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In fact, it is an expansion of this problem. But I restricted the elements of the matrix to be integers only. Obviously this probability is related to the range of random $\text{item}\in[0,n]$.

For the 2*2 matrix, I solved some probability values related to $n$ by math software violencely: enter image description here

But I don't know what the general formula is. For the 3*3 matrix, I only found 5 terms: enter image description here

But how to calculate the analytical solution of this problem?

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  • $\begingroup$ Why do you believe that this problem has an "analytical solution"? $\endgroup$ Jan 15, 2022 at 22:45
  • $\begingroup$ @BenGrossmann I'm not sure, but it might be worth trying to find out. I've been thinking about this off and on for years, but I'm so bad at math that I'm asking here for help. $\endgroup$
    – mayi
    Jan 15, 2022 at 22:48
  • $\begingroup$ It is astronomically unlikely that there is a "nice" answer to this question. Even the answer for the case of $n=1$ (for different matrix sizes) has no known analytic form. $\endgroup$ Jan 15, 2022 at 22:53
  • $\begingroup$ @BenGrossmann In my post I just don't limit the range of random values, which I think is much easier than not limiting the size of the matrix. As you can imagine, if the random values are continuous, unbounded size means that there are an infinite number of integral variables, but here I only have 4 $\endgroup$
    – mayi
    Jan 16, 2022 at 6:24
  • $\begingroup$ I suppose you have a point. At the very least, I suspect that there might be a nice general approach to the $2 \times 2$ case, if not an analytic formula $\endgroup$ Jan 16, 2022 at 6:31

1 Answer 1

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Well, the analytical solution about $\small 2*2$ case is:

$$\text{probability}(n)=\frac{2 \sum\limits _{r=1}^n \phi (r) \lfloor\frac{n}{r}\rfloor^2+(n+1) (3 n+1)}{(n+1)^4}$$

And the discussion is here. I just carried it.

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