Alternate way of computing the probability of being dealt a 13 card hand with 3 kings given that you have been dealt 2 kings

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.)


enter image description here

You're solving 2 different problems. In the above Venn Diagram,

Your First Probability is $\frac{A}{A+B+C}$

Your Second Probability is $\frac{A}{A+B}$

To understand better, I'll recommend you to solve the following version of your problem :

You're given 12 different cards out of which 11 are colored Red and 1 is colored Blue. You have to pick 11 cards from the total available 12 cards.

  1. Find the probability that there are at least 11 Red cards (or in other words all cards are red) given atleast 10 cards are known to be Red.
  2. Find the probability that there are at least 11 Red cards (or in other words all cards are red) if you pick 10 Red cards before starting the experiment.

Tags:

Probability

Conditional Probability

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