How to determine if coin comes up heads more often than tails?

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.)

I am surprised that no one has mentioned Hypothesis Testing so far. Hypothesis testing lets you to decide, with a certain level of significance, whether you have sufficient evidence to reject the underlying (Null) hypothesis or you have do not sufficient evidence against the Null Hypothesis and hence you accept the Null Hypothesis.

I am explaining the Hypothesis testing below assuming that you want to determine if a coin comes up heads more often than tails. If you want to determine, if the coin is biased or unbiased, the same procedure holds good. Just that you need to do a two-sided hypothesis testing as opposed to one-sided hypothesis testing.

In this question, your Null hypothesis is $p \leq 0.5$ while your Alternate hypothesis is $p > 0.5$, where $p$ is the probability that the coin shows up a head. Say now you want to perform your hypothesis testing at $10\%$ level of significance. What you do now is to do as follows:

Let $n_H$ be the number of heads observed out of a total of $n$ tosses of the coin.

Take $p=0.5$ (the extreme case of the Null Hypothesis). Let $x \sim B(n,0.5)$.

Compute $n_H^c$ as follows.

$$P(x \geq n_H^c) = 0.1$$

$n_H^c$ gives you the critical value beyond which you have sufficient evidence to reject the Null Hypothesis at $10\%$ level of significance.

i.e. if you find $n_H \geq n_H^c$, then you have sufficient evidence to reject the Null Hypothesis at $10\%$ level of significance and conclude that the coin comes up heads more often than tails.

If you want to determine if the coin is unbiased, you need to do a two-sided hypothesis testing as follows.

Your Null hypothesis is $p = 0.5$ while your Alternate hypothesis is $p \neq 0.5$, where $p$ is the probability that the coin shows up a head. Say now you want to perform your hypothesis testing at $10\%$ level of significance. What you do now is to do as follows:

Let $n_H$ be the number of heads observed out of a total of $n$ tosses of the coin.

Let $x \sim B(n,0.5)$.

Compute $n_H^{c_1}$ and $n_H^{c_2}$ as follows.

$$P(x \leq n_H^{c_1}) + P(x \geq n_H^{c_2}) = 0.1$$

($n_H^{c_1}$ and $n_H^{c_2}$ are symmetric about $\frac{n}{2}$ i.e. $n_H^{c_1}$+$n_H^{c_2} = n$)

$n_H^{c_1}$ gives you the left critical value and $n_H^{c_2}$ gives you the right critical value.

If you find $n_H \in (n_H^{c_1},n_H^{c_2})$, then you have do not have sufficient evidence against Null Hypothesis and hence you accept the Null Hypothesis at $10\%$ level of significance. Hence, you accept that the coin is fair at $10\%$ level of significance.

This question is one of statistics (esp. statistical inference), not probability per se. Keywords: "binomial sampling"; "confidence interval for a proportion". Asking this at http://stats.stackexchange.com will get more complete answers.

The related probability fact is: if a coin has probability $p$ of coming up heads and is tossed $n$ times, then the observed number of heads will on average be $np$, but we expect the observed number to fluctuate (if the experiment with $n$ tosses is repeated many times) around the average by an amount on the order of $\sqrt{np(1-p)}$. Keywords: normal distribution, bell curve, Central Limit Theorem, binomial distribution, convergence of binomial distribution to normal (Gaussian) distribution.