Say we flip a biased coin such that the probability of getting the same outcome in a row (head-head or tail-tail) is $p$.
What is the probability of getting three or more tails consecutively out of $n$ flips (and alternatively out of infinite number of flips).
For example: TTT-H-TTT-HH-TTTT…
I am looking for:
- The fraction of tails that are in sections of size three or more.
- The expected size of sections with tails.
Best Answer
I’ll consider the case of an infinite string of coin tosses, as it’s easier.
Start with anon’s suggestion. Let $q$ be the probability that the coin comes up tails. Then the probability of getting two tails in a row must be $q^2$, and the probability of getting two heads in a row must be $(1-q)^2$, so you know that $p=q^2+(1-q)^2=2q^2-2q+1$, and therefore $$q=\frac14\left(2\pm\sqrt{4-8(1-p)}\right)=\frac12\left(1\pm\sqrt{2p-1}\right)\;.$$ One of the solutions is $q$; the other is $1-q$, the probability that the coin comes up heads. There’s no way to decide which is which without further information.
At any stage in the process the probability of a run of $n$ tails is $q^n(1-q)$, so the expected length of a run of tails is $\sum_{n\ge 0}nq^n(1-q)$. However, this includes runs of length $0$, which you don’t want to count, so an adjustment is needed. The probability of a run of length $0$ is $1-q\,$, so the probability of a run of positive length is $q$, and for $n>0$ the probability of a run of length $n$ given that the run has positive length is therefore $$\frac{q^n(1-q)}q=q^{n-1}(1-q)\;.$$ Thus, the expected length of a non-zero run of tails is $$\sum_{n\ge 1}nq^{n-1}(1-q)=(1-q)\sum_{n\ge 1}nq^{n-1}=\frac{1-q}{(1-q)^2}=\frac1{1-q}\;.$$
For $k=1,\dots,n$ the probability that a given tail is the $k$-th of a run of $n$ tails is $q^{n-1}(1-q)^2$ (since we may ignore edge the effects at the beginning of the infinite string of tosses), so the probability that it belongs to a run of $n$ tails is $nq^{n-1}(1-q)^2$, and the probability that it belongs to a run of at least three tosses is $$\begin{align*}\sum_{n\ge 3}nq^{n-1}(1-q)^2&=(1-q)^2\sum_{n\ge 3}nq^{n-1}\\ &=(1-q)^2\left(\frac1{(1-q)^2}-(1+2q)\right)\\ &=q^2(3-2q)\;; \end{align*}$$
this is the fraction of tails belonging to runs of three or more.