10 Heads and 90 Tails

On a table in front of you there are a hundred coins with 10 bearing heads and the other 90 showing tails. The light is turned off, putting you in pitch blackness and the challenge is to arrange all of the coins into two groups so that both groups show the same number of heads when the light is turned on again. You are allowed to flip over any coins that you choose to, but you are unable to check their state by touch alone.

Answer below...

The trick to realise here is that the two groups don't have to be the same size. Instead let's split them into a big pile of 90 coins and a small pile of 10 coins. We don't know how many heads are in either pile, it couĺd be anywhere from 0 to 10. Let's say there are n heads in the large pile. That leaves 10-n heads in the small pile. If we turn over all of the coins in the small pile we get 10-(10-n) heads which simplifies to n. Therefore both piles will have the same number of heads as each other, even though we don't know how many (if any) that is.