Read up on Bernolli trials, but I can't get my head around this.
Consider a sequence of independent Bernoulli trials with p = 0 2. What
is the expected number of trials to obtain the first success? After
the eighth success occurs, what is the expected number of trials to
obtain the ninth success?
SOS!
Best Answer
For a Bernoulii trial series with success parameter $p$, the expected number of trials until the first success is: $$\begin{align}\mathsf E[N_1] & = \sum_{n=1}^{\infty} n p (1-p)^{n-1} \\[1ex] & = \frac 1 p\end{align}$$
After eight successes have occurred the expected number of trials until the ninth success is: $$\mathsf E[N_9-N_8\mid N_8] = \mathsf E[N_1]$$
Which is to say: After any number of successes the expected number of trials until the next success remains constant; there's no memory. This is because after the each success you basically start the process over again from the start.