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bonanova

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Everything posted by bonanova

  1. Let there be n red cards and n black cards.
  2. 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.
  3. 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.
  4. @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.
  5. 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.
  6. The way it's set up, he can say Stop just once.
  7. Yes. That’s a win for white
  8. 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.
  9. @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?
  10. 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.)
  11. Yup. And a bit of possibly helpful thinking: Edit: The math gets a little demanding here. Integrals and stuff.
  12. Yeah, I hate annoyances, too...
  13. OK I think I just found your first blue man.
  14. @aiemdao - Nice solve. @plainglazed - Nice puzzle.
  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. Isn't this fun? The { real numbers } between 0 and 1 comprise two infinite groups: { rationals } and { irrationals }. Rationals (expressible as p/q where p and q are integers) are countably infinite. We can order them, by placing them in an infinite square of p-q space and drawing a serpentine diagonal path. But the irrationals are not countable. The cardinality (notion of size used for infinite sets) of the rationals is called Aleph0 and for the irrationals (and reals) it's called Aleph1 or C (for continuum.) So first off, what we can do with the { numbers } between 0 and 1 depends on { which numbers } we want to deal with. Next, there's a problem that points are not objects that can be moved from place to place, as coins can. Points are more descriptions of space than of autonomous objects. If 0 and 1 are on a number line, the location midway between them is the point denoted by 0.5. It can't be removed. (We could erase the line, I guess, but that would not produce an interval [0 1] devoid of numbers, either.) But we can finesse this matter by "painting" a point. Kind of like what we did when we inscribed a number on each coin. Painting does a little less - it puts a point into a particular group or class but it does not distinguish among them. We can tell coin 2 from coin 3 and also distinguish both from a coin with no inscription. Two blue points, however, are distinguishable from unpainted points, but not from each other. But for our purpose here, painting is actually enough. Since the { rationals } are countable, we can paint them, in sequence, and if we do that at times of 1, 1/2, 1/4 .... minutes before midnight, all the rationals between 0 and 1 will be painted blue by midnight. (I don't know of a similar scheme for completing uncountable task sets, so the irrationals alas must remain unpainted.) Next, instead of asking whether all the points in [0 1] were removed, we can ask, with pretty much the same meaning, whether at this point the entire interval [0 1] has been painted blue. It should be, right? After all, between any two rationals lie countably infinite other rationals. So the paint must have covered the entire interval, right?. Well ... actually ... the answer is counter-intuitively No. And this is because even though the { rationals } are "dense" meaning there are no "gaps" between them, as noted above, the { irrationals } are even "more dense." That is, between any two irrational numbers there are uncountably many other irrationals. That means, for every blue point we created in the interval [0 1] there are uncountably many unpainted points. We can't magnify the line greatly enough to "see" individual points, but "if we could," if we were lucky enough to come across a single blue point we'd have to pan our camera over uncountably many unpainted points before we saw another blue one. In measure theory, the measure of the rationals over any interval is zero. The measure of the irrationals over [0 1] is unity. So what of our quest of emptying [0 1] of points? It's the same as our quest to paint [0 1] blue by painting all the rational numbers in that interval. (Reminding ourselves that there were too many irrationals to paint one at a time.) Turning again to measure theory, the measure of the "blueness" of the interval would be zero. That means we would not see even the hint of a faint blue haze. Back to our "empty the interval by removing the rationals" quest, Nope. [0 1] would not be empty: uncountably many points (numbers) would remain.
  17. That is what infinity does to our brains. Al retains coin 2 at step 1. But Al discards coin 2 at step 2. What is true for coin 2 is true for every coin. Every coin has a scheduled pre-midnight discard date. So "what happened to the N coins not discarded?" If N is finite, then it's not midnight yet, and the box does in fact contain coins. We have to be patient. The process has to run its course. And, specifically, {coin n+1} will be discarded at step n+1, at time t n+1, prior to midnight. So after midnight, it will be gone. Along with all the others. Every coin, identified by the number engraved on it, has a well-defined pre-midnight discard date. But then if we're Bert, who never schedules the discard of an odd coin, we'll have a ton of coins.
  18. I would dispute your first point. I can't immediately think of a scenario that permits a discard which will not eventually happen given an infinite number of opportunities. Can you provide one? We agree on the second point. My previous post answers precisely that question.
  19. @CaptainEd Wow. Not the answer I had in mind -- a much better one! Nice. Bonus (slightly modified) version. Same question, but this time no cut may end short of the opposite edge. It must go entirely through the piece being cut.
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