# Basic Stats Q - Am I Missing Something Obvious?

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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.

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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#2

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:

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.

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

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:

**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

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

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.

**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'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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#4

(Original post by

... even though the cards aren't distinct.

Anything else you can add to help me?

**Physics Enemy**)... even though the cards aren't distinct.

Anything else you can add to help me?

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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!

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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#6

(Original post by

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!

**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!

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

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.

**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.

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

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.

**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.

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

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.

**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.

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#8

(Original post by

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.

**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.

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

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.

**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.

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

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.

**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 so much for all your help!

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