Let $0\leq a<b$. Define the subset of $[a,b]^n$ by $$ X=\{(x_1,\cdots,x_{n-1})\mid b^{2n}\geq x_1^{2(n-1)}\geq\cdots\geq x_{n-1}^2\geq a\} $$
What is the probability that a uniformly distributed random-vector in $[a,b]^n$ is an element of $X$? ie: Obeys that ordering?
Comment: So far..I only can note that: If the powers of $2$ are omitted that this quantity is equal to $\frac1{(n-1)!}$.
This is difficult because of the descending powers, but it can be done inductively. First, note a couple of boundary cases:
To focus on the induction argument, let's simplify things for now by letting $a=0$ and $b=1$.
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Case $\mathbf{a = 0, \ b = 1 }$:
Take any $c \in [0,1]$, then: $$ P(c \geq x_{1}^{2} \geq 0) = c^{1/2} $$ Here's the inductive step: $$ P(c \geq x_{1}^{4} \geq x_{2}^{2} \geq 0) = \int_{0}^{c^{1/4}} P(t^{4} \geq x_{1}^{2} \geq 0) dt $$ That is, we are integrating $x_{1}$ from $0$ to $c^{1/4}$. This evaluates to:
$$= \int_{0}^{c^{1/4}} t^{4(1/2)} dt = \left[ t^{4(1/2 + 1/4)} \right]_{0}^{c^{1/4}} \cdot \frac{1}{4(1/2 + 1/4)} = c^{1/2 + 1/4} \cdot \frac{1}{3}$$
(I am purposely not simplifying the fractions, they will reveal a pattern.)
Let's do one more iteration before figuring out the inductive pattern. $$ P(c \geq x_{1}^{6} \geq x_{2}^{4} \geq x_{3}^{2} \geq 0) = \int_{0}^{c^{1/6}} P(t^{6} \geq x_{1}^{4} \geq x_{2}^{2} \geq 0) dt $$ Which evaluates to: $$ = \int_{0}^{c^{1/6}} t^{6 \cdot (1/2 + 1/4)} \cdot \frac{1}{3} dt = \left[ t^{6(1/2 + 1/4 + 1/6)} \right]_{0}^{c^{1/6}} \cdot \frac{1}{3} \cdot \frac{1}{6(1/2 + 1/4 + 1/6)} = c^{1/2 + 1/4 + 1/6} \cdot \frac{1}{3} \cdot \frac{1}{5.5}$$
Okay, so here is the pattern when $a=0$ and $b=1$: $$ P(c \geq x_{1}^{2(n-1)} \geq \ldots \geq x_{n-1}^2 \geq 0) = c^{Q_{n}} M_{n} $$ Where: $$Q_{n} = \sum_{k=1}^{n-1} \frac{1}{2k} \quad \text{and} \quad M_{n} = \prod_{k=1}^{n-1} \big( 2(k+1)Q_{k} + 1 \big)^{-1} $$ This follows inductively because: $$Q_{n} = Q_{n-1} + \frac{1}{2(n-1)} \quad \text{and} \quad M_{n} = M_{n-1} \big( 2nQ_{k} + 1 \big)^{-1} $$
Setting $c = b^{2n} = 1$, yields:
$$ P(b^{2n} \geq x_{1}^{2(n-1)} \geq \ldots \geq x_{n-1}^2 \geq 0) = b^{2nQ_{n}} M_{n} = M_{n}$$
Hence, $M_{n}$ is the answer when $a = 0$ and $b = 1$!
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Case $\mathbf{a = 0, \ b \leq 1}$: We have to scale by $b^{-1}$ every time we integrate, so:
$$ P(b^{2n} \geq x_{1}^{2(n-1)} \geq \ldots \geq x_{n-1}^2 \geq 0) = b^{2nQ_{n}} M_{n}b^{1-n} $$
This can be rewritten as the final answer: $b^{1+n(2Q_{n} - 1)} M_{n}$.
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Case $\mathbf{a \geq 0, \ b \leq 1}$: Nothing conceptual has changed, just the lower integration bound is now: $a^{1/2}$, $a^{1/4}$, $a^{1/6} \ldots$. Also, we have to scale by $(b-a)^{-1}$ every time we integrate. $$ P(c \geq x_{1}^{2} \geq a) = \left[c^{1/2} - a^{1/2}\right] (b-a)^{-1}, \quad c \in [a,b^{4}]$$ $$ P(c \geq x_{1}^{4} \geq x_{2}^{2} \geq a) = \int_{a^{1/4}}^{c^{1/4}} P(t^{4} \geq x_{2}^{2} \geq a) (b-a)^{-1} dt, \quad c \in [a,b^{6}]$$ $$ = \left[c^{3/4}/3 - a^{3/4}/3 - a^{1/2}(c^{1/4} - a^{1/4}) \right] (b-a)^{-2}$$
I don't have a closed form pattern for this case, but the inductive step is still perfectly valid. So, you could write Mathematica code to compute this out to any $n$. Once $n$ is sufficiently large that $b^{2n} < a$, the expression will become negative, and the correct probability is zero.
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Case $\mathbf{0 \leq a \leq 1, \ b \geq 1 }$: Again, the inductive approach is still valid, but now the probabilities are going to be piecewise functions.
$$P(c \geq x_{1}^{2} \geq a) = \left\{ \begin{array}{ll} \left[c^{1/2} - a^{1/2}\right] (b-a)^{-1} & c \in [a,b^{2}] \\ 1 & c \in [b^{2},b^{4}] \end{array} \right. $$ $$ P(c \geq x_{1}^{4} \geq x_{2}^{2} \geq a) = \int_{a^{1/4}}^{c^{1/4}} P(t^{4} \geq x_{2}^{2} \geq a) (b-a)^{-1} dt, \quad c \in [a,b^{6}]$$
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Case $\mathbf{1 \leq a < b }$: Because $1 \leq a$, the powers of $a$ are increasing, which affects bounds. The following math is nearly identical to the previous case just with a couple lower bounds changed. $$P(c \geq x_{1}^{2} \geq a) = \left\{ \begin{array}{ll} \left[c^{1/2} - a^{1/2}\right] (b-a)^{-1} & c \in [a^2,b^{2}] \\ 1 & c \in [b^{2},b^{4}] \end{array} \right. $$ $$ P(c \geq x_{1}^{4} \geq x_{2}^{2} \geq a) = \int_{a}^{c^{1/4}} P(t^{4} \geq x_{2}^{2} \geq a) (b-a)^{-1} dt, \quad c \in [a^{4},b^{6}]$$
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Conclusion: We have closed-form expressions for when $a=0$ and $b \leq 1$. For all of the other cases, the boundaries become messy. Fundamentally, the inductive step still works, and it can be used to compute the answer for any fixed $n$.