What's the name of this triangle?

Recall $\displaystyle(1+y)^n=\sum_{k=0}^n{n \choose k}y^k.$

Then $\displaystyle(1-x^2)^n=\sum_{k=0}^n{n \choose k}(-1)^kx^{2k}$ . Note $x^n$ has positive coefficient exactly when $n\equiv 0\pmod{4}$ .

Moreover, $\displaystyle (1-x^2)^n= (1+x)^n(1-x)^n=\sum_{k=0}^n{n \choose k}x^k\sum_{k=0}^n{n \choose k}(-1)^kx^k$ . The coefficient of $x^n$ is then

$\displaystyle\sum_{a+b=n}(-1)^{b}{n \choose a}{n \choose b}=\sum_{k=0}^n(-1)^k{n \choose k}^2$ .

Thus, $\displaystyle \sum_{k=0}^n(-1)^k{n \choose k}^2$ is positive exactly when $n$ is divisible by $4$ . Therefore, the answer is $\frac{250}{1000}=\boxed{0.25}$ .