Jump to content
BrainDen.com - Brain Teasers

bonanova

Moderator
  • Content Count

    6897
  • Joined

  • Last visited

  • Days Won

    64

Everything posted by bonanova

  1. bonanova

    Sorting out the bar bill

    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?
  2. bonanova

    making obtuse triangles

    I get fewer ...
  3. bonanova

    Coin hunt

    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.
  4. bonanova

    Coin hunt

    I can ask you to flip a coin for a 2nd time?
  5. bonanova

    Sorting out the bar bill

    @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.
  6. bonanova

    Sorting out the bar bill

    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.
  7. bonanova

    Coin hunt

    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.
  8. bonanova

    Next card is Red

    Let there be n red cards and n black cards.
  9. bonanova

    When midnight strikes

    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.
  10. bonanova

    When midnight strikes

    At an ever increasing pace Al, Bert and Charlie have been receiving into three identical boxes of limitless capacity identical pairs of silver coins engraved with the integers 1, 2, 3, 4, ... etc. These events occur on a precise schedule, each box receiving Coins marked 1 and 2 at 1 minute before midnight Coins marked 3 and 4 at 1/2 minute before midnight Coins marked 5 and 6 at 1/4 minute before midnight Coins marked 7 and 8 at 1/8 minute before midnight etc. But they were instructed at each event to remove a coin from their respective boxes and discard it. After some thought, Al decided each time to discard his lowest-numbered coin; Bert discarded an even-numbered coin; and Charlie thought what the heck and discarded a coin selected at random. Regardless of strategy, at each event the number of coins in each box grew by unity, so that after N events each box held N coins. Needless to say when midnight struck their arms were infinitely tired, but it was a small price to pay for infinite riches. But tell us, now, whether their expectations were met. Describe the contents of each box at midnight.
  11. bonanova

    Pegity - white to win

    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.
  12. bonanova

    Next card is Red

    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.
  13. bonanova

    Traffic jams

    @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.
  14. bonanova

    Traffic jams

    At 5-second intervals, 1024 automobiles enter a straight (and otherwise empty) single-lane highway traveling at initial speeds chosen at random from the interval [50, 70] miles per hour. Cars may not pass nor collide with other cars. When a slower car is encountered, a car must simply reduce its speed, and for the purposes of this puzzle we may consider the cars in such a case become permanently attached, traveling at the slower car's speed. Eventually there will be N clusters of cars. What is the expected value of N? (Equivalently, what is the expected cluster size?)
  15. @CaptainEd brilliantly answered a puzzle that I mis-worded into a much tougher one. Nice job. Here's the puzzle I had intended to post: You have just lost your 143rd straight game of checkers and have vowed never to play another game. To confirm your vow you decide to saw your wooden checkerboard into pieces that contain no more than a single (red or black) square. With each use of the saw you may pick up a piece of the board and make one straight cut, along boundaries of individual squares, completely through to the other side. You wish to inflict as much damage as possible with each cut, so you first calculate the minimum number of saw cuts needed to finish the job. And that number is ... (spoilers appreciated.)
  16. bonanova

    A big, big, big hotel

    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.
  17. bonanova

    Next card is Red

    The way it's set up, he can say Stop just once.
  18. bonanova

    Pegity - white to win

    Yes. That’s a win for white
  19. bonanova

    Green and Yellow hats again, but harder

    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.)
  20. As promised, a harder hat problem. Prisoners are seated in a circle so they can see all the others. This time the warden flips a fair coin for each prisoner and gives him a yellow or green hat, accordingly. Once all the hats have been placed, and have been seen by the others, prisoners are taken aside singly and given the opportunity to guess the color of his hat. And if instead he chooses not to guess, he is permitted to pass. Now comes the bad part. Unless at least one prisoner guesses, and all the prisoners who do guess are correct, all the prisoners will be executed. That's right, survival requires perfection from every prisoner who guesses his color. Prisoners decide on a strategy beforehand, and after the first hat is placed there is no further communication. Clearly, there can be a single "designated guesser" who ... just ... guesses a color. Half the time they all survive. But what kind of a puzzle would that be? Yes, incredibly, the prisoners can do much better. How? Maybe thinking about a three-prisoner case will answer that question. Once you're convinced they can do better than a coin toss, find their best strategy.
  21. bonanova

    Baggage on a conveyor belt

    Yup. And a bit of possibly helpful thinking: Edit: The math gets a little demanding here. Integrals and stuff.
  22. bonanova

    A big, big, big hotel

    Yeah, I hate annoyances, too...
  23. What fraction of triangles in a circle are obtuse?
×