Pool riddle; literally

[fti]

Blud's informative post on the smallest pool ball reminded me of a riddle involving pool balls.

8 pool balls. One ball is either heavier or lighter. On a balancing scale you get 3 tries to find the ball that is either heavier or lighter. How should you measure so that by the third try you can find the ball that is either heavier or lighter?

 

[fxskater]

Easy, weigh 4 against 4 whichever end goes lower those balls are eliminated, then measure 2 against 2 whichever end goes lower is eliminated then measure 1 against 1 you have your lighter ball. This is like an IQ test question.

 

[bruin70]

Easy, weigh 4 against 4 whichever end goes lower those balls are eliminated, then measure 2 against 2 whichever end goes lower is eliminated then measure 1 against 1 you have your lighter ball. This is like an IQ test question.

if you knew that the ball you were looking for either heavier or lighter, your solution would work. but what if you DIDN'T know whether the ball you're looking for is heavier or lighter. in which case the ball you're looking for could still be in either group of four.

or maybe i didn't understand the riddle correctly.

 

[efirkey]

How about this

put 2 balls on each side of the scale and the other four on the table. If the scale balances then you know the odd ball is on the table and can eliminate the four on the scale. However if the scale is not balanced then you can eliminate the four on the table.
Once you have four balls do the same thing with 1 ball on each side and eliminate 2 more balls. Once you have 2 balls left leave one of them on the scale with any of the eliminated balls on the other side. If they balance then the remaining ball is different. If the scale is not balanced then that ball is the different one.

 

[fti]

Nice one efirkey.