The prisoner paradox I am given an exercise, and I can't quite figure it out.

The Prisoner Paradox Three prisoners in solitary confinement,
  A, B and C, have been sentenced to death on the same day but, because
  there is a national holiday, the governor decides that one will be
  granted a pardon. The prisoners are informed of this but told that
  they will not know which one of them is to be spared until the day
  scheduled for the executions.
  
  Prisoner A says to the jailer “I already know that at least one the
  other two prisoners will be executed, so if you tell me the name of
  one who will be executed, you won’t have given me any information
  about my own execution”.
The jailer accepts this and tells him that C will definitely die.
A then reasons “Before I knew C was to be executed I had a 1 in 3
  chance of receiving a pardon. Now I know that either B or myself will
  be pardoned the odds have improved to 1 in 2.”.
But the jailer points out “You could have reached a similar conclusion
  if I had said B will die, and I was bound to answer either B or C, so
  why did you need to ask?”.
What are A’s chances of receiving a pardon and why? Construct an
  explanation that would convince others that you are right.
You could tackle this by Bayes theorem, by drawing a belief network,
  or by common sense. Whichever approach you choose should deepen your
  understanding of the deceptively simple concept of conditional
  probability.

Here's my analysis:
This looks like the the Monty Hall problem, but not quite. If A says I change my place with B after he is told C will die, he has 2/3 chances to be saved. If he doesn't, then I would say his chances are 1/3 to live, like when you don't change your choice in the Monty Hall problem. But at the same time, he is in a group of 2 guys, and one should die, so it is tempting to say that his chances are 1/2.
So the paradox is still here, how would you approach this. Also, I have no idea how i could make a belief network about this, so i'm interested to see that.
 A: I think you are over-thinking the problem - it is a Monty Hall problem and the same logic applies.
A: Initially there are three possibilities with equal probabilities:


*

*A will be freed (prob $1/3$)

*B will be freed (prob $1/3$)

*C will be freed (prob $1/3$)


With the promise of the message, there are four possibilities with different probabilities:


*

*A will be freed and A is told B will be executed (prob $1/6$)

*A will be freed and A is told C will be executed (prob $1/6$)

*B will be freed and A is told C will be executed (prob $1/3$)

*C will be freed and A is told B will be executed (prob $1/3$)


Conditional on "A is told C will be executed" this becomes 


*

*A will be freed and A is told C will be executed (prob $1/3$)

*B will be freed and A is told C will be executed (prob $2/3$)


So after the message A would like to swap with B (the Monty Hall problem) but cannot and so keeps the original $2/3$ probability of being executed.
A: I am not quite sure that I agree with @babelproofreader that this is a Monty Hall problem and the same logic applies.  In the Monty Hall problem, you go down and select one door.  The rules are that Monty knows where the prize is, will never 
open a door that conceals the prize, and will always open one of the unchosen 
doors (i.e. if you have chosen a door without a prize, he will not open the door you have chosen and say, "Sorry, you lose!" and send you back to your seat), 
and he will 
always offer the choice to switch to the other (unchosen unopened) door (i.e. he 
will not offer the choice only when you have chosen the door with the prize.)  In  these circumstances, if $A$ denotes the event that your initial pick is the 
the door with the prize, then $P(A) = \frac{1}{3}$.  If $B$ is the event that your final pick is the door with the prize, then


*

*if your strategy is to always stay put, then $P(B \mid A) = 1$ (since you
made the right choice in the beginning and are sticking with it) and 
$P(B \mid A^c) = 0$ (because you made a wrong choice in the beginning and
are sticking with it).  So by the law of total probability,
$$P(B) = P(B \mid A)P(A) + P(B \mid A^c)P(A^c) = 1 \times \frac{1}{3} + 0\times \frac{2}{3} = \frac{1}{3}$$

*if your strategy is to always switch, then $P(B \mid A) = 0$ (since you
made the right choice in the beginning and then switched) and 
$P(B \mid A^c) = 1$ (because you made a wrong choice in the beginning and so the remaining (unchosen unopened) door is guaranteed to have the prize).  So by the law of total probability,
$$P(B) = P(B \mid A)P(A) + P(B \mid A^c)P(A^c) = 0 \times \frac{1}{3} + 1\times \frac{2}{3} = \frac{2}{3}$$


Here, the situation is different.  There is no changing places with $B$ as in "If A says I change my place with B after he is told C will die, he has 2/3 chances to be saved."  
Added comments:  Another difference is that A has no information as to whether
the jailer knows who is going to be pardoned or whether the jailer is speaking
the truth when he says that C will be executed.  On the other hand, the jailer
is perfectly correct when he remarks that his telling A that C will be executed
has conveyed no useful information to A.  The closest analogy to the Monty Hall
problem is that after A has chosen a door, Monty opens an unchosen door to 
reveal a goat and says to A "Open your door and let's see what you got",
that is, no offer of a switch.  So A's chance of winning the prize (Monty
Hall) or being pardoned (prisoner problem) are the same: $1$ out of $3$
regardless of whether Monty opens an unchosen door to reveal a goat or not,
or the jailer tells A that C is going to be executed, or not,  exactly 
as Henry computed in detail.
A: The answer depends on how the jailer chooses which prisoner to name when he knows that A is to be pardoned. Consider two rules:
1) The jailer chooses among B and C at random, and just happened to say C in this case. Then A's chance of being pardoned is 1/3.
2) The jailer always says C. Then A's chance of being pardoned is 1/2.
All we are told is that the jailer said C, so we don't know which of these rules he followed. In fact, there could be other rules -- perhaps the jailer rolls a die and only says C if he rolls a 6. 
A: As pointed by others, the three prisoners problem is a rephrasing of Monty Hall. For more information, take a look at section 1.7 of this paper http://faculty.winthrop.edu/abernathyk/Monty%20Hall%20Problem.pdf
A: Imagine that the jailer tells A that C will definitely die.
And then he tells B that C will definitely die.
It is clear in this case that A and B has 50% each to be pardonned.
But what is the difference between the two versions?
A: Three prisoners problem is different from Monty Hall. The probability to be pardoned is actually $1/2$ for Alice, not $2/3$, but only if jailer follows "always name Bob when possible" strategy. 
Events:
$A$ - Alice is pardoned. Same for $B$ and $C$.
$J$ - jailer tells Alice the name "Bob" (as an answer to the "who will be executed"). $J^c$ - he tells name "Carl". He can't name Alice herself because of the rules.
We're interested in $P(A|J) = P(J|A)P(A)/P(J)$. Now there are two scenarios:


*

*Jailer tosses a coin before telling B or C: $P(J|A) =
\frac{1}{2}$.


$$P(A|J) = \frac{1}{2} \times \frac{1}{3} / \frac{1}{2} = \frac{1}{3}$$


*Jailer tells Bob's name whenever possible: $P(J|A) = 1$, also
$P(J|C) = 1$ and $P(J|B) = 0$. 


$$P(J) = P(J|B)P(B) +
P(J|B^c)P(B^c) = 0\times\frac{1}{3} + 1\times\frac{2}{3} = \frac{2}{3}$$
$$P(A|J) = 1 \times \frac{1}{3} / \frac{2}{3} = \frac{1}{2}$$
A: After receiving the information, that Prisoner C will die, his chances do change to 1/2, but only, because the chances that he gets that information is already 2/3 (the 1/3 possibility of prisoner C getting the pardon is eliminated)
And 2/3*1/2 is the original probability for getting freed.
More convincing is the oppositional approach:
Assume, that he is told prisoner C will get the pardon.
What are his chances not to be killed?
Everybody will acknowledge that his chances are zero, assuming the jailer doesn't lie and there is only one pardon.
This time, he has the chance of 1/1, because the chance for that information was already 1/3.
