Results 1 to 7 of 7
Like Tree4Thanks
  • 2 Post By Plato
  • 1 Post By HallsofIvy
  • 1 Post By Plato

Math Help - Dice interpretation problem

  1. #1
    Senior Member
    Joined
    Apr 2009
    Posts
    308

    Dice interpretation problem



    I understand how the question arrived at the conditional probabilities, however couldn't the question also be interpreted this way?

    Since the question never stated that the 2 dice were different, what if we assume they were not distinct and also we can assume that the 2 dice is thrown at the same time, thus P(S = 4 | M=3) = \frac{P( S=4 \cap M = 3)}{P(M=3)}

    Now to have a sum of 4 and a max of 3, all we need is to roll a 1 and a 3, there is no order here since we are throwing the dice at the same time and the dice are NOT distinct. Thus the numerator becomes 1/6 * 1/6

    To get a maximum of 3, all we need is to roll a 1 and a 3 or 2 and a 3 or 3 and a 3 (again there is no order since we are throwing the dice at the same time)

    Thus the denominator is 3*(1/6 * 1/6)

    Thus P(S = 4 | M=3) = \frac{1}{3}?

    Would the above also be a valid interpretation?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,790
    Thanks
    1687
    Awards
    1

    Re: Dice interpretation problem

    I can give a quicker explanation.
    M=3 is the space \{(1,3),~(3,1),~(2,3),~(3,2),~(3,3)\}

    Now it is clear that P(S=5)|M=3)=\frac{2}{5}.
    Thanks from usagi_killer and HallsofIvy
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Apr 2009
    Posts
    308

    Re: Dice interpretation problem

    Thanks Plato, however if the 2 dices are rolled at exactly the same time (and the 2 dices are assumed to be identical) then wouldn't (1,3) be the same as (3,1) ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,957
    Thanks
    1631

    Re: Dice interpretation problem

    Quote Originally Posted by usagi_killer View Post


    I understand how the question arrived at the conditional probabilities, however couldn't the question also be interpreted this way?

    Since the question never stated that the 2 dice were different, what if we assume they were not distinct and also we can assume that the 2 dice is thrown at the same time, thus P(S = 4 | M=3) = \frac{P( S=4 \cap M = 3)}{P(M=3)}

    Now to have a sum of 4 and a max of 3, all we need is to roll a 1 and a 3, there is no order here since we are throwing the dice at the same time and the dice are NOT distinct. Thus the numerator becomes 1/6 * 1/6

    To get a maximum of 3, all we need is to roll a 1 and a 3 or 2 and a 3 or 3 and a 3 (again there is no order since we are throwing the dice at the same time)
    Because, as you say, "there is no order here", you must also consider "a 3 and a 1" and "a 3 and a 2".

    Thus the denominator is 3*(1/6 * 1/6)

    Thus P(S = 4 | M=3) = \frac{1}{3}?

    Would the above also be a valid interpretation?
    Thanks from usagi_killer
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,790
    Thanks
    1687
    Awards
    1

    Re: Dice interpretation problem

    Quote Originally Posted by usagi_killer View Post
    Thanks Plato, however if the 2 dices are rolled at exactly the same time (and the 2 dices are assumed to be identical) then wouldn't (1,3) be the same as (3,1) ?
    Absolutely not. If you roll two dice the outcome space is a set of 36 ordered pairs.
    If you roll three dice the outcome space is a set of 6^3 ordered pairs.
    If you roll n dice the outcome space is a set of 6^n ordered pairs.

    It makes no difference if you roll n dice at once or one die n times the outcome space is the same.
    Last edited by Plato; June 27th 2012 at 09:16 AM.
    Thanks from usagi_killer
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Apr 2009
    Posts
    308

    Re: Dice interpretation problem

    Thanks to both, I understand now
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Apr 2009
    Posts
    308

    Re: Dice interpretation problem

    Hi again, I just have an extension to this problem:



    I was wondering in order to compute E(S) using the method they described, is it really as tedious as the following? (Obviously I could just apply the definition of E(S) = \sum_{s = 2}^{12} s \times p_S(s) straight away, but here I wanted to try their exercise)

    So in order to compute E(S) using the method they described we have to work out E(S|M=y) first which is given by E(S|M=y) = \sum_{s = 2}^{12} s \times p_{S|M}(s|y) = \sum_{s = 2}^{12} s \left(\frac{p_{S,M}(s,y)}{p_{M}(y)}\right)

    Now we have do that for y = 1, \cdots, 6: which means we have to manually find \sum_{s = 2}^{12} s \left(\frac{p_{S,M}(s,y)}{p_{M}(y)}\right) for y = 1, \cdots, 6

    Then we have to manually find p_M(y) for y = 1, \cdots, 6

    Then after all that sub back into E(S) = \sum_{y=1}^{6} p_M(y) E(S|M=y)

    I did all of the above and ended up with so much working, is there any shortcuts rather than applying the definitions of each expression directly?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. dice problem.
    Posted in the Statistics Forum
    Replies: 2
    Last Post: February 18th 2011, 04:45 PM
  2. Dice problem
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: January 23rd 2010, 07:10 AM
  3. Dice Problem
    Posted in the Math Challenge Problems Forum
    Replies: 2
    Last Post: November 3rd 2008, 03:22 PM
  4. Replies: 11
    Last Post: July 17th 2008, 05:37 AM
  5. Dice Problem
    Posted in the Statistics Forum
    Replies: 5
    Last Post: January 15th 2008, 05:04 PM

Search Tags


/mathhelpforum @mathhelpforum