Conditional probability for consecutive Bernoulli trials Independent trials, each of which is a success with probability $p$, are performed until there are $k$ consecutive successes. Let $N_k$ denote the number of necessary trials to obtain $k$ consecutive
successes, and show that:
$$\mathbb{E}(N_k|N_{k−1}) = N_{k−1} + 1 + (1 − p)\mathbb{E}(N_k).$$
 A: I'll give you the basic reasoning here, but you can write it out formally yourself.
Let $N_k$ be the number of trials necessary to obtain $k$ consecutive successes.
We want to show 
$$E[ N_k | N_{k-1}]= N_{k-1} + 1 + (1-p)E[N_k]$$
Firstly, given $N_{k-1}$, consider the possible values of $N_{k-1}+1$. If the trial immediately following the $N_{k-1}$'th trial is a success, then clearly $N_k = N_{k-1}+1$ with probability $p$. 
Next, suppose the trial following $N_{k-1}$ is a failure. So that we have $N_{k-1}+1$ total trails with the last one being a failure. Well, since the last one is a failure, then by the independence of each trial, under this situation we would have the conditional expected number of trials until $k$ consecutive successes as
$$N_{k-1} + 1 + E[N_k]$$
Since we already have $N_{k-1} + 1$ trials, but the last one being a failure "resets" our expectation back to $E[N_k]$.
Hence you can express the original expectation as
$$E[ N_k | N_{k-1}]= p(N_{k-1} + 1) + (1-p)(N_{k-1} + 1 + E[N_k])$$
$$=N_{k-1} + 1 + (1-p)E[N_k]$$
This answer uses the law of total expectation: $E[ E[X|Y] ] = E[X]$. Here we take $Y$ as the result of the $N_{k-1}+1$ trial, and $X$ as $N_{k}|N_{k-1}$.
A: You are conditioning on $N_{k-1}$, which means you are conditioning on the fact that you have $k-1$ consecutive successes.  Let $X \sim \text{Bern}(p)$ be the outcome of the next trial and consider the cases:


*

*If $X=1$ then you now have $k$ consecutive successes;

*If $X=0$ then you now have $0$ consecutive successes and you have to start all over again.


Hence, by application of the law-of-total probability you have:
$$\begin{equation} \begin{aligned}
\mathbb{E}(N_k|N_{k-1}) 
&= \mathbb{P}(X=1|N_{k-1}) \mathbb{E}(N_k|N_{k-1},X=1) + \mathbb{P}(X=0|N_{k-1}) \mathbb{E}(N_k|N_{k-1},X=0)  \\[6pt]
&= p \mathbb{E}(N_k|N_{k-1},X=1) + (1-p) \mathbb{E}(N_k|N_{k-1},X=0)  \\[6pt]
&= p(N_{k-1}+1) + (1-p) (N_{k-1}+1 + \mathbb{E}(N_k)) \\[6pt]
&= p(N_{k-1}+1) + (1-p) (N_{k-1}+1) + (1-p) \mathbb{E}(N_k) \\[6pt]
&= N_{k-1}+1 + (1-p) \mathbb{E}(N_k). \\[6pt]
\end{aligned} \end{equation}$$
