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?



Nilpotent infinite binary matrices
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.


