The answer is pretty clear with Bayes’ Theorem. The world in which the coin lands heads and you get the card has probability 0.0000000005, and the world in which the coin lands tails has probability 0.5. Thus you live a world with a prior probability of 0.5000000005, so the probability of the coin being heads is 0.0000000005/0.5000000005, or a little under 1 in a billion.
Given that the worst case scenario of losing the bet is saying you can’t pay it and losing credibility, you and Adam should take the bet. If you want to (or have to) actually commit to paying, then you have to decide whether you would completely screw over 1 alternate self so that a billion selves can have a bit more money. Given that $100 would not really make a difference to my life in the long run, I think I would not take the bet in this scenario.
Some properties that I notice about semistable equilibria:
It is non-differentiable, so any semsistable equilibrium that occurs in reality is only approximate.
If the zone of attraction and repulsion are the same state, random noise will inevitably cause the state to hop over to the repulsive side. So what a ‘perfect’ semistable equilibrium will look like is a system where the state tends towards some point, hangs around for a while, and then suddenly flies off to the next equilibrium. This makes me think of the Gömböc.
A more approximate semsistable equilibrium that has an actual stable point in reality will be one that has a stable equilibrium at one point, and an unstable equilibrium soon after. I think an example of this is a neutron star. A neutron star is stable because gravity pulls the matter inward while the nuclear forces push outward. With more compression however, gravity overcomes these forces and a black hole forms, after which the entire star will collapse.