OK, here's how I did it. I prefer to derive a more general result here.
First of all, it's immediately obvious that in a deck of N cards (N is even) with N/2 of one color and N/2 of the other color, if all N/2 cards of one color have been drawn by the time the (N-2)th card has been drawn, there is no guesswork entailed in predicting the last two cards (prob. of correct guess = 1). Let's call this Scenario 1.
If, in contrast (Scenario 2), exactly (N/2 - 1) of one type and (N/2 - 1) of the other had been drawn by that time, the prob. of a correct guess is only half. So the overall probability has to be somewhere between half and one.
I found it difficult to predict the respective probabilities of Scenario 1 and 2 without using a combinatorics approach. With that approach, it becomes easy.
We're given a deck of N cards, well shuffled. Array them out in order.
The total number (N) of possible distinct arrangements of these N cards is N = \frac{N!}{({(\frac{N}{2})!)}^2}
In the case where the last two cards are dissimilar (Scenario 2), (N-2)/2 cards of each sort have to be arranged to fill the first (N-2) places .
The number of ways that this can happen is n_2 = 2\frac{(N-2)!}{({(\frac{N-2}{2})!)}^2} (The multiplication by two is to take into account both the (red, black) and (black, red) possibilities).
In Scenario 1, (N/2) cards of one color and (N-4)/2 cards of the other color have to be arranged.
The number of ways for Sc. 1 is n_1 = 2\frac{(N-2)!}{(\frac{N}{2})!(\frac{N-4}{2})!}
Again, the expression is multiplied by two to consider the cases where one color predominates, and vice versa.
The respective probabilities for Sc 1 and 2 are :
p_1 = \frac{n_1}{N} = \frac{N-2}{2(N-1)}
p_2 = \frac{n_2}{N} = \frac{N}{2(N-1)}
(Alternatively, you could simply take p_2 = 1 - p_1 but this way you can check your work).
If Scenario 1 materialises, the probability of guessing the last two cards right is one (no guesswork). If Scenario 2 materialises, the probability of correctly guessing the last two right is 1/2.
So the overall probability of a correct guess is p = (1)p_1 + \frac{1}{2}p_2 = \frac{3N-4}{4(N-1)}
For a deck of 52 cards, that gives p = \frac{38}{51}.
Yes, you are completely correct. The situation is completely reversible. Reversing the draw, the first card choice is immaterial. The prob of getting a similar card with only 25 of that sort left in the deck is 25/51. Prob of dissim is therefore 26/51, and everything works out the same, with much less fuss.
