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?
 A: 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.
A: A derangement of set of $n$ uniquely labeled objects is a permutation in which each element changes its "identity."  
The number of such derangements is denoted $!n = (n-1)(!(n-1)+!(n-2))$, where $!0=1$ and $!1=0$.
So for $n=6$ the number of derangements is $265$.
The total number of permutations is, of course, $n!$, where $6! = 720$.
So the chance you do not get even one correct guess is $265/720$ or about $36.8\%$, and the chance that you get at least one correct guess is $1 - .368 = .632 = 63.2\%$, which agrees with N. F. Taussig's answer, below.
Compute the number of derangements through the definition and induction.
Example:  $!3 = (3-1)(!(3-1)+!(3-2)) = 2 (1+0) = 2$, and work up from there.
In Mathematica:
derange[n_] := (n - 1) (derange[n - 1] + derange[n - 2]);
derange[0] = 1;
derange[1] = 0;

$\left(
\begin{array}{cc}
n & derangements \\ \hline
 1 & 0 \\
 2 & 1 \\
 3 & 2 \\
 4 & 9 \\
 5 & 44 \\
 6 & 265 \\
 7 & 1854 \\
 8 & 14833 \\
 9 & 133496 \\
 10 & 1334961 \\
\end{array}
\right)$
If you have a moment, try listing all $3!=6$ permutations of $3$ objects and count the derangements ($!3 = 2$).
A good exercise is to derive the formula for derangements. Start with $n−1$ elements aligned in an "input" row, and $n−1$ elements beneath it in the "output" row.  Assume you know the number of derangements (links from an element in the input row to an element in the output row not directly beneath it). Now add an element to both the "input" and the "output". That new element must be linked to one of the $n−1$ "other" elements beneath it. Then there are two cases, either that output element is connected to the new element or it is not. These will give you: $(n−1)(!(n−1)+!(n−2))$, as you can see.
A: Suppose there are $N$ pictures that have to be matched to $N$ names, then the probability of getting exactly $k$ of them correct is given by:
$$P(k) = \binom{N}{k}\frac{ \operatorname{D}(N-k)}{N!}$$
where $\operatorname{D}(M)$ denotes the number of dearrangements of M items, which can be computed exactly by dividing $M!$ by $e$ and rounding off to the nearest integer (in the above formula, $\operatorname{D}(0)$ is an exceptional case, it has be set equal to 1) For $N$ not too small and $k$ not to close to $N$, one can replace $\operatorname{D}(M)$ by $\frac{M!}{e}$ without rounding off to an excellent approximation, this then dramatically simplifies the formula to:
$$P(k) \approx \frac{1}{e k!}$$
So, the probabilities then don't depend on $N$ to an excellent approximation. 
