Is probability determined by perspective? My question: is probability determined by perspective?  
The scenario that raised the question for me: 
Initial condition: The Monty Hall problem.  We know the contestant’s original choice of door #1 (of 3 total) is only correct 33% of the time.  After Monty Hall reveals door #3 is incorrect the contestant is asked if he would like to switch his answer to door #2.  We know he should choose to change his answer to the other unopened door (#2) which has a 66% chance of being correct.  He does so.  
However, let’s say once he decided to switch to door #2, and before either door is revealed, another contestant enters the room.  She does not have any knowledge of what has just transpired on stage.  She is offered a choice to pick which door the car behind of the 2 remains closed doors and also randomly chooses door #2.  
Are the probabilities of being correct different for each contestant?  Seemingly they are.  Contestant 1 had 3 doors to choose from, giving him a probability of 33% that door #1 is the answer.  Contestant 2 only had 2 doors to choose from, giving her a probability of 50% after choosing the same door contestant #1 choose.  
If we repeat the experiment 1000 times, what will the numbers turn out to be for door #1?  333 or 500?

Of course it is. And honestly, although I love to discuss Bayesian vs. frequentist interpretation and related issues any other day (being a decided frequentist myself), I think that we do not need any sophisticated approach here. It helps to frame it in expected values rather than probabilities. Just change the scenario to the following:

There are two doors, exactly one of them containing the car. Monty has opened door 2 for contestant A, it contains the car. Monty closes it again.
Contestant B, unaware of what happened, enters the room and is told that there is a car behind one of the doors.
Who has what chance of winning?

The single, simple fact is -- in my view -- that contestants A and B, because they have different information, have different strategies. These strategies are modeled as different random variables, one of which (A: choosing door 2, because that's where the prize is) has an expected value of 1 (if 1 = winning the car), and the other (B: choosing a door "at random", because what else to do) has expected value of 1/2.
In your scenario, the random variable / strategy "choosing door 2" of the first contestant has expected value 2/3. The random variable / strategy "choosing a door at random" has expected value 1/2.
