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In a lecture I attended today, the professor made an off-hand comment of:

"Suppose we have the set $S_n$ of permutations of $\{1, 2, ..., n\}$, which we can think of as a probability space equipped with measure $P( . )$."

I'm not sure what this means - does it mean we have a probability of picking a random permutation with some probability, or something different...?

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I assume it means that the probability of picking a given permutation is $\frac{1}{n!}$. – Qiaochu Yuan Apr 8 '11 at 15:54
up vote 3 down vote accepted

Any finite set $S$ can be equipped with a natural probability measure $P\ $ by setting, for any subset $A\subseteq S$,
$$P(A)={\mbox{number of elements in }A\over \mbox{number of elements in }S}.$$

This corresponds to selecting an item from $S$ uniformly or at random. I suspect that your professor was thinking of applying this idea to the set of permutations $S_n$.

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Yes, I think this is what he was getting at. What does P( . ) mean? Is it just another way of writing P(A), where A is any event? – Undercover Mathematician Apr 9 '11 at 7:54
@Undercover Mathematician Yes, some people use a dot as a place holder for a variable. Occasionally you see $f(\cdot)$ instead of $f(x)$, for instance. – Byron Schmuland Apr 9 '11 at 13:37

Yes, that's a consequence. The notion of measure-based probability is due to the fact that, e.g., if you were to take a probability measure on [0, 1], then you'd have nonzero probabilities associated to certain subsets of [0, 1], but not to the individual points. In a discrete case like this it's not necessary to use probability measures, and consequently it's not easy to motivate them.

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