Eight billiard balls.

 

Question:

You have eight billiard balls. One of them is defective in that it weighs more than the others. How do you tell, using a balance, which ball is defective in two weighings?

.

.

C

A

P

T

A

I

N

I

N

T

E

R

V

I

E

W

.

.

Solution: The first choice of most candidates—to weigh four balls against four balls—will not work. You will learn that one of four balls is the heavier one, but will need at least two more weighings to isolate the defective ball. The strategy that seems to work best is anarchist thinking. Work through the options, like this candidate does:

I don’t think weighing four balls against four balls will work. We would know that the heavier pan contained the defective ball. But when we split the balls in that group into two pairs and weighed the pairs against each other, we would need one more weighing to identify the defective ball. So that’s one weighing too many.

Let’s leverage the fact that whenever two pans are of equal weight, we can conclude the defective ball is not in either pan. So now for the first weighing let’s pick any three balls and weigh them against any other three balls. One outcome is that two pans balance. In that case, the defective ball must be one of the two balls we have set aside. In that case, for the second weighing, all we have to do is compare the two untested balls. The heavier is the defective ball.

The other possible outcome of the first weighing is that the pans do not balance. In that case, the defective ball must be one of the three balls in the heavier pan. For the second weighing, we set aside one ball and weigh the two remaining balls from the heavier pan against each other. If one pan is heavier, it holds the defective ball. If both pans are equal, the defective ball must be the third ball that we set aside.

As far as I know, this is the only two-step solution to this brainteaser.

Answer: Set aside two balls and weigh three balls against three balls.

 

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