Let the coins be A,B,C,D. Then: Weighing 1: AB vs CD If they are balanced, then there are an equal number of fake coins on either sides. Go to Weighing 1.1. If they are not balanced, suppose that AB is heavier, then there are more fake coins among C and D than among A and B. Go to Weighing 1.2. Weighing 1.1: A vs B If they are balanced, then A and B are of the same type of coin. Whichever they are, C and D must be of the same type too, giving either 0 or 4 fake coins. If they are not balanced, then exactly one of A and B is fake. This also means that exactly one of C and D is fake, so there are 2 fake coins. Weighing 1.2: AC vs BD If C,D are both fake and A,B are both real, this weighing gives a balanced reading. Otherwise, there are an odd number of fake coins, so this weighing cannot give a balanced reading. So in each case it can be determined whether they have exactly two fake coins or not.