Estimating a set of probabilities for sequences of independent events. Assume a line of obstacles, if dropping a marble at obstacle position $n$ it should stop with (as close to a given probability $p(n)$ as possible) somewhere before obstacle $n+m(n)$. Assuming we have $m(n)$ and $p(n)$ sampled at least some values of $n$. How can we estimate the independent probabilities of stopping (or passing) of each individual obstacle?

Let us assume the event of passing a given obstacle is independent of having passed any previous or passing any subsequent obstacles, but that they are always encountered in the same order: $$1\to 2\to 3 \to \cdots \to N$$
Found a way that works. Following the ideas in this question, we approximate the integrals with riemann sums (although we already have a discrete case with explicit product/sums). All $m(n)$ are 2 and the probabilities $p(n)$ given by the vector $\bf d$:

$${\bf v} = 2^\left[\min_{\bf v}\left\{\|{\bf Tv} - \log_2({\bf d})\|_2^2\right\}\right]$$

$${\bf T} = \left[\begin{array}{cccccccccccc}
1&amp;1&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0\\
0&amp;0&amp;1&amp;1&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0\\
0&amp;0&amp;0&amp;0&amp;1&amp;1&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0\\
0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;1&amp;1&amp;0&amp;0&amp;0&amp;0\\
0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;1&amp;1&amp;0&amp;0\\
0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;1&amp;1
\end{array}\right], \log_2({\bf d}) = \frac{1}{2}\left[\begin{array}{c}-1\\-2\\-3\\-4\\-5\\-6\end{array}\right]$$

So $$\cases{{v_1}^1\cdot {v_2}^1 = 2^{-1/2}\\{v_3}^1\cdot {v_4}^1 = 2^{-1}\\\vdots}$$

With a small isotropic Tikhonov regularizer with $\lambda = 10^{-6}$ gives us the solution: 

$$\left[\begin{array}{c}
0.840896488111804\\
0.840896488111804\\
0.707106903718765\\
0.707106903718765\\
0.594603712056721\\
0.594603712056721\\
0.500000173286739\\
0.500000173286739\\
0.420448389772112\\
0.420448389772112\\
0.353553574391631\\
0.353553574391631
\end{array}\right]$$

I we plug back in, we see several well known facts, for example $\sqrt{1/2} = 0.7071069\cdots, \sqrt{1/4} = 0.5$.

So we can build linear least squares problems encoding exponents instead of coefficients as numbers. As the system is underdetermined we happen to get the solution where pairs are equal, but bear in mind many other constraints can be put. Simultaneous overlapping constraints could probably be useful in some application..?
Found a way that works. Following the ideas in this question, we approximate the integrals with riemann sums (although we already have a discrete case with explicit product/sums). All $m(n)$ are 2 and the probabilities $p(n)$ given by the vector $\bf d$:

$${\bf v} = 2^\left[\min_{\bf v}\left\{\|{\bf Tv} - \log_2({\bf d})\|_2^2\right\}\right]$$

$${\bf T} = \left[\begin{array}{cccccccccccc}
1&amp;1&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0\\
0&amp;0&amp;1&amp;1&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0\\
0&amp;0&amp;0&amp;0&amp;1&amp;1&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0\\
0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;1&amp;1&amp;0&amp;0&amp;0&amp;0\\
0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;1&amp;1&amp;0&amp;0\\
0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;0&amp;1&amp;1
\end{array}\right], \log_2({\bf d}) = \frac{1}{2}\left[\begin{array}{c}-1\\-2\\-3\\-4\\-5\\-6\end{array}\right]$$

So $$\cases{{v_1}^1\cdot {v_2}^1 = 2^{-1/2}\\{v_3}^1\cdot {v_4}^1 = 2^{-1}\\\vdots}$$

With a small isotropic Tikhonov regularizer with $\lambda = 10^{-6}$ gives us the solution: 

$$\left[\begin{array}{c}
0.840896488111804\\
0.840896488111804\\
0.707106903718765\\
0.707106903718765\\
0.594603712056721\\
0.594603712056721\\
0.500000173286739\\
0.500000173286739\\
0.420448389772112\\
0.420448389772112\\
0.353553574391631\\
0.353553574391631
\end{array}\right]$$

I we plug back in, we see several well known facts, for example $\sqrt{1/2} = 0.7071069\cdots, \sqrt{1/4} = 0.5$.

So we can build linear least squares problems encoding exponents instead of coefficients as numbers. As the system is underdetermined we happen to get the solution where pairs are equal, but bear in mind many other constraints can be put. Simultaneous overlapping constraints could probably be useful in some application..?