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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 $p_1=1-p_2$.).

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    $\begingroup$ 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. $\endgroup$
    – mike
    Commented Aug 3, 2020 at 13:40
  • $\begingroup$ 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? $\endgroup$
    – Penelope Benenati
    Commented Aug 3, 2020 at 17:31

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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 $0.622$.
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  • $\begingroup$ 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. $\endgroup$
    – Penelope Benenati
    Commented Aug 2, 2020 at 10:17
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    $\begingroup$ @PenelopeBenenati: Re $S$ a rectangle: $S$ is a rectangle after the first step. $\endgroup$
    – Joseph O'Rourke
    Commented Aug 2, 2020 at 11:36
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    $\begingroup$ @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. $\endgroup$
    – Joseph O'Rourke
    Commented Aug 2, 2020 at 13:57
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    $\begingroup$ @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 $s_i \gg r_i$. $\endgroup$
    – Joseph O'Rourke
    Commented Aug 2, 2020 at 14:25
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    $\begingroup$ Thank you again Joseph for your comments. Based on them I have now rephrased the question, which is finally much more significant. $\endgroup$
    – Penelope Benenati
    Commented Aug 4, 2020 at 16:29

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