OK I got pretty excited, so I couldn’t wait and did the calculation myself. If my calculations are correct, the house edge in this case moves from 0.66% to 0.22%. I initially thought of royal blackjack to include Ace/King, Ace/Queen, and Ace/Jack of the same suit, which yielded 0.08% player edge. However, as shadroch mentioned, royal blackjack is probably just Ace/King of the same suit. Here are the details of my logic, let me know if there are any flaws:
There are 64 possible ways of blackjack suit combinations (i.e. Ace of Hearts and King of Hearts, Ace of Diamonds and Queen of Spades and so on.), each with a 1.6% probability. Of these 64 possibilities, 4 of them will be paid 3X, 12 of them will be paid 2X, and 48 of them will be paid 1.5X. As a result, weighted average payout will be 1.688X, instead of the usual 1.5X.
In a game of 60 hands/hour, $5 wagered/hand; total of $300 will be played. The normal return on these hands would be $298.02, with 0.66% house edge. During these 60 hands, 1.42 blackjacks will occur (60 hands x 2.37% probability). Under normal conditions, these 1.42 blackjacks would have paid $10.65 (1.42 x $7.50); however in our case these 1.42 blackjacks will pay $11.99 (1.42 x $8.44). $8.44 is calculated by multiplying 1.688 with $5 wager. Since $11.99 is $1.33 higher than $10.65, total payout during the 60 hands will go from $298.02 to $299.35. This calculates into 0.22% house edge.
Now, this joint has better blackjack games with higher minimums. There is a double deck with surrender option with $10 minimum. This game might have a break-even house edge with the promo. However, I usually am cautious with double deck games, since there is a lot of preferential shuffling going on, so I might just stick to the regular 6D shoe game and enjoy some cheaper fun.
Also, if I want to count during such a game, what could be the best way? Counting aces might be a good idea I think, since blackjacks are pretty valuable. But what should I count them against? Ace-five?