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Everything posted by bonanova

  1. @ThunderCloud you're homing in on it, but now you're a little high. @BMAD It's certainly true that if the FIRST child was a boy born on a tuesday, then it's just the prob that the second child is a boy. But ... the OP does not tell you that. That is, "one is a boy" does not imply "my oldest child is a boy." So your "second" child simply means the "other" child.
  2. Not that I know of. Tables of powers, or spreadsheet where different abc values can be simply typed in, or inspired guesswork? It's not my fav type of puzzle, but some like this type.
  3. @ThunderCloudThat's close, but a bit low.
  4. So this guy Thomas Bruss solved the general stopping problem.
  5. I ask people at random if they have two children and also if one is a boy born on a tuesday. After a long search I finally find someone who answers yes. What is the probability that this person has two boys? Assume an equal chance of giving birth to either sex and an equal chance to giving birth on any day.
  6. @plasmid Not sure why derivative failed, but you're right about the result. Nice solve.
  7. These cases are the ones I calculated as well, and led me to the right conclusion. (I had to solve it as it's not mine originally and I did not receive the solution.) It took a little insight to get me thinking along productive lines. See the spoilers in my April 1 post - btw not an April Fools post - to get there.
  8. Great job! Want to look at the reciprocal of that number and guess the exact result?
  9. @Molly Mae blazed the trail and @plainglazed nailed it.
  10. You are both on the right track, but I realize now that I mis-stated the OP. I didn't ask for what I wanted. What I wanted to get at was the average location, that is the average of all the possible ending location coordinates, more precisely, their centroid, and its distance from the origin. That's not the same as the expected distance of the ending points -- which does take sort-of serious math. My bad. I edited the OP.
  11. Al made sales calls at a number of cabins, which lay in a square field, one mile on a side. He drove his car to the first cabin, then visited the remaining cabins and returned to his car on foot, walking several miles in the process. If Al had had a map and perhaps a computer, he could have picked the shortest route to take, (traveling salesman problem). Lacking these amenities, Al simply chose to visit (after the first one) the (un-visited) cabin closest to his current location. If there were 6 cabins in all, how might they be placed, so that using Al's nearest-neighbor algorithm, and selecting the worst initial cabin, Al would be forced to walk the greatest distance; and what is that distance? Examples: 2 cabins: diagonal corners, starting at either: 2 x sqrt(2) = 2.828... miles. 3 cabins: any three corners, starting at any of them: 2 + sqrt(2) = 3.414... miles. Check out n=4 and n=5 as a warm-up.
  12. Al, Bert, and Charlie competed in a track and field event in which points were awarded for 1st, 2nd, and 3rd, place only. At the end of the day, Al had accumulated 22 points, while Bert and Charlie each garnered only 9 points. No other competitor earned points. Bert was 1st in the shot put. Who finished 2nd in the javelin throw? This is a Gold star puzzle.
  13. Sir, I bear a rhyme excelling In mystic force and magic spelling Celestial sprites elucidate All my own striving can't relate. See, I have a rhyme assisting My feeble brain, its task ofttimes resisting.
  14. You are given 5 circles, A, B, C, D, and E, whose radii are, respectively, 5", 4", 2", 2", and 1". Can you find a way to overlap circle A with portions of some or all of the other four circles so that the un-overlapped portion of A has the same area as the sum of the unoverlapped portions of the other four circles? That is, the red area is equal to the sum of the green areas. Circles B, C, D and E may overlap portions of each other as well as a portion of A.
  15. Find a, b, c. ab x ca= abca , a 4-digit number
  16. Here’s a challenge: I think this puzzle can be solved with almost no math at all. Like, think of a single “what if” question to ask that changes the conditions slightly.
  17. I have a strong feeling, and I'm working on a proof, that
  18. Which way should we evaluate this? [ [ [ [ [ x^(x/2) ]^(x/4) ]^(x/8) ]^(x/16) ]^(x/32) ].... or x^[ (x/2)^[ (x/4)^[ (x/8)^[ (x/16)^[ (x/32) .... ] ] ] ] ]
  19. One approach, among several, is to and
  20. I guess I've always been a little confused about what is being asked. Sharing is not permitted, yet an "average" result is requested. If average values are used, then two holes is enough. If there is no sharing or averaging, survival is assured only by the (very unlikely) worst case of 100 holes. If the question is what is the expected number of holes that together yields at least 100 nuts, we have an answer from simulation. Is there a way to say precisely what else might be needed?
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