Given your prefatory comment, I'm going to avoid talking about the normal curve and the associated variables and use as much straight probability as possible.
Let's do a side problem first. If on a A-D multiple choice test you guess randomly, what's the probability you get 8 out of 10 questions right?
Each problem you have a 25% (.25) chance of getting right and a 75% (.75) chance of getting wrong.
You want to first choose which eight problems you get right. That can be done in 10 choose 8 ways.
You want .25 to happen eight times [$(.25)^8$] and .75 to happen twice [$(.75)^2$]. This needs to be multiplied by the possible number of ways to arrange the eight correct problems, hence your odds of getting 8 out of 10 right is
${10 \choose{8}}(.25)^8(.75)^2$
Ok, so let's say you throw a coin 3000 times. What's the probability that it comes up heads only 300 times? By the same logic as the above problem that would be
${3000 \choose{300}}(.5)^{300}(.5)^{2700}$
or a rather unlikely 6.92379... x 10^-482.
Given throwing the coin n times, the probability it comes up heads x times is
${n \choose{x}}(.5)^n$
or if you want to ask the probability it comes up heads x times or less
$\sum_{i=0}^{x}{{n \choose{i}}(.5)^n}$
so all you have to do is decide now how unlikely are you willing to accept?
(This was a Binomial Probability if you want to read more and all the fancier methods involving an integral under the normal curve and whatnot start with this concept.)
Use the binomial distribution to get the probability of getting $k$ heads from $n$ flips:
$$p(n,k) = \binom{n}{k} \left ( \frac12 \right )^k \left ( \frac12 \right )^{n-k} = \binom{n}{k} \left ( \frac12 \right )^n$$
The probability you seek is $p(8,5)+p(8,6)+p(8,7)+p(8,8)$, or
$$\frac{\binom{8}{5}+\binom{8}{6}+\binom{8}{7}+\binom{8}{8}}{2^8} = \frac{56+28+8+1}{2^8} = \frac{93}{256}$$
Best Answer
Let $C_n$ be the $n$-th Catalan number, defined by $\frac{1}{n+1}\binom{2n}{n}$. Let's also say that you keep tossing even after you had more heads than tails, for good measure. Then there are $2^{2n + 1}$ different sequences of tosses you could've made. How many of them will at some point have more heads than tails?
We will divide this by cases in the following way: Any sequence that does at some point have more heads than tails are grouped together according to the number of tosses it took to get more heads for the first time. That means, for instance, that half of all the sequences (the ones starting with head) are in group $0$. The ones that took $3$ coin tosses are in group $1$ and so on. The ones that made it to the more-heads-than-tails-side only on the last coin flip are in group $n$.
How many sequences are in each group? Say we have the group labelled $i$. That means that for any sequence in this group, after $2i$ throws they have exactly as many heads as tails, and at no point before have they had more heads. There are $C_i\cdot 2^{2n-2i}$ tossing sequences like that. $C_i$ because that's how many ways you can reach the spot, and $2^{2n-2i}$ because that's how many ways it can go on after this, provided that the $(2i + 1)$th throw is a heads (which it is; that was the definition of group $i$).
So we're left with calculating the following sum: $$ \sum_{i = 0}^n 2^{2n-2i}C_i = \sum_{i = 0}^n \frac{2^{2n-2i}}{i+1}\binom{2i}{i} $$ and then divide it by $2^{2n + 1}$, and you have your probability.
PS. I don't know how to calculate the sum, but WolframAlpha says it's equal to $$ 2^{2 n+1}-\frac{1}{2} \binom{2(n+1)}{n+1} $$ which suggests there are $\frac{1}{2}\binom{2(n+1)}{n+1}$ sequences of coin tosses that are $2n + 1$ long and that never have more heads than tails in any initial segment.