Problem. You are practicing archery. The probability that you hit a bull's eye is 0.2. Assume that each attempt is independent of all the others. Let $T$ denote the number of tries needed to hit two bull's eyes in succession, and $\mathbb{E}[T]$ its expectation. Also let $W$ denote the RV that takes the value 1 if you hit bull's eye on your first attempt and 0 otherwise.
i. Write down an expression for $\mathbb{E}[T|W=0]$ as a function of $\mathbb{E}[T]$.
ii. Write down an expression for $\mathbb{E}[T|W=1]$ as a function of $\mathbb{E}[T]$.
iii. Compute $\mathbb{E}[T]$.
Attempted solution for i. My intuition tells me that if we miss the first time, things "start over" so that $P(T=t|W=0)=P(T=t-1)$ whenever $t\geq 3$. Hence: $$ \mathbb{E}[T|W=0]=\sum_{t=3}^\infty tP(T=t|W=0) $$ $$ =\sum_{t=3}^\infty tP(T=t-1) $$ $$ =\sum_{t=3}^\infty (t-1)P(T=t-1)+\sum_{t=3}^\infty P(T=t-1) $$ $$ =\sum_{t=2}^\infty tP(T=t)+\sum_{t=2}^\infty P(T=t) $$ $$ =\mathbb{E}(T)+\sum_{t=2}^\infty P(T=t) $$ $$ =\mathbb{E}(T)+1. $$ However, I'm uncertain of how to prove $P(T=t|W=0)=P(T=t-1)$ rigorously.
Attempted solution for ii. For $t\geq 4$ we have $$ P(T=t|W=1)=P(T=t|\text{make first and miss second}). $$ Again, using intuition, I believe that $$ P(T=t|\text{make first and miss second})=P(T=t-2) $$ hence $$ P(T=t|W=1)=P(T=t-2) $$ Now, similar to before, $$ \mathbb{E}(T|W=1)=\sum_{t=2}^\infty tP(T=t|W=1) $$ $$ =2P(T=2|W=1)+\sum_{t=4}^\infty tP(T=t|W=1) $$ $$ =2(0.2)+\sum_{t=4}^\infty tP(T=t-2) $$ $$ =0.4+\sum_{t=2}^\infty tP(T=t)+2 $$ $$ =\mathbb{E}[T]+2.4. $$
Attempted solution for iii. If my answers to (i) and (ii) are correct, then, setting $\mu=\mathbb{E}[T]$, we obtain $$ \mu=0.8\mathbb{E}[T|W=0]+0.2\mathbb{E}[T|W=1] $$ $$ =0.8(\mu+1)+0.2(\mu+2.4)=\mu+1.28 $$ which is clearly nonsense. So, obviously at least one---probably more---serious errors are involved above.
Any ideas what the trouble is?
Thanks!
$$E[T|W=0]=1+E[T]$$
That is after the first failure, the game restart.
If your first trial is a hit, there are two possibilities, you end the game by hitting it the second time or you don't and restart the game. \begin{align} E[T|W=1] &=2p + (1-p)(2+E[T))\\ &=2+(1-p)E[T] \end{align}
Another interpretation is that we need $T\ge 2$, the probability that we have to restart the game is $1-p$.
Now, using the total law of expectation.
\begin{align} E[T] &= pE[T|W=1]+(1-p)E[T|W=0]\\ &=p[2+(1-p)E[T]] + (1-p)[1+E[T]]\\ &=2p+1-p+[(1+p)(1-p)E[T]\\ &=p+1+(1-p^2)E[T]\\ &= \frac{p+1}{p^2} \end{align}