Confounding Coin Chuck Chances 
If a coin is thrown some number of times, coming up heads the last three times, what is the probability that the next (nth) throw will be heads. What is the probability that the (n+9)th throw will be the first time that four heads are thrown in a row?

I came across this recently (I've reworded it), and ended up with the answers 1/2 for part 1 and 1/32 for part b. What do you think? My main concern is that bayesion probability methods can be applied to this, but I really can't see how.
 A: Not sure I follow the meaning of step $b$.  As I understand it, we are looking at the sequence $$HHHa_na_{n+1}a_{n+2}\cdots a_{n+9}$$  and we want the probability that $a_{n+6}a_{n+7}a_{n+8}a_{n+9}$ is the first block of the form $HHHH$ to arise.  To achieve that we need a number of things:
$1.$  we need $a_{n+6},a_{n+7},a_{n+8},a_{n+9}$ to all be $H$...probability $\frac 1{16}$.
$2.$  we need $a_{n+5}$ to be $T$ (were it $H$ then $a_{n+5}a_{n+6}a_{n+7}a_{n+8}$ would be an earlier string of four $H's$...probability $\frac 12$.
$3.$  we need $a_n$ to be $T$ (else the string begins with four $H's$)...probability $\frac 12$.
$4.$  we need $a_{n+1}a_{n+2}a_{n+3}a_{n+4}$ to be anything other than $HHHH$...probability $1-\frac 1{16}=\frac {15}{16}$.
As these are independent events we get the answer by multiplying.  Hence $$\boxed {\frac 1{16}\times \frac 12 \times \frac 12 \times \frac {15}{16}=\frac {15}{1024}}$$
Note: this is about $.015$ which is roughly half of your proposed $\frac 1{32}$.
