This is annoying me, because I can't find the main source material that we used way back when to prove this. However:

http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-spring-2010/readings/MIT6_042JS10_chap18.pdf

Problem 18.5. (Do a ctrl+f) is designed to get you to exactly the right section, or:

"Problem 18.5.

I have a deck of 52 regular playing cards, 26 red, 26 black, randomly shuffled. They all lie face down in the deck so that you can’t see them. I will draw a card off the top of the deck and turn it face up so that you can see it and then put it aside. I will continue to turn up cards like this but at some point while there are still cards left in the deck, you have to declare that you want the next card in the deck to be turned up. If that next card turns up black you win and otherwise you lose. Either way, the game is then over.

(a) Show that if you take the first card before you have seen any cards, you then have probability 1/2 of winning the game.

(b) Suppose you don’t take the first card and it turns up red. Show that you have then have a probability of winning the game that is greater than 1/2.

(c) If there are r red cards left in the deck and b black cards, show that the probability of winning in you take the next card is b/(r + b).

(d) Either,

1. come up with a strategy for this game that gives you a probability of winning strictly greater than 1/2 and prove that the strategy works, or,

2. come up with a proof that no such strategy can exist."

Point (D) 2. is asked because, against intuition, the only proof that exists is one showing that no strategy can exist, that is, your odds never change. They were determined at the outset.

I can prove this (D2).

What I will show is that the odds of winning this game when there are n cards left in the deck (in other words 52-n cards have been flipped over) in an arbitrary deck are 50%.

This can be expressed as the sum from b=0 to n of the probability of there being b black cards left multiplied by the odds of winning with b black cards left in the deck.

Here's the interesting part. Consider all permutations of 52 cards and divide them up into two groups, ones which have "black" as the final card on one side and ones with "red" as the final card on the other side. You can form a one to one correspondence between these two groups because if you take each card in one group and swap each black card with a red card and each red card with a black card (in other words, inverting the color of each card), you will get a unique permutation from the other group.

Now you can see that the probability of there being b=k black cards left has to be the same as the probability of there being b=(n-k) black cards left, because for each permutation that has k black cards left, we can use that inverting function to obtain a corresponding permutation that has n-k black cards left.

Also, the odds of winning when there are b black cards left out of n cards is b/n (see part c, only I am using n instead of r+b), and the odds of winning when there are n-b black cards left is (n-b)/n. If we suppose we were playing on both situations simultaneously, the odds then of winning one of them is the sum of the odds of winning either one, since the decks are opposite each other:

(b/n) + (n-b)/n = (n-b+b) / n = n/n = 1

So now we can reduce our big summation above by pairing up those inverted scenarios:

sum from b=0 to n of the probability of there being b black cards left multiplied by the odds of winning with b black cards left in the deck

= sum from b=0 to n/2 rounded down of the probability of there being b black cards left multiplied by the odds of winning with either b black cards or n-b black cards left in the deck *

= sum from b=0 to n/2 rounded down of the probability of there being b black cards left multiplied by 1

= 1 * sum from b=0 to n/2 rounded down of the probability of there being b black cards left

= 1/2 * sum from b=0 to n of the probability of there being b black cards left

= 1/2 * 1, since the sum of the probability of every possibility has to be 1

= 1/2

* edit to note the case in the above step that if n is even, then the case where b=n/2 is also 1/2 because then n-b=b, meaning there are just as many reds as blacks left.

P.S. 40% to the OP