I pick 3 doors and the probability of success is 3/7. Monty shows 3 of the remaining 4 doors to be empty.

either
1. If I stick with my 3 doors and do not act on the new information, the information is irelevant. The probability of my success (that the prize is behind one of my original 3 doors) is still 3/7. The probability of the opposite event (that the prize is behind the one remaining door I did not choose and Monty did not show) is 1 - 3/7 = 4/7, because the probability of the prize being somewhere is 1 - certainty.

or
2. I ask myself: What is the probability that the one remaining door I did not pick is empty? The same as the probability that the 4 doors I did not pick were all empty to begin with, that is 3/7. Probability of the opposite event (that the prize is behind the one remaining door) is 1 - 3/7 = 4/7, again, because the probability of the prize being somewhere is 1 - certainty.

Always works. I make the 3 for 1 swap to increase my chances from 3/7 to 4/7.