Let P(BS) = probability of a Black Swan being involved
This makes the average payout work out to:
1-Box = $1,000,000
2-Box = $1,001,000 (1 - P(BS)) + $1,000 P(BS)
Now it seems to be that the average 2-boxer is assuming that P(BS) = 0,
which would make the 2-Box solution always == $1,001,000 which would, of
course, always beat the 1-box solution.
and maybe in this toy-problem, they’re right to assume P(BS) = 0
But IRL that’s almost never the case—after all, 0 is not a probability
yes?
So assume that P(BS) is non-zero. t what point would it be worth it to choose the 1-Box solution and what point the 2-Box solution? Lets run the math:
So, the estimated probability of Black Swan existing only has to be greater than 0.0998% for the 1-Box solution to have a greater expected payout and therefore the 1-Box option is the more rational::Bayesian choice
OTOH, if you can guarantee that P(BS) is less than 0.0998%, then the rational choice is to 2-Box.
So working the math on that
Let P(BS) = probability of a Black Swan being involved
This makes the average payout work out to:
1-Box = $1,000,000
2-Box = $1,001,000 (1 - P(BS)) + $1,000 P(BS)
Now it seems to be that the average 2-boxer is assuming that P(BS) = 0, which would make the 2-Box solution always == $1,001,000 which would, of course, always beat the 1-box solution.
and maybe in this toy-problem, they’re right to assume P(BS) = 0 But IRL that’s almost never the case—after all, 0 is not a probability yes?
So assume that P(BS) is non-zero. t what point would it be worth it to choose the 1-Box solution and what point the 2-Box solution? Lets run the math:
1,000,000 = 1,001,000(1-x) + 1000x = 1001000 − 1001000x + 1000x = 1001000 - (1002000x)
=> 1000000 − 1001000 = −1002000x
=> x = −1000/-100200
=> x = 0.000998004
So, the estimated probability of Black Swan existing only has to be greater than 0.0998% for the 1-Box solution to have a greater expected payout and therefore the 1-Box option is the more rational::Bayesian choice
OTOH, if you can guarantee that P(BS) is less than 0.0998%, then the rational choice is to 2-Box.