Probability of getting $5$ heads on $10$ (fair) coin flips? Even before attempting the problem, I immediately defaulted to an answer: $\frac{1}{2}$.
I thought that this was a possible answer since the probability of flipping a head on one flip is definitely $\frac{1}{2}$.
I then worked through the problem:
Let E be the event in the problem statement.
The total number of possible outcomes of $10$ coin flips: $N = 2^{10}$
The total number of ways in which E can result:for $n=10,r=5, nCr= 252$
$P(E) =\frac{n}{N} = .246$
How do I correct my intuition?

Let's imagine a case with $4$ rolls where we desire $2$ heads (MathJax diagrams only go so far...)
$$\newcommand{\mychoose}[2]{\bigl({{#1}\atop#2}\bigr)}$$
$$\begin{array}{ccccccccccc}  & & &   &  &  &  &  H & & & & & \\
& & &   &  &   & \swarrow &   & \searrow &  \\
& &   &   &   & H &   &   &   & T &  \\
& & &   &  \swarrow & \downarrow &   &   &   & \downarrow & \searrow \\ 
& & & H  &   & T &   &   &   & H &  & T \\
& & \swarrow  & \downarrow  &  & \downarrow  & \searrow  &   & \swarrow & \downarrow &  & \downarrow & \searrow \\
 & H  &   & T &  & H  &  \color{red}{T} &  & H & \color{red}{T} &  & \color{red}{H} & & T\end{array}$$
  $$ $$
$$\begin{array}{ccccccccccc}  & & &   &  &  &  &  T & & & & & \\
& & &   &  &   & \swarrow &   & \searrow &  \\
& &   &   &   & H &   &   &   & T &  \\
& & &   &  \swarrow & \downarrow &   &   &   & \downarrow & \searrow \\ 
& & & H  &   & T &   &   &   & H &  & T \\
& & \swarrow  & \downarrow  &  & \downarrow  & \searrow  &   & \swarrow & \downarrow &  & \downarrow & \searrow \\
 & H  &   & \color{red}{T} &  & \color{red}{H}  &  T &  & \color{red}{H} & T &  & H & & T\end{array}$$
Thus, we have a $\frac{6}{16} = \frac{3}{8}$ chance of getting two heads from four rolls by analyzing our diagram; this can also be found by taking $\frac{\mychoose{2}{4}}{2^4} = \frac{6}{16} = \frac{3}{8}$. As an addendum, note that your intuition calculation gives the chance that you get either $2$ or $1$ head from $4$ rolls ( $= \frac{8}{16} = \frac{1}{2}$ which can be easily identified from the diagram). Now just apply the logic from this example to your $10$ roll case. Also, pushing MathJax diagrams this far is a pain. You have been warned.
