# Determinant of a random matrix

Given the set $A=\{0,1\}$ of all the real numbers between $0$ and $1$, we can build the square random matrix: $$H_2=\begin{bmatrix}h_{11} & h_{12} \\ h_{21} & h_{22}\end{bmatrix}$$ where the $h_{jk}$ are random real numbers picked in $A$ in the way that every number in $A$ has the same probability to be taken. The determinant of the matrix $H_2$ is defined as: $$det(H_2)=h_{11}h_{22}-h_{12}h_{21}$$ The probability to have $det(H_2)\ge0$ is the same of $det(H_2)\le0$ and I think it's $\frac{1}{2}$. I suppose, if we have a random matrix $H_\infty$ the value of the determinant is $0$. Is it possible to find a formula for the probability of $det(H_n)\ge0$ as a function of $n$?

Thanks in advance for any suggestion.

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I would expect $P(\mathrm{det}(H_n)\geq 0)=\frac{1}{2}$. – Alex Becker Feb 28 '12 at 10:34

If you exchange two rows of your matrix, the determinant will change the sign, while this operation preserves the probability measure. Together with the fact that the probability of $\det H_n =0$ is $0$ (the measure is $0$ as it is of codimension $1$) you get that the probabiity of $\det H_n\geq0$ is $1/2$.
This means as $n\rightarrow\infty$, the probability is still $\frac{1}{2}$. Is it correct? – Riccardo.Alestra Feb 28 '12 at 12:16