|
The Unexpected Execution (Hanging) Paradox
This paradox goes by a variety of names: "Unexpected Hanging", "Unexpected Examination", "Swedish Drill", and possibly more. Basically a prisoner is told that he will be hanged next week but the day of the hanging will be a surprise. The prisoner realizes that if he wakes up Saturday morning and finds himself not dead, then he can't be hanged that day because it would not be a surprise. By induction, he then eliminates Friday and so on for every day of the week. But come Wednesday he was hanged -- much to his surprise -- as the judge promised. I think the puzzle is important since it seems to have the same characteristics of certain aspects of the Social Dilemmas and the Vagueness Problem. In particular, it is often claimed that in an Iterated Prisoner's Dilemma game if the number of plays is finite then rational players must defect on the first play due to the same backwards inductive logic as used in the Unexpected Execution Paradox. I'm thinking that the Unexpected Execution Paradox creates some doubt in that method.
An alternate formulation of this problem might help. Let us take this approach.
Consider a stack of 7 playing cards, all of which are red except one which is black. It is your job to assemble the cards in a stack face down with the black one in some position. It is my job to turn the cards over one at a time until I get to the black one.
Can you arrange the cards in the deck in such a way that at every position, I will not be able to deduce that the next card is a black one before I turn it over? That is, as I go through the stack, one at a time, I will not be able to correctly deduce that the next card is black.
You cannot put it in the bottom, 7th, position, for I can certainly deduce that it is black if I get down to the last card and I haven't seen a black one. So that rules out the 7th position. There seems to be no doubt about that. (It would seem even that the 7th card is useless and we might as well play the game with 6, but I will let that pass.)
What about the 6th position? Well when I get down to the 6th card, I can deduce that the it must be black since we have already eliminated the 7th position. So you can't use the 6th position either.
Now, I say the 5th position has exactly the same problem. We have eliminated the 6th and 7th haven't we? So if I get to the 5th card can I not deduce that it is black?
What do you think? Does this clarify the problem? Does it change the problem?
|
09-20-2002, 11:11 PM
# 1
Sponsored Links
