We throw a dice $n$ times. Each time it falls on 1 I win, each time it falls on 6 you win; otherwise nobody wins.
What is the probability that both of us win the same number of times, as a function of $n$, when $n$ is very large?
I tried to solve the question using the multidimensional central-limit-theorem. Suppose we throw a single dice. Let $X$ be a random vector of length 6, such that $X_i=1$ if the dice lands on $i$ and 0 otherwise. This is a random vector with (for $p=1/6$): $$ \mu = [p,p,p,p,p,p] $$ $$ \Sigma = \begin{pmatrix} p-p^2 & -p^2 & -p^2 & -p^2 & -p^2 & -p^2 \\ -p^2 & p-p^2 & -p^2 & -p^2 & -p^2 & -p^2 \\ -p^2 & -p^2 & p-p^2 & -p^2 & -p^2 & -p^2 \\ -p^2 & -p^2 & -p^2 & p-p^2 & -p^2 & -p^2 \\ -p^2 & -p^2 & -p^2 & -p^2 & p-p^2 & -p^2 \\ -p^2 & -p^2 & -p^2 & -p^2 & -p^2 & p-p^2 \end{pmatrix} $$
Let $S$ be a vector that is the sum of $n$ such vectors. Then, for every $i$ between 1 and 6, $S_i$ is the number of times the dice falls on $i$. When $n$ is sufficiently large, the vector $S$ is distributed like a normal random vector, with mean vector $n\mu$ and covariance matrix $n\Sigma$.
Now, the probability that $S_1=S_6$ is approximately this integral:
$$ \int_{S_1=S_6} f(S) $$
where $f$ is the probability-density-function of the normal distribution with mean $n\mu$ and covariance $n\Sigma$:
$$ f(S) = { \exp\left(-{1\over 2} (S-n\mu)^T(n\Sigma)^{-1}(S-n\mu)\right) \over \sqrt{(2\pi)^6 |n\Sigma|} } $$
I calculated $|\Sigma|$ and it is $p^6(1-6 p) = 0$, so $\Sigma$ does not have an inverse. How can I solve the integral then?
I do not need an exact solution - just an asymptotic solution for large $n$.
