Three weighings suffice. There are two different techniques for solving the problem. Let the balls be numbered 1 thru 6.

Weigh (1,2) vs (4,5), then (2,3) vs (5,6), then (3,1) vs (6,4).

All weighings involve one ball on each side of the beam balance. First weigh 1 against 2. If these are equal, then weigh 1 against 3, otherwise weigh 3 against 4. The reader may work out what the third weighing should be.

How would you divide 50 black and 50 white marbles into two piles, not necessarily of same size, so that the probability of picking a white marble as follows is maximized: we first pick one of the piles uniformly at random, then we pick a marble in that pile uniformly at random?

At a restaurant, how can Veronica choose one out of three desserts with equal probability with the help of a coin? What if the coin is biased and the bias is unknown?