6 fair coin flips: probability of exactly 3 heads 
When a certain coin is flipped, the probability of heads is $0.5$. If the coin is flipped $6$ times, what is the probability that there are exactly $3$ heads?

The answer is $\frac5{16}$. I wonder why it isn't $\frac12$. Since a fair coin flip results in equally likely outcomes, any sequence is equally likely…
I know why it is $\frac5{16}$. We divide the number of possible outcomes with exactly 3 heads by the total possible outcomes.
What bothers me is how I should think about it so I won't make a mistake anymore. Why is $\frac12$ not right? I need to get intuition.
 A: It sounds like you already have the intuition since you understand that the answer is obtained by dividing the number of outcomes with exactly 3 heads by the total number of outcomes.  From here it's a matter of understanding how to calculate these two things.
The total number of outcomes is simply $2^6 = 64$ since we're tossing a coin 6 times and each toss has only two possible outcomes.
The number of outcomes with exactly 3 heads is given by ${6 \choose 3}$ because we essentially want to know how many different ways we can take exactly 3 things from a total of 6 things.  The value of this is 20.
So the answer is $20/64 = 5/16$.
The error you made is thinking that "number of outcomes with exactly 3 heads" is equal to "half of the total number of outcomes of 6 tosses."  If this were the case then logically, "exactly 3 tails" must also be exactly half of the total outcomes.  This means that "exactly 3 heads or exactly 3 tails" must describe all possible outcomes (because each scenario joined by the "or" would have probability $1/2$) but this is clearly not the case since we can have, e.g., 1 heads and 5 tails, etc.
To put it another way... You said any sequence is equally likely.  That is correct.  But sequences containing exactly 3 heads do not make up half of the total number of sequences.  Therefore it does not follow that the probability is $1/2$.  The easiest way to see this clearly is to list every possible outcome.  But for 64 outcomes it can be tedious, so let's do it with a simpler and similar problem.
Say we want to know the probability of getting exactly 2 heads if we flip a coin 4 times.  Unless I'm misunderstanding your misunderstanding, your earlier thinking would lead you to believe the answer is $1/2$.   But if we list all possible outcomes:
HHHH
HHHT
HHTH
HHTT
HTHH
HTHT
HTTH
HTTT
THHH
THHT
THTH
THTT
TTHH
TTHT
TTTH
TTTT
We see that only 6 of them have exactly 2 heads.  $6/16 = 3/8$.  And if we do this problem the way I answered the original one, then the total number of outcomes is $2^4 = 16$ and the total number of outcomes with exactly 2 heads is ${4 \choose 2} = 6$. So we again get $6/16 = 3/8$.
A: Your incorrect thinking goes along the lines of this classic fallacy:

What's the probability of this happening? $\frac12$, either it will happen or it won't.

It is true that each sequence of heads and tails is equally likely to occur – with probability $\frac1{64}$, in this case. However, the number of those sequences having exactly three heads is not 32, but $\binom63=20$, which leads to the correct answer of $\frac5{16}$. They are two completely different things.
A: "exactly three heads" is not half of the outcomes. Try writing the outcomes down and counting them!
The probability is 
$$P=\binom{6}{3}(\frac{1}{2})^3(\frac{1}{2})^{6-3}$$
A: 
Since a fair coin flip results in equally likely outcomes, any sequence is equally likely…

... then we need to know how many of these equally-probable outcomes are possible and how many of which have "exactly three heads".
There are $2^6$ ways to make six independent "head or tails" decisions. That is $64$.
There are $\binom 6 3$ ways to arrange three heads and three tails. That is $20$
Divide and calculate.  $20/64 = 5/16$
