It's a little more complicated than that. You have to be summing everyone who is still tied to all the previous tracks. It needs to be a geometric sum formula.

[–] sabazius@lemmy.world 7 points 2 years ago* (last edited 2 years ago) (1 children)

It'll just be one fewer junctions. 2^n is always one more than the sum of 2^1+...2^(n-1)

[–] Magikjak@lemmy.world 1 points 2 years ago

I think you have to include 2^0 for that to be true?

e.g 2^0 = 1, 2^1 = 2 2^0 + 2^1 = 1 + 2 = 3, 2^2 = 4 … 7, 8 15,16 31, 32 etc.

[–] ghariksforge@lemmy.world 3 points 2 years ago* (last edited 2 years ago) (1 children)

2^33 is approximately 8.5 billion, which is roughly the population of the world.

[–] name_NULL111653@pawb.social 5 points 2 years ago (1 children)

So the 32nd person decides to either kill half of humanity and end the scenario, or give someone the power to end human kind...

[–] ghariksforge@lemmy.world 1 points 2 years ago

🤣