Probability of $m$ failed trades in series of $n$ trades This is a trading problem:

Let's say I have an automated trading system with a probability of
  success of $70\%$ on any individual trade. I run $100$ trades a year.
  What is the probability of getting $5$ or more consecutive failed
  trades?
More generally, for a probability of success $p$ on the individual
  trade, and a total of $n$ trades per annum, what is the probability of
  a series of $m$ or more consecutive failures, where $m \leq n$ and $0 \leq p \leq 1$?

I realize this can be converted into a problem about getting a run of heads with a biased coin, but I looked around and could not find a response matching my exact needs. By the way, this is a real-world problem, not schoolwork.

Let your trades be $t_1\cdots t_n$, with $t_i\in \{0,1\}$. What is the probability that this sequence contains $k$ consequtives zeroes, if the probability of zero is $p$? Well, the sequence could start at $t_1$, or $t_2,\ldots, t_{n-k+1}$. The probability that a sequence of $k$ zeroes starts at $t_i$ is $p^k$, and there are $n-k+1$ points where the sequence might start, so the first instinct is to add up these probabilities as $(n-k+1)p^k$. But that wouldn't be correct, because you haven't accounted for the fact that these sequences might overlap: you can only add probabilities if they represent disjoint events.
If you were looking for two consequtive fails, then that probability is
$$ P = (n-1)p^2 - (n-2)p^3 $$
The second term subtracts the times you counted two scenarios double. If you are looking for three consecutive fails, then
$$ P = (n-2)p^3 - (n-3)p^4 + (n-4)p^5 $$
You add the term (n-4)p^5 to compensate for the length-$5$ sequences you subtracted double (once too many) in the $(n-3)p^4$ term. See a Venn diagram if you want to know in more detail why exactly.
In general, the probability that $k$ tosses come up tails in $n$ trials (if $n\geq k$) is:
$$ P = \sum_{u=0}^{k-1}(-1)^{u}(n-k-u+1)p^{k+u} $$
