# Red, White, Blue coin weighing puzzle

Infront of you are six coins, and a sensitive balance beam.
Two of the six coins are marked with red dots. Two with white. Two with blue.
In each pair of coloured coins, one of the coins is slightly heavier (meaning there are three heavy coins, and three light coins). All the heavy coins have identical weights, and all the light coins have identical weights too. Your job is to separate them into two piles: All the heavy coins in one pile, all the light coins in another, (and to do this in the minimum number of weighings).
Other than the slight weight variance, and the dot markings, the coins are visually identical; there are no external clues as to which the heavier of each pair is. The weight discrepancy is so slight that only the balance beam is sensitive enough to differentiate.
What is the minimum number of weighings to separate the coins into two sets?

## Three weighings

It should be pretty obvious that we can do it with three weighings. We chose a colour and place one coin on each side of the balance. The heavier coin of each pair will go down. We do this three times. Simple. Can we do better?

## Two weighings

It’s possible to perform the task with just two weighings, but there is a caveat for one scenario (but, if you read the problem definition carefully, it does not violate the requirements: Our task is simply to correctly separate the coins into two piles of three, that’s it).
To help track what’s going on we can identify the heavier coins with the upper case letters RWB, and the lighter coins with the lower case letters rwb.