Calculate the probability to have $5$ consecutive $H$ in $200$ coin tosses. Also calculate the probability to have $3$ separate series of $5$ consecutive Hs in $200$ coin tosses.
I read somewhere that the solution requires Markov chains but I don't have any clue on the topic; I only know the basics on probabilities.
Is it $1 - (1 - (1/2)^5)^{200 + 1 - 5}$ ??
I would appreciate if you can help me.
 A: $\color{blue}{HINT:}$ Lets try to solve the main question using recurrence relations , you can receive detailed information about recurrence relations in the following link :
https://en.wikipedia.org/wiki/Recurrence_relation.
Lets say that the last flipping is $T$  and contain $5$ consecutive Heads ,so the number of this case is $a_{n-1}$,
or
Lets say that the last two flippings are $TH$  and contain $5$ consecutive Heads ,so the number of this case is $a_{n-2}$,
or
Lets say that the  last three flippings are $THH$  and contain $5$ consecutive Heads ,so the number of this case is $a_{n-3}$,
or
Lets say that the last four flippings are $THHH$  and contain $5$ consecutive Heads ,so the number of this case is $a_{n-4}$,
or
Lets say that the last five flipping are $THHHH$  and contain $5$ consecutive Heads ,so the number of this case is $a_{n-5}$,
or
Lets say that the last five flippings are $HHHHH$  ,so the number of this case is $2^{n-5}$
We know that the desired condition can be constructed by union of these cases. Lets say that the desired condition is $a_n$ ,then  $a_n=a_{n-1}+a_{n-2}+a_{n-3}+a_{n-4}+a_{n-5}+2^{n-5}$
This is a non-homogeneous recurrence relation , i guess you can solve it by yourself or use calculators.When you find an explicit formula ,then put $200$ into $\color{red}{n}$ .
At last divide the result by $2^{200}$ as denominator.
