Probability that sum of rolling a 6-sided die 10 times is divisible by 10? Here's a question I've been considering: Suppose you roll a usual 6-sided die 10 times and sum up the results of your rolls. What's the probability that it's divisible by 10?
I've managed to solve it in a somewhat ugly fashion using the following generating series:
$(x+x^2+x^3+x^4+x^5+x^6)^{10} = x^{10}(x^6 - 1)(1+x+x^2+\cdots)^{10}$ which makes finding the probability somewhat doable if I have a calculator or lots of free time to evaluate binomials.
What's interesting though is that the probability ends up being just short of $\frac{1}{10}$ (in fact, it's about 0.099748). If instead, I roll the die $n$ times and find whether the sum is divisible by $n$, the probability is well approximated by $\frac{1}{n} - \epsilon$.
Does anyone know how I can find the "error" term $\epsilon$ in terms of $n$?

The distribution of the sum converges to normal distribution with speed (if I remember correctly) of $n^{-1/2}$ and that error term could be dominating other terms (since you have got just constant numbers (=six) of samples out of resulting distribution). However, there is a small problem -- probability that your sum will be divisible by $n$ tends to $0$ with speed $n^{-1}$. Still, I typed something into Mathematica and this is what I've got:
In[1]:= 
    X := Sum[
         Erfc[(6*Sqrt[2]*(7n/2 - k*n+1))/(35n)]/2
        -Erfc[(6*Sqrt[2]*(7n/2 - k*n))/(35n)]/2, 
         {k, {1,2,3,4,5,6}}
    ]
In[2]:= Limit[X, n -> Infinity]
Out[2]= 0
In[3]:= N[Limit[X*n, n -> Infinity]]
Out[3]= -0.698703
In[4]:= N[Limit[(X+1/n)*n, n -> Infinity]]
Out[4]= 0.301297

The Erfc is the cumulative probability function of normal distribution,
the formula inside is to adjust for mean and variation. $X$ should be approximation of what you are looking for. What it shows (In[3] and In[4]) is that there is a typo in my formula or this does not converge to what you think it should (in fact that may be true (I am not sure), i.e. in each sixth part you are always off from center (or wherever the mean of that part value is) by constant margin). Hope that helps ;-)
