# Symmetry arguments in probability

I draw thirty cards from a regular deck (4 suits x 13 cards each). The only fact I know is:

As I draw cards from the deck, the next draw is randomly chosen from the remaining cards in the deck

How can I find the probability that the last card drawn is a spade?

I have no doubt that it's $\large \frac{1}{4}$. But I don't know how to prove it formally.

In other words, how do I prove that the deck with the above property represents a fully random shuffle?

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Let $X$ and $Y$ be independent RVs. $$f_{X|Y}(x|y) = f_X(x)$$. –  Inquest Dec 8 '12 at 2:08
@Inquest I did a poor job explaining what I know about the deck, so I updated the question. –  max Dec 8 '12 at 2:23
There are $52!$ possible orderings of the deck. The number of orderings that have a spade as the thirtieth card is $13\cdot 51!$. As we have a sample space consisting of equally likely outcomes... –  David Mitra Dec 8 '12 at 2:23

Any argument showing that the probability that that card is a spade is a certain number, would likewise show that the probability that that card is a heart is that same number. And similarly for the other two suits. They're mutually exclusive and exhaustive. So $$x+x+x+x=1.$$ Now solve for $x$.