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Nov 6, 2014 at 11:10 vote accept Irvan
Nov 1, 2014 at 8:49 history bounty ended Irvan
Nov 1, 2014 at 8:45 comment added Irvan Beautiful answer! I've been trying to compute whether the corresponding Alice strategy is $o(\log n)$: open $\log n / \log \log n$ boxes amongst the first $n(1 + 1/\log n)$ boxes, and open $\log n$ on the remaining boxes (since we the expected boxes we have to open in the worst case have to be $\Omega(\log n)$). This boils down to whether it is possible to generate a scenario in which (1) the probability that Alice goes to the remaining boxes is constant and (2) we placed a constant fraction of the red balls amongst the first set of boxes. This seems unlikely.
Nov 1, 2014 at 2:54 history edited Timothy Chow CC BY-SA 3.0
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Nov 1, 2014 at 2:45 history answered Timothy Chow CC BY-SA 3.0