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Posers and Puzzles

Posers and Puzzles

  1. Standard member talzamir
    Art, not a Toil
    13 Oct '11 10:52
    There are three dice, with different numbers on each side.

    A: 1-1-5-5-5-5
    B: 2-2-2-2-6-6
    C: 3-3-3-4-4-4


    Two players each choose one of the three dice, and roll once; high roll wins. You get to / have the misfortune to go first. Which of the three dice should you choose?

    Follow-up; is it possible to make a perfect set of dice like these, so that

    * P(A defeats B) = P(B defeats C) = P(C defeats A) = P > 1/2
    * all 18 numbers are natural numbers
    * no ties can happen; and
    * the average of the numbers on the six sides of each die is the same?

    If there are multiple such solutions, how high can P be?

    If it is not possible, why not? Perhaps it could be done with dice with a number of sides other than six?
  2. Standard member forkedknight
    Defend the Universe
    13 Oct '11 15:19 / 1 edit
    Originally posted by talzamir
    There are three dice, with different numbers on each side.

    A: 1-1-5-5-5-5
    B: 2-2-2-2-6-6
    C: 3-3-3-4-4-4


    Two players each choose one of the three dice, and roll once; high roll wins. You get to / have the misfortune to go first. Which of the three dice should you choose?

    Follow-up; is it possible to make a perfect set of dice like these, so that ...[text shortened]... not possible, why not? Perhaps it could be done with dice with a number of sides other than six?
    The worse case scenarios for each die:
    A:
    4/9 vs B

    B:
    1/3 vs C

    C:
    1/3 vs A


    One of things things are not like the others
  3. 13 Oct '11 15:30 / 1 edit
    I pick A.

    Edit: although, B probably has a slight edge over A in the long run with A having a slight edge over C and C having a slight edge over B.

    I still pick A, I'd rather have AvsB than have BvsC. No one is going to pick C first.
  4. Standard member forkedknight
    Defend the Universe
    13 Oct '11 20:18 / 1 edit
    Originally posted by Trev33

    I still pick A, I'd rather have AvsB than have BvsC. No one is going to pick C first.
    C vs A is identical to B vs C, why would you pick either one?
  5. 15 Oct '11 18:06
    Originally posted by forkedknight
    C vs A is identical to B vs C, why would you pick either one?
    If i pick C they're going to pick A and if i pick B they're going to pick C. Picking A is the only logical choice to go against their B. I have 4 sides that'll beat 4 B sides, granted B has the same but it's a lot better odds in a one off roll than any of the other possibilities for one roll.
  6. Standard member talzamir
    Art, not a Toil
    15 Oct '11 18:17
    As pointed out by forkedknight:

    A vs B: A wins with a five over two probability 2/3 x 2/3 = 4/9
    B vs C: B wins with a six over any, probability 1/3 x 3/3 = 3/9
    C vs A: C wins with any over one, probability 3/3 x 1/3 = 3/9

    so A is the best choice. I would like to have a set like this where it doesn't matter what the first player picks, the odds are always worse than 50% and always the same. With these exact dice, the odds are not identical.

    Can these dice be somehow fixed?
  7. Standard member forkedknight
    Defend the Universe
    16 Oct '11 19:34
    A:
    1-1-5-5-9-9

    B:
    3-3-4-4-8-8

    C:
    2-2-6-6-7-7
  8. Standard member forkedknight
    Defend the Universe
    17 Oct '11 20:43
    Option #2 (P = 17/36)
    A:
    1-6-7-11-15-17

    B:
    2-4-8-12-13-18

    C:
    3-5-9-10-14-16