# Odds of probabilistic outcome 1 happening at least x more times than outcome #2?

Let's take a coin flip as an example.

Let's assume we perform 10 flips.

I need to determine a formula to calculate the odds that somewhere within the 10 flips, there will be a point where heads has occurred at least 2 times more than tails.

So the final outcome could be 8 tails and 2 heads, but if the 2 heads came up in the first two events, then that outcome would still meet the criteria.

The standard binomial theorem wont work for my purposes, because if I set it to calculate the odds that I get at least 2 more heads results than tails (say 6 out of 10 heads at least), then it wont count the examples I specified above.

Mapping it out manually wont work for my purposes.

Can any clever person think of a solution?

That is, consider the random walk provided by the running variable $X(n) = \#H(n)-\#T(n)$. Here $H(n)$, for example, denotes the number of Heads you have seen after $n$ trials.
If your path is "good", i.e., if your path satisfies $X(n)≥2$ for some $n$, then consider the first time the path touches the line $X=2$. After that point, every path that ends above the line has a reflection that ends below it. Of course every path that ends with $X(10)=2$ is clearly good. Thus the answer is $$P(\max X(n)≥2)=2\times P(X(10)>2) + P(X(10)=2)$$
Direct computation then shows that this is $$2\times 0.171875+0.205078=\boxed {.548828125}$$