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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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    I've edited your title to match the body, and rephrased the body. I you intended the body to match the title, please change the title back, and change "$1$ ace and $1$ queen" in the body to "$1$ king".2012-11-02

1 Answers 1

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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.2012-11-02