Yes, losing worlds also branch, of course. But the one world where she has won wins her $2**n, and that world exists with probability 0.6**n.
So her EV is always ($2**n)*(0.6**n), which is a larger EV (with any n) than a strategy where she doesn’t bet everything every single time. I argue that even as n goes to infinity, and even as probability approaches one that she has lost everything, it’s still rational for her to have that strategy because the $2**n that she won in that one world is so massive that it balances out her EV. Some infinities are much larger than others.
I don’t think it’s correct to say that as n gets large her strategy is actually worse than Logan’s under her own utility function. I think hers is always best under her own utility function.
Yes, losing worlds also branch, of course. But the one world where she has won wins her $2**n, and that world exists with probability 0.6**n.
So her EV is always ($2**n)*(0.6**n), which is a larger EV (with any n) than a strategy where she doesn’t bet everything every single time. I argue that even as n goes to infinity, and even as probability approaches one that she has lost everything, it’s still rational for her to have that strategy because the $2**n that she won in that one world is so massive that it balances out her EV. Some infinities are much larger than others.
I don’t think it’s correct to say that as n gets large her strategy is actually worse than Logan’s under her own utility function. I think hers is always best under her own utility function.