A square appears randomly within a square of ten time its area. What is probability that the smaller square contains the larger square's center? The alignment relative to the larger square can be anywhere between 0° to 90° (0° or 90° being identical orientation to the larger square, 45° being all the way skewed), and the probability distribution of the alignment is based on percentage of total possibility given the constraint guaranteeing that it will fit within the larger square.
The largest discrete probability, for an alignment of 0°=90°, should be simple enough to figure-out‡. The smallest discrete probability, for alignment of 45°, is a bit trickier but straightforward enough.Edit:The off-kilter version would be more likely to contain the center, since it has less possible places to exist and a larger proportion of them are the center. Intuitively this should correspond directly (though not necessarily uniformly) to what I am totally clueless about, which is:

*

*how to determine the probability  distribution for between 0° to 45°,

‡ 2. how this continuum corresponds to the probability of the original problem (e.g. {1/17?} for 0° up to {2/9?} for 45° back down to {1/17?} for 90°, along finite part of some sort of curve function presumably),
and 3. how then to apply these equations formulated into the original problem to find the final probability that a square true-randomly generated wholly within a larger square with sides √(10)-times longer would surround (or contain exactly on an edge or corner) the center of the larger square.
‡ I haven't calculated the maximum or minimum ‘discrete’ probabilities (let alone any inbetween) yet. Those guesstimate values are placeholders. I am especially interested in the processes required to solve this, and appreciate any insight or clues.
 A: Let me see if I understand you problem.
You are looking at a square $S = [-R/2, R/2]^2$ of side length $R$ (for simplicity centred at the origin. I will write $C(x;\theta)$ for the smaller square of side length $r$ centred in $x\in\mathbb{R}^2$ at rotation $\theta$. You want to know what the probability is of the event that $0\in C(x;\theta)$ under the constraint $C(x;\theta)\subseteq S$.
For simplicity, we take $r \ll R$ in a suitable sense ($=$ so that everything works just fine). The first thing I would like to do is to wrap it up mathematically. For this, we define our probability space to be
$$
\Omega := \{(x,\theta)\in \mathbb{R}^2\times [0,\pi/4]\;\vert\; C(x;\theta)\subseteq S\}.
$$
Note that I only consider angles up to 45 degrees, because the situation is symmetric for angles between 45 and 90 degrees. To model "random appearance", we take the uniform distribution on $\Omega$.
First things first: Can we describe $\Omega$? For this, fix the angle $\theta$. Write $h(\theta)$ for the height ($=$ the width) of the skewed square $C(x;\theta)$. Then
$$
h(\theta) = \cos\left(\dfrac{\pi}{4} - \theta\right)\cdot\sqrt{2}\cdot r.
$$
We conclude that
$$
\Omega = \{(x,\theta)\in \mathbb{R}^2\times [0,\pi/4]\;\vert\; \Vert x\Vert_\infty \leq (R - h(\theta))/2\}.
$$
In particular,
$$
m(\Omega) = \int_0^{\pi/4} \big{(} R - r\cos(\theta)\sqrt{2}\big{)}^2\mathrm{d}{\theta} = \dfrac{\pi}{4}R^2 - 2Rr + \dfrac{2+\pi}{4}r^2,
$$
where I wrote $m$ for the Lebesgue measure on $\mathbb{R}^2$.
Now to the event we are interested in:
$$
A = \{(x,\theta)\in \Omega\;\vert\; 0\in C(x;\theta)\}.
$$
Since we took $r\ll R$ we do not need to worry about whether the squares touching $0$ are indeed included in $S$. We further simplify the problem by noticing that the size of the set of suitable angles $\theta$ for a given centre $x$ only depends on $\rho :=  \Vert x\Vert_2$. So let's fix $0\leq \rho \leq \dfrac{\sqrt{2}}{2}r$. (The upper bound is given by the half diagonal.) To get some visual intuition, take $x = (\rho, 0)$. Let us start in the most "favorable" position, i.e. at $\theta=\pi/4$. The question is by how much we are allowed to skew the square. Note that the point closest to the origin is at distance
$$
d(\theta) = \dfrac{r}{2\cos\theta}.
$$
Hence, we need
$$
\cos\theta \leq \dfrac{r}{2\rho}.
$$
This means that we have the constraint $\cos^{-1}\left(\dfrac{r}{2\rho}\right)\leq \theta \leq \dfrac{\pi}{4}$ for $\rho \geq r/2$ and bo constraint for $\rho < r/2$. This means that
$$
m(A) = \dfrac{\pi^2r^2}{8} - 2\pi\int_{r/2}^{\sqrt{2}r/2} \rho \cos^{-1}\left(\dfrac{r}{2\rho}\right)\mathrm{d}\rho = \dfrac{\pi^2r^2}{8} - \dfrac{\pi(\pi - 2)r^2}{8}.
$$
If I have not made any mistake in any of those integrals, we should get the probability
$$
\dfrac{m(\Omega)}{m(A)} = \text{something very ugly}.
$$
Edit: The following comments try to give a bit more intuitive insight into the notation.
First, note that the uniform distribution on a interval means that the probability of the event is the "length" or "size" of the event, divided by the total length of the interval. The same is going on here: Once we have understood that we want to take $(x,\theta)$ uniformly in $\Omega$, the probability of an event $A$ is the "volume" (because we operate in 3 dimensions now) of $A$, divided by the volume of $\Omega$.
Now, to obtain the volume of a set, we usually integrate, especially if the bodies are complex as in this case. But in the end, it is only "measuring the volume". The $m$ I use is just "the volume of". This "usual" volume is called Lebsegue measure, because there other ways to make sense of the word "volume", but in this case this doesn't matter.
I don't think that there is a straightforward way to obtain the explicit expression of the result without using integrals, i.e. by simply using geometric surface/volume formula, but you can always try!
For the other notation: $\Vert (x,y)\Vert_2 = \sqrt{x^2 + y^2}$ is the usual Euclidean distance, $\Vert (x,y)\Vert_\infty = \max\{\vert x\vert, \vert y\vert\}$ is the so-called infinity norm. (There are $\Vert\cdot\Vert_p$-norms for any $p\geq 1$.) When I write $r\ll R$, I mean that $r$ is a lot smaller than $R$. Here, we need this assumption when figuring out how we may describe $A$.
Also, I mainly used WolframAlpha for those horrendous integrals.
Edit 2: There is a lot easier way to calculate $m(A)$. Indeed, mjw (https://math.stackexchange.com/q/3966261) correctly points out that, for any given angle $\theta$, the centre must be in a square of side length $r$. That means that
$$
m(A) = \dfrac{\pi}{4}r^2.
$$
I corrected the above mistake.
A: Just to solve the simpler problem to get an approximate answer to check another "exact" answer.  I don't think rotating the square will give an answer too far off ...
Suppose the larger square is $\sqrt{10}$ by $\sqrt{10}$ inches, and the smaller square is $1$ square inch.  Then the center of the smaller square will be uniform within a square with side of length $\sqrt{10}-1$.
The center of the smaller square then needs to be within a square of one square inch near the center square.  Thus
$$P = \frac{1}{(\sqrt{10}-1)^2}= \frac{1}{11-2\sqrt{10}} \approx 0.214.$$
