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Probability w/ dice

  1. Jan 9, 2007 #1
    Die A has four 9's and two 0's on its faces. Die B has four 3's and two 11's on its faces. when either of these dice is rolled, each face has an equal chance of landng on top. two players are going to play a game. the first player selects a die and rolls it. the second player rolls the remaining die. the winner is the player whose die has the higher number on top.

    A) suppose you are the first player and you want to win the game. which die would you select and why

    B) Suppose the player using die A receives 45 tokens each time he or she wins the game. how many tokens must the player using die B receive each time he or she wins in order for this to be a fair game?

    i really dont know how to start this. for a, i would think that, logically, die B would be better to choose but i dont know how to prove that statistically (even if it's right in the first place)

    any help is appreciated. thanks in advance
     
  2. jcsd
  3. Jan 9, 2007 #2

    Fredrik

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    Suppose that you choose A. These are the probabilities of the different ways this can turn out, and the results:

    P(you 9, opponent 3)=2/3*2/3=4/9 . . . You win.
    P(you 9, opponent 11)=2/3*1/3=2/9 . . . You lose.
    P(you 0, opponent 3)=2/3*2/3=4/9 . . . You lose.
    P(you 0, opponent 11)=1/3*1/3=1/9 . . . You lose.

    You win 4/9 times and lose 5/9.

    Now suppose instead that you choose B, do the same calculation and compare the results.
     
  4. Jan 9, 2007 #3
    thanks, i got it
     
  5. Nov 28, 2007 #4
    i still dont get it :(
     
  6. Nov 28, 2007 #5

    Office_Shredder

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    Ok, first, spamming asking for help isn't going to get it any quicker... I was originally going to post in your other thread before realizing that you just copied this one, got annoyed and didn't post.

    Second, the concept is that since the dice rolls are independent, P(Die A=x, Die B=y) = P(A=x)*P(B=y). Then you just slug through the combinations
     
  7. Nov 28, 2007 #6
    sorry bout that
     
  8. Nov 29, 2007 #7

    uart

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    Ok how about this one with three dice.

    Dice "A" has one face marked "6" and the other five faces marked "3".

    Dice "B" has one face marked "1" and the other five faces marked "4".

    Dice "C" has three faces marked "2" and the other three faces marked "5".

    Same game play as before except player_1 gets to select his choice from the three dice while player_2 only gets to select from the remaining two. Explain why it's always advantagous to be player_2 in this game.
     
    Last edited: Nov 29, 2007
  9. Nov 29, 2007 #8

    Office_Shredder

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    If player 1 picks die A, then player 2 can pick die B, and by the same logic as above, win quite often. Try finding dice that will always beat die B and die C, and prove it using the same argument. Then no matter which die player 1 picks, player 2 can pick one that will win over 1/2 the time
     
  10. Nov 29, 2007 #9
    ohhh i get it...thanks soo much guys :)
     
  11. Nov 30, 2007 #10

    uart

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    Yep it's a classic example of the "non-transistive dice" problem.

    - Dice_B wins over dice_A by a ratio 25:36.
    - Dice_C wins over dice_B by a ratio 7:12.
    - Dice_A wins over dice_C by a ratio 7:12.

    It's a neat example of something that's non-transitive.
     
  12. Nov 30, 2007 #11

    Office_Shredder

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    Oh, I thought you were asking for help with that problem. I should have left it for someone else to do
     
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