Expected number of consecutive vowels The question is stated below:
There is a lucky draw wheel, labeled with the vowels (a, e, i, o, u) respectively. They are evenly distributed (unbiased). In this game, players have to get two consecutive spins of ‘e’ to get the prize.
Count:
(a). the expected number of times to spin to get the first two consecutive ‘e’s?
In this question, I have attempted multiple methods.
First, I tried to calculate it. I distinguished the probabilities of getting ‘e’, P(E) as 1/5, and that of getting other vowels, P(O) as 4/5. To calculate the expected value, I thought of multiplying the number of times to spin with the probability.
But I soon realized that there are infinitely many combinations, so a tree diagram is drawn to find out the patterns…
The pattern is quite unrecognizable. (E: e, O: other)
2 spins: (E, E)
3 spins: (O, E, E)
4 spins: (O, O, E, E), (E, O, E, E)
5 spins: (E, O, O, E, E), (O, E, O, E, E), (O, O, O, E, E)
…
Now I cannot think of any way to solve this program except through programming. Is there any hidden pattern here, or any method that I do not know?
Thank you!
 A: Let $E$ and $F$ denote the expected number of spins needed to get two consecutive e's if the current e-streak is $0$ and $1$ respectively. Then
$$E=1+\frac15F+\frac45E$$
$$F=1+\frac150+\frac45E$$
Solving gives $E=30$ and $F=25$. We start in state $E$, so $30$ spins are needed on average.
A: Let $s_1,s_2...$ be the expected number of spins to get $1, 2... e's$ consecutively
By the geometric distribution, we know that the expected value $s_1$ to get the first $e=5$
From there one spin takes us to our goal with $Pr= \frac 15 $, or sends us back to scratch with $Pr = \frac 45$  and we have wasted a spin, thus
$s_2 = s_1+\frac15\cdot 1+\frac45(s_2+1)$
Which yields
Expected number of spins $s_2=30$

Additional Material
An advantage of this approach is that you can easily compute  expected number of spins for,say, $s_4$, because $s_2 = 5(s_1+1)$, and parallelly you can immediately say, $s_3=5(s_2+1)$, and so on, thus
$s_3 = 5(30+1)=155$,
$s_4 = 5(155+1)=780$
And you can further see that
$s_1 =5,$
$s_2 = 5 + 5^2,$
$s_3= 5 + 5^2 +5^3$, and so on.
