Estimating weight of hidden coin Consider the following problem:
You have two coins each with it's own weight (probability of giving heads). Somebody will flip the coins for you in another room (you trust them). You can either ask them to flip both coins and they will tell you if they are both heads or not both heads. Or you can ask them to flip only the first coin (you can't only flip the second) and they will tell you if it was heads or not. 
How should you proceed to find an estimator of the weight of the second coin. You are allowed to ask the person multiple times. The main goal is to make the estimator not depend on the weight of the first coin. Is this possible? 
An aside for motivation of this problem: This is analogous to how measurements in quantum mechanics with linear loss are done. The loss is the first coin, the measurement is the second coin. We can never get rid of the loss but we can do a trivial second measurement (gives always heads) and then the real measurement. Particularly, this is the problem of measuring the polarization degree of freedom of photons from a heralded single photon SPDC source.
 A: Method 1: Comparing double and single coin flips as two binomial distributed variables (biased)
Say you flip $n$ times both with $k_n$ times double heads, and you flip $m$ times the first with $k_m$ times the heads. The unknown weights $\theta_1$ and $\theta_2$ can be estimated by the maximum likelihood estimate:


*

*The conditional probability is $$P(k_n,k_m|\theta_1,\theta_2) = {{n}\choose{k_n}} {{m}\choose{k_m}} \theta_1^{k_n}(1-\theta_1)^{n-k_n} (\theta_1\theta_2)^{k_m}(1-\theta_1\theta_2)^{n-k_m}$$

*Log-likelihood (and eliminating the binomial coefficients):
$$\log \mathcal{L}(\theta_1,\theta_2) =  k_n \log(\theta_1) + (n-k_n) \log(1-\theta_1)+ k_m \log(\theta_1\theta_2) + (m-k_m) \log (1-\theta_1\theta_2)$$

*Derivatives
$$\frac{\partial\log \mathcal{L}(\theta_1,\theta_2)}{\partial\theta_1} = \frac{k_n}{\theta_1} - \frac{n-k_n}{1-\theta_1} + \frac{k_m}{\theta_1} - \theta_2 \frac{m-k_m}{1-\theta_1\theta_2}  $$
$$\frac{\partial\log \mathcal{L}(\theta_1,\theta_2)}{\partial\theta_2} =  \frac{k_m}{\theta_2} - \theta_1 \frac{m-k_m}{1-\theta_1\theta_2}  $$

*Which is zero for 
$$\theta_1 = \frac{k_n}{n} $$ and $$\theta_2 = \frac{k_m}{m} \frac{n}{k_n} $$

This estimate does have some bias. Using simulation I found the expectation value $E(\theta_2)>\theta_2$ 
(and technically the expectation value is infinite due to the small possibility of $k_n=0$ but you could add some stopping rule that prevents $k_n$ from being zero. Since $k_m < k_n$ anyway I don't believe that this introduces too much bias, if at all)

Method 2: Performing double coin flips with number of trials depending on the amount of flips necessary to get k heads in single coin flips (unbiased ?)


*

*You flip the first coin until you got $k$ success and count the number $X$ that you need to flip. 

*Then you flip both coins a multiple $n$ times $X$ and count the number $Y$ that both coins land heads. 
Intuitively I'd say $\theta_2 = \frac{Y}{nk}$ is an unbiased estimator (with a negative aspect that $\theta_2$ can be above 1). The simulation below seems to show that there is only a small bias (also the variance/error is smaller in comparison to the first method with similar amount of coin flips). You would have to compute the conjugate distribution for the variable $Y$ to see if there is no bias.  
k=50
p1=0.5
p2=0.5
n=10
X <- rnbinom(10000,k,p1)+k
Y <- rbinom(10000,X*n,p1*p2)
mean(Y/k/n-p2)
var(Y/k/n)
plot(hist(Y/k/n))

