I was reading this explanation.
The question in concise terms is
We choose a $n$ tuple of real numbers $(x_1,...,x_n)$ uniformly randomly satisfying $x_i\geq0, ∑x_i=1.$ What is the probability that $\forall x_i: x_i<\frac{1}{2}$
The linked explanation uses the following fact as an assumption.
$P(a_1+⋯+a_{n-1} < k/2)=(\int_{0<x_1,0<x_n,x_1+...+x_n<k/2} dx...dx_{n-1}) / (\int_{0<x_1,0<x_n,x_1+...+x_n<k} dx...dx_{n-1}) = 1/2^{n-1}$
This is the elaboration given.
As you see, we consider geometrically similar sets here. Their linear sizes relate to each other as $1:2$ thus volumes of $k$-th degree will relate as $1:2^k$. We are interested in volumes of degree $n−1$ so needed number is equal to $1/2^{n−1}$. From this we get that probability of impossibility to form a polygon is equal to $n/2^{n−1}$
What does it mean by geometrically similar sets? How can we say that their linear sizes relate as $1:2$? How can we just use that knowledge without any further calculation to say that volumes of $k$-th degreee will realte to as $1:2^k$.
What is the underlying theory being used to answer all these questions. Is it measure theory? I want to rigorously study these. Could you point me to a suitable resource where I can study concepts related to these? To be precise, My Questions is What is it that I should have known so that I would have been comfortable and found the linked explanation a cakewalk?
Could the given ratio been have calculated using a perhaps more mathematically straightforward way?
There's really no need for measure theory, but the notation like $m(A)$ to denoting the size (probability measure) of event $A$ can be convenient.
One can always observe a couple of concrete examples to have a sense of what's going on.
For example, one can visually see when $n = 2$ and $n=3$ that the special threshold of $\frac12$ in each dimension ($X$ for $x_1$, $Y$ for $x_2$, and so on) cuts the whole space into $2^n$ "equal" subspaces.
One can also see that all the arguments to come based on symmetry (being "equal" or similar) will NOT work for any ratios $\sum x_i = r$ other than $r = 1/2$.
In 2-dim, there are $4$ smaller squares inside the unit square, $2$ of which are "similar" to the original "unit square with the event" (off-diagonal $x+y = 1$).
In 3-dim, there are $8$ smaller cubes inside the unit cube, $4$ of which are "similar" to the original unit square with the simplex plane $x+y+z = 1$. While the small triangle in the corner-to-origin part (the central one) is "upside down", really there are only $3$ parts that are exact scaled version of the original "unit cube with event".
A half-scaled version of the original configuration in 2-dim is of size $\frac14$, and a half-scaled config in 3-dim is of size $\frac18$ of the original.
By half-scaled we mean the linear dimension (side length) is halved. The area goes one-fourth and volume goes one-eighth.
Likewise, a half-scaled in $n$-dim is $\frac1{2^n}$ the original This is just basic geometry and should be pretty intuitive.
Note that the green part (the event), which is the edge of the solid simplex, is one dimensional lower. For example, on the right the triangle is divided into $2^{n-1}$ similar parts.
Denote the desired event as $A \equiv \{ x_i < \frac12 ~~\forall\, i\}$ and its complement event as $A^c \equiv \{\text{not all} ~x_i < \frac12\}$.
We will show that the size (measure) of $A^c$ is what you quoted "probability of impossibility to form a polygon", namely, $m(A^c) = \frac{n}{ 2^{n−1} }$.
Then by definition, $m(A) = 1 - m(A^c)$ is what we are after.
Note that $A^c$ always has exactly one of $x_i$ failing the requirement: either it is $x_1 > 1/2$, or it is $x_2 > 1/2$, or ... , or it is $x_n > 1/2$. Due to the definition of $\sum x_i = 1$, there cannot be two simultaneous $x_i > 1/2$ and $x_j > 1/2$ for $i \neq j$.
Formally, $A^c$ consists of $n$ disjoint sub-events: for $i = 1$ to $n$ $$B_i \equiv \{ x_i > 1/2,\, x_j < 1/2 ~~\forall\, j \neq i\}\, , \qquad A^c = \bigcup_i B_i$$
Due to the symmetry ($x_i$ are interchangeable), we thus know that the $B_i$ are "copies" to each other:
$$m(B_1) = m(B_2) = \ldots = m(B_n) \equiv p_B \, , \qquad m(A^c) = n \cdot p_B$$
To figure out $p_B$, one can go through the geometric argument you quoted: each piece is a half-scaled version of the original configuration thus $1/2^{n-1}$ of the original measure (total probability).
Algebraically, we can do this with a bit more rigor.
Shift the failed component $x_i$ to have a new combo within each sub-event $B_i$ (given $i$)
$$\begin{cases} w_i \equiv x_i - \frac12 & \\ w_j \equiv x_j & ~\forall j \neq i \end{cases} \qquad \text{such that} \sum_k w_k = 1-\frac12 = \frac12 ~~\text{and}~~ w_k \in (0,\frac12)~~\forall\, k = 1 \sim n$$
Now we see that each $B_i$ is the half-scaled version of the original event $$m(B_i) = \frac{ m(A) }{ 2^{n-1} }$$ in that comparing with originally for event $A$ we have \begin{align} \sum_k w_k &= \frac12 &&~\text{and} & w_k &\in (0,\frac12) &&~\forall\, k = 1 \sim n \\ \sum_k x_k &= 1 &&~\text{and} & x_k &\in (0,1) &&~\forall\, k = 1 \sim n \end{align} where one should note that each set $w_k$ are mutually independent (even with $i$ given) just like how the original $x_k$ are independent (free to vary within the specified range) up to the degree of $n-1$.
In particular, in $2$-dim the desired event $m(A) = 1 - m(A^c) = 1 - \frac{ n }{ 2^{n-1}} = 1 - \frac{ 2 }{ 2^{2-1}} = 0$ has zero measure, as it is a just a point in the left figure above.
In $3$-dim the desired event is $1 - \frac{ 3 }{ 2^{3-1}} = \frac14$ as the "central" piece in the right figure above.
Please let me know if there's anything that needs to be further clarified.