View Poll Results: How do probabilities A and B compare?
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Here's another probability problem
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From: Yorba Linda, CA
Consider a simpler situation:
There are only three cards: the ace of clubs, the ace of hearts, the king of diamonds.
Your right-hand opponent gets two cards.
Calculate the probability if he says, "I have an ace," and if he says, "I have the ace of clubs."
There are only three cards: the ace of clubs, the ace of hearts, the king of diamonds.
Your right-hand opponent gets two cards.
Calculate the probability if he says, "I have an ace," and if he says, "I have the ace of clubs."
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From: Boston
Originally Posted by magician,Oct 23 2005, 10:11 PM
Consider a simpler situation:
There are only three cards: the ace of clubs, the ace of hearts, the king of diamonds.
Your right-hand opponent gets two cards.
Calculate the probability if he says, "I have an ace," and if he says, "I have the ace of clubs."
There are only three cards: the ace of clubs, the ace of hearts, the king of diamonds.
Your right-hand opponent gets two cards.
Calculate the probability if he says, "I have an ace," and if he says, "I have the ace of clubs."
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From: Yorba Linda, CA
Originally Posted by Russian,Oct 23 2005, 08:15 PM
Either way the probablity of him having a the second ace is 1/2.
It is not 1/2 either way. It is 1/2 only in the second case.
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From: Yorba Linda, CA
Originally Posted by Russian,Oct 24 2005, 07:12 AM
I don't understand. If he declares that he has one ace out of a possible two. That leaves only one in the deck, the odds of him having that second ace are 1 out of 2. The same as when he declares he has a certain suited ace. Or so it seems.
When he says, "I have an ace," how many hands can he hold? How many of those have two aces? What's the probability?
When he says, "I have the ace of clubs," how many hands can he hold? How many of those have two aces? What's the probability?