I recently read that both player and dealer can expect to get blackjack about 5% of the time no matter how many decks are used, but that more decks increases the probability that both player and dealer will have blackjack at the same time. I haven't done any extensive research into this, so maybe someone else can offer some insights.
I think you are refering to this formula:
a * t * (1-2(a-1)*(t-1)/(n-2)/(n-3))/n/(n-1)
where a=number of aces, t=number of tens, and n=cards remaining.
from page 191 of Peter Griffin's "The Theory of Blackjack".
Of course the above formula shows the percentage of winning (not getting) a blackjack, and is not what the poster asked for - how many hands until a player gets a blackjack.
But don't worry, this book is an 'Unread Classic' and 'Obsolete', right? It doesn't contain any of the really modern cool advantage techniques like rhythmic rolling at the crap table.
Your calculation is not accounting for the effect of removal of the first card in your hand (If you get a ten as your first card, the odds are now slightly better to get an Ace as your second, since you are now drawing from a smaller deck with 1 fewer non-ace cards, for example.)
Your answer of .0473 is correct for an infinite deck (or drawing with replacement.)
Chances of getting a blackjack:
4.827% one deck
4.779% two decks
4.757% four decks
4.734% infinite deck
This decreasing likelyhood of getting a blackjack is partially why the house edge increases as number of decks increases even if the game has identical rules.
When determining BS, it is assumed that the cards comprising the hand and the dealer's upcard have been removed from the deck, in order to perform the ensuing combinatorial analysis. SD BS does NOT assume a 52-card full deck.
I think I read about the full deck method in Lance Humble's book quite awhile ago, but perhaps my memory or reading ability is not as good as it should be...With your rather extensive knowledge of the game, I'm persuaded to accept your advice.