# probability unordered sample

Suppose we have a collection of six numbers $\{1,2,7,8,14,20 \}$. What is the probability of drawing with replacement the unordered sample $\{2,7,7,8,14,14 \}$?

It seems that this probability would be $$\frac{1}{\binom{6+6-1}{6}}$$ which is $$\frac{1}{\binom{11}{6}}$$

Is that right?

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Not quite. Here's one way to get the correct answer:

Clearly there are $6^6$ ways to choose six numbers with replacement.

The set {2, 7, 7, 8, 14, 14} can be ordered 6!/(2!2!) ways - it's 6! ways to order them naively, then divide by 2! twice because we don't care about the ordering of the sevens or the fourteens.

So, there are 6!/(2!2!) different ways to obtain the desired set, and the probability of doing so is

$$\dfrac{\left(\dfrac{6!}{2!2!}\right)}{6^6}$$

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Also, if you tell us how exactly you got your original answer, we might be able to clear up a misconception or two of yours. – Lopsy Dec 28 '11 at 18:05
The number of ways of getting with replacement the unordered sample is $\binom{n+k-1}{k}$. So the probabability of getting any particular unordered sample is $1/\binom{n+k-1}{k}$. – barry Dec 28 '11 at 18:09
The problem with that logic is that not every unordered sample is equally likely: for example, you'd have to be very very lucky to get {7, 7, 7, 7, 7, 7}, as there's only one way to do it. There's many ways to arrive at {2, 7, 7, 8, 14, 14}, though, since the order doesn't matter, so the latter is a lot more likely. – Lopsy Dec 28 '11 at 18:14