I want to clarify the following problem:
I have the possibility of 8 coinflips. What is the probability that the number of tails and heads differs by three or more?
I have the following solution in mind:
$$\begin{align}\left(\binom8 8 \cdot (0.5)^8 \cdot (0.5)^0\\ + \binom8 7\cdot(0.5)^7 \cdot (0.5)^1 \\+ \binom8 6\cdot(0.5)^6 \cdot (0.5)^2 \\+ \binom8 5\cdot(0.5)^5 \cdot (0.5)^3\right)\cdot2\end{align}$$
Is this approach correct? I multiply by 2 at the end, due to the fact there are 2 options each time. Is this also correct?
This solution is incorrect because of the $\binom8 5\cdot(0.5)^5\cdot(0.5)^3$ term. This term seems to account for the case where there are $5$ heads and $3$ tails, or $5$ tails and $3$ heads, but $5-3=2<3$, so this term is extraneous.
The rest of your reasoning is correct, so removing that term will give the correct probability of: $$\begin{align}\left(\binom8 8 \cdot (0.5)^8 \cdot (0.5)^0\\ + \binom8 7\cdot(0.5)^7 \cdot (0.5)^1 \\+ \binom8 6\cdot(0.5)^6 \cdot (0.5)^2\right)\cdot2\end{align}$$ which is equal to $0.2890625$.