Estimating the number of times each of four pairs of dice was thrown PROBLEM
Suppose I have four dice: red (R), green (G), blue (B) and yellow (Y). I throw these dice several times. Each throw uses two of the four dice, and only the following pairs are allowed:


*

*RB

*RY

*GB

*GY 


After each throw I note which dies rolled a 'one'. Across the four allowed pairs there are nine possible results:


*

*Both R and B rolled a 'one' 

*Both R and Y rolled a 'one'

*Both G and B rolled a 'one'

*Both G and Y rolled a 'one'

*Only R rolled a 'one'

*Only G rolled a 'one' 

*Only B rolled a 'one'

*Only Y rolled a 'one'

*Neither of the dice rolled a 'one'


As I throw the dice, I make a tally of how many times I got each of results 1 through 8. I do not make a tally of how many times I got result 9. 
At the end of this process I give you my tallies, so that you know the number of times I got each of the first eight results. You also know the pairs that were allowed and the probability of rolling a 'one' (which is the same for all dice). However, you know neither the total number of throws I made, nor the number of times I got result 9.
Using this information, your problem is to estimate the number of times I threw each of the four allowed pairs.   
Some things to note:


*

*Order does not matter; both dice are thrown together.

*Multiple allowed pairings can have the same result; for example, both RB and RY can have result 5.


QUESTIONS


*

*Is the problem solvable?

*How would you solve this problem?

*What are some names for this type of problem that would be useful search terms if I want to look for more information?

 A: Let p be the probability of throwing  a 1 on any die. Let ni be the probability of the outcome designated by i.
The probability of the occurrences 1,2,3 and 4 are all p$^2$ as the dice are rolled independently. The probability of occurrences 5, 6, 7, and 8 are all 2p(1-p) since for example in case 5 a red must roll a 1 and the blue roll a number other than 1 or red rolls 1 and the yellow die rolls a number other than 1.
Hence E(n1)=E(n2)=E(n3)=E(n4)=Np$^2$
and E(n5)=E(n6)=E(n7)=E(n8)=2Np(1-p).
Estimate N by equating the nis with their expectations to get since
n9=N-(n1+n2+n3+n4+n5+n6+n7+n8)
and E(n1+n2+n3+n4)=4Np$^2$ 
and E(n5+n6+n7+n8) =4(2Np(1-p))
E(n9)=N-4N(p$^2$ +2p(1-p))=
N(1-4(p$^2$+2p-2p$^2$))=N(1-8p+4p$^2$)
This shows that if N were known I could estimate n9 by the nearest integer to N(1-8p+4p$^2$).  Or on the other hand if I knew n9 I could estimate N by the nearest integer to n9/(1-8p+4p$^2$).  
But if I did not know N and I also do not know n9 then for any solution I could multiply N and n9 by any integer (say 2 or 5) and get another answer.  So without additional information I cannot find a sensible and unique estimate for n9.
A: Let $p$ be the probability of any die rolling a one. Then the probability of outcomes 1, 2, 3 and 4  are all $p^2$. Let $n_1$, $n_2$, $n_3$ and $n_4$ be number of observations for outcomes 1 through 4. Let $n_{rb}$, $n_{ry}$, $n_{gb}$ and $n_{gy}$ be the values we want to estimate, which are the actual number of times each of the dice pairs was thrown. Then the best estimates are:
$$n_{rb} = \frac{n_1}{p^2}$$
$$n_{ry} = \frac{n_2}{p^2}$$
$$n_{gb} = \frac{n_3}{p^2}$$
$$n_{gy} = \frac{n_4}{p^2}$$
