It seems at first that the chance must be 1/2, as we’ve simply added a white counter and then removed it again, returning the bag to its original state.

But this is an error. After the white counter is added, the bag contains either (a) 2 white counters or (b) 1 white counter and 1 black counter. These states are equally probable. The chance of drawing a white counter from bag (a) is 1; from bag (b) it’s 1/2. So when the white counter is drawn, the chance that the bag had contained (a) 2 white counters or (b) 1 white counter and 1 black counter are proportional to (a) 1/2 × 1 (b) 1/2 × 1/2, i.e., 2 to 1. So there’s a 2/3 chance that the remaining counter is white.

In other words, the white counter that we drew is twice as likely to have come from a white-white bag than from a white-black bag.