Arthur and Bert each writes down a positive integer on a piece of paper and then shows it to Charles. Charles then writes two numbers on a blackboard, visible to Arthur and Bert: one of them is the sum of Arthur's and Bert's numbers, and the other is a random number.
After this Charles asks Arthur if he knows Bert's number. If Arthur says he doesn't know, then he asks Bert if he knows Arthur's number. If Bert says he doesn't know, Charles continues with Arthur, then if necessary with Bert and so on... until he gets a positive answer.
When will Charles get a positive answer?
(In reply to
Don't get it by Old Original Oskar!)
If A=4 and B=6, and X=10 and Y=11...
From the point of view of an observer who doesn't know what A and B is, but does know X and Y:
If A passes, then A is between 1 and 9.
If B passes (knowing this), then B is between 2 and 9. If B had
1, then he would know that Y was the random #, so he could deduce
A's number.
If A still passes (knowing that B is between 2
and 9), then A must be between 2 and 8. If A had a 1, then he
would know that Y was the random #, and could deduce B's number.
If A had a 9, then he would know that X was the random number, and
could deduce B's number.
If B still passes (knowing that A is
between 2 and 8), then B must be between 3 and 8. If B had a 2,
then he would know that Y was the random #, and could deduce A's
number. If B had a 9, then he would know that X was the random
number, and could deduce A's number.
and so, on, until B's 4th pass, which reveals to all the logicians in the room that B has 5 or 6.
A, who knew all along that B had 6 or 7, now knows that it is 6.
IT IS VERY INTERESTING TO ME THAT THE REQUIRED LOGIC IS MORE DIFFICULT
FOR A and B, BOTH OF WHOM KNOW MORE THAN THE UNINFORMED OBSERVERS ABOUT
WHAT NUMBER THE OTHER ONE MIGHT HAVE.
Edited on October 30, 2005, 7:15 pm
Edited on October 30, 2005, 7:16 pm
Edited on October 30, 2005, 7:17 pm
Edited on October 30, 2005, 9:42 pm
Edited on October 30, 2005, 9:43 pm
re: Don't get it
