<s>[INST] Summing random numbers with different domain This should be simple, but my math skills are rusty.
Given the random integers $X_0,\ldots, X_i,\ldots, X_n$, each of which is uniformly distributed on a domain $0,\ldots,J_i,\ldots,J_n$, what is the probability that the sum of the numbers will be greater than or equal to some number $Y$?
I've found a formula that works when the domain is constant (e.g. using die rolls, every integer can be only 1, 2, 3, 4, 5, or 6), but what about when the domain is different for each (e.g. one is determined by $D_6$ die and the other by a $D_{10}$ die, so $1, \ldots, 6$ and $1, \ldots, 10$)?
 [/INST] I will show you how to do the two case example, and hopefully you can then do it for a three case scenario. So, suppose that $X$ models the first dice throw, and therefore has a discrete uniform distribution on $[1,6]$, while $Y$ models the second dice throw, thus being discrete uniform on $[1,10]$. We are interested in describing $Z=X+Y$. The first observation is that the support of $Z$ is $[1+1, 6+10] = [2,16]$.
The second observation (not that obvious) is that, because of the different supports, the distribution of $Z$ will look like a trapezoid. Why? Look at the discrete convolution formula, which gives the distribution of the $Z$, assuming that $X$ and $Y$ are independent (which they are, since we look at dice rolls). The formula states that
$$
P(Z=z) = \sum_{k=0}^z P(X=k) P(Y=z-k)
$$
since when $X=k$, clearly $Y=z-k$. So:
$$P(Z=2) = P(X=1)P(Y=2-1) = \frac{1}{6}\frac{1}{10} = \frac{1}{60}\\
P(Z=3) = P(X=1)P(Y=3-1) + P(X=2)P(Y=3-2) = \frac{2}{60} \\
\vdots\\
P(Z=7) = P(X=1)P(Y=7-1) + \dots +P(X=6)P(Y=7-6) = \frac{6}{60}.
$$
We conclude that for $2\le n \le 7, P(Z) = \frac{n-1}{60}$. Now,
$$
P(Z=8) = P(X=1)P(Y=8-1) + \dots + P(X=6)P(Y=8-6) = \frac{6}{60}\\
P(Z=9) = P(X=1)P(Y=9-1) + \dots + P(X=6)P(Y=9-6) = \frac{6}{60}\\
\vdots\\
P(Z=11) = P(X=1)P(Y=11-1) + \dots + P(X=6)P(Y=11-6) = \frac{6}{60}.
$$
At this point, since $Y$ is capped at a maximum of 10, we can't form $Z=12$ when $X=1$ and therefore conclude that for $ 8\le n \le 11, P(Z) = \frac{6}{60}$. Finally,
$$
P(Z=12) = P(X=2)P(Y=12-2) + \dots + P(X=6)P(Y=12-6) = \frac{5}{60}\\
P(Z=13) = P(X=3)P(Y=13-2) + \dots + P(X=6)P(Y=13-6) = \frac{4}{60}\\
\vdots\\
P(Z=16) = P(X=6)P(Y=16-6) = \frac{1}{60}.
$$
Putting everything together we get that:
$$
P(Z=z) = 
\begin{cases}
\dfrac{z-1}{60} & 2 \le z \le 7 \\
\dfrac{6}{60}   & 8 \le z \le 11 \\
\dfrac{17-z}{60} & 12 \le z \le 16 \\
\end{cases}
$$
To do your 3 case example, pick $Z$ as described above, and pick $U$ to have any distribution you want. Then see if you can write $Z+U$ using the same type of argument.
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