TITLE: Drawing without replacement: why is the order of draw irrelevant?
QUESTION [6 upvotes]: I am trying to wrap my head around this problem:

Daniel randomly chooses balls from the group of $6$ red and $4$ green. What is the probability that he picks $2$ red and $3$ green if balls are drawn without replacement.

What I remember from my college days that the probability is found by this formula:
$$P(A)=\frac{\binom{6}{2}\binom{4}{3}}{\binom{10}{5}}=\frac{5}{21}$$
Is this correct? I am trying to understand why this works. Wouldn't probability depend on the order of balls drawn as the number of balls is changing after each draw? I get how we obtain numerator and denominator, I just feel that the probability should be dependent on the order. For example, the probability to pick red first is $\frac{6}{10}$ so the probability for the second draw becomes $\frac{5}{9}$ for red and $\frac{4}{9}$ for green. But if the first picked ball is green, the probability for the second draw becomes $\frac{6}{9}$ for red and $\frac{3}{9}$ for green. What am I missing?

REPLY [2 votes]: There is a principle called "conservation of expected evidence" that says that if you have events A and B, then when you calculate the probability of A without knowing whether B happens, the result should be the same as the expected value of the probability over the possible results of B. 
In this case, let A be the probability that the second ball is red, and B be the probability the first one is green. The principle says that P(A) = P(A|B)P(B)+P(A|~B)P(~B). That is, if you split A into two cases of A and B versus A and not B, the total probability should just be the probability of A. If you roll a die and flip a coin, the probability of getting a 1 one the die should change if you split it into P(die=1,coin=heads) plus P(die=1,coin=tails). 
We have the following values for those probabilities:
P(A) = $\frac6{10}$
P(A|B) = $\frac5 9 $
P(B) = $\frac6{10}$
P(A|~B) = $\frac 6 9 $
P(~B) = $\frac 4 {10}$
So the equation is $\frac6{10} = \frac5 9 \frac6{10}+\frac 6 9 \frac 4 {10}=\frac{30+24}{9*10} = \frac{54}{9*10}=\frac{9*6}{9*10}=\frac 6 {10}$
If you have ten cards, 6 red and 4 green, and you shuffle them, would the probability of the first one being red be any different from the probability of the second one being red?