How to estimate discrete probability distribution from a dataset of pairwise frequencies? My problem is best explained with an example..
Suppose every day I count the cars driving down my street, but because I'm a slow counter, I only count two different types of cars per day. On the first day, I count 30 trucks and 65 cars. On the second day, I count 25 trucks and 12 motorcycles. On the third day I count 19 motorcycles and 72 cars. On the fourth day I count 11 vans and 14 motorcycles. And on the fifth day I count 12 vans and 44 trucks.
While the total volume of traffic changes day to day, I expect the relative proportion of vehicle types to remain near constant.
How do I use this information to develop a discrete probability distribution for the vehicles on my road? Put another way, how can I determine the probability that the next car I observe will be type X?
I suspect there's a nice conditional probability tree solution to this, but I can't quite put my finger on it.
UPDATE
Upon more thought, I think this may be a good candidate for maximum likelihood estimation. If I assume $P(\text{car})= p_1$, $P(\text{truck})= p_2$, $P(\text{motorcycle})= p_3$, and $P(\text{van})= p_4$, then I can calculate the likelihood of my observed pairs of data given some values for $p_1$, $p_2$, $p_3$, and $p_4$.  Now I just need to define and optimize my likelihood function.. Hmm..

I have a proposition that is near to what you thought, but with an additional step of working with an exact marginal distribution.
This will be a long answer. Skip the mathematics if you are not inclined. The conclusions about the computations is explained after they are made.
The model Let $(N_t, X_{ti}, X_{tj})$ be the total number of vehicles, the number of vehicles of type $i$ and the number of vehicles of type $j$, $i\neq j$ , on the $t$-th day.
The random variable $N_t$ is latent, meaning it is not observed directly. The idea is to propose a reasonable joint distribution for this triplet, and then compute the marginal of distribution of $(X_{ti}, X_{tj})$, which will then be used to compute the proportions.
For this idea to work, we have to propose first a distribution for $N_t$. Since we are working with a countable variable, I will assume that $N_t \sim \mbox{Poisson}(\lambda)$. For the conditional distribution of $(X_{ti}, X_{tj})$, we propose
$$\mathbb{P}(X_{ti} = x_i, X_{tj} = x_j|N_t = n) = \frac{n!}{x_i!x_j!(n-x_i-x_j)!}p_i^{x_i}p_j^{x_j}(1 - p_i - p_j)^{n-x_i-x_j} \quad.$$
That is, they are a trinomial distribution with $N_t = n$ observations, where the $(1 - p_i - p_j)$ term correspond to the probability of observing vehicles of type different than $i$ and $j$.
The joint distribution can be written as
$$\mathbb{P}(N_t = n, X_{ti} = x_i, X_{tj} = x_j) = \mathbb{P}(N_t = n) \mathbb{P}(X_{ti} = x_i, X_{tj} = x_j|N_t = n) \quad.$$
We have analytical an expression for the RHS. However, we are interest in the marginal
$$\mathbb{P}(X_{ti} = x_i, X_{tj} = x_j) = \sum_{n=0}^\infty \mathbb{P}(N_t = n) \mathbb{P}(X_{ti} = x_i, X_{tj} = x_j|N_t = n) \quad.$$
Before evaluating the summation, notice that $\mathbb{P}(X_{ti} = x_i, X_{tj} = x_j|N_t = n) = 0$ when $n < x_i + x_j$. Now we evaluate the summation
\begin{align}
\mathbb{P}(X_{ti} = x_i, X_{tj} = x_j) 
&=  \sum_{n=0}^\infty \frac{e^{-\lambda}\lambda^{n}}{n!} \frac{n!}{x_i!x_j!(n-x_i-x_j)!}p_i^{x_i}p_j^{x_j}(1 - p_i - p_j)^{n-x_i-x_j}\\
&=  \frac{e^{-\lambda}p_i^{x_i}p_j^{x_j}}{x_i!x_j!}\sum_{n=x_i+x_j}^\infty \frac{\lambda^{n}}{(n-x_i-x_j)!}(1 - p_i - p_j)^{n-x_i-x_j}\\
&=  \frac{e^{-\lambda}p_i^{x_i}p_j^{x_j}}{x_i!x_j!}\sum_{k = 0}^\infty \frac{\lambda^{k+x_i+x_j}}{k!}(1 - p_i - p_j)^{k}\\
&=  \frac{e^{-\lambda}\lambda^{x_i+x_j}p_i^{x_i}p_j^{x_j}}{x_i!x_j!}\sum_{k = 0}^\infty \frac{(\lambda(1 - p_i - p_j))^{k}}{k!}\\
&=  \frac{e^{-\lambda}\lambda^{x_i+x_j}p_i^{x_i}p_j^{x_j}}{x_i!x_j!}e^{\lambda(1-p_i-p_j)}\\
&=  \frac{e^{-\lambda p_i}(\lambda p_i)^{x_i}}{x_i!}\frac{e^{-\lambda p_j}(\lambda p_j)^{x_j}}{x_j!}\\
\end{align}
After these exausting computations, we have a great result: $X_{ti}$ and $X_{tj}$ are marginally independent. Moreover, their distributions is $X_{ti} \sim \mbox{Poisson}(\lambda p_i)$ for all $i \, \in \, \{1,\ldots, k\}$, where $k$ is the number of types of vehicles.
With this, you can write the likelihood as a product of independent Poissons for each type of vehicle, where the number of observations for each will vary on how you choose the vehicle counting.
Estimation Write $\lambda_i = \lambda p_i$. Let $n_i$ be total number of days you chose to observe the $i$-th type of vehicle, and let $S_i$ be the sum of the observations of that vehicle type. The maximum likelihood estimator for $\lambda_i$ is
$$\hat{\lambda}_i = \frac{1}{n_i}S_i \quad,$$
That is, it is just the average of the observations. But we do not want to estimate $\lambda_i$, we want to estimate $p_i$. Well, we know that
$$1 = \sum_{i=1}^k p_i \quad.$$
Multiplying both sides by $\lambda$, we have
$$\lambda = \sum_{i=1}^k \lambda_i \quad.$$
By the invariance property of the MLE, we have
$$\hat{\lambda} = \sum_{i=1}^k \hat{\lambda}_i \quad.$$
But $p_i = \lambda_i/\lambda$, hence
$$\hat{p}_i = \frac{\hat{\lambda}_i}{\hat{\lambda}} \quad.$$
Therefore, we can estimate the proportions $p_i$, and also the expected number of vehicles $\lambda$!
Your example To show that this approach might work, lets compute each parameter in your data example:
\begin{align}
&\hat{\lambda}_{truck} = (30+25+44)/3 = 33\\
&\hat{\lambda}_{car} = (65+72)/2 = 68.5\\
&\hat{\lambda}_{cycle} = (12+19+14)/3 = 15\\
&\hat{\lambda}_{van} = (12+11)/2 = 11.5
\end{align}
For $\lambda$, we have
$$ \hat{\lambda} = \hat{\lambda}_{truck} + \hat{\lambda}_{car} + \hat{\lambda}_{cycle} + \hat{\lambda}_{van} = 128 \quad.$$
For the probabilities
\begin{align}
&\hat{p}_{truck} = \hat{\lambda}_{truck}/\hat{\lambda} = 0.2578\\
&\hat{p}_{car} = \hat{\lambda}_{car}/\hat{\lambda} = 0.5351\\
&\hat{p}_{cycle} = \hat{\lambda}_{cycle}/\hat{\lambda} = 0.1171\\
&\hat{p}_{van} = \hat{\lambda}_{van}/\hat{\lambda} = 0.0898
\end{align}
The probabilities are strinkingly similar to those provided by @whuber. The difference is that it is very very easy to compute it, no optimization required.
Final Analysis Here is a final overall comparison when considering this approach.
Advantages:

*

*Its very easy to compute the estimators, they are analytical;

*You probably can perform hypothesis testing, if you wish;

*You can estimate the total number of vehicle.

Disadvantages:

*

*We assumed that $N_t$ does not vary with the day, which might be false due to weekly or monthly seasonality;

*I do not know how you could check if the Poisson distribution is adequate for $N_t$ with the data at hand.

