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Everything posted by bonanova

  1. @bubbled, I would interpret it this way: Round 1: Four games => 4 winners. Round 2: Two games => 2 winners. Round 3: One game => 1 winner. Seven games in all, and there are no "middle two" rounds. If that's right, my take is
  2. @jasen, In an earlier post you said to switch boxes. (A) of course that is not permitted. (B) you can succeed even if the warden randomly re-positions them before each visit! @bubbled, if the statement is incorrect, that you say is incorrect, the puzzle has no 30% solution. We agree on that. So it's not a guessing game. It's a strategy-driven game. And the strategy does group the successes and failures to advantage for the prisoners. We're both traveling the high road, trust me. @bubbled, let's not discuss my last point right now. This is because to clarify what I was getting at would mean to
  3. I guess I misunderstood then. It sounds like the OP asks how to do it with 11, and then ask what the minimum length would be (even shorter.) What was confusing was trying to do it with 11. Good puzzle.
  4. Thalia may have it, but ... seriously? twice your mother's age?
  5. Yes, it is not a gimme that you win. The best strategy only means that among all possible strategies, it gives the best shot at winning.
  6. @harey, we have language barriers here, and I think "in the middle" in the OP is safe to take as meaning "between the groups of white and black balls." @BMAD, did you mark my previous post as the correct answer? It only found a solution for length 12, and there was an earlier indication that a shorter chute was possible.
  7. Here are three nudges. Not exactly clues. More like clarification of various points covered in the OP And here is a slight "Oops." Did I forget to mention that the boxes are externally numbered 1-100? They are, and now the OP says so.
  8. Each prisoner can open up to 50 boxes.
  9. Bingo. Can you determine the expected number of flips needed for p(H) = 1/π ?
  10. You have a coin which is known to be fair: p(H) = p(T) = 0.5 exactly. Devise a method for simulating a biased coin where p(H) = 1/π exactly. Edit: A gold star awaits a method that requires on average only two flips of the fair coin to produce an arbitrary bias.
  11. I'm not getting this. Can you do the first couple numbers step by step? How to you decided on n? Just starting with 1, 2, 3, ... ? Where does "addition" come in? Thx.
  12. I think that gives the opposite of the first flip. Pick the (slightly) lower, or bigger, hand.
  13. Lets try it together. Hidden Content Again, to obvious to be true. But what is wrong?
  14. That's the beauty of this one. You're only missing the best answer.
  15. It's time to paint the prison. The warden wants the cells cleared for the convenience of the painter. He devises a scheme by which his 100 prisoners either will be all freed or all executed. Either way, the painting will be made easier. His scheme offers a survival probability that seems terribly small. In a room he places their names, on sheets of paper, in 100 wooden boxes, numbered 1-100 and lined up on a table, one name to a box. Prisoners are led into the room one by one and are told to look inside the boxes in search of their name. But each prisoner can open at most 50 boxes. Successfu
  16. Due to temporary magical powers you find yourself undefeated in the US Open tournament, slated to play none other than Novak Djokovic for the title. Now it would not be fair to Djokovic -- and there would be no puzzle -- if you could keep your powers for the entire final match. You are informed that, at some point before the end of the match, your powers will go away, and you must finish the match using only your native skills. However, because you possess great logical skills you are granted to choose what the score will be when your powers end. In order to maximize your winning chances, what
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