Real eigenvalues by chance

Idea. Rewrite the discriminant as a sum and difference of squares of independent standard normals. The question becomes whether one squared normal is smaller than the sum of two others, and that has a neat answer.

Step 1: the discriminant. The characteristic polynomial ${\lambda }^2-(a+d)\lambda +(ad-bc)$ has real roots exactly when $(a-d{)}^2+4bc\ge 0$.

Step 2: independent pieces. Let $$u=\frac{a-d}{\sqrt{2}},v=\frac{b+c}{\sqrt{2}},w=\frac{b-c}{\sqrt{2}}.$$ These are independent standard normals (orthogonal combinations of independent normals). Since $4bc=(b+c{)}^2-(b-c{)}^2=2v^2-2w^2$, the condition becomes $$2u^2+2v^2-2w^2\ge 0⟺w^2\le u^2+v^2.$$

Step 3: the probability. $R^2=u^2+v^2$ is exponential with mean 2: $\mathbb{P}(R^2\ge t)=e^{-t/2}$. Conditioning on $w$, $$\begin{array}{rl}\mathbb{P}(R^2\ge w^2)& =\mathbb{E}\left[e^{-w^2/2}\right]={\int }_{-\infty }^{\infty }\frac{e^{-w^2}}{\sqrt{2\pi }}dw\\ & =\frac{\sqrt{\pi }}{\sqrt{2\pi }}=\frac{1}{\sqrt{2}}.\end{array}$$

So both eigenvalues are real with probability $\frac{1}{\sqrt{2}}\approx 0.707$.

Remarks. A symmetric matrix always has real eigenvalues; the risk comes from $b$ and $c$ having opposite signs. With entries uniform on $[0,1]$ the answer is 1, since then $bc\ge 0$.