Deck of 52 cards, 3 card rank/order repeat pattern probability. A standard deck of 52 playing cards is well shuffled and drawn 1 card at a time without replacement.  What is the probability that the ranks of the 3 most recently drawn cards (starting at drawn card 6) have already been seen in the exact same order in the already drawn cards?  For example, when the 6th card is drawn, we need to check if cards 4, 5, and 6 are the same exact ranks as cards 1, 2, and 3 in the same exact order.  If not, then we draw card 7 and check if cards 5, 6, and 7 have been seen in that same order in cards 1, 2, 3 or 2, 3,and 4...  So basically we are checking the last 3 cards drawn against all previously drawn triples, including overlapping triples (such as cards 2,3,4 that overlap cards 1,2,3)
So for example, using A, 2, 3, 4, 5, 6, 7, 8, 9, T, J, Q, K as the 13 possible card ranks, the following partial deck would be a pattern found scenario:
5, Q, 7, 3, K, K, 8, Q, 7, 3 because the Q, 7, 3 triple was previously seen in that exact order.
$UPDATE$: For clarification, you need not scan into the last 3 cards drawn for a match.  When I said overlapping triples I did not mean into the last 3 cards drawn, but rather overlapping triples such as scanning cards 1,2,3  then 2,3,4, then 3,4,5... from the last 3 cards drawn (start scanning after card 6 is drawn and after every other drawn card after that such as 7, 8,... 51, 52).
So my question is what is the probability this will happen at least once in a well shuffled deck?  That is, at least one triple pattern of ranks will repeat exactly in order in the same deck as the cards are drawn one at a time, as described above.
 A: Any 3 card sequence can be:
No pair, one pair, 3 of a kind.
With probabilities $\frac{52\cdot48\cdot 44}{52\cdot 51\cdot 50}, \frac{52\cdot3\cdot 48\cdot 3}{52\cdot 51\cdot 50}, \frac{52\cdot3\cdot 2}{52\cdot 51\cdot 50}$ respectively.
Verify that these sum up to 1.
If the sequence is no pair.  There is a probability that a given 3 card sequence in the deck matches rank for rank is:
$\frac {3\cdot 3\cdot 3}{49\cdot 48\cdot 47}$
If a sequence has one pair, it is less likely.
$\frac {2\cdot 1\cdot 3}{49\cdot 48\cdot 47}$
If a sequence is 3 of a kind, it is impossible to have a second sequence that matches the first.
That two sequences of 3 cards match
$\frac {52\cdot 48\cdot 44\cdot 27+52\cdot 3\cdot 48\cdot 3\cdot 6}{52\cdot 51\cdot 50\cdot 49\cdot 48\cdot 47}$
How many non-overlapping pairs sequences of 3 cards are there in a 52 card deck? ${48\choose 2}$
How to explain this?  If we think of the first sequence of sequence $A$ and the second sequence as sequence $B.$
We can burn some amount of cards before the beginning sequence $A$, and some amount of cards after the end of sequence $B$ lets call these $a,b$ respectively.   The sum of the cards burnt must be less than  or equal to 46.
The number of pairs of non-negative integers such that $a + b \le 46$ is ${48\choose 2}$ 
(remember $a,b$ could equal $0.$)
$\frac {52\cdot 48\cdot 44\cdot 27+52\cdot 3\cdot 48\cdot 3\cdot 6}{52\cdot 51\cdot 50\cdot 49\cdot 48\cdot 47}\cdot {48\choose 2} \approx 23.8\%$
Update....
As has been pointed out by the original poster, I have the expected number of duplicated sequences...
Inclusion-Exclusion
Taking a shortcut....
$p = \frac {52\cdot 48\cdot 44\cdot 27+52\cdot 3\cdot 48\cdot 3\cdot 6}{52\cdot 51\cdot 50\cdot 49\cdot 48\cdot 47}$
$p{48\choose 2} - p^2{48\choose 2}{42\choose 2} + p^3{48\choose 2}{42\choose 2}{43\choose 2} \approx 20.34\%$ which is $0.2034$.
A: Since this problem seems very difficult to count all the patterns without overcounting, I just ran a Monte Carlo type simulation of 1 billion 1 ($10^9 + 1$) random decks, stopping at the first match in the deck and counting them up.  It took more than 24 hours to complete but I just let it run on my desktop computer while I used my laptop computer to do other things like chat...
The results I got are $20.32330$% of the decks have this property.
So I am not sure what the confidence level is from this (assuming the program and random generator are correct) but I would at least feel pretty confident about $20.323$%.
I know this solution is not mathematical, but it is still useful for checking against mathematical solutions, thus helping confirm them.
In the future what I would like to do is to store the randomly generated decks in a disk file so then later I could rerun it looking for different criteria such as a quad repeating pattern instead of a triple.  Since generating the shuffled decks is the bulk of the computer time, once I do that, I can just "quickly" read it back in (a reasonable "chunk" at a time) and check for the desired patterns.
