Probability: Drawing two marbles simultaneously from the bag This is a 9. grade elementary school quiz problem.
There is a bag with 4 marbles. Two of them are yellow and the other two are green. You blindly pick two marbles at once. What is the probability that you get two yellow marbles?
I would say the answer is $\frac{1}{6}$. There is a single combination to draw two yellow marbles and 6 possible combinations how to draw two marbles. My second approach is: possibility of picking the first one yellow is $\frac{1}{2}$, picking the second one yellow is $\frac{1}{3}$, multiplying possibilities  gives me the same result $\frac{1}{6}$.
The problem is that my teacher claims that the answer is $\frac{1}{3}$. Her argument is that there are 3 possible outcomes - two yellows, two greens and a mixed one. One outcome is positive so therefore the answer is $\frac{1}{3}$. She also opposed to my second approach claiming that drawing marbles one by one is not the same as draw them simultaneously. From my opinion there should be no difference in result.
Could someone please write what is the correct answer to this problem? Does simultaneously drawing gives the same result as drawing one by one without replacement?

Finally, I think I understand the problem. Your answers help me to see it clearer. I found out that I mixed up the terminology. The problem has 6 outcomes and 3 possible events. The event that you draw the YG (mixed) combination has 4 positive outcomes while events of drawing YY or GG has a single possible outcome each. Probability is defined as a ratio between positive outcomes for specific event and all possible outcomes. Therefore the probability for drawing YY is $\frac{1}{6}$ and not $\frac{1}{3}$. By adding more green marbles you will increase the number of all possible outcomes, but the number of events will remain 3, so therefore the probability of picking YY will decrease and not stay $\frac{1}{3}$, what seems very logically. That was the catch I didn't see how to explain the teacher why I am correct. I think now I would be able to convince the teacher. Thank you all for your great help.
 A: You are correct.  The problem with the teacher's method is that the various outcomes are not equi-probable.
Since the numbers are so small, we can write out all the possible, equiprobable, draw sequences (assuming we take all four in some order).  They are:  $$YYGG\quad YGYG\quad YGGY\quad GYYG\quad GYGY\quad GGYY$$
We see that exactly one way in six works.  
Note that four out of six give mixed draws, demonstrating that the three cases your teacher points out are not equiprobable.  Indeed, your method works for the mixed case as well:  The first draw can be of either color and then the next gives a mixed draw with probabilty $\frac 23$.
A: Let's label the marbles for the sake of this question as 1 and 2 for each color $$Y_1, Y_2, G_1, G_2$$
If you choose two marble at the same time, the possible combinations are as follows:$$(Y_1, Y_2), (Y_1, G_1),(Y_1, G_2), (Y_2, G_1), (Y_2, G_2), (G_1, G_2)$$
This means there are $6$ unique outcomes, so the probability of the outcome we are looking for $(Y_1,Y_2)=\frac{1}{6}$.
In this case you are correct as your teacher is lumping the middle 4 outcomes together as one "type" of outcome which is incorrect.
I hope this helps!
A: Well, your teacher is definitely wrong and here's why. You consider 6 cases each of which has the same probability. You could write numbers 1 and 2 on yellow and same on the green marbles and count all of the outcomes as if all of the marbles are different and we take them out one by one. You would then get 12 possible cases (4 for the 1st marble and 3 for the 2nd one), two of which fit your request - both of them are yellow (Y1, Y2 and Y2, Y1) so the probability is 1/6. She, however, considers 3 cases 2 of which are equal (obviously the probability to take out 2 yellows is the same as 2 greens), but the other one (marbles are different) is more probable as it can be achieved by not 2 as each of the previous ones (Y1, Y2 or Y2, Y1) but 8 ways: Y1G1, Y1G2, Y2G1, Y2G2,... well, I guess you can finish this yourself. I don't know about your country but we in Russia have a quite famous among mathematicians anecdote on exactly this theme:  "-What is the probability to go out to the street and meet a dinosaur? -one half - either you meet it or you don't". While there can be a certain number of cases you always have to check that all of them are equally probable
A: You're completely right which makes your teacher completely wrong. I understand her intuitive approach but your analysis on 1/2*1/3=1/6 should have signal her that she's off course. 
Allow me to throw all possible permutations from the universe Y1,Y2,G1,G2 (assuming the order mattes) and show what is the probability that you get two yellow marbles? 
in this case you'll have:
('Y1', 'Y2') Good
('Y1', 'G1')
('Y1', 'G2')
('Y2', 'Y1') Good
('Y2', 'G1')
('Y2', 'G2')
('G1', 'Y1')
('G1', 'Y2')
('G1', 'G2')
('G2', 'Y1')
('G2', 'Y2')
('G2', 'G1')

Permutation result: $\frac{2}{12} = \frac{1}{6}$.
If we use combination (as suggested by @lulu) you'll get the same result: $\frac{1}{6}$.
