The Three Card Swindle (or Bertrand’s box paradox)
This piece of probability theory is probably [sic] best described in it’s form as a bar bet (or proposition wager).
It takes a little preparation but is worth it. You need three cards. Each is marked on both sides - one is red/red, one black/black, the last is red/black.
You ask your victim to choose a card blindly and place it on the table. Let’s suppose the up side is red. You bet your opponent the next round of drinks (for fun) that the other side is also red.
At this stage most people are thinking “It’s either the red/red or red/black card, so I have a 50% chance of being right”. Consequently they will take the bet. As with any bar bet it’s best to add a little theatre - a friend who used to use this would say “I know it’s a coin toss but I have the luck of the Irish” (hence the picture).
In fact the probability of the other side also being red is not 50/50. We are observing not cards but card faces. There are three possibilities:
- This is the red side of the red/black card
- This is the first red side of the red/red card
- This is the second red side of the red/red card
Thus in 2/3 of the cases the other side is also red. By betting on the hidden face being the same as the up face you will win two thirds of the time.
That works out as a fair few free drinks…