We are dealt 13 cards from a standard 52 card deck. If $A$ is the event where we are dealt at least two kings and $B$ is the event where we are dealt at least 3 kings, we want to know $P(B|A)$. I believe the correct way to do this is to calculate the probability of being dealt a hand with each number of kings separately as follows:
$\displaystyle \frac{{4 \choose 3}{48 \choose 10} + {4 \choose 4}{48 \choose 9}}{{4 \choose 2}{48 \choose 11} + {4 \choose 3}{48 \choose 10} + {4 \choose 4}{48 \choose 9}} \approx .17$.
However, it also makes sense to me that if we know we have been dealt 2 kings, it doesn't matter where in our hand they are, the $P(B|A)$ should be the same as the probability of getting dealt either 1 or 2 kings in an 11 card hand from a 50 card deck with two kings in it as follows:
$\displaystyle \frac{{2 \choose 1}{48 \choose 10} + {2 \choose 2}{48 \choose 9}}{{50 \choose 11}} \approx .4$
(Or compute $1-p$, where $p$ is the probability of getting no kings in an 11 card hand from a deck with 50 cards and only 2 kings.)
What is the issue with my logic here?
Best Answer
I'm not sure if this is an answer, but I can't fit what I want to say in a comment.
It's not a matter of what we are told, but of what is known. How we come by the information is important.
If someone, known to be truthful, looks at the hand, and we ask her, "Are there at least two Kings in the hand," and she answer "Yes," then we are certainly in the first situation, and the probability is $.17$ that there are at least three Kings in the hand.
On the other hand, if she picks up the hand and says, "The fifth and eighth cards are Kings," we really don't know enough to compute a conditional probability. Maybe she only announces the location of Kings when there are exactly two Kings in the hand. Maybe she only announce the location of red Kings.
In his "Mathematical Games" column, Martin Gardner once gave a series of three probability problems. Four people are playing bridge, so each is dealt $13$ cards. The first picks up his hand, looks at it and announces "I have an Ace." We are asked for the probability that he holds at least two Aces. (He is always truthful, and no one else has looked at his hand.)
The other two problems are the same, except that in the second case, he announces, "I have a black Ace," and in the third case he announces, "I have the Ace of Spades."
If you work out the probabilities according to the usual formula, they are all different. (If I recall correctly, they increase.) But this is paradoxical. He can always announce the color; he can always announce the suit. Why should it make any difference?
I asked a professional probabilist about this once, and his answer was essentially what I said above. We don't know enough about the circumstances under which this guy makes announcements.
Note that if he accidentally flashed a card, and we saw that it was an Ace, but couldn't tell the color or suit, or could tell the color but not the suit, or could tell the suit, then the three calculations would all be appropriate. (Or at least, I think they'd be appropriate.)