Let $A \in \mathbb{R}^{11 \times 11}$ and it's elements are form set $\{ -1,1 \}$. $\mathbb{P}(-1) = \mathbb{P}(1) = 0.5$. What is a probability to get such a matrix, that $\det A > 4000$? 


I have such an idea:


It's better to find the $\mathbb{P}(\det A \leq 4000)$:


$F\_{\det A} = \mathbb{P}(\det A \leq 4000) = \mathbb{P} \left( \sum\limits\_{\alpha=(\alpha\_1,\alpha\_2,...,\alpha\_{11})} (-1)^{\sigma(\alpha)}a\_{1\alpha\_1}a\_{2\alpha\_2}...a\_{11\alpha\_{11}} \leq 4000 \right)$.


But I have several problems:


1. $a\_{ij}$ is a random value, but not all of $a\_{1\alpha\_1}a\_{2\alpha\_2}...a\_{11\alpha\_{11}}$ are random values.
2. How to calculate $\mathbb{P}(\xi\_1+...+\xi\_{11} \leq 4000)$? I realize that $\xi\_1+...+\xi\_{11}$ is a hyperplane...


P.S. Is there any programm that can make N trials generating random matrixes?



Determinant of matrix from set {-1, 1}
Thank you, @Noam, for your idea!


**Statement 1:** Let $A \in \mathbb{R}^{n \times n}$ and $a\_{ij} \in \{-1, 1\}$, then
$\mathbb{E}(\det A) = 0$


**Proof:** It's easy to see, that because of every $a\_{ij}$ is a random value:
$ \mathbb{E}(\det A) = \mathbb{E} \left(
 \sum\limits\_{\alpha=(\alpha\_1,\alpha\_2,...,\alpha\_{n})} (-1)^{\sigma(\alpha)}
 a\_{1\alpha\_1}a\_{2\alpha\_2}...a\_{n\alpha\_{n}}
 \right) 
 = \sum\limits\_{\alpha=(\alpha\_1,\alpha\_2,...,\alpha\_{n})} (-1)^{\sigma(\alpha)}
 \mathbb{E}a\_{1\alpha\_1} \mathbb{E}a\_{2\alpha\_2} ... \mathbb{E}a\_{n\alpha\_{n}} = 0,$
as $\displaystyle \mathbb{E}a\_{ij} = (-1) \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} = 0.$ $\square$


**Statement 2:** Let $A \in \mathbb{R}^{n \times n}$ and $a\_{ij} \in \{-1, 1\}$, then
$\mathbb{E}((\det A)^2) = n!$


**Proof:**
$
 \mathbb{E}(\det A)^2 = \mathbb{E} \left(
 \sum\limits\_{\alpha=(\alpha\_1,\alpha\_2,...,\alpha\_{n})} (-1)^{\sigma(\alpha)}
 a\_{1\alpha\_1}a\_{2\alpha\_2}...a\_{n\alpha\_{n}} \right)^2 
 = \mathbb{E} \left(
 \sum\limits\_{\alpha=(\alpha\_1,\alpha\_2,...,\alpha\_{n})} (-1)^{\sigma(\alpha)}
 a\_{1\alpha\_1}a\_{2\alpha\_2}...a\_{n\alpha\_{n}}
 \sum\limits\_{\beta=(\beta\_1,\beta\_2,...,\beta\_{n})} (-1)^{\sigma(\beta)}
 a\_{1\beta\_1}a\_{2\beta\_2}...a\_{n\beta\_{n}}
 \right)
 = \mathbb{E} \left(
 \sum\limits\_{\alpha=(\alpha\_1,\alpha\_2,...,\alpha\_{n})}
 \sum\limits\_{\beta=(\beta\_1,\beta\_2,...,\beta\_{n})} (-1)^{\sigma(\alpha) + \sigma(\beta)}
 a\_{1\alpha\_1}a\_{2\alpha\_2}...a\_{n\alpha\_{n}} \cdot
 a\_{1\beta\_1}a\_{2\beta\_2}...a\_{n\beta\_{n}} \right)
$


If permutations $\alpha$ and $\beta$ are not the same, then there is $a\_{ij}$ which have power 1, and its $\mathbb{E}a\_{ij} = 0$. Then equation above can be rewritten:


$
 \mathbb{E}(\det A)^2
 = \sum\limits\_{\alpha=(\alpha\_1,\alpha\_2,...,\alpha\_{n})} (-1)^{2\sigma(\alpha)}
 \mathbb{E}a^2\_{1\alpha\_1} \mathbb{E}a^2\_{2\alpha\_2}... \mathbb{E}a^2\_{n\alpha\_{n}} = n!,
$


as $\displaystyle \mathbb{E}a\_{ij} = (-1)^2 \cdot \frac{1}{2} + 1^2 \cdot \frac{1}{2} = 1$, and the number of permutations $\alpha=(\alpha\_1,\alpha\_2,...,\alpha\_{n})$ is equal to $A\_n^n=n!$ $\square$


Then from the statements 1 and 2, we have that there are matrixes with determinant greter than 0 and $\mathbb{E}((\det A)^2) = 11!$, and correspondingly we have an **assesment**:


$\exists A: \det A \geq \sqrt{11!} \approx 6317.9 > 4000.$


The generator for matrixes in Wolfram Mathematica:



```
While[True,
 A = Table[RandomInteger[]*2 - 1, {i, 11}, {j, 11}];
 If[Det[A] > 4000, Print[MatrixForm[A]] Print[Det[A]] Break[]]
]

```

