Odds for Bad Beat Jackpot in Texas Hold em

My club is now offering a 'super beat jackpot' on top of our usual bad beat jackpot. The super beat jackpot requires that 5 players be dealt in to a hand, both players involved in the bad beat hand use both pocket cards and the losing hand must be a queen high straight flush.
What are the odds of having a queen high straight flush vs a royal flush with a minimum of 5 players (not sure if 5 players would come into calculations here) with both players using both pocket cards? What would the calculation (formula) look like?
Jody Stacey

I dont know the odds but I can tell you the only ways this is possible is the winning hand be dealt A/K suited and the loser dealt 8/9 of the same suit with the T/J/Q of that same suit on the board. Pretty rare indeed.

My club is offering a $1M jackpot for this bad beat on our 1 anniversary weekend and want to know what the odds were

Here's what I get for the original Super Beat:

There are 52 x 51 x 50 x 49 x 48 / 120 = 2,598,960 sets of community cards.
Of these, there are 45 x 44 / 2 = 990 that contain QJT of hearts but none of AK98 of hearts. Similarly, there are 990 in diamonds, 990 in spades, and 990 in clubs, or a total of 3960 sets of community hands that have QJT suited but none of AK98 of that suit.

Since each player has an equal chance of "winning," I will calculate the chance of "somebody" winning and then divide by five to get the chance of "you" winning.

Separate the hole cards into each player's "first" and "second" hole cards.
With 5 players, there are 47 x 46 x 45 x 44 x 43 / 120 = 1,533,939 sets of five cards.
One possible "winning" set is A8 in the suit and three other cards that are neither K nor 9 in that suit; there are 43 x 42 x 41 / 6 = 12,341 of those. Similarly, there are 12,341 sets with A9, 12,341 with K8, and 12,341 with K9, for a total of 49,364 sets of "first cards" that can still produce the jackpot.
At this point, there are 42 cards left in the deck; whichever of the A or K that one player has, he has to get the other, so there is 1/42 chance of that, and of the remaining 41 cards, the player with the 8 or 9 has to get the other one, so there is 1/41 chance of that.

The overall probability with five players is
990 / 2598960 x 49364 / 1533939 x 1/42 x 1/41 = about 1 in 140 million.
The probability of a particular player winning is 1/5 of that, or about 1 in 700 million.

thank you for the break down! i was hoping for this and not just a ratio

The royal flush beating 4 aces in the WSOP Main Event a few years ago was the highest bad beat type hand I have ever seen.

In your formula above, can you explain why you are dividing by 120 for the community cards

I want to try and answer this one...5!! (because 5 cards are being used as community cards)? Am I right?

Correct - in this case, we are interested only in how many different "sets" of the five community cards exist, regardless of their order. It doesn't matter what cards are in the flop, which one is the turn, and which one is the river.