# Multinomial Distributions

I'm a little bit confused about what the meaning of multinomial distributions, at least from what i've gleaned from the wikipedia page on Multinomial Distribution. In essence, a multinomial distribution is the generalized form of a binomial distribution. That is, outcomes are independent, however there are k possible outcomes, each with k success, which gives the probability mass function:

$Mult(n, p_1, p_2, p_3, ..., p_n) = {n \choose x_1, x_2, ... x_n} p_1^{x_1} ...p_n^{x_n}$

However, shouldn't evaluating only $X_1$ give a binomial distribution since the multinomial is just a generalization of it? But throwing that in, we get: $Mult(n, p_1) = {n \choose x_1} p_1^{x_1}$ which can't possibly turn into $Bin(n, p) = {n \choose x} p^x(1-p)^{n-x}$

Am I missing something in my understanding of a multinomial distribution? And if so, how do we get the binomial from the multinomial equation then?

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Your formula for the multinomial distribution is a little off: there should be $k$ different $p$s and $x$s with $\sum_{i=1}^k x_i=n$, so that the key coefficient is ${n}\choose{x_1,x_2,\ldots,x_k}$. (You can confirm this with Wikipedia.)
To get the binomial distribution, take $k=2$, $x_1=x$, $p_1=p$ and $p_2=1-p$. Then you have ${n}\choose{x_1,x_2}$$p_1^{x_1}p_2^{x_2}=$$n\choose x_1$$p^{x_1}(1-p)^{n-x_1}$, as desired.
oh, i think i see what i was missing now... since we only have $p_1$ and $p_2$, $p_2 = 1 - p_1$ by definition. Thanks! – Jaynathan Leung Nov 13 '12 at 2:50