I am familiar with the next reasoning for multinomial probability function: $P(X_1 = x_1 , X_2 = x_2,...,X_k = x_k)$ when $x_1 +x_2+...+x_k =n$ and $p_1,p_2,...,p_k$ are the probabilities to choose an element from kind $X_i$. So the probability for a specific sequence of $n$ elements $(e_1,e_2,...,e_n)$ such that the number of elements from type $X_i$ is $x_i$ for each $1\le i \le k$ is $p_1^{x_1}p_2^{x_2}...p_k^{x_k}$ and the number of sequences which fulfill those condition is ${n\choose x_1}{n-x_1\choose x_2}...{n-(x_1+x_2+...+x_{k-1})\choose x_k} = \frac{n!}{x_1!x_2!...x_k!}$ so the probability $P(X_1=x_1,...,X_k=x_k) = \frac{n!}{x_1!x_2!...x_k!} \cdot p_1^{x_1}p_2^{x_2}...p_k^{x_k} $.

However I fail to derive this probability using conditional probability, what I tried to do for the tri-nomial case, i.e $P(X_1 = x_1 , X_2 = x_2 , X_3 = x_3)$ given $x_1+x_2+x_3 = n$ and the respective probabilities $p_1,p_2,p_3$. So, $P(X_1=x_1 , X_2 = x_2, X_3 = x_3)=P(X_1=x_1 ,X_2 =x_2)$ since in case $P(X_1 = x_1 , X_2 =x_2)$ and total number of elements is $n$ then $X_3 = x_3$. Using condition probability:

$P(X_1 = x_1, X_2 = x_2) = P(X_1 = x_1 | X_2 = x_2) \cdot P(X_2 = x_2)$, for $P(X_2 = x_2)$ we get ${n\choose x_2}p_2^{x_2}(1-p_2)^{n-x_2}$ and $P(X_1=x_2 | X_2 = x_2) = {n-x_2\choose x_1}p_1^{x_1}(1-(p_2-p_1))^{n-x_2-x_1} = {n-x_2\choose x_1}p_1^{x_1}(p_3)^{x_3}$.

So multiplying gives: ${n\choose x_2}p_2^{x_2}(1-p_2)^{n-x_2} \cdot {n-x_2\choose x_1}p_1^{x_1}p_3^{x_3} = \frac{n!}{x_1 ! x_2 ! x_3 !}p_1^{x_1} p_2^{x_2} p_3 ^{x_3}(1-p_2)^{n-x_2}$.

Well it is definitely wrong, but I don't see what am I doing wrong, I think I missed calculated the conditional probailty $P(X_1 = x_1 | X_2 = x_2)$ yet I don't see what is wrong.


Among the $n-x_2$ entries which are not $X_2$-kind, the probability that a particular one is $X_1$-kind is ${p_1 \over p_1 + p_3} = {p_1 \over 1 - p_2}$. You are conditioning on that entry not being $X_2$-kind. So

$$P(X_1 = x_1 \mid X_2 = x_2) = {n-x_2 \choose x_1} ({p_1 \over 1-p_2})^{x_1} ({p_3 \over 1-p_2})^{x_3} = ({1\over 1-p_2})^{n-x_2} {n-x_2 \choose x_1} p_1^{x_1} p_3^{x_3}$$

and now when you multiply everything together you get consistent results.

Hint: next time try a very simple example. E.g. if you tried $p_1=p_2=p_3 =1/3$ and $n=2, x_2=1$, you could tell that $P(X_1=1, X_3=0 \mid X_2=1)$ should have been $1/2$ by symmetry, and therefore your original formula giving $1/3$ is wrong.


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