How to combine two independent repeated experiments with different success probabilities? Repeating an experiment (about which I asked before) with $n$ possible outcomes $t$ times independently, where all but one outcomes have probability $\frac{1}{n+1}$ and the other outcome has the double probability $\frac{2}{n+1}$, I also obtain an independent information where the outcome with double probability shows up 50% more likely than the other outcomes (i.e., with probability $\frac{3}{2n+1}$ and the others' probability is $\frac{2}{2n+1}$).
How do I combine the two results?
Alternative formulation: Given is the probability space $(\Omega\times\Omega, \mathcal{P}(\Omega\times\Omega), \mathrm{p})$ with
$$\mathrm{p}(\omega, \omega') = \left\{\begin{array}{cc}
\frac{6}{(n+1)(2n+1)} & \mbox{ if } \omega = \omega_0 \mbox{ and } \omega' = \omega_0\\
\frac{4}{(n+1)(2n+1)} & \mbox{ if } \omega = \omega_0 \mbox{ and } \omega' \ne \omega_0\\
\frac{3}{(n+1)(2n+1)} & \mbox{ if } \omega \ne \omega_0 \mbox{ and } \omega' = \omega_0\\
\frac{2}{(n+1)(2n+1)} & \mbox{ otherwise }
\end{array}\right.$$
where $\omega_0\in\Omega$ is unknown (and $|\Omega| = n$). The goal is to find $\omega_0$ given $t$ samples from $\Omega\times\Omega$.
Currently I'm counting how often each value shows up in the first coordinate and add the number of times it shows up in the second coordinate multiplied by a weighting factor of $\log_2(\frac{3}{2})$. The value with the highest number should be $\omega_0$ (if $t$ is big enough).
Is this the correct weighting factor rsp. the correct way to find $\omega_0$?
PS: I'm also thankful for anyone finding better tags for my question.
 A: Interesting problem.
Let's first generalize it and simplify the notation.  There are two multinomial distributions, one with probabilities $(p_1, p_2, \ldots, p_n)$ = $(2/(n+1), 1/(n+1), \ldots, 1/(n+1))$ and the other with probabilities $(q_1,q_2, \ldots, q_n)$ = $(3/(2n+1), 2/(2n+1), \ldots, 2/(2n+1))$.  The probabilities are in descending order: $p_1 \ge p_2 \ge \cdots \ge p_n \gt 0$ and $q_1 \ge q_2 \ge \cdots \ge q_n \gt 0$.
You make $t$ independent observations of each, with counts $k_i$ and $m_i$ ($i=1, 2, \ldots, n$), respectively.  However, you do not know the subscripts: you only have the  ordered pairs $\left((k_{\sigma(1)},m_{\sigma(1)}), \ldots, (k_{\sigma(n)}, m_{\sigma(n)})\right)$ for some unknown permutation $\sigma$ of the subscripts.
You propose identifying which of these pairs corresponds to subscript $1$ by fixing positive coefficients $x$ and $y$ and computing the statistics
$$z_i = x k_i + y m_i, \quad i = 1, 2, \ldots, n,$$
and nominating the subscript with the largest value of $z_i$.
Let's assume your loss function is simply the indicator of correctness, so that your aim is to maximize the chance that $z_1$ is the largest of the $z_i$.
To get a handle on what the optimal values of $x$ and $y$ ought to be, consider the case where both $n$ and $t$ are large.  Large $n$ allows us to ignore the slight dependency of the $k_i$ (and $m_i$), treating them as if they were independent.  Large $t$ allows us to adopt Normal approximations to the distributions of the $k_i$ and $m_i$.  These state that, to a good approximation,
$$k_i \sim N(p_i t, p_i(1-p_i)t); \quad m_i \sim N(q_i t, q_i(1-q_i)t)$$
(where the parameters are the mean and variance).  Therefore
$$z_i \sim N((x p_i + y q_i)t, (x^2 p_i(1-p_i) + y^2 q_i(1-q_i))t).$$
To maximize the chance of making a correct determination, we want to maximize the probability that $z_1 \gt z_i$ for $i \gt 1$.  Because
$$\eqalign{
z_1 - z_i \sim & N((x(p_1-p_i) + y(q_1-q_i))t, \\
&(x^2 [p_1(1-p_1) + p_i(1-p_i)] + y^2 [ q_1(1-q_1)+q_i(1-q_i)])t),
}$$
this is tantamount to maximizing the z-score,
$$z = \frac{(x(p_1-p_i) + y(q_1-q_i))t}{\sqrt{(x^2 [p_1(1-p_1) + p_i(1-p_i)] + y^2 [ q_1(1-q_1)+q_i(1-q_i)])t}}.$$
This expression takes the form
$$z = \sqrt{t} \frac{a x + b y}{\sqrt{c x^2 + d y^2}}$$
for strictly positive coefficients $a, b, c, d$ (guaranteeing $z$ will be positive, which should be obvious).  Note, too, that only the ratio $\xi = x/y$ matters, because rescaling $x$ and $y$ does not change the ordering of the $z_i$.  It therefore suffices to maximize the square of this expression,
$$z^2  = t \frac{(a\xi + b)^2}{c\xi^2 + d},$$
for $\xi \gt 0$.
This straightforward problem has the solution
$$\eqalign{
\xi = &\frac{a d}{b c} \\
    = &\frac{(p_1 - p_i)(q_1(1-q_1)+q_i(1-q_i)}{q_1 - q_i)(p_1(1-p_1) + p_i(1-p_i))} \\
    = &\frac{1/(n+1)(3(2n-2)/(2n+1)^2  + 2(2n-1)/(2n+1)^2)}{1/(2n+1)(2(n-1)/(n+1)^2 + n/(n+1)^2)} \\
    = & \frac{10 n^2 + 2 n - 8}{6 n^2 - n - 2}
}$$
for all $i \gt 1$.  Recalling the assumption that $n$ is large, we retain only the highest powers of $n$ and find (choosing $x=1$ so that $y = 1/\xi$):
$$x = 1, \quad y = 3/5 = 0.6.$$
This literally answers the question: the value $\log_2(3/2) \sim 0.585$ is not quite the right weight, although it's (surprisingly) close.
This analysis does not answer the more basic question, though: as a function of $n$ and $t$, what is the best weight?  This can be found with a similar analysis using much more painful calculations concerning the multinomial distribution in place of the Normal approximations.  I suspect, without any proof, that the formula for $\xi$ will work well even for small $n$ and small $t$.

