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?
 A: The answer you got - 0.246 is the probability of getting 'exactly' 5 heads. Your intuition gives the 'Expectation' E(x). When 10 coins are tossed, the Expectation is that you get 5 heads. Your intuition makes an average of all the cases while the solution takes only the cases where number of heads is 'exactly' 5.
Let me give you another problem- What is the probability that number of heads is greater than number of tails? This is the question which your intuition answered.
A: 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.
A: You can test your intuition by considering extreme versions of the original problem. If you had to toss a coin two million times, would your intution suggest that the chance is $1/2$ that you'll get exactly a million heads? No, because in many tosses, there is 'chance variation' that prevents you from getting exactly half heads and half tails. Also, if the chance is $1/2$ of getting exactly a million heads, then it follows that all other possibilities collectively must have a probability of $1/2$, which also doesn't sound right.
A: can also be expressed as a binomial distribution which simplifies to the same but is ${10\choose 5}({1\over 2})^5({1\over 2})^5$.  but this way if the probability was not $1\over 2$ could still calculate.
A: There are several methods to solve this, but I like the following one:
I assume that you meant to say exactly 5 heads. Then what does such a toss look like? For instance, it could be:
$$HHTHTTTHTH$$
where H means you tossed a head, and T a tail. So now all the possibilities you have are the combinations of this sequence, hence $\frac{10!}{5!5!} = 252$ possibilities. Now all the possibilities are $2^{10} = 1024$. So the probability is $$\frac{252}{1024} = \frac{63}{256} \simeq 0.246$$
I like that method. It's not the fastest one, but easily explained to anyone, I think.
A: To get exactly 5 heads and 5 tails, the score after 9 tosses has to be 5-4 one way or the other.  In this case, there is a $50\%$ chance you get to 5-5.  On the other hand if you aren't at 5-4 (one way or the other) after 9 flips, there is no chance of getting a 5-5 split.  So certainly the chance of 5-5 is less than $0.5$.
