The accepted answer conflates "the host's knowledge and motivations are unknown" with "the host's actions are uniform and random", which is at best a debatable assumption. I actually think the correct answer to this question is that there is not enough information to answer it as phrased.
For example, if I flip a coin outside of your view and ask what are the odds it landed tails, the correct answer is "I don't know", not 50%. The coin might not be fair or it might be two-headed, etc. Probability itself is defined based on distributions. If you know nothing at all about the distribution, any discussion of probability is meaningless. (As @Yakk points out in a comment, bounds on the distribution can lead to bounds on the probability. But there is nothing like that in this question.)
But I want to ignore that debate and just consider the case where the host opens a closed door at random, since a lot of people get this well-defined question wrong. Should you switch doors in this version of the game?
Just to be very clear: You will pick a door. The host will open at random one of the two doors you did not pick. Then he will give you the option of switching to the other unopened door.
Imagine playing this game 198 times with the strategy "always switch". Just for descriptive purposes, let's color the sheep red and blue.
1) In 66 instances of the game, your initial guess is correct, the host opens one of the other doors at random, you switch, and you lose.
2) In another 66 instances, your initial guess has the red sheep. In 33 of these instances, the host opens the prize door at random, you switch to the blue sheep door, and you lose. In the other 33 of these instances, the host opens the blue sheep door, you switch to the prize door, and you win.
3) In the final 66 instances, your initial guess has the blue sheep. Symmetrically with (2), in 33 of these instances the host opens the prize door and you lose. In the remaining 33 instances the host opens the red sheep door, you switch, and you win.
Adding it all up, you win in just 33 + 33 = 66 instances, or 1/3 of the time you play. This is exactly the same number of wins as if your strategy were "never switch".
Since "always switch" and "never switch" glean the exact same number of wins, there is no advantage to switching doors.
In my opinion, this is the sort of conditional probability question that most people intuitively think is being asked in the classic Monty Hall problem, which is why so many people get it wrong (where the correct answer is "always switch"). Changing the host's actions to random completely changes the problem and the solution.
Note that it does not actually matter what you are allowed to do when the host opens a door with the prize, because that did not happen in the question as stated.
A full, correct analysis is to throw out the 66 instances above where the host opens the door with the prize, because by assumption we are not in a world where that happened. In the remaining 132 instances, you win half the time and lose half the time whether you switch or not.
Here is an analogy that some have found convincing...
Suppose we play a game with an ordinary deck of 52 cards. We shuffle the cards randomly, then you pick one without looking at it. I look at the remaining 51 cards and reveal 50 (i.e. all but one) of them, none of which are the ace of spades. What is the probability that you are holding the ace of spades?
If we play this game many times, 1 time out of 52 your original card will be the ace of spades. The other 51 out of 52 times, I will need to deliberately dodge the ace in order to avoid revealing it. That second scenario is 51 times more likely, so the answer to this question is that the odds are 51-to-1 against your holding the ace.
This is exactly like the classic Monty Hall problem, but with 52 doors instead of 3. And just like that problem, we have to make a whole bunch of assumptions that are often omitted in the statement of the question; e.g. the shuffling is uniform, I always reveal 50 non-ace-of-spades cards regardless of which card you choose, etc. But subject to those assumptions, the odds are 51-to-1 against your holding the ace of spades, just as in the classic Monty Hall problem, the odds are 2-to-1 against your original door containing the prize.
Now let's change the question slightly. Suppose we shuffle and you pick a card randomly as before, but now I reveal 50 other cards without even looking at them. And suppose, just by chance, that none of those revealed cards is the ace of spades.
Granted, this is unlikely; in fact, it will only happen 1 time out of 26 that we play this game. But supposing it does happen, what then are the odds that your card is the ace? (This is what "conditional probability" is all about.)
Obviously, the card I avoided flipping over blindly is completely random, just like the one in your hand. So the answer to this modified problem is that the odds are 1-to-1.
This modified problem is exactly analogous to the modified Monty Hall question, where the host opens a door at random. And the answer is the same: If the door is opened at random, the fact that you do not see the prize tells you nothing about where the prize is, except that it is not behind that particular door.