the final roll, you have a million/6 success comes approximately with 6/6 * 5/6 * 4/6 * 3/6 * 2/6 * a million/6 which is composed of 0.01543. It skill that in the time of 1000 rolls you would be triumphant 15 cases sort of. All nicely and robust until the laptop tries it on what's noted as the monte carlo approach. it form of feels to furnish on 2000 rolls that quantity (30). that's bearing directly to the anticipated quantity. success. This answer is right. There would desire to be assorted stars in this question if 2 or 3 precise individuals can no longer do it or have hardship with it.

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  • 1 decade ago

    You can calculate the standard deviation of even a small data set, but the question becomes, is it statistically significant? and at what point (in this case how many rolls) does it become statistically significant?

    ..and the standard deviation is the square root of the sum of the squares of the deviations, your calculations cover the entire data set, not individual data points. The deviation is the only thing calculated on the individual roll.

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