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

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?

• The second probability is the probability of getting at least $3$ kings if the first $2$ cards dealt are kings. Or, if you pick two cards at random from the hand, without looking, and they both turn out to be kings, then the probability that at least one of the remaining cards is a king is about $.4$ Jun 5, 2019 at 16:25
• I see what you mean, but I think I still just don't understand why knowing where the kings are in the hand changes the probability. If I understand what you are saying, you are saying that if I just tell you that two of these 13 cards are kings, then the probability that you have 3 kings or more is .17, but if I tell you the 5th and 8th cards are kings, then the probability is .4. Jun 5, 2019 at 16:31
• An equivalent, and rather famous, problem is the Boy or Girl paradox. In particular, the observation of @saulspatz re: ace / black ace / spade ace, is equivalent to this subsection about what if you have extra, but seemingly "irrelevant" info, about the boy such as the boy being born on a Tuesday (equiv. to the ace being black / spade). Jun 5, 2019 at 17:33
• @SamEastridge yes the situations are different. Besides the answers below, here's another view. The condition "at least two are K's" is the union of 78 different conditions like "the 5th and 8th are K's" but those 78 conditions are NOT disjoint. So even though each of those 78 gives a cond. prob. for 3 K's of 0.4, it doesn't follow that the union of them does. The really simple version is: Flip a coin twice. Find prob. of 2 H's given (a) 1st is H (b) 2nd is a H (c) at least one H (d) "one coin is a H" . Answers: 1/2, 1/2, 1/3 (d) is ambiguously worded.
– Ned
Jun 5, 2019 at 20:25

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?

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.)

• This is wild. So if I tell her before I deal the hand to tell me if she has at least two kings in her hand, when she says yes after I deal the cards the probability is .17 of her having three or more kings. However, if I tell her to show me exactly two kings in her hand if she has them, then after I deal the hand and she shows me the two kings, the probability is .4? Jun 5, 2019 at 17:14
• (+1) for mention of Martin Gardner, bringing back fond memories. Jun 5, 2019 at 17:30
• @SamEastridge No, I don't see that it's any different. Showing you two Kings is the same as saying, "Yes, I have two Kings." However, if we know that she always shows black Kings when she can, and she shows two red Kings, then we know the probability that she has another King is zero. But if she picks the Kings she shows at random, then it's just like saying, "Yes." Jun 5, 2019 at 17:35
• +1 "We don't know enough about the circumstances under which this guy makes announcements." <-- This is the best one-liner explanation I have ever heard about this particular common confusion! :) Jun 5, 2019 at 18:04
• @antkam When I was first told the Monty Hall problem, I answered that there was insufficient information, for this very reason, but people heaped scorn and derision on me. Jun 5, 2019 at 18:42

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.

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.

• So, if we use Bayes Rule, we get the correct answer and your way of thinking about the problem makes sense. If you are "applying Bayes Rule," there is no way to assign A and B, as my second solution is incorrect. However, does the fact that my second solution is not an application or Bayes Rule prove that my logic is incorrect? I know that it is, but I still don't see why. Intuitively, it doesn't make sense to me that I have a higher probability of getting 3 or more kings if you tell me where in my hand the two kings are than if you don't. Jun 5, 2019 at 16:54
• Well, I didn't work out the numbers, but yes, if you tell me you have at least two kings then it is more likely that you have at least 3, then if you said nothing. Alternatively, if you announce you have four queens, that lowers the probability that you have at least three kings. If you say you have "exactly two kings", then the probability of three or more is zero. Not getting into the argument about Bayesian vs. other alternatives, if you cannot express the second method as "Bayes' rule" that should show where the inconsistency is. Have you tried to define the events?
– mjw
Jun 5, 2019 at 17:28

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.