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Let $\text{Mat}(\mathbb{N},\{0,1\})$ be the set of all maps $A:\mathbb{N}\times\mathbb{N}\to \{0,1\}$. We define a matrix multiplication for $A, B\in \text{Mat}(\mathbb{N},\{0,1\}$) and $m,n\in\mathbb{N}$ by $$(A B)(m, n) = 0 \text{ if } \sum_{i=1}^\infty A(m, i)B(i, n) < \infty \text{ and } (AB)(m, n) = 1 \text{ otherwise.}$$ We call $A\in\text{Mat}(\mathbb{N},\{0,1\})$ nilpotent if $A^N$ is the constant $0$-matrix.

If every entry in some matrix $A\in\text{Mat}(\mathbb{N},\{0,1\})$ is $0$ with a probability of $50\%$, what is the probability that $A$ is nilpotent?

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  • $\begingroup$ Is the product you're defining associative? $\endgroup$
    – Leonid Petrov
    Commented Nov 25, 2016 at 11:00
  • $\begingroup$ That's actually a very good question - I haven't figured it out yet. Do you want to ask it as a separate question here, or shall I do it? $\endgroup$
    – Dominic van der Zypen
    Commented Nov 25, 2016 at 13:41
  • $\begingroup$ I'm fine with you asking that. How about the answer below? Indeed, i.o. the terms in the sum will be equal to 1 $\endgroup$
    – Leonid Petrov
    Commented Nov 25, 2016 at 17:57

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I'm assuming you mean entries are independent and 0 or 1 with probability .5 each. I think the AA is identically 1, as $P(A_{ij} = 1, A_{jk} = 1) = \frac 14 $ as long as $ i \ne k$ and therefore happens i.o with probability 1.

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