imprisoned in a cell where there are only 3 ways out ,each of which are equally likely you are imprisoned in a cell from which there are three obvious ways to escape A,S,D. A leads you on a two-hour trip until you fall. B takes five-hours before you fall. D is locked. Each fall produces temporary amnesia and you are returned immediately after each fall. Assuming that you will always immediately choose one of the three exits from the cell with probability 1/3. On average, how long does it take before you notice D is locked.
My answer is: Let T be the waiting time till you find out D is locked, T follows a geometric distribution, the time it takes before you reach a success, in this case, before choosing D:
$$ \mathbf E(T) = \frac{1}{\frac{1}{3}}=3$$
This is where I have the problem. Can I assume that since each choice is equally likely, I can treat the choice of A or S as a Bernoulli trial where
$$ \mathbf X_t = \begin{cases}
2,  & \text{if $t^{th}$ trial is A} \\
5, & \text{if $t^{th}$ trial is S}
\end{cases} $$
Since each choice is equally likely P($\mathbf X_t$=2)=P($\mathbf X_t$=5)=$\frac{1}{2}$
and $\mathbf X_i$ is equidistributed for all i=[1,t], then $$\mathbf E(\mathbf X_i)= \mathbf E(\mathbf X_j)$$for i $\neq$ j.
Then $\mathbf E(\mathbf X_1)$= 2*0.5 + 5*0.5 =3.5
then the total waiting time is:
$$\sum_{k=1}^{T-1} \mathbf X_t $$
by the assumption above r.v X is the equidistributed so the sume becomes
$$\mathbf E(T-1)*\mathbf E(X_t)$$
with this approach my answer was 7 hours
 A: Let $E$ be the expected time taken. Then,
$$E=\frac{1}{3}\cdot 0+\frac{1}{3}\cdot(2+E)+\frac{1}{3}\cdot(5+E)=\frac{2}{3}E+\frac{7}{3}$$
So $E$ is indeed $7$.
A: I agree with your answer of $7$ hours as the expected waiting time.  I think your random variable $T$ is not the waiting time, but the number of attempts until you find the locked door.
On trials where you fall, your expected length of time is $3.5$ hours (the average of $2$ hours and $5$ hour).
You expect $2$ falls with an average time of $3.5$ hours per fall.  So your overall expected time is $7$ hours.
A: Let $A$ denote the event that the first choice falls on $A$,
Let $S$ denote the event that the first choice falls on $S$,
Let $D$ denote the event that the first choice falls on $D$,
Then:
$$\mathbb{E}X=\mathbb{E}\left(X\mid A\right)P\left(A\right)+\mathbb{E}\left(X\mid S\right)P\left(S\right)+\mathbb{E}\left(X\mid D\right)P\left(D\right)=$$$$\left(\mathbb{E}X+2\right)\frac{1}{3}+\left(\mathbb{E}X+5\right)\frac{1}{3}+0.\frac{1}{3}=\frac{2}{3}\mathbb{E}X+\frac{7}{3}$$
Leading to: $$\mathbb{E}X=7$$
