# Why doesn't order matter in this

1. Apr 10, 2012

### aaaa202

Say you have two black balls and two white in a jar and you pick 3 at random. When one is picked it is taken out of the jar.
I want to know the probability of getting for instance WWB. That must be:

P(WWB) = 1/2*1/3*1 = 1/6

Now say you want WBW. Then you get:

P(WBW) = 1/2 * 2/3 * 1/2 = 1/6

Amazingly enough the probabilities are independent on the order in which you choose your balls. This result is kind of amazing for me and not really intuitive. So far it stands more or less as a defining property of this kind of combinatorics.
Is it intuitive for you that the order really is independent? What argument would you give for this?
And lastly, perhaps even more interestingly: Can anyone prove that if you general have NA blue elements and NB red elements then the number of ways of getting a blue and b red elements is independent of order. I don't know if it's an established theorem - if so, where can I read more about it?

2. Apr 10, 2012

### HallsofIvy

Staff Emeritus
Just look at those calculations: in the first you have the fractions 2/4, 1/3, and 2/2. In the second, the fractions are 2/4, 2/3, and 1/2. Do you see that, while the fractions are different, the three numerators and three denominators are the same?
Do you see why that is true? Do you see that, because the numerators are the same and the denominators are the same, the products are the same?

3. Apr 10, 2012

### aaaa202

I see it yes. Why it's true I'm not sure.

4. Apr 12, 2012

### aaaa202

Ivy, can you please help on the intuition for it?

5. Apr 19, 2012

The order does not matter in general case.

Suppose you have $b$ black balls and $w$ white balls. The total amount of balls is $n=b+w$.
Consider a sequence {B,W,B,B,...,B,W}. The probability that this sequence occurs is:
$\frac{b}{n}$$\frac{w}{n-1}$$\frac{b-1}{n-2}$$\frac{b-2}{n-3}$...$\frac{b-(b-1)}{2}$$\frac{w-(w-1)}{1}$
Note that the numerator of this sequence is $b(b-1)...(b-(b-2))(b-(b-1))w(w-1)...(w-(w-2))(w-(w-1))$ and the denominator is $n(n-1)...(n-(n-2))(n-(n-1))$.
You should notice that whenever you want to choose $b$ black balls and $w$ white balls, the numerator and the denominator will be the same. The order of the elements of both products will be different, but you know that scalar multiplication is commutative (i.e. order does not matter).

6. Apr 19, 2012

### HallsofIvy

Staff Emeritus
The fundamental ideas are the we can multiply fractions by multiplying numerators and denominators separately and that multiplication is commutative. That is, we can rearrange the numerators and denominators as we please.