Anticipating Heads and Tails

This will show us how some clever reasoning, along with algebraic knowledge of the most elementary kind, will help us solve a seemingly “impossibly difficult” problem.

Let’s consider the following problem.

You are seated at a table in a dark room. On the table there are 12 coins, 5 of which are heads up and 7 are tails up. (You know where the coins are, so you can move or flip any coin, but because it is dark you will not know if the coin you are touching was originally heads up or tails up.) You are to separate the coins into two piles (possibly flipping some of them) so that when the lights are turned on there will be an equal number of heads in each pile.

Your first reaction is likely to be: “You must be kidding! How can anyone do this task without seeing which coins are heads or tails up?” This is where a most clever (yet incredibly simple) use of algebra will be the key to the solution.

Let’s “cut to the quick.” Separate the coins into two piles of five and seven coins, respectively. Then flip over the coins in the smaller pile. Now both piles will have the same number of heads!

That’s all! You think this is magic. How did this happen? Well, this is where algebra helps to understand what was actually done.

Let’s say that when you separate the coins in the dark room, h heads will end up in the seven-coin pile. Then the other pile, the five-coin pile, will have 5 – h heads and 5 – (5 – h) = h tails. When you flip all the coins in the smaller pile, the 5 − h heads become tails and the h tails become heads. Now each pile contains h heads!

What an awed reaction!