Again boxes with marbles and strategy 
Let’s consider 3 identical jewelry boxes, one with 2 green marbles, one with 2 blue marbles and last one with one green and one blue.
  Before opening any box we set a target, either to pick two marbles of the same color or different. 
We then randomly choose a box and then a marble (without seeing its color before choosing it). We then see the color of the marble and put it back to its box. We repeat the process by picking a box (same with the previous or different) and a marble from it (again randomly). 
Find the probability that we succeed (i.e. if we had chosen to go for “two of the same color”, we indeed got two marbles of the same color, or, if we chose to go for two marbles of different colors, we got two different. We assume that our moves are of absolute logic.    

Let's assume that we choose to find two marbles of the same color. 
We pick one box (probability $\frac{1}{3}$), say we get a green marble (is the (conditional) probability $\frac{1}{2}$??), then we must select either the same box again (hoping we will get a marble of the same color - blue probability $\frac{1}{2}$) or another box, probability $\frac{1}{2}$ for the box and probability 0 if we pick the box with the two blue marbles, 1/2 if we pick the box with 1+1 and 1 if we pick the box with the 2 green (assuming the one we chose the first time was the 1+1). 
By intuition only, I think that I must set the target to pick 2 marbles of the same color, as the chances for this are 2/3 overall.
I don't know how to continue :(
 A: The probability the first box you choose has two marbles of the same colour is $\frac23$


*

*So if you pick the same box again, the probability that your second marble is the same colour as your first marble is $\frac23 \times 1 + \frac13 \times \frac12 = \frac56$, and the probability that it is different is $\frac23 \times 0 + \frac13 \times \frac12 = \frac16$

*While if you pick a different second box, the probability that your second marble is the same colour as your first marble is $\frac23 \times \frac14 + \frac13 \times \frac12 = \frac13$, and the probability that it is different is $\frac23 \times \frac34 + \frac13 \times \frac12 = \frac23$

*If you pick the second box at random, the probability that your second marble is the same colour as your first marble is then $\frac13 \times \frac56 + \frac23 \times \frac13 = \frac12$, and the probability that it is different is $\frac13 \times \frac16 + \frac23 \times \frac23 = \frac12$, as you would expect 
So the best strategy would be, as I think you have said, to set a target of picking the same colour twice, and then to choose the second marble from the same box as the first.  The probability of achieving the target would be $\frac56$
