If diagonalizable matrices are not dense over $\Bbb R$, how common are they?

Question

The link: The diagonalizable matrices are not dense in the square real matrices says that diagonalizable matrices are dense over $\Bbb C$ but not over $\Bbb R$. If that's the case, then the logical question is, what percentage of matrices over $\Bbb R$ are diagonalizable? The answer is going to be "it depends", so let's make it concrete. We pick each element of an $n \times n$ matrix from i.i.d uniform distributions over $[-1,1]$. What is the chance we'll get a diagonalizable matrix (over the reals)?

Answer 1

For a $2×2$ matrix, I think it is $49/72$. You want real eigenvalues.
$$(\lambda-a)(\lambda-d)-bc=0\\ \lambda^2-(a+d)\lambda+ad-bc=0\\ (a-d)^2+4bc\gt0$$ For set values of $a$ and $d$, the values of $b$ and $c$ are bounded by a hyperbola. The area within the square $-1\lt b,c\lt1$ is $$2+(a-d)^2/2-(a-d)^2\ln(|a-d|/2)$$ Integrate that over the region $-1\lt d\lt a\lt 1$. Divide by $8$ which is the total volume integrated over, half of $[-1,1]^4$, and the result is $49/72$.
I suppose the fraction is less for larger matrices. The proportion for a million $3×3$ was about $13/40$ and for a million $4×4$ was about $0.1080$