(This is a cross-post from Math StackExchange <https://math.stackexchange.com/questions/609641/multinomial-distribution-sum-of-squared-probabilities>)


Let $\vec X = (X\_1, \dots, X\_k)$ be a draw from a fair multinomial distribution with $n$ trials, i.e. $P(X\_1 = x\_1, \dots, X\_k = x\_k) = \binom{n}{x\_1, \dots, x\_k} k^{-n}$


Let $\vec Y$ be an independent draw from the same distribution, i.e. $\vec X$ and $\vec Y$ follow the same law.


My questions is this: what is the probability that $\vec X = \vec Y$?


It appears that this probability, for fixed $k$, is roughly $(c\_k + o(1)) n^{-(k-1)/2}$. This makes sense; the first $k-1$ coordinates vary on the order of $\sqrt{n}$ and are "independent-ish". (The last coordinate is of course fixed by the first $k-1$). Can one compute $c\_k$?



Repeated draws from multinomial distribution
This is essentially a product of local CLTs (since one can uncover the multinomial
distribution variable by variable: decide first how many of the $n$ variables equal "1" with probability of success $1/k$ for each one and thus variance $(n/k)(1-1/k)$; Then, of the roughly $n(k-1)/k$ remaining variables, decide how many are "2" with probability of success 
$1/(k-1)$ each, and variance $(n/k)(1-1/(k-1))$, and so on). 


\**\** Edit: As pointed out by Will Nelson,
there is a factor of $1/\sqrt{2}$ in each step that comes from the fact that 
we are using the local CLT not at the mean but rather at the location of the
first multinomial $X$.


Thus, the probability is asymptotic to
$$\left(\frac{k}{4\pi n}\right)^{(k-1)/2}\cdot \frac{1}{\sqrt{\prod\_{j=2}^k (1-j^{-1})}}$$ 


For $k=2$, this is just 
coming from 
$$\left(\frac{1}{\sqrt{2\pi n\cdot (1/4)}}\right)^2\int e^{-2\cdot (x^2/(2\cdot (1/4))}dx=1/\sqrt{\pi n}$$ 


