Bingolucky

Was at a casino the other day and guy next to me looked at his first card, saw Ace of Diamonds, played in the blind without looking at other 2. Dealer flips over K clubs K hearts K spades. Dealer gets to the guy flips over Ace diamonds King Diamonds Queen of diamonds. What are the odds of this happening? (He got paid for 4 of a kind on 6 card and straight flush was an awesome hand to see!)

'Dice, verily, are armed with goads and driving-hooks, deceiving and tormenting, causing grievous woe.' -Rig Veda 10.34.4

This is a very interesting question.
Whenever dealing with card permutations, there are always 2 ways of figuring a total, one is with using sequences, the other is with out using sequences. Example : If you deal 3 cards, there are 22,100 possible 3 card combinations, however, each result has 6 separate sequences. If you deal Jc 9s 4d, that can also be dealt 9s 4d Jc, or 4d Jc 9s, etc..
In the case of this problem involving 6 cards, we MUST use sequences because if we just calculate 6-card combinations (20,358,520), some of those won't qualify even though they contain all of the cards to do so. Kc Ks Js - Ks Qs Kh may produce the 6 cards needed to have quads and a straight flush, but since they are not dealt in the proper sequence, they would produce neither in each hand.
So we must include sequences when figuring how many 6 card permutations there are, which is 14,658,134,400 and then we must calculate how many of those would produce one 3 card hand to have trips, and the other 3 card hand to have a straight flush THAT ALSO uses the 4rth card of the other hands trips. I figure that 10,368 of these six card sequences would qualify.
Therefore, we have 14,658,134,400 sequences of 6 cards that a 52 card deck produces, and 10,368 of those will produce the two three card hands we need. So, my answer would be odds of 1,413,785 - 1.
Since I am no genius like some of the people that frequent here are, I'd love to be either confirmed or corrected by someone truly qualified.