How many ways are there to pick a red ball second from a bucket of 10 blue balls and 10 red balls (no replacement)? This example is taken from my textbook:
"A bag contains twenty balls; ten of the balls are painted red and ten are painted blue. Two balls are drawn from the bag. Suppose we do not replace the first ball once it is drawn. "
"There are $20 * 19 = 380$ different ways to draw one ball and then draw a second from those that remain." This makes sense.
"There are $10*19=190$ ways to pick a ball such that the first ball is red." This also makes sense. 
"Likewise, there are $190$ ways to pick a ball such that the second ball is red." This doesn't make sense to me. How can you be certain of this probability, since it is dependent on your first pick? 
You have $20$ choices for your first pick (though $190$ isn't a multiple of $20$...), since it can be anything. After that, it has to be a red ball...but the probability of that is dependent on what color you picked previously. 
Maybe the author implied that there are $190$ ways to pick a red ball second assuming you picked a blue one first. But even then, wouldn't the answer be $10 * 10$, because you have $10$ ways to pick a blue ball first, and then $10$ ways to pick a red ball after that?
 A: There are, as you say, 100 events that are blue followed by red, and another 90 = 10$\cdot$9 events that are red followed by red, making a total of 190 in all.
The author's point, I suspect, is that there is an obvious 1-1 correspondence between events in which a red ball was picked first and events in which a red ball was chosen second  (i.e. imagining that the balls were picked in the opposite order).  Therefore, those two events are equally likely to have occurred.  This is a key idea in combinatorics -- that we can count something difficult by showing that it is equinumerous to something that is easier to count.
A: 
"Likewise, there are $190$ ways to pick a ball such that the second ball is red." This doesn't make sense to me. How can you be certain of this probability, since it is dependent on your first pick? 

Sure, the second depends on the first, but it will average out. 


*

*There are $10\cdot 9$ ways to pick a two red balls; that's $90$ ways.

*There are $10\cdot 10$ ways to pick a blue and then a red ball; that's $100$ ways.


In total there are $190$ to pick a red ball as second .  Divide by the $380$ ways to pick two balls.

By the Law of Total Probability.
$$\begin{align}\mathsf P(R_2)&=\mathsf P(R_1)~\mathsf P(R_2\mid R_1)+\mathsf P(R_1^\complement)~\mathsf P(R_2\mid R_1^\complement)\\[1ex]&=\tfrac{10}{20}\tfrac{9}{19}+\tfrac {10}{20}\tfrac {10}{19}\\[1ex]&=\tfrac {190}{380}\\[1ex]&=\tfrac{1}{2}\end{align}$$

As indeed we should anticiplate that half of the ways to select two balls will have a red ball as second -- each individual ball is exactly as likely to be the second ball drawn (there is no bias) and 10 of those 20 balls are red, so.

Now can you determine the probability that the ninth ball is red when we draw thirteen balls? …  
