Rrerku

A coin, having probability p of landing heads and probability of q=(1-p) of landing on heads. It is continuously flipped until at least one head and one tail have been flipped.

This is not part of a homework assignment. I am studying for a final and don't understand the professors solutions.

a.) Find the expected number of flips needed.

Since this is clearly geometric, I would think the solution would be:

E(N)=$Sigma_i=0^inftyip^n-1q+Sigma_i=0^niq^n-1p=frac1q+frac1p$.

However, I am completely wrong.
The answer is

E(N)=$p(1+frac1q)+q(1+frac1p)$

For example, consider we flip for heads first. Then we have:

E(N|H)=$p+pSigma_i=0^inftynp^n-1q$... I am not sure why this makes sense.

I am not entirely sure why we have an added 1 and a factored p,q. Could someone carefully explain why it makes sense that this is the right answer?

If you get a head with probability $p$ then the expected number of throws is $1+E(X)$ where $X$ is a geometric distribution requiring a tail to be thrown with probability $q$ so $1+E(X)=1+frac1q$. Similarly if you throw a tail with probability $q$ then the expected number of throws is $1+E(Y)$ where $Y$ is a geometric distribution requiring a head to be thrown with probability $p$ so $1+E(Y)=1+frac1p$. This means that the overall expected number of throws is
$$pleft(1+frac1qright)+qleft(1+frac1pright)$$
because there is a probability $p$ that the expected number of throws is given by $1+E(X)$ and probability $q$ that it is given by $1+E(Y)$.

Let $X$ be the time of the first head, and $Y$ the time of the first tail, and $W$ the first time when a head and a tail has been flipped.

You are right in assuming that $E[X]=frac1p$ and $E[Y]=frac1q$, But you are wrong in assuming that $W=X+Y$, that's simply not true, actually $W=max(X,Y)$.

A possible approach. Let $A$ be the indicator variable of the event: "first coin was a head" (hence $X=1$).

Then use $$E[W]=E[E[W | A ]] = P(A=1) E[W|A=1]+P(A=0) E[W|A=0]=\=p(E[Y]+1)+q(E[X]+1)$$