Friday's puzzle was first published in 1889 by Joseph Bertrand in Calcul des Probabilites. And it's well worth another look.
Dressing it up slightly from its original, there are three chests of drawers each with two drawers. Each chest contains two pieces of gold, two pieces of lead, or a piece of gold and a piece of lead. And each piece of metal is in a separate drawer.
You are only allowed to open two drawers.
If you strike gold with your first choice, what is the probability that the other drawer in that chest of drawers also contains a piece of gold?
If you have found gold in the drawer yo have opened then the chest can't be the one holding two pieces of lead. It must be either the two-gold chest or the gold -and-lead chest.
At this point you could decide that their is a 50% chance of the chest being one of these two options and that the probability of the second drawer in the chosen chest containing gold is also 50%.
But you would be wrong. There are three drawers containing gold - two in one of the chests and one in the other chest. It is twice as likely that you have chosen a drawer in the gold-gold chest than in the gold-lead chest, as there are twice as many. So, as you've struck gold with your first pick, the chances of the second drawer also containing gold is 2/3.
If you are familiar with the Monty Hall (two goats) problem, you might see some similarities.