In class, while illustrating the topic of conditional probability, my professor presented the following card example:

You have 3 cards that have been randomly shuffled: card1, card2, and card3. One is an ace and the other two are non-aces. We are interested in the location of the ace. Thus, the sample space is S = {card1, card2, card3}.That is, the ace can either be card1, card2, or card3. We assume that each outcome is equally likely(ie classical probability formulation). Let event

A1 = "card1 is the ace", thus P(A1) = 1/3. Let event B = "turn over card3 and it is not an ace".

My professor says that once B occurs it becomes the sample space since it becomes the full set of possibilities given that it actually occurred. He says that A1 = {1} and B = {1,2}, so now the probability of the ace being the first card is $P(A1|B) = \frac{|A1\cap B|}{|B|} = \frac{1}{2}$.

The answer to this problem made me uneasy because this problem looks very similar to the Monty Hall problem. Applying the "Monty Hall problem" reasoning to this problem would give the probability of the ace being card 1 as 1/3, since the probability of the ace being

card 2 (ie "switching your pick") would be 2/3.

Is the answer to the card example really 1/2? Are these problems the same? If not, what makes them different?