10 Stones

This makes me think of another question ....

Suppose we are told instead that $8$ of the stones have distinct weights and that $2$ have the same weight, (distinct from the other $8$ ), with all $45$ possible "equal-pairings" being equally likely. If as before $A \gt B$ and $B$ is greater than $C \to I$ , what is the probability that $J \gt B$ with this added condition? Is it less, the same, or greater than $1/5$ ?

I think that the complication here is that there is an asymmetry in the possible equal-pairings left after the initial set of weighings. We have eliminated $15$ of the $45$ , ( $A$ with 8 of the others and $B$ with 7 of the others, (and with $A$ as well but we've already counted that one)). 9 of the 30 pairings left can involve $J$ , so there is a $1/30$ chance of $J$ being paired with $A$ and the same of it being paired with $B$ . There is then a $21/30 = 7/10$ chance that $J$ 's weight is distinct, in which case there are 9 positions in the weight order it can be placed in, (as in this case 2 of $C \to I$ are paired), leading to a $2/9$ chance that it weighs more than $B$ . The final probability that $J\ \gt B$ is then

$\dfrac{9}{30} \times \dfrac{1}{9} + \dfrac{21}{30} \times \dfrac{2}{9} = \dfrac{17}{90} \lt \dfrac{1}{5}$ .

Does this sound reasonable? My concern is the $2/9$ value; it might be $2/10$ , but either way the final probability is less than $1/5$ . This did make me think of another question, but I'll leave that for another time. :)