What is the probability of dealing all $52$ cards in a standard well shuffled deck getting a single quad? I would like to know if you take a well shuffled deck of $52$ cards and then deal out all of them one at a time without replacement, what is the probability that they will be dealt such that there is exactly one quad dealt in order such as $4$ consecutive kings?  That run of $4$ has to be the only run of that length, there cannot be any others.  For example, 5, ... ,K,K,K,K,7,J,J,J,J is a "loser".  Other than computer simulation I am not sure how to solve this.

I wanted to post my algorithm for those that may want to attempt to simulate this in any general purpose language.  These are the major steps         .  Some minor details are left out for simplicity here...
Step $1$: declare and initialize an array A of size $52$ such that A[$1$] = $1$, A[$2$] = $2$... A[$52$] = $52$.
Step 2: declare an initialize and array F[$52$] of flags to false so we know which cards are already drawn/picked.  Change the appropriate flag to true when a card is picked (such as A[$22$] = true) if random number generator gives us $22$.
Step 3. Convert all $52$ cards to ranks only ($1$ to $13$)
Step 4: Count up runs of $4$.  I "cheat" when I do this by converting the cards to a string such as "K2J7777A5..." then I just scan for AAAA, 2222, 3333... KKKK.  If I detect exactly one set of quads it is a winning hand.
My first tweak is I only choose $51$ cards randomly and then just assign the one unchosen card as the last.  That sped things up by about $10$%.
My next tweak is to try a faster way to choose all $52$ cards (such as simulate a shuffle).  The language I am using has an array element delete function that will compact the remaining array so I can pick smaller and smaller random numbers that way but I don't yet know if it is faster.
