Doubt in the exercise 3.2.1 of the book

[Himanshu]

I have a doubt in the interpretation of the result of Ex. 3.2.1 (c) of the book "Probability and Statistics with examples using R". I have attached the screenshort of the problem.     

As we are first choosing a ball and then put it aside and then choosing the ball from the remaining one second time. So the sample space S is {(1,2),(1,3),(1,4),(2,1),(2,3),(2,4),(3,1),(3,2),(3,4),(4,1),(4,2),(4,3)} with the probability of each outcome = 1/12.            

P(Y=2) = P({(1,2),(3,2),(4,2)}) = 3/12 = 1/4       

We can also solve it as P(Y=2)=P(Y=2|X=1)P(X=1)+P(Y=2|X=3)P(X=3)+P(Y=2|X=4)P(X=4) = 3 X (1/3 * 1/4) = 1/4 

The result P(Y=2) = 1/4 is surprising. If we don't have the information of what we get while choosing the ball for first time, then choosing the ball for second time is like putting the ball chosen first back into the urn and choosing the ball during second time out of all the four balls. As choosing the all the balls are equally likely, So, P(Y=2) = 1/4.  

Computation-wise, the result which we arrived, I can understand. But I am not able to interpret the result physically, like how the problem reduced to choosing the ball out of 4. If we are doing the experiment and I don't know what ball I get for the first time, how likelihood of choosing the ball-2 during second time is 1/4. Intutively, I don't think the probability to be 1/4.                  

Intuitively, I am not able to interpret the result. Please help me in clarifying the doubt.   

Thanks!