Not really because:
First Step - you pick a door.
You have a 2/3 chance of picking a donkey
You have a 1/3 chance of picking the car
Step 2 - The host knows which doors contains the donkeys, so, they open the door with the donkey.
Statistically speaking, you are twice more likely to pick a door with the donkey.
If you picked the donkey, the host is forced to open the door with the other donkey.
If you swap, you have inversed your odds from being 1/3 correct to 2/3 correct.
Here's another visualisation. Below are the possible scenarios. Let's run with the assumption you always pick the first option as your door.
Car, Donkey, Donkey
Donkey, Car, Donkey
Donkey, Donkey, Car
So, if you always picked the first option, were revealed where one of the donkeys were, then swapped, you would have twice as much chance of winning the car.
David - I can see where you are coming from with this now (which is where I can now disagree ). From my understandting, you are adressing this hypothetical with a linear approach as opposed to factoring in all the possible outcomes. Is this so?