How do I find the probability of one correct guess when matching elements? If I have 6 pictures of people and 6 names. I have to match the names to the people and getting just one right results in a victory. I have no knowledge or clues to figure out which name belongs to which person, and I must use each name once. I know that my chances are not 1 in (6!) since that is the chance that I get all of them correct, I only need to get one correct. I know my chances are greater than 1 in 6 since I have a 1 in 6 chance at the beginning to meet a win condition instantly, and if I fail then I have more chances to meet the win condition.
Trying to work this out on my own, I know the first name has 1 in 6 chance to be correct when placed to a face. After it is placed, there are two possibilities, either it is correct (which is victory and I no longer need to worry) or it is not and I must continue. In which case I move onto the next name, which has a 1 in 6 chance the face it belongs to is already taken, and 5 in 6 it wasn't already taken. This is where I get stuck and I am not sure how to continue. 
How do I calculate this probability?

There are $\binom{6}{1}$ to match one of the six names to the correct picture and $5!$ ways to assign the remaining names.  However, this over counts, because some of the remaining names may be assigned to the correct picture(s).  For instance, we have counted each arrangement in which two names are assigned to the correct picture twice, once for each match.  However, we only want to count these arrangements once, so we must subtract these arrangements from the total.  There are $\binom{6}{2}$ ways to match two of the six names to the correct picture and $4!$ ways to assign the remaining names.  However, we have subtracted too much since some of these arrangements have more than two matches.  In general, there are $\binom{6}{k}$ to match $k$ of the names to a picture and $(6 - k)!$ ways to assign the remaining names.  By the Inclusion-Exclusion Principle, the number of ways the names could be assigned to the pictures so that at least one of the names matches the picture is 
$$\binom{6}{1}5! - \binom{6}{2}4! + \binom{6}{3}3! - \binom{6}{4}2! + \binom{6}{5}1! - \binom{6}{6}0!$$
Dividing this number by the $6!$ possible ways to assign names to pictures gives the probability that at least one name has been correctly matched.
In connection to this problem, you may want to read about derangements, permutations which leave no object in its original position.
