Chances of winning - multiple lanes I assume this is stats 101 level of question.
If you have PC that is set to "randomly" pick a number between 1 and 100 each time a button is pushed, the chances of getting a 50 is 1 in 100, right?
So if you have two PCs doing the exact same thing, and you can press a button on either, but you can only pick one PC, what are the chances of getting a 50?
Also
if you have one person at each PC, each pressing the respective button, what are each of their chances that one of them will get a 50?
 A: 
If you have PC that is set to "randomly" pick a number between 1 and 100 each time a button is pushed, the chances of getting a 50 is 1 in 100, right?

Assuming by "randomly pick" you mean "with equal probability", then yes.

So if you have two PCs doing the exact same thing, and you can press a button on either, but you can only pick one PC, what are the chances of getting a 50?

Yes. Law of total probability ($\Pr(A)=\sum_i \Pr(A| B_i)\Pr(B_i)$). Let's call getting 50 a success ("S") and let $C_A$ and $C_B$ be the events that you picked computer A or B respectively.
$P(S)=P(S|C_A)\cdot P(C_A)+P(S|C_B)\cdot P(C_B)$
$ \quad= \frac{1}{50}\cdot P(C_A)+\frac{1}{50}\cdot (1-P(C_A))$
$ \quad= \frac{1}{50} (P(C_A)+ (1-P(C_A)))$
$ \quad= \frac{1}{50}$

Also if you have one person at each PC, each pressing the respective button, what are each of their chances that one of them will get a 50?

Depends if you mean "exactly one" or "at least one". If the PC's work independently, then
a) $P(\text{At least one }S) = 1-P(\text{no } S) = 1-(\frac{49}{50})^2 = 0.0396$
(i.e. just under 2/50)
b) 
$P(\text{Exactly one }S) = P(\text{success on }A, \text{fail on }B) + P( \text{fail on }A, \text{success on }B)$
 $\quad = \frac{1}{50}\cdot \frac{49}{50}  + \frac{49}{50}\cdot \frac{1}{50}$
 $\quad = 0.0392$
Alternate method:
$P(\text{Exactly one }S) = P(\text{At least one }S)-P(\text{Both }S)$
 $\quad = 0.0396 - \frac{1}{50}\cdot \frac{1}{50}$
 $\quad = 0.0392$
