If n is even, there is a parity constraint that prevents this: the sum of the kasha at even-numbered seats in one iteration is the same as the sum of the kasha in odd-numbered seats the next iteration. However, within each of the two groups, they asymptotically approach having the same amount. Overall, each person approaches alternating between two values, the same two values for each person, but the even-numbered people are out of phase with the odd-numbered people.

I don’t understand the math behind the proof of the original puzzle, but here (for both cases), you can consider the result if all the kasha starts in one bowl. Then the pattern is initially that every other person has an empty bowl, and the others have C(s,i)/2^s of the total kasha, until s is large enough that the kasha starts to overlap, and then the numerator starts being sums of binomial coefficients where i assumes a set of values which are the same modulo the number of people in the loop (where, for the even case, there are two loops containing half the people each, and for the odd case there is a single loop). The result starting from any other state is the superimposition of multiple instances of this, rotated and multiplied by the starting values. Once everybody in the loop has some kasha, you can ignore the minimum amount of kasha, and treat it as if that amount is constant, and the remainder is circulating. This leads to the asymptotic approach to even distribution of the kasha (within each loop).

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