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Question

What is the probability that a five-card poker hand contains "four of a kind hand"?

For More information about poker's four of a kind hand,please refer here

My Attempt

$\text {Rank of card-: Jack,King,Queen,Ace}$$,9,8,...1$

In four of a kind hand , $4$ cards are of the same rank( and obviously of diiferent type)

$\Rightarrow$ Now in a set of $52$ card,we have $13$ ranks of $4$ kind(spade,heart,diamond,club)

$\Rightarrow$ four of a kind hand says $4$ cards are of the same rank.

So Among $13$ ranks ,select $1$ of the rank i.e $\binom{13}{1}$,

and the remaining last card can be any thing

I am not getting how to move forward ,please help me out.

Thanks in advance

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Yes, the remaining card can be anything. But you still need to count it. In how many ways can that happen?

You could approach it in the same manner. First count the number of ways to choose the rank. Then count the number of ways to choose one of the suits. Alternatively, ignore all that and just choose one of the remaining cards. This can be done in $\binom{48}{1} = \binom{12}{1}\binom{4}{1} = 48$ ways. Then remember to divide by $\binom{52}{5}$.

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$$\frac{13\times(52-4)}{\left( \begin{array}{cc}{52}\\{5}\end{array}\right)}\simeq 0.024\%$$

First, we have 13 possibilities of "four of a kind". Once that's picked, we have 48 cards remaining, and any one may be picked. So there're $13\times 48$ possible combinations. Dividing this by the total possible hands, i.e. $\left(\begin{array}{cc}52 \\ 5\end{array}\right)$, gives the answer.

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