Expected Value of Bills 
A chest contains 5 envelopes that each contain one bill. One envelope
  contains a $\$5$ bill, two envelopes contain a $\$20$ bill, and the
  remaining two contain a $\$100$ bill. You randomly pick two envelopes
  from the chest. You then open one of the envelopes in your hand and
  reveal a $\$20$ bill. 
At this point, you have the option of making a swap. You can either
  open the other envelope in your hand and keep the bill inside, or you
  can pick one of the three remaining envelopes from the chest and keep
  the bill there. Do you make a swap?

Is my approach correct?
Assume you make the swap, and calculate the expected value of the increase in earnings. 
There is a $\frac{1}{10}$ chance the two bills in your hand are both $\$20$. By switching, you gain $\frac{1}{3}(205) - 20 = \frac{145}{3}$ on average. 
There is a $\frac{2}{10}$ chance the two bills in your hand are a $\$20$ and a $\$5$. By switching, you gain $\frac{1}{3}(220) - 5 = \frac{205}{3}$ on average.
Similarly, there is a $\frac{4}{10}$ chance the two bills in your hand are a $\$20$ and a $\$100$. By switching, you gain $\frac{1}{3}(125) - 100 = \frac{-175}{3}$ on average.
Therefore, by switching you earn $\frac{1}{10} \cdot \frac{145}{3} + \frac{2}{10} \cdot \frac{205}{3} + \frac{4}{10} \cdot \frac{-175}{3} < 0$ on average, so you shouldn't switch. 
 A: How can it matter whether you swap or not?  You might as well have drawn one envelope, opened it to find $\$20$, then picked another.  You know nothing about the other envelope you hold nor about the ones you don't.
A: Your approach is not correct. 
Prior to selecting two envelopes and opening one, there was indeed a $\frac2{10}$ chance that when selecting two envelopes you would choose envelopes with contents $\$5$ and $\$20.$
But there was only a $\frac1{10}$ chance that you would do that and then open the $\$20$ envelope.
In fact, you initially have a $\frac1{10}$ chance to select envelopes with $\$5$ and $\$20$ and then open the one with $\$20,$
a $\frac1{10}$ chance to select the two envelopes with $\$20$ and then open one of them, 
and a $\frac2{10}$ chance to select envelopes with $\$100$ and $\$20$ and then open the one with $\$20.$
Putting those values in place of the incorrect ones in your expected value calculation, the result is zero, not negative. 
Technically, there is a second error: you should be using probabilities conditioned on the fact that you did open an envelope with $\$20,$ not using prior probabilities. After all, your probabilities added to only $\frac7{10},$ whereas they describe all possible ways to observe an event that you would already know had happened when you made your decision to swap or not. 
The conditional probabilities come out to 
$\frac14,$ $\frac14,$ and $\frac12,$ 
that is, all the terms of the expected-value sum are multiplied by $\frac52.$
The result is still zero. 

Another way to look at this is that selecting two envelopes at random and then randomly selecting one of them to open is tantamount to selecting one envelope at random from the original five (and then opening it) and then selecting another envelope. Opening the first envelope has reduced the possible outcomes to just four (two of value $\$100,$ one $\$20,$ and one $\$5$), and every remaining envelope (including the one that was chosen but not opened) has the same likelihood to be any of those four outcomes. 
It’s like playing a variant of the Monty Hall game where Monty doesn’t know where the car is.
A: Let's first suppose your other bill is the $\$5$ bill. There is a $\frac14$ chance of this.
If you switch, you are guaranteed to make more than if you don't.
Let us now suppose you have both $\$20$ bills. There is also a $\frac14$ chance of this. 
If you switch, it is worth it $\frac23$ of the time.
If you have a $\$100$ bill, of which there is a $\frac12$ chance, switching is not worth it $\frac23$ of the time, and makes no difference $\frac13$ of the time.
Multiplying and summing the probabilities, we have the switch being worth it $\frac{5}{12}$ of the time, not worth it $\frac{5}{12}$ of the time, and no difference made by switching or not $\frac16$ of the time. On account of this last probability where it doesn't matter, I'd say the switch is overall worth it.
A: As was correctly pointed out in another answer, swapping yields no gain or loss - as long as the envelop that you opened was chosen at random from the two in your hand, which is not made entirely explicit in the original question. That's because in this case, you have effectively selected an envelop at random, opened it to reveal a $20$ dollar bill, and then picked another envelop at random - the expectation of what's in that second, unopened envelop is exactly the same as for the remaining $3$.
However, suppose that a knowledgeable third party tells you to open what he knows to be the "richest" of the two envelops in your hand. In that case, finding a $20$ dollar bill should definitely prompt you to switch your second envelop for one of the remaining $3$ (easy to see: in the unopened envelop you have in hand, you certainly have no more than $20$ dollars, while there are at least $205$ dollars spread between the remaining $3$ envelopes).
