Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.


Mr G
Posts:
1
Registered:
2/21/13


Puzzle in the new Ian McEwan book
Posted:
Feb 21, 2013 2:00 PM


Hello,
I came across a problem in the new Ian McEwan that I can't quite figure out. It's basically a modification of the Monty Hall problem, this is the actual scenario from the book but don't worry  no spoilers.
A man is in a hotel standing in front of three doors, and he knows his wife is with her lover behind one of them. He's about to break down the first of these doors when suddenly a couple open the second door to leave their room. The man then decides that his probability of choosing the first door was only 1/3 and decides to break down the third door instead.
However, McEwan later dismisses this change of decision as erroneous. He argues that because the couple left the second room of their own accord, the occurrence was random and not the Monty Hall problem, since there's no host who knows a priori about the doors and who will always open a door that leads to an empty room.
So in this case, McEwan argues that it wouldn't matter which door the man chose since the probabilities are 1/2 and 1/2, and not 1/3 and 2/3.
I don't quite understand the difference, and was wondering if anyone can shed some light on this?
Thank you.



