If you draw two cards in consecutively in a standard deck of 52 cards, what is the probability of getting black on the second draw? This is my thought process:
$P(2^{nd}\ \text{black}) = P(2^{nd}\ \text{black} \mid 1^{st}\ \text{red}) P(1^{st}\ \text{red}) + P(2^{nd}\ \text{black} \mid 1^{st}\ \text{black}) P(1^{st} \ \text{black}) $ 
$\frac{26}{51}*\frac{26}{52} + \frac{25}{51}*\frac{26}{52} = \frac{1}{2} $
Is this correct?
 A: Yes, it is correct. 
Sanity check: By symmetry, the likelihood to get a black is equal to the probability og getting a red at any draw position.
A: Yes you are right.
The intuition behind why your answer is symmetrically ${1\over 2}$ is that if you first draw a card and close your eyes over its color (e.g. throw it away), you are in a full ambiguity or lack of information about the color of the second one. Hence the probability is the same. 
This is also the same case for any number of color types and drawing the second card.
A: The "hard way":
To get a black card on the second draw you must get either "red, black" or "black, black".
1) "red, black":  There are 52 cards in the deck, 26 or them red.  The probability the first card drawn is red is 26/52= 1/2.  Given that, there are now 51 cards in the deck, 26 of them black.  The probability the second card drawn is black is 26/51.  The probability of "red, black" is (1/2)(26/51)= 13/51.
2) "black, black:  There are 52 cards in the deck, 26 or them black.  The probability the first card drawn is red is 26/52= 1/2.  Given that, there are now 51 cards in the deck, 25 of them black.  The probability the second card drawn is black is 25/51.  The probability of "red, black" is (1/2)(25/51)= 25/102.
The probability of either "red, black" or "black, black", since those are mutually exclusive, is 13/51+ 25/102= 26/102+ 25/102= 51/102= 1/2!
A: You are correct... unless the colour of the first card is known, in which case it would be either:
$\dfrac{25}{51}$ if the first card was black, or
$\dfrac{26}{51}$ if the first card was red
