PDF of 'blind' multinomial distribution

Question

I want to describe the distribution where different coloured balls are drawn from a bag with replacement (so far, I know this is the multinomial). However, the observer only knows that he got different coloured balls, he can't say which colour they were specifically.

So if there were red, blue, and green balls in the bag and he reports a sampling of (3, 2, 1) then he could have sampled (3 reds, 2 blue, 1 green) or (3 blue, 2 red, 1 green) or (3 green, 2 blue, 1 red) or ...

I need to calculate the probability of observing a specific numerical vector from this distribution, given the composition of balls in the bag. At the moment I'm treating the samples as unordered, so P(X=(1,2,3) == P(X=(3,2,1)), and summing the individual multinomial probabilities over the permutations of X. Is this correct?

I feel like this distribution (and the without replacement sampling version of it) must be described somewhere in the literature but I don't know where to look. I'm also curious about the 'ordered' version, where reporting X = (2,3,1) implies that the colour sampled 2 times was the first colour found in the sampling.

Answer 1

With three distinct colors there would be $3!=6$ possible permutations, so indeed you would have to sum the probabilities over this 6 possible identifications ... or so it would seem. But in reality it could be more complex, and you didn't specify enough information. If this is really a multinomial setup, as your tag multinomial indicates, you would have to know the composition of the urn. Lets say the composition of the urn is $R=3$ red balls, $B=6$ blue balls, $G=5$ green balls, then the multinomial probabilities would be unequal, with largest probability $=6/14$ for blue. Then after observing that $X=(2,3,1)$ (after six draws with replacement from the urn), you would have to calculate the posterior probability, using Bayes, for each possible color specification. Those six posterior probabilities would have to be used as weights when summing the probabilities.

So, in short, this is more complicated than you envisioned. I will not attempt those calculations here, and do not know of references. In the case of sampling without replacement, of course, some color identifications might be impossible after the fact.