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Math Help - rolling two die and cardinality

  1. #1
    Senior Member Danneedshelp's Avatar
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    Exclamation rolling two die and cardinality

    When rolling two fair 6-sided dice, it is well known that one can list the sample space of this experiment by using direct products. Answer the following questions:

    Q1: What is the cardinality of the set of all possible sums in this experiment?
    Q2: What is the cardinality of the set of all possible products in this experiment?
    Q3: What is the cardinality of the set of all possible quotients in this experiment?
    Q4: What is the cardinality of the set of all possible values of the first die outcome modulo the blue die outcome in this experiment?

    Firstly, there are 36 outcomes to this experiment. Each outcome can be represented by an ordered pair (i,j). Thus, when I use the letter "i" I am referring to the number on the face of the first die and when I use the letter "j" I am referring to the number on the face of the second die.

    A1: I am looking for the cardinality of the set A={x|x=i+j}. I am assuming the problem is asking for unique solutions. We know the smallest product is 2 and largest is 12. First, all ordered pairs with a first coordinate equal to 1 can be added to their second component giving us the sums 2 through 7. We then look at all ordered pairs with a first coordinate equal to 2 and see the only sum we haven't seen yet happens when the second coordinate is 6. I continue this and find that,

    |A| = 11.

    A2: I am looking for the cardinality of the set B={x|x=i*j}. I got that

    |B| =18

    A3: I am having trouble with this one. I first though I'd write out the set of unique ordered pairs (i.e., if I already have (1,2), I delete (2,1) for the set). I originally did this for the first problem. The set had 21 elements. However, the quotient from (1,2) is different from (2,1). So, I figured I could multiply 21 by 2 to get 42 possible quotients. I am not sure if I am looking at this correctly.

    A3: I am most confused by this problem. I started out by writing out x=i(mod j). So, I need to look at each value of j=1,2,...,6. So, I would start with the first case; that is, j=1. Now, I have to run through each i=1,2,...,6. So, I am looking for the equivalence classes of each i. In mod 1, the equivalence class for i=1 is the set {1,2,3,4,5,6} which includes all outcomes.

    I am not sure if I am approaching this correctly. I am not too familiar with mod operation.

    Thanks for the help.
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Danneedshelp View Post
    When rolling two fair 6-sided dice, it is well known that one can list the sample space of this experiment by using direct products. Answer the following questions:

    Q1: What is the cardinality of the set of all possible sums in this experiment?
    Q2: What is the cardinality of the set of all possible products in this experiment?
    Q3: What is the cardinality of the set of all possible quotients in this experiment?
    Q4: What is the cardinality of the set of all possible values of the first die outcome modulo the blue die outcome in this experiment?

    Firstly, there are 36 outcomes to this experiment. Each outcome can be represented by an ordered pair (i,j). Thus, when I use the letter "i" I am referring to the number on the face of the first die and when I use the letter "j" I am referring to the number on the face of the second die.

    A1: I am looking for the cardinality of the set A={x|x=i+j}. I am assuming the problem is asking for unique solutions. We know the smallest product is 2 and largest is 12. First, all ordered pairs with a first coordinate equal to 1 can be added to their second component giving us the sums 2 through 7. We then look at all ordered pairs with a first coordinate equal to 2 and see the only sum we haven't seen yet happens when the second coordinate is 6. I continue this and find that,

    |A| = 11.

    A2: I am looking for the cardinality of the set B={x|x=i*j}. I got that

    |B| =18

    A3: I am having trouble with this one. I first though I'd write out the set of unique ordered pairs (i.e., if I already have (1,2), I delete (2,1) for the set). I originally did this for the first problem. The set had 21 elements. However, the quotient from (1,2) is different from (2,1). So, I figured I could multiply 21 by 2 to get 42 possible quotients. I am not sure if I am looking at this correctly.

    A3: I am most confused by this problem. I started out by writing out x=i(mod j). So, I need to look at each value of j=1,2,...,6. So, I would start with the first case; that is, j=1. Now, I have to run through each i=1,2,...,6. So, I am looking for the equivalence classes of each i. In mod 1, the equivalence class for i=1 is the set {1,2,3,4,5,6} which includes all outcomes.

    I am not sure if I am approaching this correctly. I am not too familiar with mod operation.

    Thanks for the help.
    A1 agree

    A2 agree

    A3 certainly can't be 42 since 42 > 36. I recommend looking for duplicates, 1/2 = 2/4 = 3/6, and 1/3 = 2/6. Actually I guess those are the only ones, multiply by 2 for reciprocals, take care you don't accidentally get off by 1 or some such. Edit: Sorry also 1/1 = 2/2 = ... of course. Edit: Ah and 2/3 has a duplicate.. well anyway it's your job to find them

    A4 mod operation gives "remainder when divided by". Think about it you will see the set is {0,1,...,5}
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    MHF Contributor undefined's Avatar
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    Forgot to mention that the reasoning in A1 is a bit more than necessary. Once you realise that all sums (you made a typo by writing products) between 2 and 12 inclusive are possible, you know the answer is 12-2+1 = 11.
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  4. #4
    Senior Member Danneedshelp's Avatar
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    I am still not sure what to do with the 4th part of the problem. Are all my operations going to be mod6, since there are six outcomes?
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  5. #5
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Danneedshelp View Post
    I am still not sure what to do with the 4th part of the problem. Are all my operations going to be mod6, since there are six outcomes?
    Note: I assumed blue die means second die.

    1 mod 1 = 0
    2 mod 1 = 0
    ...
    1 mod 2 = 1
    2 mod 2 = 0
    3 mod 2 = 1
    4 mod 2 = 0
    ...
    1 mod 3 = 1
    2 mod 3 = 2
    3 mod 3 = 0
    1 mod 3 = 1
    ...
    you will get the most from mod 6
    1 mod 6 = 1
    ...
    5 mod 6 = 5
    6 mod 6 = 0

    Make sense?
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  6. #6
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    On Q2. |B| = 18. I did it by enumerating the possible distinct outcomes.
    But is their a smart way count this? For e.g what if the die had 100 faces.

    Have been thinking about this but couldn't find a better way - Any pointers please?
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  7. #7
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by aman_cc View Post
    On Q2. |B| = 18. I did it by enumerating the possible distinct outcomes.
    But is their a smart way count this? For e.g what if the die had 100 faces.

    Have been thinking about this but couldn't find a better way - Any pointers please?
    Looks like there is no known nice formula, just algorithms and estimation (lower and upper bounds, asymptotic behaviour).

    id:A027424 - OEIS Search Results

    For getting exact answers, 100 faces is trivial for a naive computer program, I think it's even so for 10,000 faces on a low-end computer and 100,000 faces on a nicer system with a bit of a wait and 1.25 GB RAM to spare, but for a million faces something more clever would need to be found (unless you have a lot of computing resources and/or time on your hands), perhaps by using floor/integer division, maybe even integer factorisation/primes, not sure, interesting question.
    Last edited by undefined; September 2nd 2010 at 06:24 AM. Reason: slight clarification
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