I was recently trying to do some mental mathematics and found myself stumped when trying to handle what ought to be a simple problem. My issues essentially boiled down to the question that I've asked, although for the sake of making the numbers easy I suppose that we can stick with the discrete uniform distribution with P(X=x)=1/6, i.e. the distribution that most often be used in problems related to six-sided fair dice.

So, with all of this is mind, is there any "easy" way to even mentally do something as simple as find the distribution of the sum of two i.i.d discrete uniform random variables, each with P(X=x)=1/6? What if I was trying do something more advanced like finding the probability that the sum of 6 fair six-sided dice is greater than or equal to 18?

  • $\begingroup$ Getting the distribution of a sum of iid random variables is not difficult, but not extremely simple (and not be done mentally). See here and references, esp dartmouth.edu/~chance/teaching_aids/books_articles/… $\endgroup$
    – leonbloy
    Sep 9, 2017 at 14:08
  • $\begingroup$ I think your comment might be the best answer that I'm going to get. So far I've seen no signs of an easy way to do this sort of stuff mentally. $\endgroup$
    – J. Min
    Sep 13, 2017 at 11:31

2 Answers 2


I picture the pdf mentally. A single uniform has a pdf whose graph looks like a rectangle. Convolve it against itself and you get a thing whose graph looks like a triangle. If all we are doing is adding two identically distributed rvs I stop and try to puzzle out the numerical coordinates of the vertices of the triangles; this calculation is slighty different in the continuous case. If we are adding two independent but non identically distributed uniform rvs (such as a die roll plus a coin flip) we get a trapezoid instead of a triangle, and the coordinate puzzle is a tiny bit harder.

If we convolve three uniforms I try to imagine the convolution integral of a rectangle sliding against a triangle, and mentally tote up the area of the instersection at various offset slides. To begin with, when the two figures don't intersect much, the area obviously grows quadratically. Then when the lead edge of the rectangle passes the apex of the triangle the formula switches. Finally when the trailing edge of the rectangle passes the triangle's apex, we are back in a quadratic regime again. At this point I realize the behavior is probably quadratic in the middle range, too, and that the answer must be a piece-wise quadratic approximation to the bell shaped curve.

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    $\begingroup$ We're in the realm of speculation here. I think this is interesting speculation. (+1) $\endgroup$
    – BruceET
    Sep 9, 2017 at 19:48

I doubt there is a really simple and useful answer to your question. But it's fun to speculate. Here is an approach using means and variances that might sometimes be easy enough to use.

If $X$ is the first random variable and $Y$ is a second one, then $E(X+Y) = E(X) + E(Y).$ So if you know the mean of the first random variable is $3.5$ and the mean of the second is $0.5$ then the mean of the sum is $4.$

The next part is harder, but sometimes useful. (Especially for normal random variables, but I know you asked about discrete ones.)

If the random variables are independent, then $Var(X+Y) = Var(X)+Var(Y).$ Then if you also knew that the variance of the first is about 2.9 and the variance of the second is about .25, then the variance of the sum is about 3.15, and the SD of the sum is about 1.8.

Finally, you might use the Empirical Rule to guess very roughly that about 95% of the values of the sum are within $4 \pm 3.6.$

Note: For a roll of a die and toss of a coin, quantiles .025 and .975 of the total are 1 and 7, respectively. The graph is from a very simple simulation. I sometimes use simulations to visualize probability results.

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