Let's say I have 3 objects that can belong to some type (color, size, whatever you like). However, I'm not certain (drunk or blind, lol) whether or not each pair of objects belongs to the same type. The probability $p_{12}$ measures this uncertainty, namely, tells me how frequently the objects 1 and 2 are indeed of the same type.
My question is: what are the probabilities (which should sum up to 1) that all of the objects are of different type, encoded as $P_{123}$, two of the same type but the third is not, encoded as $P_{(12)3}$, $P_{1(23)}$ and $P_{(13)2}$, and that all of them are of the same type, i.e., $P_{(123)}$.
The problem here is that this "sameness" is actually an equivalence relation, which means that if 1~2 and 2~3, then definitely 1~3, which should affect the probabilities.
This can not be determined from the given data.
To see this, consider two scenarios.
$I$. Suppose that each object is one of three types, independently and uniformly. In that case, $P_{ij}=\frac 13$ for $i\neq j$. In that case, the probability that all three match is, of course, $\frac 19$.
$II$. Suppose there are only two types, $H, T$ and that the objects might be $HHT, HTH, THH$ with equal probability. In that case we again have $P_{ij}=\frac 13$ for $i\neq j$. But now it is impossible for the three to match.
If it troubles you that we have three types in the first scenario but only two in the second, rewrite the second to have cases $AAB, BBC, CAA$, for example. That version has the nice property that, as in $I$, each object is equally likely to be any of the three types (though, of course, the probabilities are far from independent).