Probability of drawing 5 spades, 3 hearts....

[Addez123]

TL;DR

What are the odds of getting 5 spades, 3 hearts, 3 diamonds, 2 clubs when drawing cards at random (Without putting them back).

The odds are $$\frac {13C5 \times 2 \times 13C3 \times 13C2} {52C13}$$

This is correct according to my book. The follow up question then becomes, what if you can pick 5 of any suit, 3 of any other suit, another 3 from the remaining 2 suits and 2 from the last untouched suit?

The solution would be to calculate
4C1 * 13C5 for the first suit, then 3C1 *13C3 for second suit etc..
Basically same answer as first question except times: 4C1 * 3C1 * 2C1 * 1C1 = 24

Textbook says the correct aswer is same answer as first question * 12, not 24.
That is what I need help with, why 12?

 

[hutchphd]

Because 3H then 3D is the same as 3D then 3H.

 

[Addez123]

Because 3H then 3D is the same as 3D then 3H.

Am I doing completely wrong by doing 4C1 * 3C1..?
Because if I divide by the amount of ways that 4 suits can be arranged, aka 4!, I get 1.
That's not even close to 12 anymore..

 

[etotheipi]

The odds are $$\frac {13C5 \times 2 \times 13C3 \times 13C2} {52C13}$$

Shouldn't that second term be $13C3$?

In any case, @hutchphd is right, you are double counting. Think of it like this; there are 3 types of cards remaining, you want to choose two of them to take 3 cards from and one of them to take 2 cards from. How many ways can this be done?

 

[Addez123]

It should be 13C3, I think I ment to write (13C3)^2

Ok so first the odds of picking any suit is 1. So we don't do 4C1 (altho it still sounds like the logical expression).
Instead we start at 13C3 * 3C1, then 13C3 * 2C1 then last one is 13C2 * 1C1.
We end up with
$$3C1 * 2C1 * 1C1 = 6 $$
Which isn't 12 either :/

 

[etotheipi]

It should be 13C3, I think I ment to write (13C3)^2

Ok so first the odds of picking any suit is 1. So we don't do 4C1 (altho it still sounds like the logical expression).
Instead we start at 13C3 * 3C1, then 13C3 * 2C1 then last one is 13C2 * 1C1.
We end up with
$$3C1 * 2C1 * 1C1 = 6 $$
Which isn't 12 either :/

You're right about the 13C5 x 13C3 x 13C3 x 13C2 part, that doesn't change. You've just got to work out how many ways we can assign the different suits to 5/3/2 cards.

There are 4 choices for the suit from which we draw 5 cards. Let's suppose this is spades. Now you have 3 suits remaining, hearts, diamonds and clubs.

3 cards	3 cards	2 cards
Hearts	Diamonds	Clubs
Hearts	Clubs	Diamonds
Diamonds	Clubs	Hearts

That's not 3! ways, that's only 3 ways. It's because the other 3 ways in all of the possible permutations are obtained by switching the elements of the first 2 columns, which results in an identical scenario. It would be incorrect to double count them!

 

[Addez123]

Ah ok!
So because 3 cards are picked twice, it shouldn't be counted twice since it doesn't matter if I pick 3 diamonds first then 3 spades or 3 spades then 3 diamonds. True that!

So the solution is:
$$4C1 * 3C1 = 12$$

Correct?

 

[etotheipi]

Yes, I'll agree it's perhaps not obvious at first! But then again the whole point of combinatorics is to make you feel lost and confused - or at least that's my experience... !

There's a few different ways you could think about it. It essentially just comes down to "in how many ways can we assign one suit to '4', two suits to '3' and one suit to '2'". You could imagine a little table (it's sometimes useful!) and try to fill in the gaps. Without accounting for any of the funny business, we'd get 4! permutations. Now, any distinct arrangement corresponds to two of these permutations (obtained by switching the 3's columns), so we just need to divide the 4! by two.

 

[WWGD]

Just a quick note. Odds and probability are not the same thing. You can go from one to the other but they are not the same thing.