pick card from a shuffled deck of cards until an ace appears.  on
average, how many cards are required to produce an ace?  we need the
theoretical value and the explanation please.

Are you reinserting the selected card and reshuffling each time or
just removing cards until you get an ace? It makes a difference.

also, are you drawing from the top of the deck or from anywhere, I was
working out from top of the deck which I think is more work then
needed.

no reinsertion, just picking cards until i get an ace.  i'm also
drawing from the top of the deck.  thank you

Calculate the following:
(51 C 3)+2(50 C 3)+ 3(49 C 3)+ ...+49(3 C 3)
-------------------------------------------
                 (52 C 4)
The answer I got using Maple was 53/5.

Note that if it's a shuffled deck it doesn't matter where you're
drawing from, as long as you're not reinserting. (But don't use this
argument in poker games!)

More explicitly, the probability of getting the first ace on the n'th try is
(48/52)*(47/51)*(46/50)*...*((50-n)/(54-n))*(4/(53-n))
= 4 . 48! (52-n)! / [52! (49-n)!]
= 4.(52-n).(51-n).(50-n) / (52.51.50.49)
where n goes from 1 to 49.

To get the expected number of cards to choose, we need to sum
4n.(52-n).(51-n).(50-n) / (52.51.50.49) for n from 1 to 49.
You can derive a fifth-order polynomial for this: the sum of numerator
of the above expression as n goes from 1 to k is
k (k+1) (-4k^3 + 759k^2 -51249k + 1299994) / 5
(you can prove this by induction if you like).
Evaluating this for k=49, or simply pasting cells into a spreadsheet
and adding them up, will give you a numerator of 68872440; cancelling
out common factors with the denominator leaves you with 53/5, as
reported by etotheipi-ga.