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?

(In reply to re: Simply stated (solution) by levik)

If you pay your money before he draws the card, the chances remain the same as in any of the other games. If you pay after you see that it's got a red back, this is a different situation from the other puzzle.

The original puzzle was symetrical: your bet and your odds would be the same if the side you saw were black instead of red.

In this version, whenever the blue-backed ten is drawn, you will pass. Since you will only play when one of the two red-backed gards are drawn, they are the only cards to worry about. You have a 50:50 chance

Posted by TomM on 2002-07-16 00:56:54