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Given a traditional deck of $52$ cards, find the probability that a $5$-card hand contains four cards of the same rank.

So far, we only dealt with the notion of a finite sample space and that each elementary event is equally probable. So I defined the sample space

$$ \Omega := \{\omega : \omega = (a_1,a_2,a_3,a_4,a_5), a_j \neq a_k, j\neq k, a_i \in \{1,\dots,52\}\}$$ since the sampling is without replacement. Now I am trying to figure out $$|\{\text{exactly four cards of the same rank}\}|$$ since then we have $$P(\{\text{exactly four cards of the same rank}\}) = \frac{|\{\text{exactly four cards of the same rank}\}|}{|\Omega|}$$ Does anyone has a nice way of calculating the cardinality of the above set? I tried to split it up into several disjoint subsets, but I do think this gets quite complicated.

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    $\begingroup$ Well, there are $\binom {52}5$ hands, and there are $13\times 48$ possible five card hands containing four-of-a-kind so... $\endgroup$ – lulu Feb 24 '17 at 23:02
  • $\begingroup$ There actually are two ways to count the sample space. You can distinguish between two hands with the same five cards arranged in a different sequence ("counting permutations"), or you can consider those hands the same ("counting combinations"). To consider them the same, you can make the condition $a_1<a_2<a_3<a_4<a_5.$ I think combinations are easier to work with for a problem like this. (The previous comment counts combinations.) $\endgroup$ – David K Feb 24 '17 at 23:24
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There will be 13 different cases, one for every existing rank.

$${13 \choose 1} = 13$$

For every case you have that one of the cards can be whatever other card with a diferent rank, $52-4 = 48$, and the rest need to be one of the 4 cards of the same rank no matter the order in which they are picked. Therefore:

$${48 \choose 1} {4 \choose 4} = 48 $$

And in total that leads to $13\cdot48 = 624$ possible cases.

To calculate the probability you divide the favorable cases with the total possible states, i.e. all the possible 5 cards combinations taken randomly out of the maze:

$${52 \choose 5} = \dfrac{52\cdot 51\cdot 50 \cdot 49 \cdot 48}{5!} = 2,598,960$$ $$\dfrac{624}{2,598,960} \approx 0.00024 $$

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It's easy to see that there are 2 ranks (1 with 4 cards and 1 with 1). There are $2!\frac{13!}{11!2!}=(13)(12)$ possibilities to choose the ranks. Then there are $4$ possibilities to choose the lone card's type (i.e. spade, heart, ...). So there are $(13)(12)(4)=(13)(48)$ possibilities. Multiply this by $5!=120$ to arrange (so far we have only chosen). We obtain $(13)(48)(120)=(52)(48)(10)(3)$. Divide this by the sample space $(52)(51)(50)(49)(48)$. Now we obtain $$\frac{1}{(17)(5)(49)}=\frac1{4165}\text.$$

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  • $\begingroup$ $156\times 4 = 624$. $\endgroup$ – David K Feb 24 '17 at 23:22

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