Standard deck of cards, full straight flush probability question Assume a standard $52$ card deck (no jokers) well shuffled.  I was wondering how many cards would need to be randomly drawn from it (without replacement) on average to get about a $50$% chance of getting all $13$ cards in exactly one suit.  For example, getting all $13$ of the diamond cards.  How do you solve for this type of problem?
For clarification, a "winning" draw would be when one of the four suits is completely filled (all $13$ ranks of that suit are in the drawn cards).  Any of the $4$ suits can be filled then the # of cards in the draw would be # of interest.  There will be no cases where more than $1$ suit is completely filled in the hand.  So for example, if the first $13$ cards drawn are all hearts, then hand stops at $13$ cards.  If $12$ cards of each suit are drawn ($48$ cards total), then the hand would stop at the $49$th card draw (thus completing one of the suits in full).
 A: I can't think of a straight up combinatorial argument without splitting it into too many cases, so I'll use the inclusion-exclusion principle.
Number the suits $1,2,3,4$ (this is for commodity of notation). Let: 
$$A_k=\text{the draw contains 13 cards of suits k}$$ 
Then the desired event is $A_1 \cup A_2 \cup A_3 \cup A_4$
Let $n$ be the size of your draw. Then there are ${52 \choose n}$ possible draws. 
Then $P(A_1)={{52-13 \choose n-13} \over {52 \choose n}}$, since you need to choose $n-13$ cards which are not of suit 1.
Similarly: $P(A_1 \cap A_2)={{52-26 \choose n-26} \over {52 \choose n}}$
$P(A_1 \cap A_2 \cap A_3)={{52-39 \choose n-39} \over {52 \choose n}}$
The others are similar. Compute them to obtain an expression which depends on $n$ and try out different values. 
Obviously some of the probabilities might be $0$ depending on $n$.
A: I ran a computer simulation of $1$ million decisions and got about $45.33$ cards on average are needed to get a winning hand (exactly $1$ of the $4$ suits all filled).  It might be fun to play a game with a naive person betting them even money (dollar for dollar) that they cannot get all $13$ cards of one suit drawing from $3/4$ of the deck ($39$ cards). The answer is somewhat surprising because worst case is $49$ cards.  That is, if you draw $49$ cards you are guaranteed to fill at least one suit completely.  $45.33$ is not much less than $49$ so the average case is close to the worst case and "far away" from the best case of $13$.
