Yeah, that seems key. Even if the probability that Linda will eventually get 0 money approaches 1, that small slice of probability in the universe where she has always won is approaching an infinity far larger that Logan’s infinity as the number of games approaches infinity. Some infinities are bigger than others. Linear utility functions and discount rates of zero necessarily deal with lots of infinities, especially in simplified scenarios.
Linda can always argue that in every universe where she lost everything, there’s more (6 vs 4) universes where her winnings were double what they would have been had she not taken that bet.
Linda can always argue that in every universe where she lost everything, there’s more (6 vs 4) universes where her winnings were double what they would have been had she not taken that bet.
It’ll actually look more like this: [60 worlds where Linda has won each bet:40 where she has lost] → [36:64] → [21.6:78.4] → etc. If you’re invoking manyworlds branching, note that the losing worlds also continue to branch, so the ratios will still be what I wrote
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.
Yeah, that seems key. Even if the probability that Linda will eventually get 0 money approaches 1, that small slice of probability in the universe where she has always won is approaching an infinity far larger that Logan’s infinity as the number of games approaches infinity. Some infinities are bigger than others. Linear utility functions and discount rates of zero necessarily deal with lots of infinities, especially in simplified scenarios.
Linda can always argue that in every universe where she lost everything, there’s more (6 vs 4) universes where her winnings were double what they would have been had she not taken that bet.
It’ll actually look more like this: [60 worlds where Linda has won each bet:40 where she has lost] → [36:64] → [21.6:78.4] → etc. If you’re invoking manyworlds branching, note that the losing worlds also continue to branch, so the ratios will still be what I wrote
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.