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Suppose that there are $k$ random discrete integers $x_1, x_2, \dots, x_k$ that are each uniformly chosen from their respective sample space sets. If it helps, let us denote each random discrete integer variable's respective sample space set as $s_1, s_2, \dots, s_k$.

Then, you want to find the probability of $\sum\limits_{i = 1}^{k} x_i \le n$ for some $n \in \mathbb{Z}$.

How would you find this probability, and is there any good intuition behind it? (I have only a basic knowledge of statistics.)

Niche example: What is the probability that the sum of $3$ standard dice is less than $n$?

Rodrigo de Azevedo
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Gigi Bayte 2
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    This is probability question. I doubt it has relation to statistics. – kludg Feb 15 '23 at 10:22
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    The case of $3$ standard dice has been adressed here and more generally here. For your general problem, can the $s_i$ be arbitrary sets of integers? (If they're intervals, more could be said.) – joriki Feb 15 '23 at 10:27
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    If $n$ were negative, then the probability is obviously equal to zero because the sum of integers can never be smaller than zero. So you could consider also $n \in \mathbb{N}$. – WarreG Feb 15 '23 at 10:31
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    Assuming independence, write the probability generating function for each, take the product of these (i.e. the pgf for the sum) and expand, remove the terms with exponents greater than or equal to $n$, and then substitute $x=1$ – Henry Feb 15 '23 at 10:47
  • @joriki Thanks for those links! Also, I was assuming that $s_i$ could be arbitrary sets of integers, but I would be interested in the conclusions that could be drawn if each of $s_i$ consists of a contiguous integer interval. – Gigi Bayte 2 Feb 15 '23 at 11:18
  • @Henry That seems like a good approach, but I guess I'm just having trouble connecting the dots/visualizing that solution. Could you perhaps provide an example? – Gigi Bayte 2 Feb 15 '23 at 11:19
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    Pgf for one fair die: $\frac{1}6x^1+\frac{1}6x^2+\frac{1}6x^3+\frac{1}6x^4+\frac{1}6x^5+\frac{1}6x^6$. Cube that for the sum of three fair dice and you get something starting $\frac{1}{216}x^3+\frac{1}{72}x^4+\frac{1}{36}x^5+\frac{5}{108}x^6+\frac{5}{72}x^7+\cdots$. If you want the probability of being less than $n=7$, you drop the higher terms and then let $x=1$ so $\frac{1}{216}+\frac{1}{72}+\frac{1}{36}+\frac{5}{108}=\frac{5}{54} \approx 0.0926$ – Henry Feb 15 '23 at 11:28
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    I'm not sure whether what I could contribute would deserve to be called "conclusions" :-) There's a more general form of the "balls in bins with limited capacity" solution (described here and linked to in the second answer linked to above) that gives the answer by inclusion–exclusion, but if Henry's approach with probability generating functions (which doesn't require the sets to be intervals) is suitable for you, that might not be helpful. Let me know whether you want me to post it as an answer. – joriki Feb 15 '23 at 11:42
  • @Henry That makes sense. Thank you! – Gigi Bayte 2 Feb 15 '23 at 12:27
  • @joriki Ah, that is very informative. Thanks! I'd still upvote it if you made it into an answer. Perhaps you can put that information plus what Henry put into a comprehensive answer? – Gigi Bayte 2 Feb 15 '23 at 12:29

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