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.

Joriki provides the elegant solution ... but also shows the value of checking theoretical work via simulation. The discussion got me playing with simulating it in Mathematica, and thought I would post some code.
A pack of cards can be represented by the set:
cards = Table[Range[1, 13], 4] // Flatten 

Sampling without replacement, and dealing out a hand of 52 cards   i.e. RandomSample[cards] ... and then counting all the cases where there are four identical cards in a row {x_,x_,x_,x_}, and doing it 10 million times:
4Row = ParallelTable[Count[Split[RandomSample[cards], SameQ], {x_,x_,x_,x_}], {10^7}];

All done. The empirical distribution is given by:
Tally[4Row]


{{0, 9976557}, {1, 23415}, {2, 28}}

In other words, out of 10 million drawings, there was:


*

*exactly 1 sequence of 4 identical cards 23,415 times (out of 10 million)

*exactly 2 sequences of 4 identical cards just 28 times (out of 10 million)
The above Monte Carlo simulation (10 million times) takes about 42 seconds on my Mac. I am sure there are faster ways of doing it, but that's another question :)
