logical deduction - Four coins (plus one) and a balance

Someone gives you 4 apparently identical golden coins.

You know that among them there could be exactly one fake coin, and that a fake coin hasn't the same weight of a golden one.

In your pocket you have a fifth - surely genuine - golden coin, and in front of you a balance with two pans.

What is the minimum number of weighings to determine if there is a fake coin between the four you've been given, and if it's heavier or lighter than a golden one?

Answer

How many weighings you need:

2

Call the four coins A,B,C and D, and the true gold coin G. You start by weighing

AB | CG

1. 
   

   
   

   If this balances, then it is only possible that D is defective. Weigh D against G to find out for sure.

   
   

   
   


2. 
   
   

   
   

   If it tips towards AB, there are three possibilities: either A or B is heavy, or C is light. To figure out which one, weigh A against B. If they don't balance, you learned which of them was heavier. If they do balance, you learn C was light.

   
   

   
   


3. 
   

   
   

   If it tips toward CG, this is very similar to the previous case. Just switch the words "heavy" and "light" in that paragraph.

   
   

   
   

Furthermore, it can't be done in just one weighing. There are nine possible situations (each coin could be heavy or light, or all could be same), and a weighing has only three outcomes. By the pigeonhole principle, there will be some outcome which would result from from two different situations, so a weighing could not distinguish between those situations.