The classical answer to this problem is yes, because the chances of winning the prize are twice as high when the player switches to another door than they are when the player sticks with their original choice. This is because upon the original choice, the player has only a 1/3 chance of choosing the door with the prize; this probability does not change when a door opens with a goat. Hence the chances of winning the prize are 1/3 if the player sticks to their original choice, and thus 2/3 if the player switches.
Instead of one door being opened and thus eliminated from the game, it may equivalently be regarded as combining two doors into one, as a door containing a goat is essentially the same as a door with nothing behind it. In essence, this means the player has the choice of either sticking with their original choice of door, or choosing the sum of the contents of the two other doors. Clearly, the chances of the prize being in the other two doors is twice as high.
It may also be easier to appreciate the result by considering a hundred doors instead of just three, with one prize behind only one of the doors. After the player picks a door, 98 doors are opened with goats behind them. Clearly, there's now a very high chance (precisely 99/100) that the prize is in the other door not opened.

Yeah, double post... I'm forewarned... 