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Everything posted by bonanova

  1. One approach, among several, is to and
  2. 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?
  3. @harey, some guidance about completing a formula would be appreciated. I'm interested in the "surprise."
  4. Yes you cand do that. The elevation of the two men as they ascend and descend are continuous functions of time. If at one time one is greater and at a later time the other is greater there must be a time when they are equal. Since they both stay on the path, and if the the slope of the path never changes sign, then they meet at that time. If the slope does change sign, then multiple points on the path will have the same elevation and there can be multiple times when the elevation of the two men are the same. They meet at one of those times.
  5. Good start -- That's shorter than 3, which comes from any three of the square's edges.
  6. Al and Bert are among 100 passengers assigned to one hundred seats on an airplane. Al was first to board, and Bert was last. Strangely, the first 99 passengers ignored their boarding passes and took random unoccupied seats. Bert liked the seat he was assigned and is not happy with the situation. If he's lucky, his seat is unoccupied and there's no problem. Otherwise, he insists the passenger erroneously occupying it move to his own assigned seat. The displaced passenger must then move, possibly displacing another person. This process continues until all passengers are seated. What is the probability that Al must move?
  7. Recently we considered the shortest roadway that connects the four corners of a square. Here we seek the shortest set of line segments, one attached to each of a square's corners, that need not connect with each other. Instead, what we ask of the line segments is that is they will block any ray of light attempting to pass through the square.
  8. Consider a random walk in the plane where each step is taken, beginning at the origin, in either in the positive x or positive y direction, i.e. either east or north, each choice being made by the flip of a fair coin. The length of each step is 1/2 the length of the previous step, and the first step has length √2. After infinitely many steps have been taken, what is your expected distance from the origin? Edit: Ignore the original text in pink. Instead, What is the distance to the origin of the centroid of the possible termination points? You find the centroid of a set of points by averaging respectively their x- and y- coordinates. First correct answer wins, but style points will be awarded as well.
  9. @plasmid Nice solve. The puzzle itself was more a math exercise than a puzzle -- that part was thinking it through and setting up the calculation. I thought is was interesting in that you can sort of envision the setup and know that it had to be e.g. greater than 1/3, but maybe not as great as 2/3, so the question was would it be greater than 1/2?
  10. Yes there is a rule, L'Hopital's rule. Basically you can just replace functions by their derivatives to resolve indeterminate values. Or, you can just evaluate expressions and see how they behave: It's intuitive that the exponential dominates for large x if you think of say x / e3x, instead of x e-3x.
  11. I guess we can compute expectation value as well:
  12. The King has decreed that his daughter the Princess shall marry the most wonderful Prince in all the land. One hundred suitors have been selected from their written applications, and on a certain day the King arranges for them, in turn, to interview the Princess. Each suitor must either be chosen or eliminated on the spot. If the Princess does not choose any of them, she will marry the last Prince to speak with her. You have been chosen as the Royal Advisor to the Princess and tasked with implementing her best strategy to choose the Most Wonderful Prince of the realm. You devise an evaluation scheme by which the princess can assign a unique "wonder number" to each prince as she meets him. The strategy then is to have the Princess reject, but record the highest score of, the first N princes that she meets. The Princess will then choose the first Prince that she subsequently interviews whose score exceeds that recorded score. That's it. The puzzle is basically solved. Except, of course, to decide on the optimal value of N. It requires some thought. If N to too high, the most wonderful prince is likely to be eliminated at the outset, and she ends up with the last guy. If N is too small, the Princess will likely settle for a fairly undistinguished prince. What value of N optimally balances these two risks? What is the probability that the Most Wonderful Prince will be chosen? Disclaimer: I recall this puzzle being posted before, with different flavor text. And it's somewhat of a classic. To give it a fair play here, I'll ask not to post any links and not to just give the answer if you know it, at least not without "showing your work."
  13. There is survival in numbers.
  14. The cars in front of the slowest car are the remaining cars. (Each slowest car captures those behind it.)
  15. I think this puzzle is exactly the same as the traffic jam puzzle I posted recently, although that's not obvious at first glance. @plainglazed found a solution in which he formed clusters of cars by recursively locating the slowest of a group of cars, assuming on average it was in the center of the remaining cars. This corresponds to "grabbing" on average one-half of the remaining marbles from the bag. Picture the marbles in a line and, grabbing a random percentage of them starting from one end. This has to end up having the same number of marble grabs and car clusters. It leads to a logarithmic answer, but to the wrong base -- it should be the natural loge, not log2, which gives too large an answer. Instead of decreasing the number by 1/2 each grab, the remaining number is decreased by 1/e each grab. Same must go for locating the slowest of the remaining cars. @plasmid found a solution that leads to Sum { 1/k }, which as you point out is ln { n } + gamma, and is here confirmed by simulation. What I can't find is a corresponding analysis for the marbles problem (although plainglazed's approach is applicable to both problems) that will lead to that same sum, instead of yours, where both sums appear to be correct. This has been fun to think about. .
  16. That agrees with simulations I ran, which gave, approximately, Nice solve.
  17. You're right. And it's not much of a puzzle after all. You have to allow Albert to lean forward as he runs to make this puzzle at all interesting. But even then, the obvious solution is to have Albert lie horizontally and crawl at infinite speed. Only the top of his head gets wet then. I ran into this puzzle a few years back and "solved" it, more interestingly but also more incorrectly, by multiplying his front and top areas respectively by sin theta and cos theta where theta was determined by his speed compared to the speed of the rain, and some other stuff. It was nonsense.
  18. @Thalia You are so right. Thanks. Locking this thread.
  19. You are wearing gloves while trying to retrieve the marbles contained in a bag. Because of the gloves, you're doing a pretty poor job of it. With each grab, you are only able to retrieve a random number of marbles, evenly distributed between 1 and n, the number of marbles currently in the bag. With 30 marbles initially in the bag, how many grabs do you expect it will take to retrieve them all?
  20. Twenty coins lie on a table, with ten coins showing heads and the other ten showing tails. You are seated at the table, blindfolded and wearing gloves. You are tasked with creating two groups of coins, with each group showing the same numbers of heads (and tails) as the other group. You are only permitted to move or flip coins, and you are unable to determine their initial state. What's your plan?
  21. For the purposes of this puzzle, consider our old friend Albert to have the shape of a rectangular paralellepiped (when he was born the doctor remarked to his mother, I don't explain them, ma'am I just deliver them) just meaning a solid having (six) rectangular sides. At a recent physical exam, Albert was found to be 2 meters tall, 1 meter wide and .20 meters thick (front to back.) He maintains his geometric rectitude by never leaning forward when he walks or runs. So anyway, Albert, alas, has found himself caught in a rainstorm that has 1000 raindrops / cubic meter that are falling at a constant speed of 10 meters / second, and he is 100 meters from his house. Just how fast should Albert run to his house so as to encounter as few raindrops as possible?
  22. If you chose to answer this question completely at random, what is the probability you will be correct? 25% 50% 0% 25%
  23. If 5+3+2 = 151022 9+2+4 = 183652 8+6+3 = 482466 5+4+5 = 202541 Then 7+5+2 = ______ ?
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