Probability of drawing a 13 card straight from a deck Suppose you have a deck of standard 52 card deck.  You shuffle the deck and draw (13 + k) cards where k is a non-negative intger.  
What is the probability that you have a 13 card straight (Ace, 2, 3, ... Queen, King) amongst the cards drawn?  Note that cards in a straight can be of different suits.  
Thanks,
 A: This is an inclusion-exclusion question. 
Let $A_j$ be the set of such draws that do not contain rank $j$. Let $A$ be all such draws. Then you want: $|A\setminus(A_1\cup\cdots\cup A_{13})|/|A|$. 
Now: $|A|=\binom{52}{13+k}$.
And we have:
$$\begin{align}|A\setminus(A_1\cup \cdots\cup A_{13})|=|A|-&(|A_1|+|A_2|+\cdots + |A_13|)\\&+(|A_1\cap A_2|+|A_1\cap A_3|+\cdots+|A_{12}\cap A_{13}|)\\
&-\cdots
\end{align}$$
Now the number of elements in $A_{i_1}\cap A_{i_2}\cap A_{i_m}$, with the $i_j$ distinct, is always equal to $\binom{52-4m}{13+k}$. So we get:
$$|A\setminus(A_1\cup \cdots\cup A_{13})|=\sum_{m=0}^{13}(-1)^m\binom{13}{m}\binom{52-4m}{13+k}$$
That's not going to be any fun to calculate.
A: Using generating functions we get
$$f(z) = (-1+(1+z)^4)^{13}$$
where the thirteen factors  correspond to the thirteen different ranks
and we subtract one because there  must be at least one card from each
rank. We then obtain
$$[z^{13+k}] (-1+(1+z)^4)^{13}
\\ = [z^{13+k}] 
\sum_{q=0}^{13} {13\choose q} (-1)^{13-q} (1+z)^{4q}
\\ = \sum_{q=0}^{13} {13\choose q} (-1)^{13-q} {4q\choose 13+k}.$$
This may be written as (reverse index)
$$\sum_{q=0}^{13} {13\choose q} (-1)^{q} {52-4q\choose 13+k}.$$
Observe that
$$(1+z)^4 = {4\choose 0} + {4\choose 1} z +
{4\choose 2} z^2 + {4\choose 3} z^3 + {4\choose 4} z^4$$
and hence this term represents the possibilities of choosing $p$ cards
from the four possible suits of each rank.
