Ordered v.s. unordered and distinct v.s. identical in probability For example:
There are two green marbles and two red marbles in a bag. Two marbles are chosen at random, what is the probability that the two marbles which have been chosen are of the same colour?
Ordered + distinct: 
Marbles = $G1, G2, R1, R2$ 
Event space = $\{|G1~G2|, |G2~G1|, |R1~R2|, |R2~R1|\} = 4$ ways 
Sample space = $^4P_2 = 12$ ways 
Probability = $\dfrac{4}{12} = \dfrac{1}{3}$ 
Unordered + distinct: 
Marbles = $G1, G2, R1, R2$ 
Event space = $\{|G1~G2|, |R1~R2|\} = 2$ ways 
Sample space = $^4C_2 = 6$ ways 
Probability = $\dfrac{2}{6} = \dfrac{1}{3}$ 
Ordered + identical: 
Marbles = $G, G, R, R$ 
Event space = $\{|G~G|, |R~R|\} = 2$ ways 
Sample space = $\{|G~G|, |G~R|, |R~G|, |R~R|\} = 4$ ways 
Probability = $\dfrac{2}{4} = \dfrac{1}{2}$ 
Unordered + identical: 
Marbles = $G, G, R, R$ 
Event space = $\{|G~G|, |R~R|\} = 2$ ways 
Sample space = $\{||G~G|, |G~R|, |R~R||\} = 3$ ways 
Probability = $\dfrac{2}{3}$
Which approach is the correct one? And more importantly, why are the other approaches wrong?
 A: The first two are correct, the third and fourth are wrong.  The third and fourth are wrong because there are two different green marbles in actuality.  It can't be right to assume they're really the same.  I can't think offhand of a situation where either the third or fourth calculation is appropriate.
Both of the first two calculations are acceptable.  The only difference really is whether you think of grabbing two marbles at once, or taking one, then the other.  But since you don't put the first marble back, these two models are equivalent, and you can solve it either way.
A: All of the methods can be done to find the probability.
Your problem in the "identical" approaches is that you are considering each choice as equally likely, whereas this is not the case. Using the unordered identical example, picking $GG$ has a probably of $\frac16$, but the probability of picking $GR$ is not (it's $\frac23$) There's a higher chance of picking one of each ball rather than picking two of the same color.
So you would want to add the probabilities of $GG$ and $RR$ which will come out to the correct answer of $1/3$.
It can be done similarly for the ordered identical approach.
