What is the probability of getting $3$ triples before a single quad in a $52$ card deck? Suppose I have a well shuffled standard deck of $52$ cards.  I draw them one at a time (without replacement) and sort the chosen cards by rank so that it is easy for me to see pairs, triples, quads....  I want to know what is the probability of seeing exactly three triples (such as $5,5,5$, $3,3,3$, $K,K,K$) before seeing a quad (such as $10,10,10,10$).  Note that the triples and quad need not be in any order, they can have other "irrelevant" cards drawn in between them.
Just to clarify, the triple and quads do not need to be consecutive cards.  For example, if the first $6$ cards drawn are $3,5,6,5,K,5$, then that is a triple for the $5$s.
Note that when we either see $3$ triples or one quad, we are done with that hand.  No other cards will be drawn.  So for example, if you have $3,3,3,5,5,7,7,7$ so far (and all the other drawn cards are singles, then another $3$ or $5$ draw will immediately stop the hand but for different reasons.  We will either have quad $3$s at that point or $3$ triples.  A drawn $7$ would also stop this sample hand.
A quad winner would be something like $K,K,K,7,7,7,3,3,K$ with any "irrelevant" single or double (paired) cards interspersed such as $2,J,9...$.
Just for "extra" clarity, a quad does NOT also count as a triple.  A triple becomes a quad and loses its triple status if the $4$th card of that same rank is drawn in the same hand.
For even more clarity (I didn't know this question would be ambiguous), I am $NOT$ asking for $3$ triples followed by a quad before seeing the $4$th triple.  I am saying all you have to see is $3$ triples without seeing a quad and you can stop that hand and it is counted as a winner.  Sorry for any confusion. I thought it was clearly stated but now that I reread the title it is a little confusing but not if you read the examples it shouldn't be.
 A: Note: This is an answer to the original question, which seemed to be asking for the probability of getting at least three triples before a quad, not exactly three triples.

The probability of completing a third triple without having completed a quad on draw $k$ is
$$
\frac{\sum_{m=0}^{\min(10,\left\lfloor\frac{k-9}2\right\rfloor)}\binom{12}2\binom43^2\binom32\binom{10}m\binom{10-m}{k-9-2m}\binom42^m\binom41^{k-9-2m}}{\binom{51}{k-1}}\;,
$$
since, given the card drawn, we can select $k-1$ from the remaining $51$ by selecting $2$ of $12$ ranks, and in each $3$ of $4$ suits, for the triples already completed, $2$ of $3$ suits for the double being completed to a triple, $m$ of $10$ ranks for which doubles have been drawn and $k-9-2m$ of the remaining $10-m$ ranks for which singles have been drawn, along with the corresponding choices of suits. Summing this for the possible values $8\le k\le29$ yields
$$
\frac{2556075271418192}{3583273791308225}\approx71\%\;.
$$
Here's code that computes the sum and checks it with a simulation.
