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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
up vote 2 down vote accepted

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 $1/2$.

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This means as $n\rightarrow\infty$, the probability is still $\frac{1}{2}$. Is it correct? – Riccardo.Alestra Feb 28 '12 at 12:16

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