Probability of $k$ changeovers when a coin is thrown $n$ times with $p=0.6$ E.g. 2 changeovers from 3 coin flips: HTH. So I have seen the answer to this question with p=0.5. What would be the answer to $P(k)$ for $k=1, \ldots, n-1$ when $P(Head)=0.6$?
 A: Update. (The original answer was wrong.)
We are considering $\{H,T\}$-strings of length $n\geq1$, and the number $K$ of changeovers in these strings. Let $p$ denote the probability for $H$ and $q=1-p$ denote the probability for $T$. We group the strings according to the number $h$ of $H$s occurring therein; let $t=n-h$ be the number of $T$s in these strings. All strings in such a group have the same probability $p^h\,q^t$.
There is exactly one string with $t=0$, namely $H^n$; its probability is $p^n$. Similarly there is exactly one string with $h=0$, namely $T^n$; its probability is $q^n$. It follows that $$P[K=0]=p^n+q^n\ .$$ All other strings have $h\geq1$ and  $t\geq1$, hence $K\geq1$. Let $h\geq1$ and $t=n-h\geq1$ be given. 
How many strings of type $(h,t)$ are there with exactly $k=2j-1\geq1$ changeovers? Since there are $2j-1$ changeovers we  have an alternating sequence of $j$ nonempty blocks of $H$s and $j$ nonempty blocks of $T$s (this requires $h\geq j$ and $t\geq j$). To define the sizes of the successive $H$-blocks we have to insert $j-1$ separators  into the string $H^h$, and to define the sizes of the successive $T$-blocks we have to insert $j-1$ separators into the string $T^t$. This can be done in
$${h-1\choose j-1}{t-1\choose j-1}$$
ways. Since we may  choose whether to begin with an $H$- or a $T$-block  we have to multiply this by $2$ in the following final formula:
$$P[K=2j-1]=2\sum_{h=j}^{n-j} p^h q^t{h-1\choose j-1}{n-h-1\choose j-1}\qquad(j\geq1)\ .$$
How many strings of type $(h,t)$ are there with exactly $k=2j\geq2$ changeovers? Since there are $2j$ changeovers we have have an alternating sequence of $2j+1$ nonempty blocks of $H$s and  $T$s. If we begin with an $H$-block there are $j+1$ of them, separated by  $j$ $T$-blocks; and if we begin with a $T$-block there are $j+1$ of these and $j$ $H$-blocks in between. It follows that we can realize the $2j$ changeovers in
$${h-1\choose j}{t-1\choose j-1}+{h-1\choose j-1}{t-1\choose j}={n-2j\over j}{h-1\choose j-1}{t-1\choose j-1}$$
ways. The final formula therefore looks as follows:
$$P[K=2j]={n-2j\over j}\sum_{h=j}^{n-j} p^h q^t{h-1\choose j-1}{n-h-1\choose j-1}\qquad(j\geq1)\ .$$
(Note that $2j=n$ enforces $P[K=2j]=0$.)
