Posers and Puzzles

Posers and Puzzles

  1. Standard memberXanthosNZ
    Cancerous Bus Crash
    p^2.sin(phi)
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    28 Mar '06 10:301 edit
    Postulate:

    Given a mixed box of chocolates containing n distinct chocolates, r of which are wrapped, there is a greater than r/n chance that the last remaining chocolate will be wrapped.

    Assumptions:
    n > r
    r != 0
    n,r are natural numbers.
  2. Joined
    12 Mar '03
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    28 Mar '06 11:18
    I don't know any natural number r for which r!=0.

    If you mean n > r >= 0, then I assume your postulate refers to human behaviour. For me, it depends on how tasty each type is, and who is to pick the last wrapped resp. unwrapped.

    Of course, only Belgian chocolates will do 😀
  3. Joined
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    28 Mar '06 13:04
    Originally posted by Mephisto2
    I don't know any natural number r for which r!=0.
    42 is a natural number...? And 42 != 0?
  4. Joined
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    28 Mar '06 14:21
    I think he meant that 0 is not part of the set of natural numbers. Although there really is no general agreement as to whether it should be included or not.
  5. Joined
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    28 Mar '06 14:29
    Originally posted by Balla88
    [b]I think he meant that 0 is not part of the set of natural numbers. b]
    Yes, I think youre right.

    Sometimes "!=" is pronounced "not equal to" (used by programmers), and in this case "r != 0" mening "r not equal to zero" says that r can be any natural number except zero. For example 42.
  6. Joined
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    28 Mar '06 14:395 edits
    I see the problem (therefor my edits): the "is not equal' sign is not supported by all editors (although it is on my keyboard).
  7. Standard memberPBE6
    Bananarama
    False berry
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    28 Mar '06 16:56
    Originally posted by XanthosNZ
    Postulate:

    Given a mixed box of chocolates containing n distinct chocolates, r of which are wrapped, there is a greater than r/n chance that the last remaining chocolate will be wrapped.

    Assumptions:
    n > r
    r != 0
    n,r are natural numbers.
    If you have n distinct chocolates, there are n! possible arrangements in total. To find out how many have a distinct wrapped chocolate at the end, we designate one to be put at the end and arrange the rest, so the number of possible arrangements here is (r choose 1)*(n-1)!. The probability that the chocolates will be selected in an order represented by the second group is then:

    P = (r choose 1)*(n-1)! / n!

    = r*(n-1)! / n!

    = r / n

    There must be some trick I'm missing.
  8. Standard memberleisurelysloth
    Man of Steel
    rushing to and fro
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    28 Mar '06 17:13
    People like to see what they're getting? 😕
  9. Joined
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    28 Mar '06 17:19
    Aren't the wrapped chocolates usually the ones with alcohol inside? Maybe the majority of chocolate eaters doesnt like alcohol in chocolates. Especially during daytime when they have to work etc and shouldnt get drunk yet. 😀

    But I also think it's the fact that its wrapped. People are just too lazy.
  10. SubscriberAThousandYoung
    iViva la Hispanidad!
    tinyurl.com/y2c6j2t3
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    29 Mar '06 00:27
    Originally posted by crazyblue
    Aren't the wrapped chocolates usually the ones with alcohol inside? Maybe the majority of chocolate eaters doesnt like alcohol in chocolates. Especially during daytime when they have to work etc and shouldnt get drunk yet. 😀

    But I also think it's the fact that its wrapped. People are just too lazy.
    Wrapped chocolates look like they'll last longer so ppl will eat the others first.
  11. Account suspended
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    29 Mar '06 03:52
    Life is like a ... either a chicken or an orange.

    lol 🙄
  12. Account suspended
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    29 Mar '06 03:53
    Actually, life is like a lemon.
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