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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 $\frac {13}9-\frac{40}{9 \pi}\approx 0.0297338$

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You should describe things better, ¿how are the sizes? ¿Is the 1x1 square in the fusrt square the same size as the 1x1 square in the 3x3 grid?

If the answer to that is yes, then the answer to the problem is 0, as there's a continuum of possibilites.

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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

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