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bonanova

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Everything posted by bonanova

  1. At first blush I'm amazed that you came up with the number .6094, Hi, by the way, and welcome to the Den, because it's not a familiar number, and it agrees closely with simulations. Also given an assumption that a single point, repeatedly placed, will cluster at 1/4 or 3/4 (the center points of either half of the line). Simulations show that it will cluster at 1/2. But then 1/4 and 3/4 would average to 1/2 so maybe that doesn't matter. Very interesting result.
  2. I think it's interesting that the sum of the squares of the means doesn't match the mean of the sum of the squares. Maybe that shouldn't be interesting? I think it's instructive. Taking the mean is usually the last step. That guides us first to determine the most relevant quantities. Edit: And now I have to somehow correct my first post. The numbers I gave that sum to ~.5 are misrepresented as the squares of the probabilities.
  3. I spent a long time on this puzzle. I liked it a lot, (thanks, @BMAD), and I like probability puzzles in general. Precisely for the reason @bubbled told us.
  4. @bubbled, our numbers agree, including sum of the squares of means being 0.426... I also find that the mean of the sum of the squares is 0.500.
  5. A crystal goblet contains water to the brim. It has a conical shape, with a 5-inch vertical height and a 30-degree angle at the base. I mischievously want to spill as much of the water as possible. To accomplish this I have a supply of spherical lead weights of any radius I desire. Which single lead weight shall I drop into the goblet?
  6. CaptainEd, are these numbers the expected (average) lengths of the longest arc (interval) for each value of n? Concerned about using the round-the-corner length, since the (sorted) n uniform points will tend to cluster around i/(n+1), i=1, ... n which will make r-t-c lengths twice as long as the other intervals. the i=0th point is eliminated when you join the 0 and 1 endpoints. For n uniform points on [0,1) the intervals are properly (after sorting the points lowest to highest) n1, n2-n1, n3-n2, n4-n3 ... 1-nmax. Note the first interval, n1, is formally n1 - 0.
  7. What do you get for prob (middle arc) and prob (shortest arc)?
  8. Hi Jelly, and welcome to the Den. I searched on the Web a bit, and did not find puzzles of this type, but they may be out there. If anyone does find them, a link can be posted to this Forum: http://brainden.com/forum/forum/9-other-mind-boggling-stuff-on-web/
  9. I did more simulations. The expected middle length remains slightly larger than the geometric mean of the other two. If I impose a 2.3 ratio I get 0.6150, 0.26775, 0.11641 Slightly larger, slightly smaller, and about the same, respectively, compared to simulated values. For what it's worth, the subtended angles are 221.7o, 96.389o, 41.908o
  10. Hidden Content Hidden Content Hidden Content Expanding on my post of Sunday at 8:09 AM:
  11. @plasmid, nicely done. @bubbled, my spoiler 3 was a complete brain fart.
  12. Here's another try at distinguishing our ways of looking at this problem.
  13. What ... my description of selecting random points does not overwhelmingly convince you? How about this.
  14. OK, and I have a suspicion that my answer is too easy and there's more to it.
  15. Totally! And, nice find on that last one.
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