My student made the interesting problem and asked me how to solve it. I believe the Bayesian approach is promising, and got a kind of solution. But, I'm not confident of my solution because it's too complicated considering the simpleness of the problem.

There are two questions:

  • Is this problem equivalent some famous one?
  • Could you please suggest how to solve it?

The problem my student made


There are two boxes. Each box contains $N^{c}$ coins. The coins are labeled with numbers $1,2,3,\cdots,N^{C}$. The probabilities of getting heads or tails are NOT necessarily 0.5 (distorted cons).

Also, $p^{A,i}$ is NOT necessarily $p^{B,i}$, where $p^{\eta,i}$ is the probability of getting head when tossing $i$-th coin in the box $\eta \in \{A,B\}$.

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Data 1

Alice tossed each coin in both of the box A and B with $N^{\eta,i}$ times, where $N^{\eta,i}$ is how many times she tossed the $i$-th coin in the box $\eta$. She recored how many times head or tail had been obtained. So, the observed data is like $D_1=\{N^{A,1}_{head}, N^{A,1}_{tail},N^{A,2}_{head}, N^{A,2}_{tail},\cdots,N^{B,1}_{head}, N^{B,1}_{tail},N^{B,2}_{head}, N^{B,2}_{tail},\cdots\}$. (Of course, $N^{\eta,i}_{head}+ N^{\eta,i}_{tail} = N^{\eta,i}$)

Data 2

Then, Bob selected randomly the box A or box B and tossed all of coins inside the selected box. He recorded head or tail of each coin and passed the result to Alice as the data 2. So, the data 2 is like $D_2=\{x^{1}, x^{2}, \cdots,x^{N_c}\}$, where $x^{i}$ is head or tail of $i$-th coin.


Using the two kind of data, Alice is supposed to infer the posterior probability that Bob chose the box A. Formally, she calculates $P(Box=A|D_1, D_2)$.

My answer

probability of each coin

Using the data $D_1$ the probability of getting the head of each coin can be calculated as below.

By the bayese theorem, the probability $p^{\eta, i}$, by which one gets the head of the $i$-th coin in the box $\eta \in \{A, B\}$ is $Beta(\alpha=1+N^{A,i}_{head}, \beta=1+N^{A,i}_{tail})$, where $Beta(\alpha, \beta)$ is the beta distribution.

Likelihood of $D_2$

With Monte Carlo simulation, you can calculate the likelihood $P(D_2|\eta, p^{\eta, i})$.


With MCMC, you can calculate $P(\eta|D_1, D_2) = P(\eta|D_1, p^{\eta, i})$ with a setting of that the likelihood is set as $P(D_2|\eta, p^{\eta, i})$ calculated at the above step, and the prior $P(A) = P(B) = 0.5$.


1 Answer 1


Towards reducing the complexity of your notations, let us consider the case there is a single coin in each box, with probability of head $p_A$ and $p_B$, respectively.

The learning experiment means observing $N^A$ heads and $M^A$ tails, versus $N^A$ heads and $M^A$ tails, for boxes A and B, respectively. Starting with a flat prior on both $p_A$ and $p_B$, the resulting posterior is then $$p_A\sim\text{Be}(1+N^A,1+M^a)\qquad p_B\sim\text{Be}(1+N^B,1+M^B)$$ which becomes the prior for the testing experiment.

Having observed $x\in\{\text{head},\text{tail}\}$, rewritten as $y=\mathbb{I}_{\text{head}}(x)\in\{0,1\}$, the marginal distribution of $Y$ is $$m_A(y)=\int_0^1 p_A^y (1-p_A)^{1-y} \text{Be}(p_A;1+N^A,1+M^A)\text{d}p_A$$ for box A and $$m_B(y)=\int_0^1 p_B^y (1-p_B)^{1-y} \text{Be}(p_B;1+N^B,1+M^B)\text{d}p_B$$ for box B. Those integrals are available in closed form: \begin{align*}m_A(y)&=\dfrac{B(1+N^A+y,1+M^A+1-y)}{B(1+N^A,1+M^A)}\\ &=\dfrac{(2+N^A+M^A)!}{(1+N^A+M^A)!}\Big/\dfrac{(N^A+y)!(M^A+1-y)!}{N^A!M^A!}\\ &=\dfrac{2+N^A+M^A}{y(N^A+y)+(1-y)(M^A+1-y)} \end{align*} and $$m_B(y)=\dfrac{2+N^B+M^B}{y(N^B+y)+(1-y)(M^B+1-y)}$$ Therefore,$$\mathbb{P}(A|y)=\dfrac{m_A(y)}{m_A(y)+m_B(y)}$$ The extension to the case described in the question is straightforward since the $N^c$ coins are independent. Therefore the marginal distribution of $D_2$ is just the product of the marginals over all coins.


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