Find the conditional probability that a ball of certain color is chosen from an urn among several urns containing variable number of colored balls. There are $n$ urns of which the $i$-th urn contains $i − 1$ red balls and $n − i$ blue balls. You pick an urn at random and remove two balls at random without replacement. Find the probability that (a) the second ball is blue; (b) the second ball is blue given the first ball is blue.
This is a repeated question, and I found its solution here in Find probability of specific ball getting selected on second turn  .  
I am having some problems with the solution. According to the solution, the probability is $2/3$.
Consider $n=3$ . Then the first urn has $2$ blue balls, second has $1$ red and $1$ blue ball, and the third has only $2$ blue balls
Hence probability of choosing a blue ball the second time, given the first one is blue is : $1/3\cdot1$ [ Since $2$ blue balls, one already taken, the other one is obviously blue]  $+ 1/3\cdot0$ [ There was just one blue ball, the other was red. The blue ball was picked, hence only red ball remains]  $+ 1/3\cdot0$ [ No blue balls] . Hence the probability must be $1/3$. 
What is the problem in my logic?
 A: There are three blue balls, all equally likely to be the first ball chosen.  In two cases, the second ball chosen will also be blue, and in one case, it will be red.
EDIT What you say in your objections is true, but it has nothing to do with the case.  Instead of balls, let us says that we have one urn with two gold coins, one with a gold coin and a silver coin, and one with two silver coins.  The coins are the same size and cannot be distinguished by touch.  You draw a coin at random from a randomly chosen urn, and it's gold.  What is the probability that the other coin in the urn you have chosen is gold?  According to your logic, it $1/2.$  
Now I tell you that the gold coin in the urn with the silver coin is dated $1900,$ and the two gold coins in the same urn are dated $1901$ and $1902.$  You draw a gold coin, and place it on the table, and tails is showing, so that date cannot be seen.  Now, what is the probability that the other coin in the urn is gold?  This is the question being asked.  The answer is that the other coin is gold if when we flip the coin we've drawn, the date is $1901$ or $1902$ but not if it's $1900.$
Do you really want to maintain that it's equally likely to be $1900$ as it is to be $1901$ or $1902?$      
A: 
There are three urns.  The first contains two blue balls, the second contains one blue ball and one red ball, and the third contains two red balls.  You pick an urn at random and remove two balls at random without replacement.  Find the probability that the second ball was blue given that the first ball was blue.

Our sample space consists of all sequences of draws in which the first ball is blue.  Let $U_i$ denote the event that urn $i$ was drawn.  Let $B_i$ denote the event that the $i$th ball drawn is blue.  Then
\begin{align*}
\Pr(B_1) & = \Pr(U_1)\Pr(B_1 \mid U_1) + \Pr(U_2)\Pr(B_1 \mid U_2) + \Pr(U_3)\Pr(B_1 \mid U_3)\\
         & = \frac{1}{3} \cdot 1 + \frac{1}{3} \cdot \frac{1}{2} + \frac{1}{3} \cdot 0\\
         & = \frac{1}{3} + \frac{1}{6}\\
         & = \frac{1}{2}
\end{align*}
Our favorable case is that the second ball is blue given that the first ball is blue.
\begin{align*}
\Pr(B_2 \mid B_1) & = \Pr(U_1)\Pr(B_1 \mid U_1)\Pr(B_2 \mid B_1 \cap U_1) + \Pr(U_2)\Pr(B_1 \mid U_2)\Pr(B_2 \mid B_1 \cap U_2)\\
                  & \quad + \Pr(U_3)\Pr(B_1 \mid U_3)\Pr(B_2 \mid B_1 \cap U_3)\\
                  & = \frac{1}{3} \cdot 1 \cdot 1 + \frac{1}{3} \cdot \frac{1}{2} \cdot 0 + \frac{1}{3} \cdot 0 \cdot 0\\
                  & = \frac{1}{3}
\end{align*}
Thus, the conditional probability that a blue ball is selected on the second draw given that a blue ball was selected in the first draw is 
$$\frac{\Pr(B_2 \mid B_1)}{\Pr(B_1)} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}$$
which is consistent with Joriki's answer to the linked question.
In your calculations, you found the probability that both the first and second balls are blue, but you have to divide this by the probability that the first ball is blue.
