Counting outcomes of flipping coins

I know this is an extremely basic question, but I have a slight misunderstanding (?) regarding this question:

How many possible outcomes are there when flipping two coins?

At first glance, this is a really easy problem: 2 x 2 = 4!

But if I list all the possible outcomes:

{H, H}
{H, T}
{T, H}
{T, T}


I noticed that there are really three "distinct" outcomes:

{H, H}
{H, T}
{T, T}


Is it wrong to consider {H, T} and {T, H} the same? Why or why not?

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The answer to this question might be of interest: math.stackexchange.com/questions/19850/… – Elliott Jun 3 '11 at 7:56

It depends on how you decide to count them. You could say either. But, if your three events are two heads, one heads and one tails, or two tails - they do not have the same probability. But if your events are HH, HT, TH, or TT - they all have the same probability.

Ultimately, one can define one's events and sample space however one wants. But we usually design it with some sort of problem in mind.

Does that make sense?

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And for $n$ coins, these two different sample spaces have size $n+1$ and $2^n$ respectively. Here $n=2$. – Did Jun 3 '11 at 5:44

You can understand it by assuming that 1 coin is red and the other is black. In this case there are exactly 4 outcomes when you flip these 2 coins, that is HH, HT, TH or TT.

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I think you missed the point of the question. If you will denounce considering them the same, give a reason. Renaming them changes mostly semantics. Why can't they both be black? – mixedmath Jun 3 '11 at 9:20

that's because it is the first time or first flip of the first coin and the next flip of the second coin so the both cannot be the same as you suggested T,H and H,T. They are not the same.

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