Consider the case of an infinite (or finite $n$) string of coin tosses, and let $q$ and $1-q$ be the probabilities that the coin comes up tails and heads, respectively. (For simplicity, we can take $q=\frac12$ so that the fraction of tails and heads are the same.)
Let $p$ be the probability that, if a given toss is tails, it will be followed by tails. (If $q = \frac12$, this is the same probability as for getting heads after heads, might as well be a different probability though.)
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: $TTT-H-TTT-HH-TTTT\dots$
We can have half tails and half heads in total (when $q=\frac12$). $p$ does not affect the fraction of tails and heads in the limit, but affects how they are clustered. So when $p=1$ we have only 2 separate 'sections' such that one section is composed of only tails and the other runs of only heads. (ex: $TTTTT \dots HHHHH$). For $p=0$, it will be like $THTHTHTH\dots$
I am looking for:
- The fraction of tails that are in sections of size three or more.
- The expected size of sections with tails.
I'd be very thankful if you could help me with it!
Best Answer
To calculate the probability of getting 3 tails in a row in $n$ tosses, you could construct a Markov chain with the states $H$, $T_1$, $T_2$ and $T_3$, and the transition probabilities:
$$P(H \to H) = P(T_1 \to T_2) = P(T_2 \to T_3) = p$$ $$P(H \to T_1) = P(T_1 \to H) = P(T_2 \to H) = 1-p$$ $$P(T_3 \to T_3) = 1$$
Then calculate the probability of ending up at the absorbing state $T_3$ after $n-1$ steps, starting from state $H$ with probability $\frac12$ and from $T_1$ with probability $\frac12$ (to account for the first toss).
Edit: To answer your later questions, the length of consecutive sequences of tails is geometrically distributed with parameter $1-p$. Thus:
The probability that a randomly chosen section of tails has length $k$ is $p^{k-1}(1-p)$, and so the fraction of tails in sections of length $k$ is proportional to $kp^{k-1}(1-p)$ (see below for the actual probability). Thus, the fraction of tails in sections of $n$ or more is $$\frac{\sum_{k=n}^\infty kp^{k-1}(1-p)}{\sum_{k=1}^\infty kp^{k-1}(1-p)} = \frac{\sum_{k=n}^\infty kp^k}{\sum_{k=1}^\infty kp^k} = \frac{p^n(n-np+p)/(1-p)^2}{p/(1-p)^2} = p^{n-1}(n-np+p).$$
The expected length of a randomly chosen section of tails is $\dfrac{1}{1-p}$.
However, the fraction of tails in sections of $k$ tails is $kp^k\frac{(1-p)^2}{p}$, and so the expected length of the section a randomly chosen "tails" toss belongs to is in fact $$\sum_{k=1}^\infty k^2p^k\frac{(1-p)^2}{p} = \frac{p+1}{1-p} = \frac{2}{1-p}-1.$$