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Specific question about what sample spaces are exactly

  1. Dec 28, 2014 #1

    s3a

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    1. The problem statement, all variables and given/known data
    A card is drawn at random from an ordinary deck of 52 playing cards (which means there are 0 jokers).

    Give the sample space.

    2. Relevant equations
    A set that consists of all possible outcomes of a random experiment is called a sample space, and each possible outcome of the aforementioned random experiment is called a sample point.

    3. The attempt at a solution
    My confusion:
    Is

    S = {A♥, A♠, A♦, A♣, 2♥, 2♠, 2♦, 2♣, 3♥, 3♠, 3♦, 3♣, 4♥, 4♠, 4♦, 4♣, 5♥, 5♠, 5♦, 5♣, 6♥, 6♠, 6♦, 6♣, 7♥, 7♠, 7♦, 7♣, 8♥, 8♠, 8♦, 8♣, 9♥, 9♠, 9♦, 9♣, 10♥, 10♠, 10♦, 10♣, J♥, J♠, J♦, J♣, Q♥, Q♠, Q♦, Q♣, K♥, K♠, K♦, K♣}

    the only way to describe the sample space (ignoring order of the elements, since a set with the same exact elements but in a different order are equal), or not?

    Is it possible to give another set S_2 that consists of unordered pairs such as what is shown below?:

    S_2 = {(A,♥), (A,♠), (A,♦), (A,♣), (2,♥), (2,♠), (2,♦), (2,♣), (3,♥), (3,♠), (3,♦), (3,♣), (4,♥), (4,♠), (4,♦), (4,♣), (5,♥), (5,♠), (5,♦), (5,♣), (6,♥), (6,♠), (6,♦), (6,♣), (7,♥), (7,♠), (7,♦), (7,♣), (8,♥), (8,♠), (8,♦), (8,♣), (9,♥), (9,♠), (9,♦), (9,♣), (10,♥), (10,♠), (10,♦), (10,♣), (J,♥), (J,♠), (J,♦), (J,♣), (Q,♥), (Q,♠), (Q,♦), (Q,♣), (K,♥), (K,♠), (K,♦), (K,♣)}

    Assuming everything I said up until now is correct, are A♥ and (A,♠) each a (correct/valid) sample point?

    In short, I want to ask if the set can be a "regular" set with one element at a time, or if it can also be a set of unordered pairs (such as the above set S_2, where the ordering of the pairs and the ordering of the elements in the pairs both do not matter - because that is what it means to be a set with unordered pairs)?

    Any input would be GREATLY appreciated!
     
  2. jcsd
  3. Dec 28, 2014 #2

    haruspex

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    Yes, they're equally valid representations.
    Both sets consist of 52 elements. There's nothing special about S_2 at that level. They're both 'one element at a time'.
    Your S_2 is a set of ordered pairs. The components of each pair are of distinct types, so it makes little sense to write them in the other order.
    You can express S_2 as the Cartesian product of two sets, which would be a little neater than writing out all 52 elements.
     
  4. Dec 28, 2014 #3

    Ray Vickson

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    Both are correct, and both describe the same thing, but in different ways. Your ordered pairs are of the form (rank, suit), but you could also list them as (suit,rank), to give you a third (but still equivalent) sample space. These give different sets mathematically, but there is no practical difference between them. It is just the case that in any single problem you should choose the convention you want to use and stick to it from start of the problem to its finish---don't mix them up in the same problem.

    Personally, I would either label the suits as H,D,S,C or 1,2,3,4 (with a documented translation between number and suit), and the rank as perhaps going from 1 to 13 (Ace = 1, King = 13) so I would list them as {(1,H),(1,D),... (13,S),(13,C)}. The unordered pairs would be 1H, 1D,..., obtained just by erasing the parentheses and the comma. Alternatively, just number the cards from 1 to 52 in some documented and specified way, and take the sample space as {1,2,3, ...,52}.
     
  5. Dec 30, 2014 #4

    s3a

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    Thanks for the answers. :)

    So, is the reason why a sample space set is ordered (as opposed to unordered), if it is a relation, simply because the number of elements has to be equal to the number of possible outcomes?
     
  6. Dec 30, 2014 #5

    Ray Vickson

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    I do not understand your question, but let me expand, anyway.

    There are several different ways to represent a sample space, and some of them may be more convenient than others, depending on context. In any case, for finite sample spaces the number of elements must equal the number of distinct outcomes, but what "distinct outcomes" means is context-dependent. For example, look at two tosses of a coin (or one toss of two coins at once). One way to represent the sample space is S1 = {HH,HT,TH,TT}, where each outcome represents toss1, toss2. For a fair coin each outcome has probability 1/4. You could answer just about any question about tossing a (fair) coin twice using sample space S1. However, suppose we are interested only in the number of Heads, not their order of occurrence. We could represent this via a sample space S2 = {0,1,2}, representing the numbers of possible Heads. Now we need different probabilities: P(0) = 1/4, P(1) = 1/2, P(2) = 1/4. You could answer any question about numbers of heads/tails (but not their order) using sample space S2. S1 is more versatile, allowing you to answer a wider range of questions, but in some problems S2 is more convenient (albeit, more restricted in scope).
     
  7. Dec 30, 2014 #6

    haruspex

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    The sample space is unordered. In your {(suit, pip)} notation, the elements of the sample space are ordered pairs.
     
  8. Jan 8, 2015 #7

    s3a

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    Sorry for the delayed response!

    Okay, so, basically, the ordering of the elements (where, in the case of S_2, a pair is considered to be an element of the set) in both of the sets S and S_2 don't have to be ordered, but the ordering of the elements in the pairs of S_2 does matter (assuming the set is a binary relation), right?

    In other words, the ordering of the elements in S simply does not matter at all, and, in the case of S_2, after having defined the sets that are being related and which set is being related to which other set, I could have (A,♥), (A,♠) be in the order (A,♠), (A,♥) instead but not (♥,A), (♠,A), for example, right? (I emphasized the "after having defined the sets that are being related and which set is being related to which other set" part, because I could have defined it such that (♥,A), (♠,A) is acceptable and (A,♥), (A,♠) is not.)
     
  9. Jan 8, 2015 #8

    Ray Vickson

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    Basically, the answer to your question "right?" is YES. When constructing the sample space in preparation to solving a problem, you are free to use whichever of the two conventions (A,H) or (H,A) appeals to you---or whichever is more convenient for a possible algorithm you might want to use---but once you have made a choice you should stick to it from start to finish of the problem. Of course, in another problem you might choose the opposite convention---again, as long as you don't change it in the middle of the problem's analysis.
     
  10. Jan 8, 2015 #9

    s3a

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    Alright, thank you (both) a lot! :)
     
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