Even Cooler Problem Involving e

[madnak]

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I'm guessing about 30% of the time and it doesn't matter how many girls. If you let 20-50% of them go by then take the 1st better-then-the-group one.

The times you haven't let the best slip through, you are protected by chances of the 2nd best being in the 1st group , or the 3rd best, etc which drags you into the winning 30% range.

very rough in-my-head calcs 'guess'.

luckyme

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If you cut at the 50% mark (I think this is most efficient, but I can't prove it), then there's a 50% chance that the hot one is in the first 50% and you lose. But in the other 50% of cases where the hot one is in the second half, this means there's a >50% that the second-hottest chick is in the first half. And if the second-hottest is in the first half while the hottest is in the second half, then the hottest will be chosen.

Thus, there is a bare minimum 50%*50%=25% chance that you'll hit. But you'll also hit in many cases where the hottest and second-hottest are both in the latter group. So the chance is considerably higher than 25% but considerably lower than 50%. My answer in white below:

<font color="white">The title of this thread suggests that the answer has something to do with e. Since 1/e (~36% range) fits my observations, I'm guessing that as n approaches infinity, the success rate approaches 1/e.</font>