Origional Question:
If you are picking two cards in sequence from a pack of 52 cards,
what is the probability that the second card you picked is a spade?
Note: This is saying that before the two cards are taken out, we have a complete deck (no jokers). It also doesn't say what suit the first card is.
The Question is asking that, before you pick either of the cards, what is the probability that the second card you picked is a spade?
NOT: after you pick the first card, what is the probability that the second card you picked is a spade?

As Guv -and others - put it (correctly):
P(spade in nth postion of 51-card deck) = (1/4)*(12/51) + (3/4)*(13/51)
Note: 51 card deck because one card has already been drawn.

If you think this is incorrect, I have used another reasoning:
The probability must be 25%, because:
(Where P(heart) = probability of the nth card being a heart)

1) P(heart) + P(spade) + P(club) + P(diamond) = 1 = 100%
2) P(heart) = P(spade) = P(club) = P(diamond) - ** see below
=> P(heart) = P(spade) = P(club) = P(diamond) = 0.25 = 25%
** P(heart) = P(spade) = P(club) = P(diamond) because:
The deck is a complete deck, with the same number of cards in each suit. As there is no force (i.e. weight) making the appearance of a particular suit more likely than any other, then statement 2 must be true.

I apologise for saying you can't make a computer program for finding the probability. You can, just not one that randomly deals cards. All you have to do is make one that deals 52 cards, labelled H, D, S, C (only use suits as numbers aren't important), 13 of each, into 52 places, and then get the computer to count how many of these combinations have a S in the 2nd place.