I think that the answer given is wrong. It is incorrect to assign equal probabilities to B, W1 and W2. B and W1 have half the probability of W2.

If you ran the experiment 100 times you would expect that in 50 cases the unknown counter was drawn first and in the other 50 cases the added white one. In the first 50 cases you would expect to draw a white counter 25 times. In the remaining 50 cases it would be 50. So the expectation is 75/100 ie 3/4, not 2/3.

No, this does not make sense. The original counter is either this or that. Once it is here, you draw it with the probability of 50%. You can't split this probability into one for a possible black counter and 25% for a possible white counter.

In addition, the question is not about drawing the original counter, but the white one.