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Message: <blockquote data-quote="MerakSpielman" data-source="post: 1354178" data-attributes="member: 7464">OK, here's the version of the treasure chest dillema I remembered, from a book on paradoxes and probability: From Aha! Gotcha by Martin Gardner, pages 100-101  Operator: Step right up, folk, See if you can guess which shell the pea is under. Double your money if you win. After playing the game a while Mr. Mark decided he couldn't win more than once out of three times. Operator: Don't leave, Mac. I'll give you a break. Pick any shell. I'll turn over an empty one. Then the pea has to be under one of the other two, so your chances of winning go way up. Poor Mr. Mark went broke fast. He didn't realize that turning on empty shell had no effect on his chances. Do you see why? After Mr. Mark has selected a shell, at least one of the remaining two shells is certain to be empty. Since the operator know where he put the pea, his act of doing so adds no useful information for Mr. Mark to resvise his estimate of the probability that he has picked the right shell. You can demonstrate this with an ace of spades and two red aces. Mix the cards and deal them face down in a row. Allow someone to put a finger on a card. What is the probablitly he or she has picked the ace of spades? Clearly it is 1/3. Suppose, now, you peek at your two cards and turn over a red ace. You can argue, like the game operator, as follows. Only two cards are face down. The ace of spades is as likely to be one of them as the other. Therefore the probability that the ace of spades has been picked seems to have gone up to 1/2. Actually it reamins 1/3. Because you can always turn over a red ace, turning it adds no new information that affects the probability. You can puzzle your frinds with the following variation. Instead of peeking a the two unselected cards to made sure you turn a red ace, allow the person whose finger is on a card to turn over one of them. If it should be the ace of spades, the deal is declared void and the game is repeated until the reversed card is a red ace. Does this procedure raises the probability of picking the ace of spades? Oddly enough, it raises it to 1/2. We can see why by taking a sample case. Call the positions of the cards 1,2,3. Let's assume the person puts a finger on card 2, then turns over card 3 and it is a red ace. There are six equally possible ways the six cards can be delt. 1. S H D2. S D H3. D S H4. D H S5. H S D6. H D S If the third card had been the ace of spades, the deal would have been declared void, therefore cases 4 and 6 never enter into the problem. We rule them out as possible cases. Of the remaining four (1, 2, 3, 5), card 2 (on which the finger rests) is the ace of spades in two cases. Therefore the probability card 2 is the ace of spades is indeed 2/4 = 1/2. The result is the same regardless of which card the person chooses, and which card is exposed as the red ace. Had Mr. Mark been allowed to pick the shell to be turn over, and had it been empty, his chances of being right would have gone form 1/3 to 1/2. end.  I highly recomend this book. The artwork is downright silly, but the reasoning is solid and understandable.</blockquote>

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