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?
 A: The probabilities are Door 1: $1/3$ and Door 2: $2/3$ for both contestants, it's simply that one of them knows which doors bears $1/3$ and the other doesn't.
One thing that will make things clearer is the following: 


*

*Assume that the contestant who arrives late is told what happened before, then asked what are the probabilities for each door. The correct answer is "one of them bears $1/3$, the other one $2/3$, but I don't know which is which.

*Now assume that the second contestant is only told that one door has a prize and the other nothing. He's told nothing more about what happened or how it was decided what door would be the right one. He is then asked what are the probabilities for each door. The correct answer is I have absolutely no clue.
There is a fundamental difference between "I don't know" and "$50/50$". It could very well be that the organisers always put the car behind Door 2, in which case the probabilities would be $0$ and $1$. You can't make up probabilities if you don't know the protocol.

Edit: one last point is that there is a difference between the probability that the car is behind Door 2, and the probability that the second contestant will get the car. If the second contestant knows nothing, and picks a door by flipping a fair coin, then indeed they will get the car with probability $0.5$. This doesn't depend on the probability for each door to be the correct one.
A: This is an ongoing point of philosophical debate, generally subsumed under the name "interpretations of probability". Joseph Butler famously wrote the following in the introduction to his The analogy of religion, natural and revealed, to the constitution and course of nature (1736). Here, he frames probability as essentially a tool for beings in the world dealing with limited and possibly flawed perspective:

Probable Evidence, in its very nature, affords but an imperfect kind
  of Information; and is to be considered as relative only to Beings of
  limited Capacities. For nothing which is the possible object of
  Knowledge, whether past, present, or future, can be probable to an
  infinite Intelligence; since it cannot but be discerned absolutely as
  it is in itself, certainly true, or certainly false: But to us,
  Probability is the very Guide to Life.

Wikipedia provides the following summary of current interpretations. Note in particular that the "Subjective" interpretation states that probability is fundamentally about "Degree of belief", while the others do not:



*

*Wikipedia: Probability Interpretations

*Stanford Encyclopedia of Philosophy: Interpretations of Probability
A: I'll assume the standard Monty Hall assumptions (host is required to reveal a goat and offer the option to switch, and the contestant knows this).
Otherwise, as usual, there is no definitive answer to the question.

She does not have any knowledge of what has just transpired on stage.

It is not at all obvious that this is true in the situation you describe.
Of course, if the second contestant has never heard of the Monty Hall problem,
has never even been told the rules of the game show, perhaps does not even know that it is a game show, then if you tell her she must pick one of the two closed doors she has a $50\%$ chance to pick the car.
But then she will not know she has a $50\%$ chance to pick the car,
and she will have no reason to say there is a $50\%$ chance that there is
a car behind door number $2.$
If you explain the rules of the game show before bringing on the second contestant, then she can see that the contestant has already chosen a door, that it was not door number $3,$ and that Monty opened door $3.$
She does not know whether the first contestant decided to switch.
She may not even know that the other contestant has already made his final choice.
Knowing what the second contestant knows, and knowing one of the following additional pieces of information, she would find the following probabilities that the car is behind door number $2$:


*

*The first contestant originally choose door $1$ and switched to door $2.$ Then the probability is $\frac23.$

*The first contestant originally choose door $2$ and was offered a chance to switch but chose not to. Then the probability is $\frac13.$

*The first contestant originally choose door $2$ and has not yet been offered a chance to switch. Then the probability is $\frac13.$


Of course, we know that only the first "additional piece of information" could be true, but if the second contestant cannot determine that from the state of the room when she enters, in order to assess the probability that the car is behind door $2$ she should estimate the probability of each of those three possible circumstances. The probability that the car is behind door $2$ can then be computed via Bayes' Theorem.
She could give $\frac13$ weight to each of the three cases, but they are not symmetric so I see no reason she should give them equal weight.

As for the implied question in your question's title,
yes, probability estimates should be conditioned on what you know,
and people with different knowledge often will infer different probabilities
even if they both reason completely correctly.
A: The other answers have adequately explained this variation of the Monty Hall problem, so I'm going to focus on the question:

My question: is probability determined by perspective?

The answer is: It depends.
Broadly speaking, there are two schools of thought about what a "probability" is and how it should be used.  The frequentist believes that probability describes something like a roulette wheel, which can be spun many times but always behaves the same (i.e. it always produces the same distribution of results).  When we say $P(X) = \frac{1}{6}$, we mean that, if a given event is tried many times, in the long run the outcome $X$ will happen one sixth of the time.  This convergence is guaranteed by the law of large numbers.  Probability is thus an objective fact about the universe, and not something subject to a person's perspective.  In this worldview, the probabilities in the Monty Hall problem are always 1/3 and 2/3, regardless of whether we know which is which.
The Bayesian, on the other hand, sees probability as a degree of belief.  You might think of this like in a court of law: In order to convict the defendant of some crime, we need to be 99% certain that the defendant committed the crime.  Seeing incriminating evidence may raise our subjective belief in the defendant's guilt, while exculpatory evidence would lower it, both according to Bayes' theorem.  When we're making a verdict, we ask ourselves whether the defendant has at least a 99% probability of having committed the crime.  In the frequentist worldview, this is a nonsensical question; either the defendant is truly guilty or the defendant is truly innocent, and the probability is accordingly either 100% or 0% (we just don't know which).  Similarly, in the Bayesian worldview, the Monty Hall problem is nonsensical unless you specify the person whose worldview we are following and their subjective prior probabilities for each door.  Bayesian reasoning, then, could give you a 50-50 split for your hypothetical second contestant, but only if she started with 33-33-33 priors, and then only if there is no other evidence allowing her to distinguish between the two remaining doors.
It is also important to recognize that the frequentist and Bayesian approaches are not mathematically distinct as both probabilities are subject to the same mathematics (i.e. each system admits both the law of large numbers and Bayes' theorem).  What differs is how the math is applied to the real world.  Because the frequentist deals in objective probability, they cannot tell you "the probability that candidate X wins the election."[1]  Because the Bayesian deals in subjective probability, they cannot tell you much of anything without a set of prior probabilities,[2] which are necessarily tied to a particular observer at a particular point in time and space.  Ultimately, both systems inevitably require making certain assumptions about how your data relates to the real world.  So you should examine those assumptions with care before blindly accepting the result which they have produced.

[1]: The election happens only once; it doesn't make sense to ask how often candidate X wins.  Imagining the election being re-held many times doesn't work either because elections are deterministic, so the same people will vote or not-vote in the same ways every time, and you will get the same result.  Instead, you have to engage in a far more roundabout investigation of the likely level of errors in the polls, which gives a less obviously meaningful value as its final output.
[2]: In cases like Monty Hall, some set of priors is typically "obviously correct" (e.g. "All three doors are equally likely to conceal the car").  However, this still has to be explicitly stated as an assumption of the Bayesian method.  Many circumstances, including elections, have no obviously correct set of priors (though betting markets may be a good first step).  In cases like the court of law, it may be desirable to begin with a set of priors which is "obviously wrong" (we must assume the defendant is probably innocent, even though most criminal defendants are probably guilty).
A: If assume that Contestant #2 always picks the same door as Contestant #1 did, then the choice made by Contestant #2 is depending on the information of what happened before, violating one of your premises.
If you do not specify such a rule and make #2's choice random, then the two contestants will pick the same door only 50% of the time, justifying their different win rates.
A: Probability is determined by perspective if the perspectives have different information.
In an omnipotent context, the probability of everything is either 0 or 1.
In an oblivious context, the probability is evenly distributed amongst the possible options.
For each piece of information provided on the scale between "oblivious" and "omnipotent", the probability will tend either towards 0 or towards 1. In your example, the original guesser has one more piece of information than the latecomer, so the probabilities are different.
A: Probability inherently asks "What is the probability of $X$, given this set of information".
For the example you give, the probability given one set of information is different to the probability, given a smaller set of information.
The greater the gamut of information, provide there is no known bias to the info, the closer the probability will be to reality.  Given perfect knowledge (i.e. the actual outcome), all probabilities are either $0$ or $1$.
A: 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.
A: 
Is probability determined by perspective?

The answer is undoubtedly yes, but I don't think the Monty Hall problem illustrates the possibility of different perspectives well. Rather I like Bertrand's Paradox.

Consider an equilateral triangle inscribed in a circle. Suppose a chord of the circle is chosen at random. What is the probability that the chord is longer than a side of the triangle?   

Bertrand then gave three answers depending on different probability distributions. The whole point was that for the problem to have a particular solution then the probability distribution for the chord length distribution must be specified in the problem statement since different distributions lead to different answers. 
