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bonanova

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  1. bonanova's post in Chronicles of the Redrum vs. Ecidius Prank War: The Purple Pool Party was marked as the answer   
  2. bonanova's post in A society that prizes girls was marked as the answer   
    Since it's solved, I won't spoiler this.

    The answer is not complex. Every birth, regardless of anything that is said, done, legislated, prohibited, presupposed, calculated, imagined or hoped for, has equal chances of being a boy or girl.

    If you flip a coin an odd number of times, then, yes, the number of heads and tails cannot be equal. But the question is for large numbers of conceptions and births - for a society - where, even though the numbers of boys and girls may not be exactly equal at every moment - each birth changes the running total - the chances of an excess of boys exactly equals the chances of an excess of girls. The question asks whether a birth control strategy can affect the society's overall gender balance in a systematic way. The answer is no, it can not.

    No calculus needed.
  3. bonanova's post in Probability of guessing was marked as the answer   
  4. bonanova's post in A Side Bet was marked as the answer   
  5. bonanova's post in Ants On a Ruler was marked as the answer   
    Hi sergyegi, and welcome to the Den.


  6. bonanova's post in No Jokers was marked as the answer   
  7. bonanova's post in Weighing in a Harder Way was marked as the answer   
    Jkyle1980's solution is correct, and much less verbose than what follows.

    You can distinguish among 3**N cases in N weighings.
    There are 54 possible cases in this puzzle [one of 27 coins is heavy or light].
    So three weighings [27 cases] won't do it, but four [81 cases] can.

  8. bonanova's post in Hats on a death row!! One of my favorites puzzles! was marked as the answer   
  9. bonanova's post in HOW DID IT HAPPEN??? was marked as the answer   
    I think this answer
    doesn't work: when you get to the South pole, how do you run West?
    But this answer:
    does: for example, any point on the circle (1 + 1/2pi) miles from the South Pole.
    After going South 1 mile, you're (1/2pi) miles from the Pole,
    which allows you to run West 1 mile [1 lap of a 1-mile circumference circle]
    and be able to go a mile North to the starting point.

    As Martini noted, there is an infinite number of starting distances:
    1 + 1/2Npi miles North of the South pole where N is any positive integer.
    N is then the number of circular laps in your westerly mile.

    e.g. N=5280 - you'd run 5280 laps around a 1-foot circumference circle.

    Here's a counter question - why can't N be negative?
    i.e. start closer than a mile - you could still do N laps
  10. bonanova's post in Hole in a sphere was marked as the answer   
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