# Probability of # of suits dealt given that your hand must have one ace and one king.

Suppose you are dealt a $5$-card hand from a standard deck, and that your hand has $1$ ace and $1$ queen. What is the probability that your hand has only one suit? two suits? three suits? four suits?

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The discrepancy between the title and the body of the question creates ambiguity. The title seems to imply that the king's suit is included in the suit count, whereas the body seems to imply that it isn't. Also, the title appears to ask for a conditional probability, whereas the body appears to ask for the probability of the intersection of events, without conditioning. Please clarify. – joriki Nov 1 '12 at 22:51
Write clearly and in complete sentences. Phrase your question as a question. – ncmathsadist Nov 1 '12 at 23:04
I wondering about the conditional probability. Clarification: what is the probability that the number of suits in the hand is 1 given the hand has a king. What is the prob that the number of suits in the hand is 2 given the hand has a king. What is the prob that the number of suits in the hand is 3 given the hand has a king. What is the prob that the number of suits in the hand is 4 given the hand has a king. – Elliot Nov 1 '12 at 23:07
Intuitively it seems like it shouldn't matter at all... you've got some ranks in your hand, who cares if a king is one of them? Or do you mean that your hand contains exactly one king (not two or more)? – mjqxxxx Nov 1 '12 at 23:54
@Elliot: That kind of clarification should be incorporated into the question so that people don't have to delve into the comments to understand the question unambiguously. – joriki Nov 2 '12 at 3:58

Revised and extended:

There are $\binom{48}4$ hands containing a given king and no other. The probability of having a $k$-suited hand, given that you have exactly one king, is the same as the probability of having a $k$-suited hand, given that you have a particular king and no other, say the heart king.

Suppose that you have the heart king and no other.

• There are then $\binom{12}4$ ways to choose the rest of the hand so that all $5$ cards are hearts. Thus, the probability that you have a one-suited hand, given that you have the heart king, is $$\frac{\binom{12}4}{\binom{48}4}=\frac{11}{4324}\approx0.002544\;.$$

• One of the suits in a two-suited hand is hearts. There are $3$ choices for the second suit. There are $24$ non-kings left in these two suits, so there are $\binom{24}4$ ways to choose $4$ cards from them, but $\binom{12}4$ of them result in a one-suited hand. Thus, there are $3\left(\binom{24}4-\binom{12}4\right)$ ways to choose the other $4$ cards to get a two-suited hand. The probability of having a two-suited hand, given that you have exactly the heart king, is therefore

$$\frac{3\left(\binom{24}4-\binom{12}4\right)}{\binom{48}4}=\frac{30,393}{194,580}=\frac{3377}{21,620}\approx0.156198\;.$$

• In a three-suited hand the suits may be distributed $2,2,1$ or $3,1,1$. Suppose first that they are $2,2,1$. If there are $2$ hearts, one of them is the king, and there are $12$ ways to choose the other. There are $3$ ways to choose the other $2$-card suit, $\binom{12}2$ ways to choose $2$ cards from it, $2$ ways to choose the third suit, and $12$ ways to choose a card from it, for a total of $$12\cdot3\cdot\binom{12}2\cdot2\cdot12=57,024$$ hands. If there is only one heart, it’s the king; there are $\binom32=3$ ways to choose the other two suits, and $\binom{12}2^2$ ways to pick $2$ cards from each, for a total of $$3\binom{12}2^2=13,068$$ hands. Now suppose that the distribution is $3,1,1$. If there are $3$ hearts, one is the king, the other $2$ can be chosen in $\binom{12}2$ ways, the remaining $2$ suits can be chosen in $\binom32=3$ ways, and there are $12^2$ ways to choose one card from each, for a total of $$\binom{12}2\cdot3\cdot12^2=28,512$$ hands. Otherwise there are $3$ ways to choose the $3$-card suit, $\binom{12}3$ ways to choose $3$ cards from it, $2$ ways to choose the third suit, and $12$ ways to choose one card from it, for a total of $$3\cdot\binom{12}3\cdot2\cdot12=15,840$$ hands. That makes a grand total of $114,444$ three-suited hands and a probability, given that you have the heart king, of $$\frac{114,444}{194,580}=\frac{12,716}{21,620}\approx0.588159$$ of having a three-suited hand.

• In a four-suited hand there are $2$ cards of one suit and one of each of the other three suits. There are $12^4$ ways to choose one non-king from each suit to go with the king to make a four-suited hand with two hearts. For each of the $3$ other suits there are $12^2\binom{12}2$ ways to choose two cards from that suit and one from each of the other non-heart suits. Thus, there are $12^4+3\cdot12^2\binom{12}2=49248$ four-suited hands in which the heart king is the only king, and the corresponding probability is $$\frac{49,248}{194,580}=\frac{5,472}{21,620}\approx0.253099\;.$$

As a quick check,

$$\frac{11}{4324}+\frac{3377}{21,620}+\frac{12,716}{21,620}+\frac{5472}{21,620}=\frac{21,620}{21,620}=1\;.$$

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I'm not quite seeing how to get my remaining probabilities. I tried a couple different methods for x=3 and x=4 but I ended up over counting both times. – Elliot Nov 2 '12 at 2:39
I would have upvoted the original answer, but I don't understand what the reasoning is behind answering all four questions completely. Now Elliot has no way left to check whether he's understood your answer. – Noah Snyder Nov 2 '12 at 9:02
@Noah: Read his comment: the original answer clearly wasn’t enough for him. Given the amount of detail that it contained, I think it likely that my choice is between finishing the job in hopes of giving him something solid to use as a basis for future problems, and a long exchange of incremental extensions and quite possibly growing frustration for both of us. If he and I were sitting down together, I’d go for at least a few more iterations, but this medium makes that harder, especially for this kind of problem, with its typically somewhat lengthy explanations. In this case I judged that ... – Brian M. Scott Nov 2 '12 at 9:11
... giving him more to work with for the future was the better course. And it’s certainly not true that he has no way left to check whether he’s understood the answer: it’s not as if this is the only counting problem he’ll encounter that uses these ideas! Moreover, if in fact he has not, he may well recognize this and ask for further clarification $-$ which I will be happy to try to give. – Brian M. Scott Nov 2 '12 at 9:14
Certainly you're allowed to use your own judgement, but let me propose two other alternatives that might be worth considering in the future. You could give explain a second example that wasn't taken from his homework, or you could suggest that this is not the right medium for the sort of help he needs and suggest that he talk to someone in person if possible. (Of course if he says why that's not possible, then you can take another approach.) – Noah Snyder Nov 2 '12 at 9:19