Has anyone compared these two (I call them the MHP and TCP, respectively)? I certainly have.
The short answer is that, as you have worded them, they are different. But a more common wording of the TCP - which is different than yours but too often interpreted as being the same - they are both variations of Joseph Bertrand's famous Box Problem. Other similar problems include the Three Prisoners, Three Pancakes, and/or Three Cards. You can look them up on the internet, but they are all essentially the same thing. There is even a similar situation in card games like Bridge, where experts use what they call the Principle of Restricted Choice to help estimate who has a missing card.
What is common in all these situations, is that there can be more information in what we learn than is immediately obvious. Because its importance isn't obvious, it can be either omitted in the problem statement, like you did in your MHP, or strongly worded to prevent it, like you did in your TCP. I call it "hidden information." What makes it important, is if it could have taken a different form in the situation we are asked to provide a probability for, than in others.
This may be easiest to explain with the MHP. Your statement of the problem left out the fact that we actually see which doors are chosen at each point. Say, as in Marilyn vos Savant's version, that the contestant chose Door #1 and the host opened Door #3. I'm not saying these specific door numbers matter - they don't - but the fact that the host theoretically could have opened door #2 does matter. So "Opened #3" is the hidden information, and we have to consider the cases where he could have opened #2 instead.
If the prize is behind Door #2, he would have been forced to open Door #3; that is, a 100% chance. If the prize is behind Door #1, he must choose between two equivalent doors, and there only is a 50% chance he would have opened Door #3. It is the ratio of these two numbers - 100% to 50% - that makes it twice as likely that we are in the first case instead of the second!
Note that most solutions you will see say that switching has a 2/3 chance because the 1/3 chance that your original choice had the prize can't change. THIS IS WRONG; or at least, incomplete. This probability can change based on what door we see opened, but only if the host is biased when he makes the choice. What if, when he can choose between #2 and #3, he chooses #2 100% of the time? Then the only way he will ever open #3 is if the car is behind #2 and the 1/3 probability has changed to 0. It changes to 1/2 if he opens #2. But note that the weighted average of these two results - that is, the probability if you ignore the source of this hidden information - has to remain 1/3.
No matter how it is worded, the TCP is usually interpreted as you stated it. But it is not often worded so strongly. Martin Gardner (Scientific American's "Mathematical Games" columnist) popularized it in 1959, with the problem statement "Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?" He answered "1/3," but later acknowledged that the problem was ambiguous. He received many letters pointing out that the gender you know is hidden information. If you don't know how you learned the fact "at least one is a boy," you have to consider whether it would have been possible to learn "at least one is a girl."
If you decide it would have been possible, the answer to the TCP is 1/2: there was a 25% chance that Mr. Smith has two boys and you could only learn about a boy, and while there was a 50% chance he has a boy and a girl, there is only a 25% chance that he has a boy and a girl AND the gender you would know is "boy."