Can we find an unbiased estimator of $\pi$? Suppose we want to estimate $\pi$. We adopt the following strategy. We generate $n$ iid observations $Z_1,Z_2,...,Z_n$ from the unit disk $D=\{(x,y):x^2+y^2<1\}$ and let $R=\{(x,y):-\dfrac{1}{\sqrt{2}}\leq x,y\leq \dfrac{1}{\sqrt{2}}\}$ be a square inside $D$. Let $T$ be the number of times a generated observation falls inside $R$. Let $I=\max\{1\leq i\leq n:Z_i\in R\}$ and define $I=0$ if no observation falls inside $R$.
I am supposed to find an unbiased estimator of $\pi$ using $T$ and $I$. I have made certain observations. It is clear that $T\sim Bin(n,\dfrac{2}{\pi})$. I have also found that $P(I=k)=P(Z_k\in R,Z_j\notin R,j>k)=\dfrac{2}{\pi}(1-\dfrac{2}{\pi})^{n-k}$ if $1\leq k\leq n$ and $P(I=0)=(1-\dfrac{2}{\pi})^n$.
$E(I)$ is turning out to be a pretty nasty thing to compute. How can we use $T$ and $I$ to compute a function $f(T,I)$ with $E(f(T,I))=\pi$?
 A: I assume  $\{Z_i\}_{i=1}^{n}$ are i.i.d. vectors  uniformly distributed over the unit disc in $\mathbb{R}^2$.
If you are seeking something more than my $f(T,I)=\pi$ answer, then I think you are on a wild goose chase.  That is because you suggest an answer can be given for any positive integer $n$, which would include $n=1$. 
In the case $n=1$ we have $T=I\sim \mbox{Bernoulli}$ with $P[T=1]=p=2/\pi$. If you want a function $f(T)$ such that $E[f(T)]=\pi$, you must find $f(0)$ and $f(1)$, such that
$$ (1-2/\pi)f(0) + (2/\pi)f(1) = \pi$$
Thus
$$ (\pi-2)f(0) + 2f(1)=\pi^2 $$
Now if both $f(0)$ and $f(1)$ are rational numbers, it would contradict the fact that $\pi$ is transcendental!   
So we are forced to make irrational choices for $f(0)$ and/or $f(1)$, in which case why not just choose $f(0)=f(1)=\pi$? 

More generally, for any positive integer $n$ we seek a rational-valued function $f(t,i)$ such that 
$$ \sum_{t=0}^n \sum_{i=0}^n f(t,i)(2/\pi)^t(1-2/\pi)^{n-t} c(t,i)=\pi$$
where $c(t,i)$ is the (integer) number of ways to have $t$ successes with the last success at index $i$.  Multiplying this equation by $\pi^n$ gives
$$ \sum_{t=0}^n \sum_{i=0}^n f(t,i) 2^t(\pi-2)^{n-t}c(t,i) = \pi^{n+1}$$
Hence the equation
$$ \sum_{t=0}^n \sum_{i=0}^n f(t,i)2^t(x-2)^{n-t}c(t,i)=x^{n+1}$$
is a polynomial equation with rational coefficients for which $\pi$ is a root, again contradicting the fact that $\pi$ is transcendental!

On the other hand I observe: 
i) With an infinite number of observations $\{Z_i\}_{i=1}^{\infty}$, the expected number of observations until the first one lands in the square is $1/p = \pi/2$. 
ii) If we define a larger square $\{(x,y):-1\leq x,y\leq 1\}$ that contains the unit disc, and generate a random vector that is uniformly distributed over that square, then the probability this vector lies in the unit disc is $\pi/4$. 
