Little explorations with HP calculators (no Prime)

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(12-28-2018 09:42 PM)brickviking Wrote:  I contend (again, without a rigid mathematic proof, 'cos I'm not that bright) that you can't actually arrange the balls specially to 'guarantee' that both Anna and Bertha get their required proportions of Christmas baubles.

You can arrange them like that. It's actually pretty easy.

To start, pick the first 70 balls such that 50 of them are blue and 20 are white. That takes care of the case when Anna's chosen range starts at the first one.
Now, let's look at what happens when the range is shifted by one. The n-th ball (where n specifies where the range started before shifting) is put back into the line, and the (n+70)-th ball is removed from it. We want to keep Anna's counts of blue and white balls stable, so just make the n-th and (n+70)-th ball have the same color. Done!
In short, this means the arrangement has to be a pattern of 50 blue balls and 20 white ones repeated twice, which is consistent with the requirement that the entire line shall consist of 100 blue balls and 40 white ones.

As a bonus, this still works when Anna's range is allowed to wrap around. That's not exactly hard to achieve though, because Anna's selection is the inverse of Berta's - when Anna's range wraps, Berta's doesn't, so you could just swap the names.