The events "the number of heads is $0$," "the number of heads is $1$," "the number of heads is $2$." and so on, are very far from being equally likely.
For a fair coin, the probability of $k$ heads in $2n$ tosses is $\binom{2n}{k}\cdot \frac{1}{2^{2n}}$. This is because any particular sequence of heads and/or tails has probability $\frac{1}{2^n}$, and there are $\binom{2n}{k}$ sequences of length $2n$ that have $k$ heads and the rest tails.
As an extreme example, the probability of $0$ heads in $20$ tosses is $\frac{1}{2^{20}}$, because there is only $1$ pattern of tosses that yields $0$ heads. The probability of $10$ heads is $\binom{20}{10}\frac{1}{2^{20}}$, very much larger, since $\binom{20}{10}$ is large.
Computation will show that $8$ heads and $12$ tails is a fair bit less likely than $10$ and $10$.
The binomial coefficients $\binom{2n}{k}$ climb from $k=0$ to $k=n$, and then decrease symmetrically.
I don't know your level in Mathematics so I will write it as clearly as possible. There might be some English errors, feel free to edit the post.
Suppose that your current score is $m$ and you still have to flip $n$ coins. What is the probability of getting a final score higher than $k$ ?
Mathematical definition of the score
It is crucial to determine what is the mathematical deifnition of the score here. After flipping a coin, either you get $-1$ or $1$ with probability $p=0.5$ (assume more generally that $p\in[0,1]$.
Noticing that leads to model a point as a Bernoulli variable of parameter $p$. The score is then a sum of Bernoulli variables with parameter $p$. In the following, we assume that each flip is done independently of the others.
Simplifying the problem
By an independence argument, it is easy to see that your problem is equivalent to:
Assuming the current score to be 0, what is the probability that after flipping $n$ coins, the score will be higher than $k-m$
WLOG, let denote $h:=k-m$.
Case $h>n$. It is not difficult to see that the probability, starting from 0, to be higher than $n$ after flipping $n$ coins is 0.
Case $h<-n$. The same reasonning leads to same conclusions.
Case $-n\leq k \leq n$. This case is not trivial as the others. As we said before, the score is a sum a i.i.d. Bernoulli variables of parameter $p$. Let notice the usefull theorem below.
Theorem Let $X_1,...X_i$ be a sequence of i.i.d variables following a Bernoulli distribution of parameter $p\in [0,1]$. The random variable defined as $S_i=X_1+...+X_j$ follows a Binomial distribution of parameters $p$ and $i$.
At this stage it is almost done since (you are lucky) binomial distribution has been studied for long time ago. A closed formula exists and is given by:
Proposition Let $S_i$ be a Binomial variable of parameter $p$ and $i$. For any $j$ such that $-i\leq j \leq i$, we have:
$$\mathbb{P}(S_i \geq j) = \sum_{\ell = j}^i \binom{i}{\ell}p^\ell (1-p)^{i-\ell}.$$
Thanks to this proposition, starting from $0$, the probability that we get a score higher than $k$ (with $-n\leq k \leq n$) is:
$$\sum_{\ell = k}^n\binom{n}{\ell}p^\ell(1-p)^{n-\ell}.$$
Solution to your problem
By shifting the score again, the probability of getting a score higher than $k$ after flipping $n$ coins and starting from a score of $m$ is:
$$
\begin{cases}
\displaystyle 0 \text{ if } k-m>n \text{ or } k-m<-n,\\
\displaystyle \sum_{\ell = k-m}^n \binom{n}{\ell}p^\ell(1-p)^{n-\ell} \text{ if } -n\leq k-m\leq n.
\end{cases}
$$
Conclusion
The key takeway are:
- Model properly the situation your are dealing with
- Try to link it with known objects (distributions, ...)
- Try to seek for the desired properties (this forum, books, research articles, etc...)
- Try simplify when possible (here, you shift the score to have a simpler but equivalent problem).
Hope it helps!
[Edit]
Other scores
In this section, I briefly introduce situations where the score not $1$ or $-1$.
In the initial set up, you get either $+1$ or $-1$ at each flip. You can assume that at each step, you get a score that is "randomly chosen" on $\mathbb{R}$. By "randomly chosen", I mean that a each step, you get a score that follow a specific distribution (gaussian, exponential, etc...). We still assume the i.i.d. property.
1. Gaussian score
You can imagine that at each step, you will get a score that follow a normal distribution, denoted $\mathcal{N}(0,1)$. the score is no longer an integer but a real number.
Theorem If $S_n:=X_1,...X_n$ are $n$ i.i.d. gaussian variables, the variable $S_n$ follows the distribution $\mathcal{N}(0,n)$.
This theorem is interesting because in this case, the probability for the score to be higher than a certain level can be expressed with the error function $\text{erf}$.
To conclude the gaussian case, if the current score is at level $a\in\mathbb{R}$, the probability of getting a score higher than $b\in\mathbb{R}$ after flipping $n$ coins is:
$$\frac{1-\text{erf}\left(\frac{b-a}{\sqrt{n}}\right)}{2}.$$
This can be proved by a bit of algebras.
2. Exponential case
You can also assume that the score you get follows an exponential distribution. At each step, you get a positive number.
Theorem If $S_n:=X_1,...X_n$ are $n$ i.i.d. exponential distribution of parameter $\lambda>0$, the variable $S_n$ follows a gamma distribution $\Gamma(n,\lambda)$.
The case is also interesting since the cumulative distribution of this distribution is expansively studied in the literrature.
The probability you are looking for is then:
$$1-\frac{\gamma(n, \lambda(b-a))}{\Gamma(n)}$$
where $\gamma$ is the incomplete gamma function.
Simple non-i.i.d. situation
You can also imagine a situation where the flips are not independent. To this aim, you can model a sequence of $n$ flips as a sequence of $n$ gaussian variable. To describe the correlation, you need to know the correlation matrix, usualy denoted $\Sigma$.
The distribution followed by the by the sum of gaussian variables (here the score) follows a multivariate normal distribution. You can then apply some well known properties on this disbtribution to find the desied propbability.
I know that it is quite brief. Detailing all of these situations would require a lot of properties. It is more an overview of what you can do without knwowing hard probabilities.
Best Answer
If you flip $n$ coins, the number of Tails follows a binomial distribution that you can approach with a normal distribution of parameters $\mu=\frac n2$ and $\sigma=\frac{\sqrt{n}}{2}$.
With this approximation, $95\%$ of the time the amount of Tails will fall within $1.96\sigma$ of $\mu$, so in the interval $[\frac {n-1.96\sqrt n}{2},\frac {n+1.96\sqrt n}{2}]$ and in these cases the ratio T/H will be between $\frac {n-1.96\sqrt n}{n+1.96\sqrt n}$ and $\frac {n+1.96\sqrt n}{n-1.96\sqrt n}$.
You want therefore to solve $$\frac {n-1.96\sqrt n}{n+1.96\sqrt n}\geq\frac{45}{55}$$ which is equivalent to $$10n\geq 196\sqrt n$$ and $$\sqrt n \geq 19.6$$ or finally $$n\geq 385$$
If you flip a fair coin at least $385$ times, you are $95\%$ certain to end up with a ratio $T/H$ between $\frac{45}{55}$ and $\frac{55}{45}$.