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?


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 $1902?$

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    $\begingroup$ But the problem is, once we have chosen an urn, we cannot change the urn . So if we chose the first urn, only then can we get two blue balls, otherwise, we cannot. Isn't this argument correct? $\endgroup$ – noobcoder Sep 1 '18 at 15:02
  • $\begingroup$ What you say is true, but it isn't relevant to the problem. You're answering the question, "What is the probability of picking the urn with two blue balls?" not the question given. $\endgroup$ – saulspatz Sep 1 '18 at 15:11
  • $\begingroup$ No. I understand that. But if we chose the other two urns, we'd have a probability of 0 choosing a second blue ball, given that the first ball is blue. And if we choose the first urn, we choose the second ball blue with a probability of 1. Which means , it boils down to choosing the urn with the two blue balls. I mean that is what I think. $\endgroup$ – noobcoder Sep 1 '18 at 15:31

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