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Math Help - Probability and Dice Loading

  1. #1
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    Probability and Dice Loading

    Is it possible to load three dice so that the first will probably show a higher number than the second, the second will probably show a higher number than the third, and the third will probably show a higher number than the first?

    Loading dice means you change the probability that each number will be rolled, but they all still add up to one.

    Please only give me hints. Don't tell me how to solve the problem. Thanks.
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  2. #2
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    Quote Originally Posted by seuzy13 View Post
    Is it possible to load three dice so that the first will probably show a higher number than the second, the second will probably show a higher number than the third, and the third will probably show a higher number than the first?
    Let me confirm this.

    If you have 3 dice, d_1, d_2, d_3 such that the numbers showing once rolled are n_1, n_2, n_3 then you are trying to find where

     n_1> n_2> n_3> n_1 ? I don't think this is possible.
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  3. #3
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    Quote Originally Posted by pickslides View Post
    Let me confirm this.

    If you have 3 dice, d_1, d_2, d_3 such that the numbers showing once rolled are n_1, n_2, n_3 then you are trying to find where

     n_1> n_2> n_3> n_1 ? I don't think this is possible.
    Sort of, but not quite. Dice one has a probability of showing a higher number than dice two, etc. So if dice one has, for instance, a 1/2 probability of getting a six and dice two has a 1/4 probability of getting a 5 and a 1/4 probability of getting a 4 (the rest of the probabilities being evenly distributed over the remaining numbers 1-6), then dice one will probably show a higher number than dice two. So is there a way to continue this such that the last dice will probably be higher than the first dice?

    It seems like I wouldn't be asked this question if it weren't possible.
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  4. #4
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    Notation: n_i > n_j means "at least a 50% chance that n_i is bigger than n_j

    proposition: n_1 > n_2 > n_3 > n_1 is impossible

    remove the middle steps, and just think about 2 dice to start with.
    n_1 > n_2 > n_1
    You should be able to show that this is impossible.

    Now add 1 more die:
    n_1 > n_2 > n_3 > n_1


    You can probably generalise your method from the 2 dice problem. i also suspect that its possible to show that the 2 dice version is a necessary condition for the 3 dice version. Either of these approaches would complete the proof


    I hope that isn't too much info!
    Last edited by SpringFan25; May 24th 2010 at 03:40 PM.
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  5. #5
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    Quote Originally Posted by SpringFan25 View Post
    Notation: n_i > n_j means "at least a 50% chance that n_i is bigger than n_j

    remove the middle steps, and just think about 2 dice to start with.
    n_1 > n_2 > n_1
    You should be able to show that this is impossible.

    Now add 1 more die:
    n_1 > n_2 > n_3 > n_1

    Try and see a way that your 2 dice version of the problem is a necessary condition for the 3 dice version to work. Since you have already established that 2 dice version is impossible, this will complete your proof.
    I know it really does seem impossible, and if I were working with ordinary numbers instead of probability distributions, I know I could easily prove that, but is it possible that this is different?

    Ah, well, I'll think about it a little more. Thanks.
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  6. #6
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    An alternative, and probably faster approach is to tackle your proposition directly. Try and show that:

    n_3 > n_2
    and
    n_2 > n_1

    is sufficient condition for
    n_3 > n_1


    This would make the proposition impossible.
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  7. #7
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    I think you're looking for nontransitive dice.
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  8. #8
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    Quote Originally Posted by undefined View Post
    I think you're looking for nontransitive dice.
    Yes, that's definitely it. Thanks.
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  9. #9
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    that explains why i couldn't finish those proofs then
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