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Basic Stats Q - Am I Missing Something Obvious? Watch

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    Hey,

    I've been doing a bit of stats lately and don't get one concept. Here's a Q:

    5 Cards dealt w/o replacement from a pack of 52. Find Prob that 3 of the 5 are hearts. The book explains it like this:

    There's 52C5 sets of 5 cards.
    Theres 13C3 sets of 3 hearts and 39C2 sets of 'non-hearts'.
    So the no. of sets of 5 cards with 3 hearts is (13C3)x(39C2)
    So P(3 hearts in a set of 5) = (13C3)x(39C2) / 52C5

    Am I missing something obvious? To me, the 52 cards aren't distinct, so you aren't going to get 52C5 sets of 5. I could have a billion cards, half blue half red, and choose 2, but I don't get 1billionC2 combos. It'd be blueblue, bluered or redred. They seem to do the same thing regarding 13C3; to me you only get one combo, a 3-set of hearts. The concept of different combinations of 3-hearts doesn't make sense to me; all these sets are the same.

    Any help? There's another Q with 47 sweets; 22 red, 15 blue and 10 green, and the answers are achieved using this type of working too. But it doesn't make sense to me because we don't have 47 distinct sweets, or different sets of only-blue sweets etc (in analogy to different sets of 3 hearts)

    Thanks.
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    The problem with your approach is that if you treated the cards as identical apart from whether they have a heart and you don't worry about order then your workings would look something like:

    There's 6 sets of 5 cards (0 hearts, 1 heart, ..., 5 hearts)
    There's 1 set of 3 hearts and 1 set of 2 non-hearts
    So the number of sets of 5 cards with 3 hearts is 1

    But then when you calculate the probability, it doesn't work. You can't just do 1/6 because these 6 sets of heart/non-heart combinations are not equally likely. In case you don't believe me:
    Spoiler:
    Show
    In a lottery you either have a winning ticket or a non-winning ticket. That doesn't mean your chances of winning are 1/2


    By treating the different cards as distinct, they're able to say that there are 52C5 sets of 5 cards and each of these sets are equally likely so they can use that to calculate a probability.
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    (Original post by ttoby)
    The problem with your approach is that if you treated the cards as identical apart from whether they have a heart and you don't worry about order then your workings would look something like:

    There's 6 sets of 5 cards (0 hearts, 1 heart, ..., 5 hearts)
    There's 1 set of 3 hearts and 1 set of 2 non-hearts
    So the number of sets of 5 cards with 3 hearts is 1

    But then when you calculate the probability, it doesn't work. You can't just do 1/6 because these 6 sets of heart/non-heart combinations are not equally likely. In case you don't believe me:
    Spoiler:
    Show
    In a lottery you either have a winning ticket or a non-winning ticket. That doesn't mean your chances of winning are 1/2
    I never said it could be 1/6, because it depends on the relative amount of hearts and non-hearts in the pack. There's more chance of selecting non-hearts, so yes the 6 sets are not equally likely. 5 non-hearts and 1 heart, is evidently more likely than 5 hearts and 1 non heart, even without calculation. But my point still stands; the cards aren't distinct.

    Regarding your lottery analogy; yes you either have a winning ticket or a non-winning ticket. But the non-winning tickets heavily outweigh the winning ones, so the probabilities will reflect that. Just like non-hearts outweighing the hearts.

    (Original post by ttoby)
    By treating the different cards as distinct, they're able to say that there are 52C5 sets of 5 cards and each of these sets are equally likely so they can use that to calculate a probability.
    I can see that's what they're trying to do, and no doubt it works, but it doesn't seem well justified to me. Again look at my example with a billion balls, half red and half blue. The only combos are redred, redblue and blueblue.

    I'm struggling to wrap my head around the idea that they'll just treat the cards as distinct for the sake of calculating probability, then work out the various 'unique' sets, then divide to get probabilities ... even though the cards aren't distinct.

    Anything else you can add to help me?
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    (Original post by Physics Enemy)
    ... even though the cards aren't distinct.

    Anything else you can add to help me?
    The cards are distinct.
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    To add, I can see how this is a good trick for the sake of calculating probability:

    Treating the cards as distinct allows us to have 52C5 unique 5-sets, and 13C3 unique 3-sets of hearts, and 39C2 unique 2-sets of non-hearts. 13C3x39C2 gives us the number of 5-sets with 3 hearts and 2 non-hearts. Dividing this by 52C5 gives us the relative frequency of obtaining such a set out of all the sets possible, and thus it's probability.

    I guess, as this trick seems a little counter intuitive at first, how else could one do the Q? That got me thinking. I have a pack of 52 with 39 non-hearts and 13 hearts. I want to know the probability of selecting 5 cards such that 3 are hearts, and 2 are non-hearts.

    Heart -> Heart -> Heart -> Non -> Non ... is one way
    Heart -> Heart -> Non -> Heart -> Non ... is another way

    etc.

    And there's many ways of getting 3 hearts and 2 non-hearts (5! / 3!2! ways infact). Each of these ways has it's own assigned probability; it's conditional since each card taken affects the probability of the next card taken, and so on.

    If you work out the probability for each way, then sum them, you should get the probability of 3 hearts, 2 non-hearts. But this is clearly a longer method!
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    (Original post by Get me off the £\?%!^@ computer)
    The cards are distinct.
    I don't see how. In any case, this method is also used for the sweets Q, where there's only 3 colours of various proportion, making up 47 sweets. They use the same trick of pretending they're all distinct at first, and then proceeding with the combinations. I can see that it's a good trick though and works well. If I did it via the alternative method I suggested, I'd have to work out 10 probabilities (1 for each of the 10 ways of pulling out 3 hearts and 2 non-hearts out of 5 draws from the pack) and then sum them.

    EDIT: On the sweets Q I tried my method of finding the conditional probabilities for each of the ways, then summing, and it works perfectly. You just end up summing the probabilities assigned to each permutation to get your answer. The counting trick works very nicely though and is much quicker. :-)
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    (Original post by Physics Enemy)
    To add, I can see how this is a good trick for the sake of calculating probability:

    Treating the cards as distinct allows us to have 52C5 unique 5-sets, and 13C3 unique 3-sets of hearts, and 39C2 unique 2-sets of non-hearts. 13C3x39C2 gives us the number of 5-sets with 3 hearts and 2 non-hearts. Dividing this by 52C5 gives us the relative frequency of obtaining such a set out of all the sets possible, and thus it's probability.

    I guess, as this trick seems a little counter intuitive at first, how else could one do the Q? That got me thinking. I have a pack of 52 with 39 non-hearts and 13 hearts. I want to know the probability of selecting 5 cards such that 3 are hearts, and 2 are non-hearts.

    Heart -> Heart -> Heart -> Non -> Non ... is one way
    Heart -> Heart -> Non -> Heart -> Non ... is another way

    etc.

    And there's many ways of getting 3 hearts and 2 non-hearts (5! / 3!2! ways infact). Each of these ways has it's own assigned probability; it's conditional since each card taken affects the probability of the next card taken, and so on.

    If you work out the probability for each way, then sum them, you should get the probability of 3 hearts, 2 non-hearts. But this is clearly a longer method!
    That is a valid method. In fact it's the method I would have used if I hadn't seen the answers. What I would have done for that is calculated the probability of getting heart, heart, heart, non, non (in that order) then multiplied it by 5! / (3!2!) because each ordering of 3 hearts/2 non is equally likely.

    To get the probability of 3 hearts then 2 non-hearts, think about how many cards of each type are left in the pack as you pull out each card, so you should get the probability as (13/52)*(12/51)*(11/50)*(39/49)*(38/48). You can then calculate this and multiply it by 5! / (3!2!) to get your final answer.

    With your billion balls you do need to do a similar sort of trick. The probability of getting red,red is not the same as the probability of getting red,blue in that order. This is because once you've chosen your red ball, there are 0.5billion-1 red balls left and 0.5billion blue balls left so you are slightly more likely to have your next ball as blue rather than red. Of course in this case since there are so many balls you might consider making approximations, e.g. by saying that once you have chosen one red ball, there are an equal number of red and blue balls left. But with 52 cards, you can't take this approach.
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    (Original post by ttoby)
    That is a valid method. In fact it's the method I would have used if I hadn't seen the answers. What I would have done for that is calculated the probability of getting heart, heart, heart, non, non (in that order) then multiplied it by 5! / (3!2!) because each ordering of 3 hearts/2 non is equally likely.
    I didn't realise the 10 probabilities are equal. Care to explain why? I tried a couple below; they´re equal, but the fractions are different:

    Heart, Heart, Heart, Non, Non : P = (13/52)*(12/51)*(11/50)*(39/49)*(38/48)
    Heart, Non, Heart, Non, Heart : P = (13/52)*(39/51)*(12/50)*(38/49)*(11/48)

    etc.

    EDIT: I guess you always have 13, 12, 11, 39, 38 in the numerator and 52, 51, 50, 49, 48 in the denominator. Doesn't matter which way round, works out the same with multiplication. Is there a more intuitive reason though that you spotted?

    (Original post by ttoby)
    To get the probability of 3 hearts then 2 non-hearts, think about how many cards of each type are left in the pack as you pull out each card, so you should get the probability as (13/52)*(12/51)*(11/50)*(39/49)*(38/48). You can then calculate this and multiply it by 5! / (3!2!) to get your final answer.
    Yes. But I don't see why the probabilities work out to be the same for all 10 ways. See above. Maybe I missed something obvious.

    EDIT: I see it with regards to the numbers in the numerator/denominator always being the same, but did you spot it differently?

    (Original post by ttoby)
    With your billion balls you do need to do a similar sort of trick. The probability of getting red,red is not the same as the probability of getting red,blue in that order. This is because once you've chosen your red ball, there are 0.5billion-1 red balls left and 0.5billion blue balls left so you are slightly more likely to have your next ball as blue rather than red. Of course in this case since there are so many balls you might consider making approximations, e.g. by saying that once you have chosen one red ball, there are an equal number of red and blue balls left. But with 52 cards, you can't take this approach.
    Yep. I tried the sweets Q using my approach and summing all the ways, and I got the same answers as using the book's clever combinations trick. So both ways work; I just found the 'lets say they´re distinct then use the teams argument´to be a little counter-intuitive at first.
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    (Original post by Physics Enemy)
    I didn't realise the 10 probabilities are equal. Care to explain why? I tried a couple below; they´re equal, but the fractions are different:

    Heart, Heart, Heart, Non, Non : P = (13/52)*(12/51)*(11/50)*(39/49)*(38/48)
    Heart, Non, Heart, Non, Heart : P = (13/52)*(39/51)*(12/50)*(38/49)*(11/48)

    etc.

    EDIT: I guess you always have 13, 12, 11, 39, 38 in the numerator and 52, 51, 50, 49, 48 in the denominator. Doesn't matter which way round, works out the same with multiplication. Is there a more intuitive reason though that you spotted?


    Yes. But I don't see why the probabilities work out to be the same for all 10 ways. See above. Maybe I missed something obvious.

    EDIT: I see it with regards to the numbers in the numerator/denominator always being the same, but did you spot it differently?


    Yep. I tried the sweets Q using my approach and summing all the ways, and I got the same answers as using the book's clever combinations trick. So both ways work; I just found the 'lets say they´re distinct then use the teams argument´to be a little counter-intuitive at first.
    Since each card is just as likely as any other card to be chosen then any arrangement of heart,heart,heart,non,non is equally likely. The cards are stacked in a random order so there's no reason for one type of card to be more likely to appear earlier than later.

    You could use the lottery analogy here. Ignoring the bonus ball, if you take some combination of balls coming out of the machine then each possible ordering of the balls is equally likely because all of the balls are identical apart from the number printed on them.
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    (Original post by ttoby)
    Since each card is just as likely as any other card to be chosen then any arrangement of heart,heart,heart,non,non is equally likely. The cards are stacked in a random order so there's no reason for one type of card to be more likely to appear earlier than latter.
    I see your point. Say I put a number on each card (1 to 52). Cards 5, 6, 7, 8, 9 are the heart, heart, heart, non, non under consideration. Drawing 5 then 6, 7, 8, 9 is equivalent to any other order; each numbered card has the probability 1/52.

    Of course, that's why they say 'Find P(3 hearts, 2 non-hearts)' rather than average probability; it implies the order doesn't alter the probability.

    (Original post by ttoby)
    You could use the lottery analogy here. Ignoring the bonus ball, if you take some combination of balls coming out of the machine then each possible ordering of the balls is equally likely because all of the balls are identical apart from the number printed on them.
    Thanks. I can see how 6, 7, 8, 9, 10, 11 coming out of the machine is the same as that of any other order. It wouldn't make sense otherwise; they'd then have to award prizes weighted to order. 11, 10, 9, 8, 7 may be worth £50 million.

    Thanks so much for all your help!
 
 
 
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