What theorem can be used to explain this occurrence?

Question

I'm not highly versed in research-level mathematics. I do conduct research in cellular biology. I was wondering if you could help me find a term that can be referred to when discussing the following concept:

Say you have 2 circles, one within the other. If you throw an object, as the size of the object decreases the probability of it lying entirely in the inner circle increases.

This is rather obvious, but is there a theorem/axiom/postulate or even a field of mathematics I can reference when discussing this idea.

Answer 1

[Edit 1: it seems I decided to call the object a disk for some reason. Considering a non-disk will mess up thought 1 but not thought 2.]

[Edit 2: If I understand you right, then see my answer below. But perhaps I understand you wrong. If you are conditioning on the object lying entirely in the outside circle, then my reasoning below isn't right, but it could probably be fixed?]

No. But here are two and a half thoughts to help?

Thought 1: If the inner circle has radius $R$ and the disk has radius $r$, then the thing lies entirely in the inside circle if and only if the center lies in a circle of radius $R-r$. So as $r$ decreases, the probability increases.

Thought 2: This one's more subtle, but it's based on the same thinking as above. Consider we throw the disk. I don't see where it landed (but perhaps you do). I then think about shrinking the disk but keeping the center in the same spot it landed. If I'm hoping that the thing is entirely inside the inner circle, I might as well shrink the disk because I've got literally nothing to lose [either it's already inside, in which case shrinking will still keep it entirely inside; or it's not already entirely inside, in which case I have nothing to lose].

This second idea is like a "coupling" idea, you could say. It's not an axiom or theorem so much as a way to argue about certain probabilities by splitting the random process into easy to understand pieces (here, we first throw the disk, then we think about changing the size of the disk).

Thought 2.5: We do thought 2, but then ask me if I want to make the inner circle bigger, which is effectively the same as making the object smaller. Same reasoning as before.

If you want to call it something I would perhaps say that one event is contained in the other [thinking along the thought 2 lines]. And then you could say it's because probability is monotone (this is essentially an axiom of probability). But this sounds too fancy for anybody who needs any reference in the first place.

[All the same, you could say "by monotonicity of probability" if you're itching to say something]

Answer 2

I realized I cannot fit my answer in the comment space.

As several people pointed out, the concept of "size" is too vague for a mathematician. Consider first the simplest case when the two objects are coins. In this case the concept of "size" in unambiguous: size is the radius of the coin. So we have two coins $S$ (small) and $L$ (large) of radii $r<R$. Suppose that we have two concentric circles circle in the plane of radii $\underline{r} <\underline{R}$. Assume their common center is the origin of the plane. $\newcommand{\ur}{\underline{r}}$ $\newcommand{\uR}{\underline{R}}$ The location of a given coin is specified by its center.

The concept of randomness is very important in these issues and this is where it gets tricky. I will first assume that you're comparing coins with centers located in the bigger circle.

The coin $S$ will fit inside the small circle if its center is located at a distance $\ur-r$ from the origin. Thus the probability that such a coin fits inside the small circle is

$$ p_s=\frac{\pi (\ur-r)^2}{\pi\uR^2}=\frac{(\ur-r)^2}{\uR^2}. $$

Similarly, the probability that the larger coin $L$ fits inside the smaller circle is

$$ p_L= \frac{(\ur-R)^2}{\uR^2}. $$

Since $r<R$ we have $p_s<p_L$ as you have correctly guessed.

Now let me change my assumption. Suppose that I only look at coins that fit inside the bigger circle. A large coin fits inside the big circle if its center is located at a distance $\uR-R$ from the origin, and a small coin fits inside the big center if its center is at a distance $\uR-r$ from the origin. The probability that a small coin that fits inside the big circle also fits inside the small circle is

$$ p_s= \frac{(\ur-r)^2}{(\uR-r)^2}. $$

Similarly, the probability that a large coin fits inside the small circle is

$$ p_L= \frac{(\ur-R)^2}{(\uR-R)^2}. $$

It is less obvious, but again, $p_s<p_L$.

Suppose now that we have two irregularly shaped coins $S$ (small) and $L$. I will say that $L$ is larger than $S$, if we can rigidly move $S$ inside $L$. This involves rotating and and translating the irregular coin $S$ until it fits a region previously occupied by $L$.

Suppose that we look at all the positions of $L$ and $S$ inside the big circle. We denote by $p_L$ the fraction of those positions corresponding to $L$ being inside the small circle. Define $p_s$ in a similar fashion, as the fraction of those positions of $S$ corresponding to $S$ being inside the small circle. Your claim is that $p_S<p_L$. I am inclined to believe that it is true, but I cannot think of an argument.