Probability of 0% correct choices If I have 12 boxes, in order of #1 to #12, and I close my eyes and stick a known label (A to L) on a random box, what is the probability that I get zero correct labels on the corresponding boxes?
In other words, after sticking all the labels on the boxes, not a single box has the correct label on it.
 A: The answer to your question is $\approx 0.3678794$.
Your question in its general form can be seen as a simple application of the concept of derangement (not the psychological condition) were we want to find the probability that a randomly selected permutation is a derangement. To directly quote Wikipedia on this: "a derangement is a permutation of the elements of a set, such that no element appears in its original position." Therefore what you want is simply the ratio of possible derangements over the total number of possible permutations of a given sequence (of labels in your case). This ratio converges to $\frac{1}{e}$ quite quickly and it can be also approximated for a given $n$ by $\sum_{i=0}^{n} \frac{(-1)^i}{i!}$ (or simply sum((-1)^(0:n) / factorial(0:n)) in R). 
It is easy to evaluate this expression for $n=12$ and get $0.3678794$. The Wikipedia page on random permutation statistics has some further derangement-associated fun facts.
Just to put some code behind this I append a very simple script doing exactly the random assignment you described and then plotting the estimated simulated probabilities and the asymptotic ratio $1/e$.
oneExperiment <- function(K){
  myLabels = sample(letters[1:K], replace = FALSE, size = K );
  trueLabels = sample(letters[1:K], replace = FALSE, size = K)
  # print(myLabels); print(trueLabels)
  all( myLabels != trueLabels)
}

set.seed(55)
dPr = sapply(2:23, function(L) { 
       mean(sapply(1:(2^16), function(x){ K = L;  oneExperiment(K) } ) ) } )
plot(2:23,y = dPr, panel.first = grid(), ylab = 'Prob. of Derangement')
abline(h = 1/exp(1)) # Add asymptotic limit line


