I came up with a mental problem in probability and I found out that it's either trivial or more complex than what I thought.
The problem is this:
Take a coin of diameter $D$ and radius $R$, and toss it.
Cover the table with a rectangular tablecloth made of small squares of side $\ell$, with $\ell \geq D$.
What is the probability for the coin to land exactly inside a little square?
If the problem has a trivial solution then I expect it to be this: the probability is the area of the coin over the area of the square:
$$P = \frac{\pi R^2}{\ell^2}$$
But here I would a sort of $\theta$ function such that (in case this hypothesis is missing) the probability is zero when $D>\ell$ hence:
$$P = \frac{\pi R^2}{\ell^2}\theta(\ell - 2R)$$
But here many problems arise: first of all my convention demands that $\theta (\ell - 2R) = 1$ for $2R = \ell$, because as improbable it is, there could be a golden toss by which the coin lands exactly perfectly within the square the side of which is identical to the diameter.
But now I strongly believe I shall consider the whole number of squares of the tablecloth.
But more squares = more probability?
I'm stuck on an either trivial or complicated reasoning.
Any clarification?
You need only think about the square (area $\ell^2$) containing the center of the coin. The coin is entirely within that large square just when its center is in the small square with side $\ell - D$ (area $(\ell - D)^2$) centered in the large square. Assuming the center is uniformly distributed, the probability that the coin is entirely inside is $$ \frac{(\ell - D)^2}{\ell^2}. $$