Probability of a square landing within squares on a grid

A 2 x 2 square is tossed randomly on a grid of 3 x 3 squares. What is the probability that the 2 x 2 square falls completely within one of the 3 x 3 squares?

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You need to be careful to specify how the landing point is chosen randomly. In this case it is natural to consider that the center point of the $2 \times 2$ square is chosen uniformly in a $3 \times 3$ square, then the rotation angle $\theta$ of the side is chosen uniformly in $[0, \frac \pi 2)$, and ask the probability that the $2 \times 2$ square does not protrude outside the $3 \times 3$.
If you follow this procedure, the horizontal and vertical extent of the $2 \times 2$ square is $2(\cos \theta + \sin \theta)$. For a given $\theta$ the center must be in a region of size $3-2(\cos \theta + \sin \theta)$ square, so the chance is $\frac{(3-2(\cos \theta + \sin \theta))^2}9$. Averaging over $\theta$, Alpha gets $\frac {13}9-\frac{40}{9 \pi}\approx 0.0297338$
I don't see mention of a $1 \times 1$ square in the question. If the thrown square is smaller than the grid, as I believe is intended, the probability is greater than zero. –  Ross Millikan Jan 11 '13 at 23:48