# 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