Four different aces are dealt, face up, one apiece, to you and three other players. They are shuffled and redealt, this time face down.

Before looking, the chance is 75% that you have an ace different from your original card.

Question 1) But what if the first player looks at his card and, without showing the card, truthfully reveals that it is not his original ace? What is the chance that you also have a different ace, given that one player's card is known to be different?

Question 2) And what if the second player also looks at his card and, without showing the card, also truthfully reveals that it is not his original ace? What is the chance that you also have a different ace, given that two players' cards are known to be different?

Question 3) Finally, what if the third player also looks at his card and, without showing the card, also truthfully reveals that it is not his original ace? What is the chance that you also have a different ace, given that all three other players' cards are known to be different?

According to my math, There are 18/24, 14/18, 11/14, and then 10/12 for the probability of drawing a different ace. After the first round, 6 possibilities are eliminated. After the second, 4 possibilities are eliminated, and after the third, 2 possibilities.

Posted by Joe on 2005-02-09 15:47:08