Mdthbs

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?

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 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.

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.)

Your first answer is a correct application of Bayes Rule. If, as you say, $A$ is the event "at least 2 kings" and $B$ is the event "at least 3 kings", then

$$P(B|A)=frac{P(A|B) cdot P(B)}{P(A)}$$

$$P(A|B) = 1$$

$$P(A) = frac{begin{pmatrix} 4\2 end{pmatrix}begin{pmatrix} 48\11 end{pmatrix}+begin{pmatrix} 4\3 end{pmatrix}begin{pmatrix} 48\10 end{pmatrix}+begin{pmatrix} 4\4 end{pmatrix}begin{pmatrix} 48\9
end{pmatrix} }{begin{pmatrix} 52\13 end{pmatrix}}$$

$$P(B) = frac{begin{pmatrix} 4\3 end{pmatrix}begin{pmatrix} 48\10 end{pmatrix}+begin{pmatrix} 4\4 end{pmatrix}begin{pmatrix} 48 \9 end{pmatrix} }{begin{pmatrix} 52\13 end{pmatrix}}$$

$$P(B|A) = frac{P(B)}{P(A)} = frac{begin{pmatrix} 4\3 end{pmatrix}begin{pmatrix} 48\10 end{pmatrix}+begin{pmatrix} 4\4 end{pmatrix}begin{pmatrix} 48\9 end{pmatrix} }{begin{pmatrix} 4\2 end{pmatrix}begin{pmatrix} 48\11 end{pmatrix}+begin{pmatrix} 4\3 end{pmatrix}begin{pmatrix} 48\10 end{pmatrix}+begin{pmatrix} 4\4 end{pmatrix}begin{pmatrix} 48\9 end{pmatrix} } $$

Your second answer, if you were to try to "apply Bayes Rule" would show the falicy. What are your events "$A$" and "$B$" in this case?

You can also answer the question with the $1-p$ type argument, but again, you need to consider what the events are, and compute Bayes' rule accordingly.

They're different situations with different outcomes. You calculated the probabilities correctly, though.

The first situation, you have all of your cards face down in front of you. The dealer puts on some X-ray glasses and tells you, "You have at least two kings in your hand." This eliminates all hands with no kings or exactly one king. But you already have all of your cards in front of you.

You constructed the number of hands for two, three, and four kings, and calculated the probability.

The second situation, though, is equivalent to the dealer being sweet on you, pulling out two kings out of the deck and handing them to you. Then she deals eleven more cards to you. You already have two kings dealt to you, because you're choosing from the fifty remaining cards, two of which are kings.

If you object to the way I described the second situation, that's fine, but it matches how you calculated the probability the second time. Solving a problem equivalent to the one you want to solve works.

But ... I would say that your first way is closer to the spirit of how the problem was intended.

Comment: Parallel to @mjw's answer (+1), an exact computation in R, using the hypergeometric CDF phyper.

a = 1 - phyper(1, 4, 48, 13); a

[1] 0.2573349

b = 1 - phyper(2, 4, 48, 13); b

[1] 0.04384154

b/a

[1] 0.1703676

Approximate results from simulation in R, based on a million 13-card hands (from which we can expect about three decimal places of accuracy):

set.seed(605)  # for reproducibility

deck = rep(1:13, each=4)  # for simplicity Kings are 13s

nr.k = replicate(10^6,  sum(sample(deck, 13)==13) )

mean(nr.k[nr.k>=2]>=3)  # read [ ] as 'such that'.

[1] 0.1703104

mean(nr.k >= 3)/mean(nr.k >=2)

[1] 0.1703104

In R: The object nr.k is a vector of length $10^6.$ There are several 'logical vectors' consisting of TRUEs and FALSEs.
One of them is nr.k >= 3. The mean of a logical vector is its proportion of TRUEs. Thus mean(nr.k >= 3) approximates the probability of getting at least 3 Kings.