Probability of consecutive heads with biased coin How would I calculate the probability of rolling 3 or more heads consecutively out of total 525600 flips of a biased coin whose probablity of heads is 0.01? I am able to approximate this through simulation, but I need a faster calculation. With simulation I get about 0.4. The sequence T,H,H,H,H,H,H,T counts as a single occurence.
 A: Referring to the answers to the question "Expected number of tosses for two coins to achieve the same outcome for five consecutive flips,"
the expectation of the number of times you will need to flip a coin
with probability $p$ of heads before you see $n$ consecutive heads
for the first time is
$$
E[X_{p,n}] = \frac{(1/p)^n - 1}{1 - p}.
$$
For $p = 0.01$ and $n = 3$ this comes out to $1\,010\,100$.
On average, I'd expect to see three consecutive head about once in
every $1\,010\,100$ flips.
(I say I expect this "about" once because you might consider a
single head flipped after the first occurrence of three consecutive heads
to be another occurrence of three consecutive heads.
Since there is only a $0.01$ chance to continue the streak of heads,
however, this will have only a small effect on the answer.)
Since you have only $525\,600$ flips,
we might expect $525\,600/1\,010\,100 \approx 0.520$ occurrences of 
three consecutive heads.
But if $Y$ is the number of times we get three consecutive heads,
then $0.520 \approx P(Y=1) + 2P(Y=2) + 3P(Y=3)+\cdots$,
whereas we actually want $P(Y=1) + P(Y=2) + P(Y=3)+\cdots$.
A better estimate (though not exact) would be
$$
1 - \left(1 - \frac{1}{1\,010\,100}\right)^{525\,600} \approx 0.405683,
$$
which seems to agree with your simulation.
To get this, I made the simplifying assumption that each flip
(even the first one!) has a $1/1\,010\,100$ chance to complete a
run of three consecutive heads.
(Treating the first two flips the same as the others has very little effect
on the final result.)
A: The usual one-step Markov decomposition yields that the generating function of the first time $T$ when three consecutive heads appear is $$E(s^T)=x^3s^3f(s)^{-1}$$ with $x=.01$ the probability of heads and $$f(s)=1-(1-x)s-x(1-x)s^2-x^2(1-x)s^3$$ In the limit $x\to0$, one sees that $f(1)\sim x^3$ and $f'(1)\sim-1$ hence the root $s^*$ of smallest modulus of $f$ is such that $s^*-1\sim x^3$ and $f(s)\approx s^*-s$ hence $$E_0(s^T)\approx x^3\sum_{n\geqslant0}s^{n+3}(1+x^3)^{-n-1}$$ which yields $$P_0(T⩾n)≈x^3\sum_{k⩾n−3}(1+x^3)^{-k-1}=(1+x^3)^{-n+3}$$
Thus, the probability to observe at least once three consecutive heads before time $t=525600$ is approximately
$$1−(1+x^3)^{−t}=1−1.000001^{−525600}\approx.408799$$
(Using the true root of $f$ yields the approximate value $.405684$ instead.) 
