What I understand happened in Prisoner's Paradox is that the very field of probabilistic distribution of pardons has shrinked from 3 to 2. It's like binary search. You cheated your way by knowing where _not to look_.
Another simple analogy: you can increase the value of rational number both by increasing the numerator and decreasing denominator. In this case you decreased denominator.
Monty Hall requires a slightly bigger leap of logic.
Actually, here's a more directly linked problem: http://en.wikipedia.org/wiki/Three_Prisoners_problem
This isn't the same as Monty Hall. Monty Hall only increases the odds after switching doors. This actually seems like a trick meant to fool people who know about the Monty Hall problem but don't really understand it.
I re-read all three problems after you replied and this is what I understood. Talking about Three Prisoners' problem since the article's wording doesn't include the very crucial ingredient for easier understanding: coin flip on warden's side.
For A, it doesn't really matter if B or C were executed. In fact, he knew that either B or C are going to be executed for sure. He knew that at least one person who is not him will be executed.
Warden just named the sure-to-be-executed-guy. Didn't change what he previously knew about his chances, since he still doesn't know for sure whether the-other-not-named-guy is going to be executed.
However, the distribution changed for C. Now A still knows that there 1/3 chance of him living but since we know for sure that the-guy-who-warden-named (B or C) will be killed the-guy-who-warden-didn't-name-and-who-isn't-A (C or B) will have 1 - 1/3 = 2/3 probability of living.
Prisoner A rejoices that his own chance of survival has improved from 1/3 to 1/2. But how is this possible? The guard has given him no new information. Has he?
No, Prisoner A still has a 1 in 3 chance of [edit: not] dieing. If he asked if Prisoner B had a pardon then you could argue his odds became 1/2. But, now he still only knows that at least one of his cellmates is going to die.
So when the guard says B is sure to die he is actually opening door number 2 to show no prize is there.
Before this event the odds of survival for each prisoner were 1/3, 1/3, 1/3.
The odds that the guard would choose prisoner B as the non-surviving one (dead that is) in case prisoner A/B/C is the surviving one are: 1/2, 0, 1.
That is if A was pardoned then the guard can chose either of B or C so the odds of pointing out B are 1/2. The odds of pointing out B as dead when he is pardoned are 0 and the odds of pointing out B as dead when C is pardoned are 1 (because he cannot indicate A as dead because the guard is not allowed to tell A about his own fate). Hence the final odds for each prisoner must add up to 1 so through proportionality they are 1/3, 0, 2/3.
So prisoner A didn't find out anything new about his chance of being the pardoned one.
That prisoner should immediately switch name-tags with the other prisoner who's fate is in question :)