Doubt while constructing sample spaces While making sample spaces for studying probability, why do we count each arrangement differently?
For example, let's say I want to make a sample space of the number of possible outcomes if three coins are tossed.
We count HTH and HHT and THH  separately though all of them represent the same thing. Does this make any sense? What's the need to count the same thing again and again since each one of them contains two heads and one tail.
If we don't count them differently, will that change the answer?
I'm asking this question in general not about this particular case only.
Thanks in advance!
 A: One possible reason for counting them separately is that it makes computing certain probabilities easier. Suppose you wanted to compute $P(\text{two heads out of three flips})$. You could treat HHT, HTH, and THH as merely variations of a single result, but it turns out that that makes computing that probability harder.
If we simply count the number of heads in the result, for instance, there are four possible results:

*

*Zero heads out of three flips

*One heads out of three flips

*Two heads out of three flips

*Three heads out of three flips

The problem, however, is that these are not equally probable results. We can't conclude, as might be tempting, that $P(\text{two heads out of three flips}) = 1/4$.
On the other hand, if you count all the possible sequences of flips, those are equally probable. Ultimately, this is because the three flips are independent, so that whether the first flip comes up heads or tails has no impact on whether the second flip does or whether the third flip does. Each subsequent flip just cuts the various results' probabilities into two equal pieces, so to speak.
At any rate, we can now compute the desired probability by simple counting: There are eight possible sequences—HHH, HHT, HTH, THH, HTT, THT, TTH, TTT—each of which have probability $1/8$. Of these, three—HHT, HTH, and THH—satisfy the condition of two heads, so
$$
P(\text{two heads out of three flips}) = \frac38
$$
A: Practical (not theoretical) probability connects mathematics with the real world.  The consequence of this is that there are always going to be some questions that cannot be answered on the basis of mathematics alone, but have to be answered by looking at what happens in the real world.
You can regard HHT, HTH, THH as all being the same if you like, but then the next question is, what is the probability of this (single) outcome?  If you say, there are $4$ possibilities (no heads, $1$ head, $2$ heads, $3$ heads) and so the probability is $1/4$ each, then this is a statement about the real world and you should check it by doing a real world experiment.  For example, you could toss your three coins $80$ times, and you would expect to get each outcome $20$ times, or something close to that.  Try it!
Another idea to think about: in some situations, it is definitely not reasonable to say that HHT, HTH, THH are all the same.  For example, I am playing a game where I toss three coins.  Heads on the first toss wins me \$10, heads on the second toss wins me \$100, heads on the third toss wins me \$1000000.
