Sure. Not being a mathist, I only have my engineering intuition here.
An intuition is that:
- As the number of participants increases, the probability one of them has a reason to force-close must increase. Assuming there is some reasonable average (mean) probability of force-close at any moment of time,
P(c). Then for N participants, the probability of force-close at any moment of time is1 - (1 - P(c))^n. - More concretely: HTLCs are the most common reason for force-closing. In particular, once you participate in public routing, some fraction of your incoming HTLCs can be held hostage by your outgoing HTLCs, for a time. If the outgoing HTLCs are not resolved, you need to drop them onchain. If it resolves onchain, great, but if the outgoing HTLC is not confirmed onchain, or you, by complete accident, are forced offline (e.g. clumsy human foot kicking power supply), then your incoming HTLC is also dropped onchain.
- Dropping an HTLC onchain usually means the hosting mechanism is also dropped onchain, i.e. force-closed.
- Force-close due to HTLC timeout affects all participants in the mechanism; usually, the mechanism cannot continue operating offchain if any hosted HTLC has to be dropped onchain due to the accepting node going offline.
Thus, unless a multi-party mechanism can drop individual HTLCs onchain and continue operating, then N-party mechanisms are more brittle than 2-party channels.
It is probably measurable how much uptime a random node gets (just try to connect to a bunch of them, divide the successes by the number of attempts), and from there, we can assume that probability that an HTLC will time out due to timeout based on the probability of a random node being offline. We can then compute the brittleness of N-party mechanisms as we vary N.