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?
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

$$1-\displaystyle\frac{\left\lfloor\displaystyle\frac{4!}{e}+\frac{1}{2}\right\rfloor}{4!} =1 -\frac{9}{24} = \boxed{\,\frac{5}{8}\,\,} $$
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.