You seek the probability that one from the four aces is the first from the $40$ non-royal cards encountered.
There will be $0$ to $12$ royal cards in the deck before the first of the $40$ remaining cards is encountered. Four of these $40$ cards is an ace, so the probability that an ace is encountered first among the non-royal cards is clearly $1/10$.
Consider the sample space Ω to be the set of all possible sequences of 52 cards drawn from the deck (without replacement). So we do not remove the Jacks, Queens and Kings from the deck.
We do not remove them, but we may safely ignore them.
Still, if we insist on not doing so, there are $52!/40!$ ways to select places in the deck for the 12 royal cards and arrange them. For each of these arrangements, there are $40!$ ways to arrange the remaining cards. Among these arrangements there are $4\cdot 39!$ in which one from the four aces are placed before the other thirty nine cards.
Thus there is a total of $52!$ ways to arrange the cards in the deck, among which there are $52!\cdot 4/40$ ways to arrange the cards so an ace occurs before the first non-royal.
Hence the probability that an ace is encountered before any other non-royal is simply $1/10$.