Suited blackjacks

A few weeks ago, I was playing Spanish 21 at Primm Valley Casino. They were having a promotion where the player is given a card, and when he gets a suited blackjack, the card is signed. When the player gets all four suits, he gets to spin a wheel for a prize. Lady Luck was on my side that night, and I had three of the suited blackjacks within ten minutes. It took a couple hours, but I finally filled in the diamonds spot and won a dinner in their steak house. (A very good dinner, BTW.)
Now the question that's been racking my brain ever since: is the player more or less likely to get a suited blackjack in Spanish 21 (no 10s in the deck) as opposed to regular blackjack? On the one hand, a player is more likely to get a blackjack in regular blackjack, since there are more 10 value cards. On the other hand, if a Spanish 21 player gets an ace, there are fewer cards in the deck that are not a ten value card of the matching suit. Any thoughts from the math guys?

I don't know the odds, but Red Lion casino in Elko, Nevada had a similar promotion when I was driving through, except it was $200 for hitting all four suited blackjacks. I played two hands for about 4 hours, got three, couldn't get the fourth, and quit down $250.

In a regular 6-deck game you have 312 cards of which 24 are Aces and 96 are '10 value'. So, the probability of receiving Ace/10 or 10/Ace is:-

24/312 X 96/311 X 2 = 4.749% and we divide by 4 to get the suited number = 1.187% or roughly 1 in 84 hands.

In a Spanish 6-deck game you have 288 cards of which 24 are Aces and 72 are '10 value'. So, the probability of receiving Ace/10 or 10/Ace is:-

24/288 X 72/287 X 2 = 4.181% and we divide by 4 to get the suited number = 1.045% or roughly 1 in 96 hands.

Thanks for that well explained answer. I couldn't even think of how to make the equation. So the chances of getting all four suited blackjacks is a small fraction of 1%, now I feel even luckier!