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Suppose we have a well shuffled deck of 52 cards and we want to divide those cards evenly among four non distinct players. We're asked to calculate the probability that all of them receive exactly one ace.

Without the added constraint of four of the players being non distinct this chance would have been easily calculable by using the multinomial theorem like so:

$$n_a = \frac{4!\cdot48!}{(12!)^4},\, N=\frac{52!}{(13!)^4},\, P=\frac{n_a}{N}=\frac{4!\cdot48!\cdot(13!)^4}{52!\cdot(12!)^4}$$

where $n_a$ are the outcomes we're interested in and $N$ is the complete outcome space but with the added constraint I'm stuck. Initially, I thought that this could be solved by multiplying $n_a$ with $4!$ but this won't fly since this would result in a higher probability than 1. Could someone point me in the right direction?

Martijn
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    Since we are interested in the probability, the clause specifying the players to be unlabelled is immaterial. Your answer is correct. You could see http://math.stackexchange.com/questions/1345413/is-there-an-alternative-intuition-for-solving-the-probability-of-having-one-ace for a number of approaches. – true blue anil Oct 12 '16 at 17:21

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My professor explained to me today that if you want to apply the multinomial theorem with $n$ non-distinct objects divided over $k$ non-distinct groups it goes like this:

$$\binom{n}{n_1, n_2, \dots, n_k} \cdot \frac{1}{k!} = \frac{n!}{n_1! \cdot n_2!\cdots n_k! \cdot k!}$$

The reasoning for this is that there are $k!$ ways of ordering the different groups and since they are non-distinct, we need to divide out the duplicate outcomes we were left with without the compensation.

The reason I was left with the correct answer, was because these extra terms cancel each other out in the final $P=\frac{n_a}{N}$ formula resulting in a correct answer. This doesn't, however, include a correct reasoning for the answer.

Martijn
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