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Apr 13, 2017 at 12:19 history edited CommunityBot
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Aug 18, 2010 at 0:48 answer added Aaron Meyerowitz timeline score: 1
Aug 17, 2010 at 7:38 history edited Casebash CC BY-SA 2.5
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Aug 17, 2010 at 7:15 answer added Artem Kaznatcheev timeline score: 0
Aug 17, 2010 at 5:41 answer added Gerry Myerson timeline score: 1
Aug 17, 2010 at 4:29 comment added Gerry Myerson I think it's an interesting question if you fix $k$ and imagine $n$ large compared to $k$. For $k=1$ it's trivial; you may need $\lceil n/2\rceil$ boxes, you never need more. But for $k=2$ it's already not clear (to me) what the answer might be. I can see where you may need $(n/2)+1$, I can't see if you may need more.
Aug 16, 2010 at 20:46 comment added Casebash @Steve, @Dan: Updated the question
Aug 16, 2010 at 20:45 history edited Casebash CC BY-SA 2.5
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Aug 16, 2010 at 14:51 comment added Steve Kass @Casebash: Not sure what your comment is saying. Dan is correct. Your question isn't clear, so Dan may not have answered it. Do you mean "Given a known distribution, what is the number B (which depends on the distribution) so that every subset of B boxes contains at least half of each type, and B is minimal (some B-1 boxes fail to contain half of each type)?" If this is the question, there are some distributions for which B=n (everything in one box and n-1 empty boxes), and some distributions for which B = n/2 (identical boxes). What is your question?
Aug 16, 2010 at 10:43 comment added Casebash @Dan: The boxes can contain mixed types
Aug 16, 2010 at 10:43 history edited Casebash CC BY-SA 2.5
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Aug 16, 2010 at 10:41 comment added Dan Brumleve If k = n and each box contains a different kind of object then we need all boxes.
Aug 16, 2010 at 10:22 history asked Casebash CC BY-SA 2.5