Take into account that in order for that hand to be the best hand, the dealer must also flip over a card that matches the suit that your best hand is for.
Edit: the original question had two potential methods, one of which was correct and one of which was not. For posterity, here are the two methods, paraphrased from the original question.
Method one (correct):
5 cards are chosen without replacement and without the order mattering from a deck of 24 (9, 10, J, Q, K, A in four suits). Then a sixth card is chosen. The 5 cards must be JQKA in one suit and J in the other suit of the same color. The sixth card must be 9 or 10 of the JQKA suit.
There are 42,504 possible hands of 5 cards. There are four possible "best hands" corresponding to the four suits. Once one has fixed a best hand, there is a 2/19 chance of the sixth card being one of the two chosen ones.
Therefore the probability is (4/42504) * (2/19), approximately 9.90619 x 10^-6.
Method 2 (incorrect):
Suppose you want the first two cards to be jacks and then the remaining cards to be QKA of a corresponding suit. Perform each draw sequentially.
You have a 4/24 chance of getting a jack with your first draw. Then you have a 1/23 chance of getting the complementary jack on the second draw. Then you have a 6/22 chance of getting one of the possible remaining cards (queen, king, or ace of either jack suit). Once this happens, the suit is determined and there are only two possible remaining cards, so there is a 2/21 chance of getting one of the possible remaining cards at this stage. Then to get the last necessary card is a 1/20 chance. Finally, there is 2/19 chance that the sixth draw is as required.
4/24 x 1/23 x 6/22 x 2/21 x 1/20 x 2/19 = 9.90619 x 10^-7
The first method is correct.
The problem with the second method is that it assumes that the first two cards in the deal are jacks and the third, fourth, and fifth cards are the non-jacks. One could fix this by multiplying the probability by 5C2 (you are choosing which two of the 5 dealt cards are the jacks).
There is no problem with ordering within the two jacks because there is no assumption about which jack came first. There is no problem with ordering within the other three cards because there are no assumptions about which one was which.
Multiplying by 5C2 = 10, one obtains exactly the answer as via Method 1.
As pointed out in a comment, the rules (and thus the probability) are different for the dealer. The dealer must add the sixth card to her or his hand and then discard one of her or his six cards.
The sixth (face-up) card must then be a jack, queen, king, or ace of one fixed suit and the dealer's original hand must be the other three of these cards, the complementary jack, and an arbitrary other card. The chances of the sixth card being a jack, queen, king, or ace are 16/24. Then there are precisely 19 cards left to be the arbitrary card in the dealer's original hand (and the other four cards are fixed). Then there are 19*16 deals that result in the dealer having the best possible hand. The total number of deals (including the sixth card as a separate choice) is again 42504 * 19. Then the probability for the dealer is 16/42504, approximately 3.76 * 10^(-4), quite a bit more likely than for the other players.