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Probability dice problem

  1. Sep 3, 2011 #1
    1. The problem statement, all variables and given/known data

    How many ways can you roll six dice so that at least 2 numbers are the same? At least 3? At least 4? At least 5?

    2. Relevant equations



    3. The attempt at a solution

    :cry: I've used every equation in the chapter and filled page after page with numbers all moved around. I'm drowning! Please help. I hate probability.
     
  2. jcsd
  3. Sep 3, 2011 #2

    micromass

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    Hi Arcana! :smile:

    I'd say this problem is a little too annoying to tackle immediately.
    Can you solve first the following for me:

    In how many way can you roll 2 dice such that 2 numbers are the same???

    Solve this first, then we'll analyze what you did.
     
  4. Sep 3, 2011 #3

    I like Serena

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    Hi ArcanaNoir! :smile:

    Let's start with the first.
    You need to consider the negation of the outcome to make the problem a lot easier.

    What is the negation of: "at least 2 numbers are the same"?
     
  5. Sep 3, 2011 #4
    6/36
    because 1,1 or 2,2, etc.
     
  6. Sep 3, 2011 #5

    micromass

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    OK, so you're finding tuples with entries between 1 and 6.

    So for the first one, you need to find 6 tuples (a,b,c,d,e,f) such that we have at least 2 numbers that are the same.

    Following ILS's lead: what is the negation of that.
     
  7. Sep 3, 2011 #6
    All numbers are different? Which is nPr(6,6) = 6!
    Right? So, okay, for at least 2 are the same it's [tex] 6^6-6! [/tex] and then over [itex]6^6[/itex] for the probability.

    I'm really getting messed up for at least 3 and at least 4.
     
  8. Sep 3, 2011 #7
    I'm confused.
     
  9. Sep 3, 2011 #8

    micromass

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    Did my tuples analogy confuse you? :frown: I thought it made it easier...
    I'll probably leave the thread to ILS, then. He can explain things much better than me :biggrin:
     
  10. Sep 3, 2011 #9
    Well yes, the tuples sort of confused me. Can you be more specific and less vague? lol... I need a shove more than a nudge here.
     
  11. Sep 3, 2011 #10

    I like Serena

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    For the next one, first you need to find what the chance is for exactly 2 numbers that are the same.
    After that you can find the answer again with negation.

    So going back to micromass's example, suppose you have 3 dice. :wink:
    How many possibilities for exactly 2 numbers to be the same?
     
  12. Sep 3, 2011 #11
    I don't know how to do that without tabling the possibilities.
     
  13. Sep 3, 2011 #12

    vela

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    You might find it easier to start at the other end. How many ways are there to roll 6 dice so all 6 numbers are the same?

    To figure out "exactly 5 numbers the same", you have to figure out how many ways you can roll 5 dice so all 5 numbers are the same, the number of ways you can roll the die that's different, and then how they can be arranged, e.g. yxxxxxx, xyxxxx, etc.

    Then "at least 5" is "5 all the same" plus "6 all the same".
     
  14. Sep 3, 2011 #13

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    All right, zooming in.
    Suppose we roll the dice one by one.

    What is the chance for the first 2 dice to be the same, and the 3rd die to be different?
     
  15. Sep 3, 2011 #14

    vela

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    That might not be a bad idea. If you write out the combinations in an orderly way, you should be able to identify what factors contribute to the total, and then generalize from there.
     
  16. Sep 3, 2011 #15
    Exactly 5 is 6 times 5= 30

    It's really 3 and 4 that I can't get, and it's because I don't know what formulas to use. The other numbers I did by tabling.
     
  17. Sep 3, 2011 #16
    :frown: I don't know.
     
  18. Sep 3, 2011 #17

    vela

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    Does the order matter? I'm assuming it does, in which case you're missing a factor of 6.
    Well, the factor of 5 above is the total number of ways to roll one die so that it doesn't match the value of the other dice. In the case where you have 4 matching dice and two different, how many combinations are there for the two dice? You know that each can take on one of five possible values.
     
  19. Sep 3, 2011 #18
    There's a formula where you find the number of ways you can arrange the letters in PEPPER,
    and you do 6!/(1!2!3!).

    So, for exactly 4 numbers the same, would I do 6!/(4!1!1!)?

    [edit] never mind I see that's not going to work.
     
  20. Sep 3, 2011 #19

    I like Serena

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    All right.
    I'll make the table for you (first 2 dice the same, 3rd different).

    Code (Text):

                    chance
    1 1  2,3,4,5,6  (1/6)(1/6)(5/6)
    2 2  any not 2  (1/6)(1/6)(5/6)
    3 3  any not 3  (1/6)(1/6)(5/6)
    4 4  any not 4  ...
    5 5  any not 5
    6 6  any not 6
     
    Total chance is: 6 x (1/6)(1/6)(5/6)

    I might also say:
    First die: any number - chance is 6/6
    Second die: the same number - chance 1/6
    Third die: any different number - chance 5/6

    The product is (6/6)(1/6)(5/6).

    Does that make sense?

    If so, then what would the chance be for 4 dice with the first 2 dice to be the same and the other 2 different?
     
  21. Sep 3, 2011 #20
    crap

    halp.

    I really want to know how to do this without tabling, but every question people ask, I can only answer by tabling.
     
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