Cutting dna using restriction endonuclease
Welcome to Biology.SE. I think this question would be better on-topic on math.SE. 

Assuming the chromosome is either 4 bases long or infinitely long, then the answer is simply $\left(\frac{1}{4}\right)^4$. However, chromosomes are not infinitely long so the result should be (negligibly) lower than $\left(\frac{1}{4}\right)^4$. If chromosomes are only 3 nucleotide long, then of course, the probability will be $0$. If it is 4 bases long the probability is $\left(\frac{1}{4}\right)^4$. If it is 5 bases long the probability is $\frac{2}{4^5} = \frac{1}{2}\left(\frac{1}{4}\right)^4$.

Consider this numerical approximation, to ensure the result is correct with this simple one liner in R

n = 1e7
length(gregexpr('abcd', paste(letters[sample(1:4,n,replace=TRUE)], collapse=""), perl=TRUE)[[1]])/n
[1] 0.0039041


which is really close to $\left(\frac{1}{4}\right)^4 ≈ 0.00390625$. 

So, in a genome of 3 giga bases, that represents an expected $\frac{3\cdot 10^9}{4^4} = 11,718,750$ (or ~11 millions) such locations where Alu 1 can cut.