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Say I want the probability that a five card poker hand contains exactly two kings. This would be

$$\frac{{4\choose 2}{48 \choose 3}}{52\choose 5}$$

Now if I drop the $48 \choose 3$, which represents the 3 non king cards, what can the probability $\frac{4\choose 2}{52\choose 5}$ be taken to represent?

Is it the number of hands containing at least 2 kings?

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3 Answers 3

up vote 3 down vote accepted

No, it is not true.

It may represent the probability that your hand contains 2 kings and 3 prefixed cards (not kings). For instance, the probability that your hand contains any 2 kings and the ace of spaces, the ace of diamonds, and the ace of clubs: this probability equals

$$\frac{\binom{4}{2}\binom{1}{1}\binom{1}{1}\binom{1}{1}}{\binom{52}{5}}=\frac{\binom{4}{2}}{\binom{52}{5}}.$$

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It would be the probability of a hand containing exactly two kings and three specified non-kings, e.g. the ace, 2 and 3 of spades.

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Cool, that is good to know thanks –  sonicboom Oct 28 '12 at 20:09

In this context the fraction $$\frac{\binom42}{\binom{52}5}$$ has no very natural interpretation. The original fraction $$\frac{\binom42\binom{48}3}{\binom{52}5}$$ is another story: the numerator is the number of $5$-card hands containing exactly two kings, and the denominator is the number of $5$-card hands, so the fraction is the probability of being dealt a hand containing exactly two kings.

Just as there are $\binom42\binom{48}3$ hands with exactly two kings, there are $\binom43\binom{48}2$ hands with exactly three kings and $\binom44\binom{48}1$ hands with exactly four kings. Thus, the number of hands with at least two kings is $$\binom42\binom{48}3+\binom43\binom{48}2+\binom44\binom{48}1\;,$$ and the probability of being dealt such a hand is $$\frac{\binom42\binom{48}3+\binom43\binom{48}2+\binom44\binom{48}1}{\binom{52}5}\;.\tag{1}$$

Note: The number of hands with no kings is $\binom40\binom{48}5$, and the number with exactly one king is $\binom41\binom{48}4$, so the number with at most one king is $$\binom40\binom{48}5+\binom41\binom{48}4\;.$$ Thus, the number of hands with at least two kings is $$\binom{52}5-\left(\binom40\binom{48}5+\binom41\binom{48}4\right)\;,$$ the total number of possible hands minus the number having fewer than two kings. Thus, we could also have computed the probability in $(1)$ as

$$\frac{\binom{52}5-\left(\binom40\binom{48}5+\binom41\binom{48}4\right)}{\binom{52}5}=1-\frac{\binom40\binom{48}5+\binom41\binom{48}4}{\binom{52}5}\;.$$ This has a perfectly good intuitive significance: it’s $1$ minus the probability of getting a hand with fewer than two kings.

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Very nice answer thanks –  sonicboom Oct 28 '12 at 20:10

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