Probability of intersecting a rectangle with random straight lines

We are given a rectangle $$R$$ with sides lengths $$r_1$$ and $$r_2$$, contained in a square $$S$$, with sides lengths $$s_1=s_2\ge r_1$$ and $$s_2=s_1\ge r_2$$. $$R$$ and $$S$$ are axis-aligned in a cartesian plane $$P$$. With the following recursive random process, we select straight lines orthogonal to the sides of $$R$$ (and $$S$$), until $$R$$ is cut.

At each time step, we select one of the two axes of $$P$$ with probability $$\tfrac12$$. Let $$a$$ the axis selected. Thereafter a straight line $$L$$ is selected uniformly at random from the ones cutting $$S$$ and orthogonal to $$a$$. Let $$S'$$ and $$S''$$ be the two parts of $$S$$ generated by the cut of $$L$$. These two random steps are repeated until $$R$$ is cut by $$L$$, and each time $$R$$ is not cut, $$S$$ is transformed by removing its part (either $$S'$$ or $$S''$$) that does not contain $$R$$.

Question: Given the coordinates of the vertices of $$R$$ providing its position within $$S$$, what is the probability $$p_i$$ that it is eventually cut (at the end of the random process) by a line orthogonal to its sides with length $$r_i$$ for $$i\in\{1,2\}$$?

(For the sake of clarity, we obviously have $$p_1=1-p_2$$.).

• If you put the rectangle in the lower left hand corner, as in J O'Rourke's first picture, the process of cutting either side by itself is the well-known stick breaking process. The number of tries until you cut the side on the y axis is about -log(r_1/s_1), and similarly for the side on the x axis. It seems to me that the probability of hitting one side first is likely to depend on those quantities. I haven't tried to make a proof out of this.
– mike
Aug 3, 2020 at 13:40
• Thank you for your comment @mike. I see your point. Would you have any suggestion about what happens then when the rectangle is not positioned in a corner of the square? Aug 3, 2020 at 17:31

I wonder if the probability is dependent only on $$r_i$$, or also dependent on the placement of $$R$$ within $$S$$? In these two examples,

it takes an average of $$2.70$$ steps to reach slicing $$R$$ on the left, but $$3.16$$ steps on the right.

I realize I'm ignoring your condition that $$s_i \gg r_i$$.
Added 4Aug2020. I include below some simulation data that might help a theoretical investigation. Here are two examples where $$R = 0.2 \times 0.1$$ in a unit square $$S$$.

On the left, after one million trials, the probability that the long side of $$R$$ is sliced was $$0.591$$. On the right, the probability was $$0.622$$.
• Thank you for your comment Joseph. In my opinion the required probability is independent of the placement of $R$ within $S$, and is independent of the number of time steps to reach slicing R, but I might be wrong. Aug 2, 2020 at 10:17
• @PenelopeBenenati: Re $S$ a rectangle: $S$ is a rectangle after the first step. Aug 2, 2020 at 11:36
• @PenelopeBenenati: I think I finally understand your question. $R$ will eventually be cut. You are seeking the probability that it is cut on the long side vs the short side, the $r_1$ side vs the $r_2$ side. Aug 2, 2020 at 13:57
• @PenelopeBenenati: Ignoring the $s_i \gg r_i$ condition, simulations suggest: (1) the conjecture is not precisely true; (2) the probability depends on the position of $R$ within $S$. But (a) this is from simulations, and (b) still your conjecture may hold for $s_i \gg r_i$. Aug 2, 2020 at 14:25
• Thank you again Joseph for your comments. Based on them I have now rephrased the question, which is finally much more significant. Aug 4, 2020 at 16:29