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3 cards (Posted on 2002-05-11) Difficulty: 3 of 5
A dealer offers to you to play a game. He shows you three two-sided cards: one with both sides red, one red and black and the other black and black. He puts them in a hat, and randomly (no tricks here) takes out a card and puts it on the table.

You both see only one side of the card. At this point he says that if the bottom side is the same as the top, he will take your money. If the other side is different, you double it. He explains that by now one of the cards is ruled out - if you're seeing red, the card cannot be a double black card, and vise versa - so you have a 50/50 chance of winning.

Is this a fair game? Why or why not?

See The Solution Submitted by levik    
Rating: 3.5385 (13 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
A-priori probability | Comment 16 of 18 |
Understanding the deception here is akin to understanding a-priori probability. Before a card is drawn you have a 1 in 3 chance of winning. No matter which card is drawn you will still play the game. Even though you have new information after the draw (one of the cards cannot be the visible one), you are not making any decisions based on this new information, therefore you cannot change your previous odds of winning.
  Posted by Andy on 2003-11-20 11:22:59
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