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What is the probability of getting 4 Kings if I draw 9 cards from a pack of 52 without replacement?

Now the way I see it, I do $\tfrac 4{52}\tfrac 3{51}\tfrac 2{50}\tfrac 1{49}{}^9C_4$.

But here is my question, shouldnt we also multiply it with $4!$ since there are $4!$ ways in which the kings can be picked King $\heartsuit$, King $\spadesuit$, King $\diamondsuit$, King $\clubsuit$, but this order can be any other order so shouldn't that be figured into the calculation?

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Firstly your 9 card hand has 4 kings. There is only 1 way that 4 kings can be picked from 4 kings$$\dbinom{4}{4}=1$$ This leaves 5 cards to be picked from the remaining 48 $$\dbinom{48}{5}=1712304$$

Total possible 9 card hands $$\dbinom{52}{9}=3679075400 $$

The number of 9 card hands that contain 4 kings divided by the number of all possible 9 card hands gives the probability of a 9 card hand containing 4 kings $$\frac{1712304}{3679075400}\approx. 0.000465416936 $$ This equates to more than a 2000 to 1 chance of this happening

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Hint: how many possible, distinct $9$ card hands are there in total? How many distinct hands are there with four kings? Divide one by the other, and you have your probability.

This kind of problem / solution is called hypergeometric probability, by the way. In case you want to google for more resources.

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You seem to be calculating the probability for drawing $4$ from $4$ kings when drawing $4$ from $52$ cards, multiplied by the ways to select which $4$ from the $9$ places in the hand to put the kings, and ... ignoring what the other cards might be.

That's not really sensible.


You simply want the probability for drawing $4$ from $4$ kings, and $5$ from $48$ non-kings, when drawing any $9$ from all $52$ cards.

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