Probability problem: what is the probability one card from each suit will be represented when 6 cards are dealt?

Question

What is the probability one card from each suit will be represented when 6 cards are dealt?

Answer 1

Let $N_s , N_h, N_d $ and $N_c$ be the events that we get no Spades, Hearts, Diamonds or Clubs respectively when dealing 6 cards. Thus the event $X := N_s \cup N_h\cup N_d \cup N_c$ Is such that at least $1$ suit is unrepresented. By the inclusion exclusion principle we have that $\mathbb{P}(X) = \mathbb{P}(N_s) + \mathbb{P}(N_h) + \mathbb{P}(N_d)+ \mathbb{P}(N_c) - \mathbb{P}(N_s \cap N_h) - \mathbb{P}(N_s \cap N_d) - \mathbb{P}(N_s \cap N_c) - \mathbb{P}(N_h \cap N_d) - \mathbb{P}(N_h \cap N_c) - \mathbb{P}(N_d \cap N_c) + \mathbb{P}(N_s \cap N_h \cap N_d) + \mathbb{P}(N_s \cap N_h \cap N_c) + \mathbb{P}(N_s \cap N_d \cap N_c) + \mathbb{P}(N_h \cap N_d \cap N_c) - \mathbb{P}(N_s \cap N_h \cap N_d \cap N_c)$

This might look horribly complicated but its actually very simple. Here is what it is doing: Add all the $\textbf{singletons}$, Subtract all the $\textbf{doubletons}$, Add all the $\textbf{triplets}$ and Subtract all the $\textbf{quadruples}$. And from symmetry we know that all the singletons are equal so are all the doubletons so are all the triplets and so is the quadruple. Let us solve for them all now:

$\textbf{Singletons:}$
The singletons, $\mathbb{P}(N_i)$, are simply the probability there are none of a particular suit. Which occur simply with probability: $\frac{39}{52} \cdot \frac{38}{51} \cdot \frac{37}{50} \cdot \frac{36}{49} \cdot \frac{35}{48} \cdot \frac{34}{47} = 0.160258358663. $ This is the same as 39 chose 6 over 52 chose 6

$\textbf{Doubletons:}$
The doubletons are the event that there are neither of 2 suits appearing in 6 cards. $\mathbb{P}(N_i \cap N_j) = $ having 26 cards available to make up 6 over the total number of combinations = $\frac{C^{26}_6}{C^{52}_6}$

$\textbf{Triplets:}$
The event $\mathbb{P}(N_i \cap N_j \cap N_k)$ is the event that none of 3 suits appear. i.e only 1 suit and occurs with probability $\frac{C^{13}_6}{C^{52}_6}$

$\textbf{Quadruple:}$
$\mathbb{P}(N_s \cap N_h \cap N_d \cap N_c)$ is the only quadruple which is clearly 0 as it is stating no spades, no hearts, no clubs and no diamonds.

Hence $\mathbb{P}(X) = 4\cdot \frac{C^{39}_6}{C^{52}_6} -6\cdot\frac{C^{26}_6}{C^{52}_6} +4\cdot\frac{C^{13}_6}{C^{52}_6}= 0.573517917805 $ This is the value that at least 1 suit is
$\textbf{Un}$represented and hence the answer to your question is $1-0.573517917805 = 0.426482082195 \approx 0.426 $