Probability of candidate voting tie This question seemed simple at first glance, but I quickly realized it was not.
You have a group of 25 candidates with 1 vote each; they cannot vote for themselves.
What is the probability of a 25-way tie (that each voter casts their vote for in such a way that all 25 receive only 1 vote)?
Edit: More details.  Voters are not aware of each others' votes.  The reasoning behind each vote cast for a person is because that voter believes they are most deserving of the title/award.
 A: Yes, it's not trivial because this is about derangements. Total number of choices is $24^{25}$. For everyone to get one vote, we need a permutation, but no one is allowed to vote on himself/herself. This is a derangement, and the number of possible derangements for $n$ is shown with $!n$. So, the answer is $$\frac{!25}{24^{25}}$$
The number of derangements is calculated several ways as laid out in the wikipedia entry. One of them is the simple recursive formula
$$!n=(n-1)(!(n-1)+!(n-2))\rightarrow a_n=(n-1)(a_{n-1}+a_{n-2})$$
where $a_n$ is the number of derangements.
According to this series, we have results up to $a_{23}$, but you can easily calculate $a_{25}$.
We can test this for smaller numbers, i.e. for $n=6$, the probability is $$\frac{!6}{5^6}=\frac{265}{15625}\approx 0.017$$
An example matlab program to demostrate this is as follows:
n = 1000000;
c = 0;
m = 6;  % num people

ids = 1:m;
for i = 1:n
    votes = randi(m-1,1,m);
    inds = votes >= ids;
    votes(inds) = votes(inds) + 1;
    if length(unique(votes)) == m
        c = c + 1;
    end
end

disp(c / n)

The output is:
0.0170

