The difference is that I'm starting out by assuming that *all* the probabilities T(i,j) of arriving at the point (i,j) are given for i,j >= 0, in advance. I'm now seeing that the intended problem was evidently to provide in advance only the probabilities along a fixed NW-SE diagonal D_n = {(i,j) | i+j = n}. (I made my hypotheses clearer in my post following the quoted one.) Apologies for misunderstanding the problem. --Dan Jim wrote: << I still assert that the probabilities are underdetermined, even in the case n=3. Consider a situation board in which balls get routed from (0,0) to (1,0) w.p. (with probability) 2/3, from (1,0) to (2,0) w.p. 1/2, and from (0,1) to (1,1) w.p. 0. In the Galton board picture, that's 2/3:1/3 1/2:1/2 0:1 The output distribution is 1/3:1/3:1/3. On the other hand, the biased Galton board 1/3:2/3 1:0 1/2:1/2 also gives output distribution 1/3:1/3:1/3. So the n output probabilities do NOT uniquely determine the n(n-1)/2 biases. My guess is that Dan is solving a different problem. Can anyone reconcile his analysis with mine and explain why we came to such different conclusions?

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