Extending the Socialist millionaire problem to three parties
A tripartite version of EKE would appear to work.  Here's how it works:


We select a pairing friendly elliptic curve, with a point $G$ that generates a prime order subgroup.
We define an invertible keyed hash function $F_k(H)$ that maps (based on the key) a point in the subgroup to a point in the subgroup (and so that, for two different keys $k_1, k_2$, we don't know the discrete log of $F_{k_1}(H)$ to the base of $F_{k_2}(H)$


So, what Alice does is take her networth $n_A$, and selects a random value $a$; she computes $A = F_{n_A}(aG)$ and publishes it.

Similarly, Bob and Carol also publish their $B = F_{n_B}(bG)$ and $C = F_{n_C}(cG)$.

Alice computes $e( F_{n_A}^{-1}(B), F_{n_A}^{-1}(C) )^a$ and publishes that.  Bob and Carol do the same computations and publish their values.

If $n_A = n_B = n_C$, this joint computation is $e( F_{n_A}^{-1}(B), F_{n_A}^{-1}(C) )^a = e(bG, cG)^a = e(G,G)^{abc}$, and hence they will publish the exact same value.

If those three values are not the same, for example, if $n_A \ne n_B$, then $F_{n_A}^{-1}(B) = F_{n_A}^{-1}(F_{n_B}(bG))$ is effectively a random point, and so Alice will generate a value that is distinct from what anyone else generates.



On review, this idea doesn't quite work, if we have $n_A \ne n_B$ and $n_B = n_C$, then B and C will compute the same value.  I believe that, from the problem statement, we want to jointly compute the value $n_A = n_B \land n_B = n_C$; we end up leaking more than that...

I'll leave this answer open in case someone can fix it...