Just when I thought I had seen every wrinkle on the Monty Hall problem, Raymond Smullyan has to go come up with another one. Here’s an excerpt, from his book *The Riddle of Scheherazade and Other Amazing Puzzles*:

“And now,” said Scheherazade, “I have a paradox for you. There are three boxes labeled A, B, and C. One and only one of the three boxes contains a gold coin; the other two are empty. I will prove to you that regardless of which of the three boxes you pick, the probability that it contains the gold coin is one in two.”

“That’s ridiculous!” said the king. “Since there are three boxes the probability is clearly one in three.”

“Of course it’s ridiculous,” said Scheherazade, “and that’s what makes it a paradox. I will give you a proof that the probability is one in two, and your problem is to find the error in the proof – since the proof must obviously contain an error.”

“All right,” said the king.

“Let’s suppose you pick Box A. Now, the coin is with equal probability in any of the boxes, so if Box B should be empty, then the chances are fifty-fifty that the coin is in Box A.”

“RIght,” said the king.

“Also, if Box C is empty, then again the chances are fifty-fifty that the coin is in Box A.”

“That’s right,” said the king.

“But at least one of the boxes, B or C, must be empty, and whichever one is empty, the chances are fifty-fifty that the coin is in Box A. Therefore, the chances are fifty-fifty, period!”

“Oh my!” said the king.

What is the solution to the paradox?

Feel free to hash that out in the comments.