# Combinatorics formula seemingly overcounting

Recently I've been searching the formula to compute the probability of obtaining certain cards combinations. My researches came across a formula, and I tried to extend it to a more general use. However, my extended version does not seem to be correct, as it likely overcounts. My first formula basically calculates the probability of obtaining l times different k of a kind amongst n drawn cards. Formula is the following : $$\frac{\begin{pmatrix}13\\l\end{pmatrix}\cdot{\begin{pmatrix}4\\k\end{pmatrix}}^l\cdot\begin{pmatrix}13-l\\n-kl\end{pmatrix}\cdot4^{n-kl}}{\begin{pmatrix}52\\n\end{pmatrix}}$$ Seemingly this is correct, probabilities appear realistic to me and I've never came across probabilities superior to one. To help you understanding this formula, here is an example : we would like to obtain 3 different two of a kind amongst 8 drawn cards, which means $l=3$, $k=2$, and $n=8$. We get : $$\frac{\begin{pmatrix}13\\3\end{pmatrix}\cdot{\begin{pmatrix}4\\2\end{pmatrix}}^3\cdot\begin{pmatrix}10\\2\end{pmatrix}\cdot4^2}{\begin{pmatrix}52\\8\end{pmatrix}}$$ So here's how it works : $\begin{pmatrix}13\\3\end{pmatrix}$ is the number of possible triple two of a kind ranks combinations. ${\begin{pmatrix}4\\2\end{pmatrix}}^3$ is the number of ways the the three two of a kind can be drawn (depending on if the cards are clubs, spades, diamonds, or hearts). $\begin{pmatrix}10\\2\end{pmatrix}\cdot4^2$ is the possible combinations for the 2 remaining cards, and finally $\begin{pmatrix}52\\8\end{pmatrix}$ is the number of possible combination of 8 cards (which you maybe guessed). For this very example, the probability is about $0.0591$.

The reason why I wanted to extend this formula is basically the formula I explained above only computes the probabiliy of getting exactlly l k of a kind, but no other special combination. So I tried to create a formula which could basically give the probabilities of getting at least the specified cards combinations, and any other random cards (which can contain some more combinations). Here was my idea : $$\frac{\begin{pmatrix}13\\l\end{pmatrix}\cdot{\begin{pmatrix}4\\k\end{pmatrix}}^l\cdot\begin{pmatrix}52-kl\\n-kl\end{pmatrix}}{\begin{pmatrix}52\\n\end{pmatrix}}$$ While $\begin{pmatrix}52-kl\\n-kl\end{pmatrix}$ was intended to compute the number of different possible combinations for the remaining $n-kl$ specified cards, it appears that this formula leads to probabilities superior to one. I believe it counts two combinations of different orders as two different combinations, but I can't figure out where does it. So after so much text here's my real question : where is this wrong ?

• Your formula for exactly three different two of a kind is correct. Your extension to at least three different two of a kind overcounts. Mar 28, 2016 at 20:09
• As expected yes ... Should I try using a different structure instead of the single 52-kl C n-kl ? Mar 28, 2016 at 20:12
• I am not sure about what probability you are looking for. It it is three different two of a kind or $4$ different two of a kind, calculate the probability of each and add. Mar 28, 2016 at 20:16
• I'm not so sure about your first formula, what if $n = 52$? Mar 28, 2016 at 20:19
• Answer to André : I am trying to find the probability of the following event : we choose l and k to determine how many two, three, or four of a kind do we want at least, and n the number of cards we draw, which means out of the wanted two, three or four of a kind we can have anything (including some more two, three or four of kind since we want at least l k of a kind. Hopefully I explained it well. Answer to Adam : according to all I've been calculating with the formula using www.desmos.com, it looks like have 13 four of a kind (basically the whole) deck is certain for $n=52$. Mar 28, 2016 at 20:24