jmc

Let $p(x)$ be the given polynomial. Notice that $$p(1)=1+(2a)+(2a-2)-(4a+3)-2=0,$$so $1$ is a root of $p(x).$ Performing polynomial division, we then have $$p(x)=(x-1)(x^3+(2a+1)x^2+(4a-1)x+2).$$Notice that $$p(-2)=1\cdot (-8+4(2a+1)-2(4a-1)+2)=0,$$so $-2$ is a root of $p(x)$ as well. Dividing the cubic term by $x+2,$ we then have $$p(x)=(x-1)(x+2)(x^2+(2a-1)x+1).$$Therefore, we want to find the probability that the roots of $x^2+(2a-1)x+1$ are all real. This occurs if and only if the discriminant is nonnegative: $$(2a-1{)}^2-4\ge 0,$$or $(2a-1{)}^2\ge 4.$ Thus, either $2a-1\ge 2$ or $2a-1\le -2.$ The first inequality is equivalent to $a\ge \frac{3}{2},$ and the second is equivalent to $a\le -\frac{1}{2}.$ This shows that all values of $a$ except those in the interval $\left(-\frac{1}{2},\frac{3}{2}\right)$ satisfy the condition. This interval has length $2,$ and the given interval $[-20,18],$ which contains it completely, has length $18-(-20)=38,$ so the probability is $$1-\frac{2}{38}=\boxed{\frac{18}{19}}.$$