Spoiler warning: This is a fun puzzle -- solution follows; don't read on if you want to try it yourself :-)
The dwarves, being rather pensive types, reflect deeply upon their predicament. Having been imprisoned for a long (infinite?) time, they realize that they can only liberate themselves externally if they first liberate themselves internally. They mustn't let their long captivity rob them of their inner freedom. Despite having been incarcerated for as long as they can remember, they realize that they can only escape if they truly believe that they have a choice. And so they embrace the axiom of choice.
Using the axiom of choice, they choose (beforehand) one representative from each class of (countably) infinite binary sequences under the equivalence relation that relates all sequences that differ only in finitely many places. Then each dwarf determines the equivalence class of the actual hat assignment and guesses that s/he is wearing the hat specified by the representative chosen for that class.
Label the couples A, B, C so that the six people are A$_{1}$, A$_{2}$, B$_{1}$, B$_{2}$, C$_{1}$ and $C_{2}$.
Now, consider your 6 seats: _ _ _ _ _ _
There are 6 choices for who can sit in the first seat; without loss of generality, say it's A$_{1}$.
There are 4 choices for who can sit in the second seat (anyone else but A$_{2}$); without loss of generality, let's say this is B$_{1}$.
Our table now looks like this: A$_{1}$ B$_{1}$ _ _ _ _
For the 3rd seat, we just can't have B$_{2}$, so we can either have A$_{2}$ or one of the C's; I'll split this into 2 cases.
Case 1: A$_{2}$ sits in the third seat.
A$_{1}$ B$_{1}$ A$_{2}$ _ _ _
Now the fourth seat can't be B$_{2}$, otherwise the C's will sit next to each other in the last two seats. So the fourth seat must be one of the C's; there are 2 choices for this. Then the fifth seat must be B$_{2}$ and the sixth must be the other C.
Thus there are $6*4*1*2*1*1 = 48$ ways for Case 1 to happen.
Case 2: One of the C's sits in the third seat; there are 2 possibilities for this. Let's say it's C$_{1}$
A$_{1}$ B$_{1}$ C$_{1}$ _ _ _
Then there are 2 choices for the fourth seat: A$_{2}$ or B$_{2}$. I'll treat these as subcases.
Case 2.1: A$_{2}$ sits fourth.
A$_{1}$ B$_{1}$ C$_{1}$ A$_{2}$ _ _
For the fifth seat, we have 2 choices, and then 1 for the sixth.
Thus $6*4*2*1*2*1 = 96$ possibilities for this case.
Case 2.2: B$_{2}$ sits fourth.
A$_{1}$ B$_{1}$ C$_{1}$ B$_{2}$ _ _
For the fifth seat, there is only one possibility: A$_{2}$, since otherwise A$_{2}$ would be in the sixth seat, which is adjacent to the first (the table is circular). Of course there is only one possibility for the sixth.
Thus there are $6*4*2*1*1*1 = 48$ ways for Case 2.2 to happen.
There are then $48+96+48 = 192$ possible seating arrangements. The probability of this happening is $\frac{192}{6!} = \frac{192}{720} = \frac{4}{15} \approx 0.267$
Best Answer
Consider the number of coins held by dwarves sitting in even-numbered seats (seats starting at the "rich" dwarf and counting every other seat going around the circle). What happens when a dwarf is chosen to give away coins?
If the dwarf is in an even seat, then the number of coins owned by the even seats goes down by 2.
If the dwarf is in an odd seat, then his two neighbors (both even seats) gain one each, so the total goes up by two.
Since these are the only ways by which coins can move, the end result is that the number held by even-numbered seats can only go up or down by 2's.
Now, we start with all 10 coins in even-numbered seats. (The "rich" dwarf is at an even seat.) If we consider the case where everybody has 1 coin, that means there's 5 coins in even seats. But this is impossible, since we cannot get to 5 by adding/subtracting 2's from 10.
Thus it is not possible to evenly distribute the coins using the method described.