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I read a probability book that says:

Let us roll 2 symmetrical cubes. Determine the probability of an event that a different number on both cubes has fallen?

The solution start with writing of $\Omega$. $$\Omega = \{ (i,j):i,j=1,2,3,4,5,6 \}$$ My question is why Omega isn't filled with set of $\{i,j\}$ because if I have 2 the same dice why I count event (3,4) and (4,3)?

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The space is determined, in part, by whether the dice are distinguishable or not. Once you set that, the rest is easy.

If the dice are indistinguishable, then the outcomes are: $(1,1), (1,2), \ldots , (6,6)$ and of course these are not equally likely, as governed by simply binomial distribution. For instance, it is twice as likely you get $(1,2)$ than $(1,1)$.

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  • $\begingroup$ Oolala, are you saying that if the dice are visually indistinguishable, then (3,4) is same probability as (4,4) ? $\endgroup$ – Kostas Nov 23 '19 at 21:04
  • $\begingroup$ @Kostas: No... of course not. Note that not all outcomes in an outcome space need have the same probability. $\endgroup$ – David G. Stork Nov 23 '19 at 21:20
  • $\begingroup$ How are you going to determine those probabilities if not by saying that P(12) = P'(1,2) + P'(2,1) where P' is probabilities calculated with distinguishable dice? $\endgroup$ – Kostas Nov 23 '19 at 21:28
  • $\begingroup$ You can mathematically incorporate indistinguishability, generally by dividing by the multiplicity. But even if you do first consider the distinguishable case first, you needn't adhere to it once you have the final space that is relevant. $\endgroup$ – David G. Stork Nov 23 '19 at 22:01

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