Results 1 to 3 of 3

Math Help - Dice

  1. #1
    Junior Member
    Joined
    Mar 2010
    Posts
    40

    Dice

    Hi,

    We throw three dice: A(1-1-1-2-2-3), B(1-2-3-4-5-6) and C(4-4-4-4-5-6), what are all different possibilities, ans what are the possibilities to have the number 1 exactly twice???
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Apr 2010
    Posts
    57
    Just to clarify, only dice B is a standard 1-6 dice? The rest have 3 or 4 sides with the same number?

    Under that assumption, the answer to the second half is

    (1/2) * (1/6) = (1/12) = 8.33% chance that you will have two 1's (the third die is irrelevant in this case)

    As for the first question, it's impossible to answer since you haven't clarified what counts as a possibility. For example, if the 3 dice end up being {1} {2} {4}, is that the same as {2} {1} {4}?

    Assuming that 1 2 3 = 3 2 1 etc., there are 32 unique combinations total.

    Last edited by shenanigans87; April 29th 2010 at 12:31 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,707
    Thanks
    626
    Hello, bhitroofen01!

    We throw three dice: A(1-1-1-2-2-3), B(1-2-3-4-5-6) and C(4-4-4-4-5-6).
    What are all different possibilities?
    There are 54 possible outcomes.

    . . \begin{array}{cccccc}<br />
(1,1,4) & (1,2,4) & (1,3,4) & (1,4,4) & (1,5,4) & (1,6,4) \\<br />
(1,1,5) & (1,2,5) & (1,3,5) & (1,4,5) & (1,5,5) & (1,6,5) \\<br />
(1,1,6) & (1,2,6) & (1,3,6) & (1,4,6) & (1,5,6) & (1,6,6)\end{array}

    . . \begin{array}{cccccc}<br />
(2,1,4) & (2,2,4) & (2,3,4) & (2,4,4) & (2,5,4) & (2,6,4) \\<br />
(2,1,5) & (2,2,5) & (2,3,5) & (2,4,5) & (2,5,5) & (2,6,5) \\<br />
(2,1,6) & (2,2,6) & (2,3,6) & (2,4,6) & (2,5,6) & (2,6,6) \end{array}

    . . \begin{array}{cccccc}<br />
(3,1,4) & (3,2,4) & (3,3,4) & (3,4,4) & (3,5,4) & (3,6,4) \\<br />
(3,1,5) & (3,2,5) & (3,3,5) & (3,4,5) & (3,5,5) & (3,6,5) \\<br />
(3,1,6) & (3,2,6) & (3,3,6) & (3,4,6) & (3,5,6) & (3,6,6) \end{array}




    What are the possibilities to have the number 1 exactly twice?
    If you mean probability, then shenanigans87 is absolutely correct!


    There is basically one scenario:

    . . \begin{array}{ccccc}P(\text{1 on Die A}) &=& \frac{3}{6} &=& \frac{1}{2}\\ \\[-3mm]<br /> <br />
P(\text{1 on Die B}) &=& \frac{1}{6} \\ \\[-3mm]<br /> <br />
P(\text{any on Die C}) &=& \frac{6}{6} &=& 1 \end{array}

    Therefore: . P(\text{two 1's}) \;=\;\frac{1}{2}\cdot\frac{1}{6}\cdot 1 \;=\;\frac{1}{12} \;=\;8\tfrac{1}{3}\%

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Dice
    Posted in the Statistics Forum
    Replies: 3
    Last Post: May 12th 2010, 01:40 PM
  2. Replies: 0
    Last Post: December 29th 2009, 08:08 PM
  3. dice
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 30th 2009, 01:08 PM
  4. Dice
    Posted in the Statistics Forum
    Replies: 1
    Last Post: February 27th 2009, 02:41 AM
  5. dice
    Posted in the Statistics Forum
    Replies: 2
    Last Post: October 17th 2007, 12:17 PM

Search Tags


/mathhelpforum @mathhelpforum