Let’s flip 0s and 1s, so that we have one 0 and n = 9 1s, and the goal is to make the 0 bucket as high as possible. It’s convenient to think of the numbers as temperatures, and the averaging operations as heat exchange.

At any point in time, let a_1 >= … >= a_n be the temperatures of the “1” buckets, and b be the temperature of the “0” bucket. Let k be such that a_k >= b >= a_{k+1}.

By mixing b with a_k, then a_{k-1}, then a_{k-2}, etc, (the strategy described in previous comments) we can raise the temperature of b to T* = b / 2^k + \Sum_{i=1}^k a_k / 2^k.

A case-by-case analysis (examples below) should show that no moves can ever increase T*, so the strategy from the previous comments is the best we can do (I haven’t really done all the cases, but I think this approach should work). For example:

– Moving epsilon heat from a_k to b keeps T* constant.

– Moving epsilon heat from a_i to a_j (i <= j <= k) reduces T*.

– Moving epsilon heat from a_i to b (i <= k) reduces T*.

– Moving epsilon heat from b to a_i (k < i) reduces T*.

– Etc.

Finally, note that a given state can have multiple values of k, but T* is still well-defined. For example, if a_j = b, both j and j – 1 are possible values for k. However, T* has the same value:

T*_{k=j} = b / 2^j + \Sum_{i=1}^j a_i / 2^i = b / 2^j + a_j / 2^j + \Sum_{i=1}^{j-1} a_i / 2^i = b / 2^{j-1} + \Sum_{i=1}^{j-1} a_i / 2^i = T*_{k=j-1}.

]]>And Ben, your variation is interesting.

]]>But I wonder if the intended problem involves the additional constraints that you may only average *adjacent* numbers. E.g., after one step you could have 0.5 0.5 0 0 0 0 0 0 0 0, and after two steps you could have 0.5 0.25 0.25 0 0 0 0 0 0 0.

In this case, it seems we shall approach 0.1 in the limit, but giving a closed formula for the leading value after n steps is not so obvious!

]]>I don’t think averaging any of the other values is helpful at any time, since it either involves averaging two values that are both bigger than the “1” slot (which does nothing), or it destroys a zero (which is counterproductive).

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