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An Ace for a Pair (Posted on 2005-03-01) Difficulty: 3 of 5
Playing five-card stud poker with two friends one night, one of them, Kevin, accidentaly drops one of his cards on the table, the ace of hearts. My other friend, Nick, laughs and says, "I also have at least one ace in my hand." I have no reason not to believe him. Now, I do not have any aces in my hand, but I do have a pair of kings. Which of my friends is more likely to have at least a pair of aces (that is to say, at least one more ace) in his hand?

See The Solution Submitted by DJ    
Rating: 3.7500 (4 votes)

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Solution I think this is it. | Comment 1 of 12

The one whose ace you saw is more likely to have another ace.

There are 52 cards all together.  Four of them are aces.  We know that Kevin has the one you saw and that Nick has at least one other.

What's the probability that Kevin has no other aces?

We know that the cards in your hands are not aces, and Kevin doesn't have these. We don't know what cards Nick has except that at least one is an ace. There are 52-7=45 cards that could be the first of Kevin's unseen cards, of which 2 are aces, so the probability that first other card of Kevin's is not an ace is 43/45; and that the second is not an ace given that the first was not, is 42/44; the third, 41/43 and the fourth, 40/42, giving the probability of no second ace as 41*40/(45*44) = 82/99, or probability of a second ace as 17/99.

How about Nick?

Of Nick's cards, the a priori probability that he would not have gotten at least one ace, given that an ace is in Kevin's hand and none in yours, would be similarly calculated: the probability that his first dealt card would not be an ace is 43/46 (43 non-aces in the 46 cards that are not in your hand and not the ace that Kevin dropped). Continuing from there, the probability he'd have no aces is 43*42*41*40*39/(46*45*44*43*42)=533/759 or about 0.7022397891963109354, and the probability he'd have at least one ace would then be 226/759.  We need to take the probability of at least two aces and divide by this, to get the probability he has at least one other ace.

We can calculate the probability he has exactly one ace: The probability the first dealt card was an ace is 3/46 and that the subsequent ones were not, as 43*42*41*40/(45*44*43*42). We have to multiply by 5 to allow for the fact that any one of the cards could be the ace, so it comes out in all as 5*3*43*42*41*40/(46*45*44*43*42) or 205/759. So the probability of more than one ace is (226-205)/759 or 21/759.

Then the probability of more than one ace given there is at least one, is 21/226 (the two numerators over the same denominator), or about 0.0929203539823008849, which is smaller than Kevin's over 17% probability.


  Posted by Charlie on 2005-03-01 19:48:11
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