Gene,

I have a methodology that gives me the answer with 3 uses of the balance 50% of the time but requires 4 uses of the balance the other 50% of the time.
Are you sure this can be done???? 

For my first use of the scales, I place a total of 6 balls on the scales (3 on each side). If they balance, then none of the six balls on the scales is the odd ball. If they do not balance, then none of the six balls that were left off of the scales is the odd ball. Thus, I have reduced the possible suspects to 6 after one use of the scales.

Is this the right starting point???

Second, I place three balls that have been eliminated as suspects on one side of the scales and three balls that are still possibles on the other. Two outcomes:

1. If the scales do not balance, I know that one of the three suspects on the scales is the odd ball and I know whether it is heavier or lighter than the others. I can determine which is the odd ball on the third use of the scales by placing only one remaining suspect on each side of the scales. If they balance, then the suspect left off is the odd ball. If they do not balance, I know which one is odd because I already know whether the odd ball is heavier or lighter.  

2. If the scales balance on my second use of the scales, then I have three remaining suspects (the balls left off the scales) but do not know whether the odd ball is heavier or lighter. I cannot figure out how to determine both the heavier/lighter question plus identify the odd ball in just one more use of the scales. I need two more uses. 

Am I close at all????


I hope that you had a good trip back. Stay in touch.

Paul