smc

We consider an individual one-by-one block. If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius $d$, the area covered by the circles should be $0.5$. Because of this, and the fact that there are four circles, we write $$4\cdot \frac{1}{4}\cdot \pi d^2=\frac{1}{2}$$ Solving for $d$, we obtain $d=\frac{1}{\sqrt{2\pi }}$, where with $\pi \approx 3$, we get $d\approx \frac{1}{\sqrt{6}}\approx \frac{1}{2.5}=\frac{10}{25}=\frac{2}{5}$, and from here, we see that $d\approx 0.4⟹\boxed{0.4}.$ $\text{Note:}$ To be more rigorous, note that $d<0.5$ since if $d\ge 0.5$ then clearly the probability is greater than $\frac{1}{2}$. This would make sure the above solution works, as if $d\ge 0.5$ there is overlap with the quartercircles. $\text{- Emathmaster}$