TITLE: Fair coin tosses: Probability of $\geq 4$ consecutive heads
QUESTION [5 upvotes]: I know that there are some related questions, but they seem to be overkill for this small exercise. 
I have 10 (fair) coin tosses and am interested in the probability that I have at least 4 consecutive heads.  
So I had a lot of different ideas, but I think many of them do not work too well. 
1) Markov chain: But then, somehow, we need to keep track of the number of coins we have already tossed, so one "loses" the Markov property. 
2) count all possibilities: we have $2^{10}$ possibilities in total and can subtract the ones that have at most 3 consecutive heads. But seems to be nasty. 
3) recursive equation? Let $p_{i,j}$ for $i \leq j$ be the probability that we have at least $j$ consecutive heads. But also this seems to be not that easy.. 
So this question is asked in a book in the first chapter, so it shouldn't be too hard?

REPLY [2 votes]: Another answer is you could take the formula for getting $5$+ consecutive out of $10$ random draws and then just add in the case $4$s ($139$ of those).
The formula for $5$+ consecutive out of 10 random draws is $p^5(6-5p)$ which for p=$0.5$ gives us $7/64$ which is $112/1024$.  
Another answer is you could take an "almost correct" formula for getting $4$+ consecutive out of $10$ random draws but you'd have to subtract out the $5$ occurrences where they are either actually case $5$s or "collide" like in my other enumerated answer to this same question.  
The "almost" formula for $4$+ consecutive out of 10 random draws is $p^4(7-6p)$ which for p=$0.5$ gives us $4/16$ which is $256/1024$ then subtract out the $5$ we double counted with this formula (see my enumerated answer elsewhere) and we get the correct $251$ (out of $1024$).