Given a coin which comes up Heads with probability $p$, let $N_p$ denote the number of times the coin is tossed until the first Heads. Argue that $$N_p = \begin{cases} 1 & \text{with probability } p\\ 1+\tilde{N}_p & \text{with probability } 1-p \end{cases}$$ where $\tilde{N}_p$ is an independent random variable with the same distribution as $N_p$. Using this, show that $\mathbb{E}[N_p] = 1/p$. Also use a similar argument to find $Var(N_p)$.
I find the expectation fairly straightforward: By rule of iterated expressions, we have: $$E[E[N_p|\tilde{N_p}]] = E[N_p] = E[p\times 1 + (1-p)\times (1+\tilde{N_p})]$$ $$= p\times p + (1-p) + (1-p)\times E[\tilde{N_p}] = 1 + \tfrac 1p - 1 = \tfrac 1p$$ However, I am having trouble with finding $V(N_p)$. There is an identity where $$V(N_p) = E[V(N_p|\tilde{N_p})] + V(E[N_p|\tilde{N_p}])$$ but I do not know how to handle $V(N_p|\tilde{N_p})$. By definition this should be $$V(Y|X=x)=\sum_{x\in X}(y-E(Y|X=x)^2P(Y|X=x)$$ What is little $y$ here? Is it simply $1$ and $1+N_p$? What does $E(Y|X=x)$ look like? This is a number, whereas $E(Y|X)$ is a random variable whose actualizations is $E(Y|X=x)$, but I am not sure what these actualizations look like.How do I find $V(Y|X)$ with $V(Y|X=x)$?
You should condition on the outcome of the first toss (call this $Y$), not on $\tilde{N}_p$. Then,
$$V\left(N_p\,|\,Y\right) = \left\{ \begin{array}{c c} 0 & Y=1\\ V\left(N_p\right) & Y=0 \end{array} \right.$$
and, similarly,
$$\mathbb{E}\left(N_p\,|\,Y\right) = \left\{ \begin{array}{c c} 1 & Y=1\\ 1+\mathbb{E}\left(N_p\right) & Y=0 \end{array} \right.$$
Combining these (and having solved for $\mathbb{E}\left(N_p\right)$ before) should give you an equation that has $V\left(N_p\right)$ on both sides, but you can rearrange terms and solve for it.