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If there are multiple possible approximations, I'm looking for the most basic one.

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You can approximate it with the multivariate normal distribution in the same way that binomial distribution is approximated by univariate normal distribution. Check Elements of Distribution Theory and Multinomial Distribution pages 15-16-17.

Let $P=(p_1,...,p_k)$ be the vector of your probabilities. Then the mean vector of the multivariate normal distribution is $ np=(np_1,np_2,...,np_k)$. The covariance matrix is a $k \times k$ symmetric matrix. The diagonal elements are actually the variance of $X_i$'s; i.e.$ np_i(1-p_i)$, $i=1,2...,k$. The off-diagonal element in the ith row and jth column is $\text{Cov}(X_i,X_j)=-np_ip_j$, where $i$ is not equal to $j$.

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I guess my statistical sophistication is not enough to connect the dots in this answer. If I have the sample size n and the probabilities P, how do I calculate the mean vector and the covariance matrix of the multivariate normal distribution? – ericstalbot Aug 17 '12 at 20:35
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Check out the 2nd reference. – Stat Aug 17 '12 at 21:06
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Stat, so that this answer can stand by itself (and be resistant to link rot), would you mind giving a summary of the solution? – whuber♦ Aug 17 '12 at 21:32
Let P=(p_1,...,p_k) be the vector of your probabilities. Then the mean vector of the multivariate normal distribution is np=(np_1,np_2,...,np_k). The covariance matrix is a k X k symmetric matrix. The diagonal elements are actually the variance of X_i's i.e. np_i(1-p_i), i=1,2...,k. The off diagonal element in the ith row and jth column is Cov(X_i,X_j)=-np_ip_j, where i is not equal to j. – Stat Aug 17 '12 at 22:38

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