Well, clearly it's a suboptimal strategy in the following way. If after seeing N-2 candidates, I still haven't seen one better than the K'th, I know I only have two left. If the next one is above the median seen so far, I'm probably certainly doing better to pick that one and pass on the last, than to wait for the last. And of course a similar argument applies for N-3, and so on. Unless I'm missing something? On Tue, Jun 17, 2014 at 5:06 PM, Bernie Cosell <bernie@fantasyfarm.com> wrote:
I was just reading a PDF about the secretary problem and all of the analyses I've ever seen seems to start with the assumption that the optimal strategy will necessarily be of the form "examine the first K candidates and then take the first candidate better than any you've seen so far", and if you haven't picked any candidate before that you MUST pick the N'th. And the math goes from there to derive the usual N/e value for K.
What got me curious [and this has actually drifted in and out of my pondering for some time now] is whether there's a proof that includes proving that the *assumption* is correct, also. I could see a strategy, for example, where you examine K candidates and then take the *second* best [that is, wait until you get one better than K, and then take the one better than that second one]. There might be other strategies more subtle or something beyond my ken. I don't see how to prove the assumption: that that's the best *possible* selection strategy. ???
/Bernie\
-- Bernie Cosell Fantasy Farm Fibers mailto:bernie@fantasyfarm.com Pearisburg, VA --> Too many people, too few sheep <--
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