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bonanova

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Everything posted by bonanova

  1. If we place four matches in the form of a square, they form 4 right angles. If we place them like a hash-tag (#) they form 16 right angles. If someone removes one match, can we still form 12 right angles? (No bending or breaking of the matches is allowed.)
  2. The answer probably does change with the nature of the distances.
  3. Regarding the 5 probabilities after 12 ant moves:
  4. Case by case results, inviting algorithm falsification, while limiting lengths to 10.
  5. Naively.I might want the first n-1 coins to be heads if the valuable one was tails in the nth position. I'd have you keep flipping them until they showed heads. But /// I watched the video, so ... nah.
  6. I can ask you to flip a coin for a 2nd time?
  7. @Pickett It may not impact your solution, but the OP did not intend to say that all of us drank something. "Class of drinkers" was not meant to preclude anyone from drinking nothing. Meaning there are eight "classes" of drinkers. Sorry for that. I have this thing about insisting that zero is a number, rather than a denial. Also, I get a different answer. I'll check my analysis against yours to see why.
  8. It means that all of the teetotalers drank neither wine nor beer. Let's say also that the others did consume alcohol. One class of friends may have drunk nothing, and perforce not used a glass.
  9. Thinking. Meantime I'll call your 10 minutes of cat flight and raise you 25 minutes of Roomfull of Teeth, as they explore new uses for the human voice, which you might actually enjoy.
  10. A bunch of friends went to the sports bar and got a group rate on the drinks: $5/glass for wine, $2/glass for beer, and $1/glass for water. When we left, the waiter asked me to sort out the bill. There was enough uncertainty in what people remembered that I could not be precise. So we happily just threw in enough to cover the bill, which came to $293 and we went home. But it got me thinking. None of us had multiple glasses of the same beverage. The waiter said 106 glasses were used, once each. 18 of us did not drink water. 39 people had wine. I was certain that 9 of us were teetotalers. If I had known the sizes of just three classes of drinkers I could have figured out the bill, as it was, I could not. But it did occur to me that if those who drank beer and water but not wine were as many as possible, and if those who drank only wine were half as many as that, I could say the smallest number of us who drank all three. Can you?
  11. Let there be n red cards and n black cards.
  12. I just re-read this, and my reply wasn't totally responsive. But let's see whether Charlie even makes us address this question. Remember, Charlie doesn't get any more coins to deal with than Albert did. If we believe Albert's box (can) get emptied, we can't say discarding Charlie's coins is an impossible task. We simply have to show that none of Charlie's coins is able to "hide" indefinitely, in hopes of hearing the clock strike while still being in the box. Charlie's coins are not inscribed, so we have to treat them collectively. We can't trace the destiny of any individual coin, as we could in Albert's case. Can we still show they are all eventually discarded, and not face the task of depleting an infinite set along the way? We claim that we can do that. We can refer to each of Charlie's coins as "one of the coins that entered Charlie's box at event n that was also not discarded at event n." The n coins in his box at any such time are indistinguishably equivalent. Their "fate" is countably infinite events with non-zero probability of being discarded. We have shown in a preceding post that each of Charlie's coins has a survival probability of k/n where k is the event when the coin entered his box and n is the current event number. k is fixed. n becomes infinite. The probability of the coin that entered Charlie's box at event k still being in his box after midnight is limit (n->inf) k/n = 0. Since this holds for each k it holds for each coin that entered Charlie's box. Using other words, every coin that is in Charlie's box at midnight is not a coin that entered Charlie's box. A contradiction. Also, at no event time did Charlie have an infinite number of coins.
  13. Pencils down. Discussion time over. For the believers in the utility of clever play, choose a convenient number of cards (e.g. pick a small number and enumerate the cases) and quantify the advantage that can be gained. For the nay-sayers, give a convincing argument that all the factors already discussed (permission granted to peek at spoilers) exactly balance each other out.
  14. @plainglazed (first) and @plasmid (with a formula) both have it. Both posts show how to get the answer, with the formula garnering "best answer" designation. Nice job both.
  15. Precisely. If every room houses a green man, new customers have no place to sleep. This does not work: { g1 g2 g3 ... gi .... b1 } <=> { r1 r2 r3 ... ri .... rinf+1 } Do we want another guest? We can't assign gi to ri. This does work: { b1 g1 g2 g3 ... gi .... } <=> { r1 r2 r3 r4 ... ri+1 .... } gi is no longer in ri.
  16. The way it's set up, he can say Stop just once.
  17. Yes. That’s a win for white
  18. OK so I was a kid once, back in the 40s and 50s, and we had this game called Pegity. It had a board with a 16x16 array of holes, and in turn each of 2-4 players inserted a peg of his own color into a hole. The object was to get 5 pegs of your color in a row: vertically, horizontally or diagonally. Not every game was won: similar to tic-tac-toe, you could run out of holes. But with only two players, there was usually a win. For simplicity let's shrink to a 9x9 board, mark the holes like in chess (A1, E7, etc.) and say that after O and X have each made three moves we have this position, with O to move: 1 2 3 4 5 6 7 8 9 A + + + + + + + + + B + + + + + + + + + C + + + + + + + + + D + + + + + + + + + E + + + + O + + + + F + + + O + O + + + G + + + + X + + + + H + + + X + X + + + I + + + + + + + + + With best play by both players, how soon can O win? Use chess notation (naming the holes, in two columns) to list the moves.
  19. @CaptainEd shuffles a standard deck of playing cards and begins dealing the cards face up on the table. At any time @plainglazed can say Stop and bet $1 that the next card will be red. If he does not interrupt, the bet will automatically be placed on the last card. What is the best strategy? How much better than 50% can @plainglazed do?
  20. That is amazing. And you've shown that 1 fewer than a power of 2 is the optimal value for n. Can you reduce it to only q memorized strings, and the success rate to (q-1)/q, where q is the highest power of 2 less than or equal to n? (Purposely leaving this un-spoilered, cuz this is a tough problem.)
  21. Yup. And a bit of possibly helpful thinking: Edit: The math gets a little demanding here. Integrals and stuff.
  22. Yeah, I hate annoyances, too...
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