A ilttle while back on this blog I proposed that probability is fundamentally an epistemological matter. If someone had exact knowledge of the location of all of the particles in a "fair coin" as well as all of the relevant force vectors in a particular toss, then for that person the odds of heads coming up would not be 1/2 but 0 or 1. John G. in the comments section argued that while my contention is true, it is trivially so.
I think not, and as a paradigmatic case of why not, I offer the Monty Hall problem. In the TV show Let's Make a Deal, a contestant would choose among three closed doors. Then, often, Monty Hall (the host) would call for one of the unchosen doors to be opened, revealing the "booby prize" of, say, a goat. Then he would present the contestant with the option of switching her pick to the other, still-closed, door. Most people's intuition is that there is no statistical advantage to either sticking or switching -- the odds the contestant initially faced of any door concealing the top prize were 1/3, and that's still the case. But it turns out that that intuition is incorrect, and that the contestant interested in maximizing her odds of winning should always switch. Here's the explanation of why that is so from Wikipedia:
"The chance of initially choosing the car is one in three, which is the chance of winning the car by sticking with this choice. By contrast, the chance of initially choosing a door with a goat is two in three, and a player originally choosing a door with a goat wins by switching. In both cases the host must reveal a goat. In the 2/3 case where the player initially chooses a goat, the host must reveal the other goat making the only remaining door the one with the car.
"More formally, when the player is asked whether to switch there are three possible situations corresponding to the player's initial choice, each with probability 1/3:
"The player originally picked the door hiding goat number 1. The game host has shown the other goat.
"The player originally picked the door hiding goat number 2. The game host has shown the other goat.
"The player originally picked the door hiding the car. The game host has shown either of the two goats.
"If the player chooses to switch, the player wins the car in the first two cases. A player choosing to stay with the initial choice wins in only the third case. Since in two out of three equally likely cases switching wins, the probability of winning by switching is 2/3. In other words, players who switch will win the car on average two times out of three.
"The solution would be different if the host did not know what was behind each door..."
The alteration in the "objective" probabilities faced by the contestant hinges on the fact that Monty Hall knows what is behind each door, and is not picking randomly. It is the state of his knowledge that alters the contestant's odds.
Since many, many people have had a very difficult time accepting the correct solution to this problem, it suggests to me that the dependence of probability on states of knowledge is far from being an immediately obvious relationship.