I apologize if this turns out to be a simple question with an overly long premise.
I'm playing a game of chance where players are distributed into 5 categories. Call them A, B, C, D, E. I know exactly who is in each category.
The game essentially works like this - each day, every player is given a ticket corresponding to their category. And each day, there are 100 empty buckets on the game board. Every player independently chooses which bucket to place his ticket in. At the end of the day, a ticket is drawn randomly from each bucket, and those drawn are winners for the day.
However, the probability of a player's ticket being drawn is weighted according to category. One category A ticket has a much higher chance of being drawn than a category E ticket, but that exact weight is unknown.
At the end of each day, I know everything about what happened that day except for these weights - I know exactly who placed tickets in each bucket, the categories of each player, and the winner of each bucket.
I do not have the weights assigned to each category of players, and this is what I am trying to calculate.
In other words, I have multiple days of data that looks like this:
Bucket | # of A | # of B | # of C | # of D | # of E | Winner Category I | 2 | 6 | 12 | 36 | 90 | *B* II | 0 | 1 | 5 | 10 | 25 | *D* ...
Given this, how do I calculate the weighted probabilities for the different categories?
This is really more of a problem of statistics (rather than probability).
Let $W_A, W_B,\dots,W_E$ be the corresponding weights and $D$ be your observed data. One approach would be to fix a value for, say, $W_A$ $($for instance, $W_A = 1)$ and let the other weights be unknown. Then, for each value of $\mathbf w = (1, w_b,w_c,w_d,w_e)$ calculate
$$f(\mathbf w) = \mathbb P \left.\Big(D\,\,\middle|\,(W_A,W_B,W_C,W_D,W_E) = \mathbf w\Big)\right.,$$
and pick some value of $\mathbf w$ that maximizes $f$.
This way, we're taking a maximum likelihood approach. Does this make sense to you?
For each bucket $\beta$ and class $l \in \{a,b,c,d,e\}$, let $n_{l,\beta}$ denote the number of tickets of class $l$ in bucket $\beta$. Then, the probability that the winning ticket is from class $j \in \{a,b,c,d,e\}$ is
$$p_{j,\beta}(\mathbf w) = \mathbb P(\text{$j$ = class of winning ticket from bucket $\beta$}\,|\,\mathbf w) = \frac{n_{j,\beta}\,w_j}{\sum_{l \in \{a,b,c,d,e\}}\,n_{l,\beta}\,w_l}.$$
If we denote by $k(\beta)$ the observed class of the winning ticket for bucket $\beta$, then independence gives us that
\begin{align} f(\mathbf w) &= \prod_{\text{bucket $\beta$}}\,\left( \frac{n_{k(\beta),\beta}\,w_{k(\beta)}} {n_{a,\beta}+n_{b,\beta}\,w_b+n_{c,\beta}\,w_c+n_{d,\beta}\,w_d+n_{e,\beta}\,w_e} \right) \\&= \prod_{\text{bucket $\beta$}}\,{\left( 1+\frac{\sum_{l\neq k(\beta)}\,n_l\,w_l}{n_{k(\beta),\beta}\,w_{k(\beta)}} \right)}^{-1} .\end{align}
Write
$$g(\beta; w_a,w_b,\dots,w_e) = {\left( 1+\frac{\sum_{l\neq k(\beta)}\,n_l\,w_l}{n_{k(\beta),\beta}\,w_{k(\beta)}} \right)}^{-1} = p_{k(\beta),\beta}(\mathbf w),$$
so that $f(\mathbf w) = \prod_{\text{bucket $\beta$}}\,g(\beta;\mathbf w)$. Then
\begin{align} \frac{\partial f}{\partial w_j}(\mathbf w) = \sum_{\text{bucket $\beta$}}\,\left( \left(\prod_{\text{bucket $\gamma \neq \beta$}}\,g(\gamma;\mathbf w)\right) \cdot \frac{\partial g}{\partial w_j}(\beta;\mathbf w) \right). \end{align}
On the other hand, a direct calculation shows that
$$ \frac{\partial g}{\partial w_j}(\beta;\mathbf w) = \left\{ \begin{array}{ccc} \displaystyle g(\beta;\mathbf w)^2 \cdot \frac{\sum_{l\neq k(\beta)}\,n_{l,\beta}\,w_l} {n_{k(\beta),\beta}\,{w_{k(\beta)}}^2}&;&j = k(\beta)\\ \\ \displaystyle -g(\beta;\mathbf w)^2 \cdot \frac{n_{j,\beta}} {n_{k(\beta),\beta}\,w_{k(\beta)}}&;&j\neq k(\beta) \end{array} \right. $$
Thus, we have that
\begin{align} \frac{\partial f}{\partial w_j}(\mathbf w) = f(\mathbf w)\cdot\left[ \sum_{\substack{\text{bucket $\beta$}\\j=k(\beta)}}\left( g(\beta;\mathbf w) \cdot \frac{\sum_{l\neq j}\,n_{l,\beta}\,w_l} {n_{j,\beta}\,{w_j}^2} \right) - \sum_{\substack{\text{bucket $\beta$}\\j\neq k(\beta)}}\left( g(\beta;\mathbf w) \cdot \frac{n_{j,\beta}} {n_{k(\beta),\beta}\,w_{k(\beta)}} \right) \right] .\end{align}
Letting
$$ h_j(\beta;\mathbf w) = \left\{ \begin{array}{ccc} \displaystyle \frac{\sum_{l\neq j}\,n_{l,\beta}\,w_l} {n_{j,\beta}\,{w_j}^2}&;&j = k(\beta)\\ \\ \displaystyle -\frac{n_{j,\beta}} {n_{k(\beta),\beta}\,w_{k(\beta)}}&;&j \neq k(\beta) \end{array} \right. $$
simple manipulations show that
$$h_j(\beta, \mathbf w) = \frac{\delta_{j,k(\beta)} - p_{j,\beta}(\mathbf w)} {w_j\cdot p_{k(\beta),\beta}(\mathbf w)},$$
where $\delta_{ij}$ is the Kronecker delta. We can hence simplify the derivative with
\begin{align} \frac{\partial f}{\partial w_j}(\mathbf w) &= f(\mathbf w)\cdot\left( \sum_{\text{bucket $\beta$}} g(\beta;\mathbf w) \cdot h_j(\beta;\mathbf w) \right) \\&= \frac{f(\mathbf w)}{w_j}\cdot\left( \sum_{\text{bucket $\beta$}} \delta_{j,k(\beta)} - p_{j,\beta}(\mathbf w) \right) .\end{align}
Assuming all weights are positive, $f>0$. This implies that
\begin{align} \nabla f(\mathbf w) = 0 &\iff \Big(\forall j \in \{b,c,d,e\}\Big)\,\sum_{\text{bucket $\beta$}} \delta_{j,k(\beta)} - p_{j,\beta}(\mathbf w) = 0 \\&\iff \Big(\forall j \in \{b,c,d,e\}\Big)\, \sum_{\text{bucket $\beta$}} p_{j,\beta}(\mathbf w) = \#\{\,\text{buckets $\beta$ with $j=k(\beta)$}\,\} .\end{align}