I am asked to find the probability of getting HHHTHHTT or (111010011) combination in infinite coin tosses sequence where probability of getting heads is $p$.
So I thought that if it starts with the first toss it would be simply $p^3(p-1)p^2(p-1)^2=p^5(p-1)^3$. But it does not need to start with the first toss, so what then? How to take this into consideration?
Best Answer
Look at the string as concatenation of strings of length $8$ and call these strings the special strings.
The probability that the first special string is not HHHTHHTT is a number $r<1$.
Then the probability that none of the first $n$ special strings is HHHTHHTT is $r^n$.
If $s$ denote the probability that none of the special strings is HHHTHHTT then $s\leq r^n$ for every $n$.
This for $n=1,2,3,\dots$ and $r<1$ so this can only be true if $s=0$.
That means that the probability that at least one of the special strings is HHHTHHTT is $1$.
Then of course the same is true if we look at all strings of length $8$ (not only the special ones).
It can also be proved for every integer $n$ that the probability that not more than $n$ strings of length $8$ will be HHHTHHTT is $0$, and from this it follows that the probability that an infinite number of strings of length $8$ will be HHHTHHTT equals $1$.