It all started with this

Q: There are 8 balls. Four of them weigh X grams each, and the other four weigh Y grams each. Your task is to find two balls having different weights. You have a balance scale with two pans on which you can compare the weight of any number of balls. What is the minimum number of weighings?

My answer:


Split them into two equal groups of four balls each and weigh them . Two things are possible.

1.The Two groups are Unequal The two groups are unequal. This gives two possibilities: (case 1) xxxy /// xyyy (case 2) xxxx /// yyyy

In either case you take the heavier group and split it into two equal groups and weigh it. Possible results are:

(a) the groups are unequal, which means it must have been case 1 --> xy /// yy, so you take both from the lighter side and voila (b) the groups are equal, which means it must have been case 2 --> yy /// yy, so you take one of these four balls and one of the four balls you had discarded earlier and voila

2.The Two groups are Equal The two groups are equal, in which case you can represent them like this: xxyy /// xxyy. So you take all four from one side and weigh them. This is second weighing . Then you will either get: (a) both equal, which means xy /// xy, and you just take both from one side and voila (b) unequal, which means xx /// yy and you just take one from each side and voila

Now we have invented a question out of this:

Lets say I want to be lucky and without weighing I want to pick two balls of unequal weight.

1.) What is the probability that I pick two balls of unequal weight?

2.) If I have two strategies, a.) two pick one ball of a certain weight first and then pick another ball of the other weight and b.) I pick two balls at the same time will the probabilities change in cases a and b or remain the same and what are the answers?


The simplest way to solve 1) is to note that whatever ball we pick first, the probability the next one is of different weight is $\frac{4}{7}$. 2) If you don't like picking first and second, we can pick out by eye one ball, then another, and lift them from the bag simultaneously. This makes no difference to the probability.

A more complicated way to solve the problem is to imagine that all the balls have ID numbers engraved on them. There are $\binom{8}{2}$ equally likely ways to choose $2$ balls. There are $(4)(4)$ ways to choose $2$ balls of different weights. Thus our required probability is $$\frac{(4)(4)}{\binom{8}{2}}.$$ Calculate. This simplifies to $\frac{4}{7}$.

  • $\begingroup$ 1.) is expected. But for 2.) doesnt actually randomly picking one after the other affect the overall probability $\endgroup$ – Slartibartfast Feb 4 '14 at 10:01
  • $\begingroup$ No it doesn't. This is because not only are all (unordered) pairs of choices equally likely, all ordered pairs are equally likely. $\endgroup$ – André Nicolas Feb 4 '14 at 14:49

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