Buttonwood (a dude); Distinguishing skill from luck; In The Economist; 2012-12-14.
It’s a book review of: Michael J. Mauboussin; The Success Equation: Untangling Skill and Luck in Business, Sports, and Investing; Harvard Business Review Press; 2012-11-06; 320 pages.
I don’t have any quibble with the book (yet), but with of the review wherein I don’t agree with the mindpuzzle in the middle:
Statement: Jack is looking at Anne but Anne is looking at George. Jack is married, but George is not.
Question: Is a married person looking at an unmarried person?
- (A) Yes
- (B) No
- (C) Cannot be determined.
The claim on the answer is controversial:
Like 80% of those tested, [the author] went for C because we don’t know the marital status of Anne. But if Anne is married, she is looking at the unmarried George, while if she is unmarried, then the married Jack is looking at her. So the answer is A.
If I recall, you’re not allowed to do that in all logical systems. I’m pretty sure you gotta go with C … using Computational Matrimonial Calculus (CMC) we can see that you just can’t say anything about poor Anne. You can’t unify like that in the computational logical systems. There’s the problem.
GNU Prolog 1.3.1
By Daniel Diaz
Copyright (C) 1999-2009 Daniel Diaz
| ?- [user].
compiling user for byte code...
looksatmarried(A, :- looksat(A, , married(A), married(B).
anyone :- married(A), not(married(), looksatmarried(A, .
user compiled, 7 lines read - 1356 bytes written, 15848 ms
(1 ms) yes
| ?- anyone.
| ?- married(X).
X = jack
| ?- not(married(X)).
X = george
| ?- married(anne).
| ?- not(married(anne)).
Anne is in a state of both being married and not being married. So there. I feel so frosh.