What is the probability that at least one card is in the right box 
Four Christmas cards are randomly placed in their recipients' boxes
  (one in every box). What is the probability that at least one card is
  in the right box?

So, if we number the boxes with $1,2,3,4$ and let the first card should be placed in the box with number $1$, the second in the box with number $2$, etc. If only one card is in the right box, we have $4$ possibilities; if only two cards are in the right boxes, we have $6$ possibilities. I am having troubles with what happens when three cards are in the right boxes. Do we have only $1$ way or $4$ ways?  
 A: The probability that at least one card is in the right box is $1$ minus the probability that no cards are in the right boxes. The latter is a derangement. 
There are $n!$ total arrangements, and you are interested in the probability of not getting a derangement, which is $$1-\displaystyle\frac{\left\lfloor\displaystyle\frac{n!}{e}+\frac{1}{2}\right\rfloor}{n!}$$
Your problem is the case $n=4$, for which we have
$$1-\displaystyle\frac{\left\lfloor\displaystyle\frac{4!}{e}+\frac{1}{2}\right\rfloor}{4!} =1 -\frac{9}{24} = \boxed{\,\frac{5}{8}\,\,} $$
A: Your calculations are off, because you only focus on the cards in the right box, and forget to take  into account the cards that have to end up in the wrong box.
Now, you seem to be trying to count the number of ways to get at least $1$ card in the right box by adding up the number of ways to get exactly $1$ in the right box, exactly $2$, exactly $3$, and exactly $4$. You don't use the word 'exactly' yourself, but this is certainly implied by your very method of breaking the problem down the way you do. Well, you can certainly do this as a strategy, so so far so good.
However, you then say:

If only one card is in the right box, we have $4$ possibilities

No.  There are $4$ possibilities to pick a box with the right card, but for each of those, there are $2$ ways to distribute the $3$ remaining cards for the $3$ remaining boxes so that the other cards are all in the wrong box. For example, both distributions $1342$ and $1423$ have $1$ in the right place, but the other three in the wrong box. So, you get $4 \cdot 2 = 8$ ways to get exactly $1$ card in the right box.

if only two cards are in the right boxes, we have $6$ possibilities. 

This turns out to be correct, but I bet you were still just thinking about the number of ways to pick the two 'correct' boxes ... you still need to think about in how many ways the other two cards can end up in the wrong box .. but with 2 cards and 2 boxes left there is only one way to get them both wrong. So, it is $6 \cdot 1 = 6$ ways to get exactly 2 cards in the right box.

I am having troubles with what happens when three cards are in the right boxes. Do we have only $1$ way or $4$ ways?  

Neither. There are $0$ ways: if three cards are in the right box ... then the fourth is as well!  So, here is the perfect example of how not thinking about how the remaining cards have to be in the wrong box will get you the wrong answer. 
Now, of course, there is exactly $1$ way to get all $4$ cards in the right box, and in that case there are no 'wrong' cards to worry about.
So, the number of ways to get at least $1$ card in the right box is $4 \cdot 2 + 6 \cdot 1 + 1 = 8+6+1=15$, meaning that the probability to get at least $1$ card in the right box is $\boxed{\frac{15}{24}}$
A: One of the more fundamental approaches that students first being exposed to derangements (but maybe even before being told the name of the objects they are counting) will come up with will be to use inclusion-exclusion.
Let $X_1,X_2,X_3,X_4$ be the event that the first, second, third, and fourth boxes contain the correct letter respectively.
We are tasked with calculating $Pr(X_1\cup X_2\cup X_3\cup X_4)$, the probability that at least one of the boxes contains the correct letter.
We expand this via inclusion-exclusion as:
$$Pr(X_1\cup X_2\cup X_3\cup X_4) = Pr(X_1)+Pr(X_2)+Pr(X_3)+Pr(X_4)-Pr(X_1\cap X_2)-Pr(X_1\cap X_3)-\dots-Pr(X_3\cap X_4)+Pr(X_1\cap X_2\cap X_3)+\dots+Pr(X_2\cap X_3\cap X_4)-Pr(X_1\cap X_2\cap X_3\cap X_4)$$
We calculate each individual term in the above.  By symmetry, we need only calculate the four probabilities $Pr(X_1),Pr(X_1\cap X_2),Pr(X_1\cap X_2\cap X_3)$ and $Pr(X_1\cap X_2\cap X_3\cap X_4)$
We find that the probability that the first letter is in the correct box will be $\frac{1}{4}=\frac{3!}{4!}$ as given the first letter is in the correct box there are $3!$ ways to arrange the remaining letters without any consideration as to whether any are correct or not.
Similarly, we find $Pr(X_1\cap X_2)=\frac{2!}{4!}, Pr(X_1\cap X_2\cap X_3)=\frac{1!}{4!}$ and $Pr(X_1\cap X_2\cap X_3\cap X_4)=\frac{0!}{4!}$.  (Remember: $0!$ is equal to $1$)
Grouping the terms together, we have then
$$Pr(X_1\cup X_2\cup X_3\cup X_4) = \dfrac{\binom{4}{1}3!-\binom{4}{2}2!+\binom{4}{3}1!-\binom{4}{4}0!}{4!} = \dfrac{15}{24}$$

Indeed, what we just calculated is one of the ways to derive the value for $!n$, the number of derangements of an $n$ element set, i.e. the number of permutations in $S_n$ with no fixed points.  We have:
$$!n = n!-\sum\limits_{k=1}^n \left(-1\right)^{k+1}\binom{n}{k}(n-k)!$$
A: If 3 cards are in the correct box, then the fourth card must be in the remaining box. There is only 1 way to do this.
