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Dice with weighted sides, looking for odds of a sum:faster way

  1. Oct 17, 2007 #1
    Dice with weighted sides, looking for odds of a sum:faster way plz

    1. The problem statement, all variables and given/known data

    6 side die is weighted so prob of rolling 2,3,4,5,6 is same and rolling 1 is 3x rolling a 2. If die is thrown twice, what's odds of getting a sum of 4?

    2. Relevant equations



    3. The attempt at a solution
    x+x+x+x+x+3x represents the "chances" of rolling a 2,3,4,5,6, and 1. So 2,3,4,5,6 individually have a 1/8 chance of rolling each time. 1 has a 3/8 chance of hitting each time.

    a)For sum of 4 from 2 rolls, I need to get 1,2,3 on either rolls. Chance of getting that on first roll is 1/8 + 1/8 + 3/8 = 5/8.

    If I get a 1 or 2 on first roll, then my 2nd roll can only be one result, so 5/8*1/8=5/64. Then now it's weird. If I get a 3, I multiply 5/8 by 3/8 right? o_O Answer is 7/64 though and for 3, it'd be 15/64.

    Also, 7/64 doesn't match the 5/64 I got for getting a 1 or 2 on 2nd roll.

    b)I made a chart. I got the correct answer this way, but can someone show me a faster way?

    http://s000.tinyupload.com/index.php?file_id=15919977748640328011
     
  2. jcsd
  3. Oct 17, 2007 #2

    Hurkyl

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    That can't be right. 5/8 is the odds of getting a 1, 2, or 3 on the first roll... but you said you were considering the case where you got a 1 or 2 on the first roll.


    Incidentally, why bother breaking the problem into "first roll" and "second roll"? Why not just consider each possible pair of rolls as outcomes?
     
  4. Oct 17, 2007 #3
    Incidentally, why bother breaking the problem into "first roll" and "second roll"? Why not just consider each possible pair of rolls as outcomes?

    You mean 5/8 * 5/8? That wouldn't equal 7/64. My chart works out to 7/64, but I want to find a less meticulous way.
     
  5. Oct 17, 2007 #4

    Hurkyl

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    Well, most of the outcomes in {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)} do not have a sum of 4. So clearly the answer is not the probability of getting any one of these outcomes.
     
  6. Oct 17, 2007 #5

    Hurkyl

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    I can finally see your chart. I'm not sure what the problem with that approach is -- unless you mean writing the entire chart. Of course you don't need to write down all 36 outcomes; you only need to write down the 3 relevant ones.
     
  7. Oct 17, 2007 #6
    Just wondering if there was an equational way?
     
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