19/08/19 10:20:19.02 eIAyJJOc.net
>>377
補足
下記 P：The Riddle→Q：The Modificationが導かれる
対偶 Qの否定→Pの否定が導かれる
Q：The Modificationは、実際現代確率論・確率過程論に矛盾し、否定される
よって、Pの否定が導かれ、The Riddleは不成立
QED （＾＾；
URLﾘﾝｸ(mathoverflow.net)
Probabilities in a riddle involving axiom of choice asked Dec 9 '13 at 16:16 Denis
(抜粋)
The question is about a modification of the following riddle (you can think about it before reading the answer if you like riddles, but that's not the point of my question):
The Riddle:
We assume there is an infinite sequence of boxes, numbered 0,1,2,…. Each box contains a real number.
No hypothesis is made on how the real numbers are chosen.
You are a team of 100 mathematicians, and the challenge is the following:
each mathematician can open as many boxes as he wants, even infinitely many, but then he has to guess the content of a box he has not opened.
Then all boxes are closed, and the next mathematician can play.
There is no communication between mathematicians after the game has started, but they can agree on a strategy beforehand.
You have to devise a strategy such that at most one mathematician fails. Axiom of choice is allowed.
The Anwser: 略
The Modification:
I would find the riddle even more puzzling if instead of 100 mathematicians, there was just one, who has to open the boxes he wants and then guess the content of a closed box.
つづく