1 box then 2 box = $1,001,000
1 box then 1 box = $1,000,000
2 box then 2 box = $1,000
2 box then 1 box = $0
$2,002,000 divided by 4 is $500,500. Effectively you’re betting a million dollars on two coinflips, the first to get your money back (1-box on the first day) and the second to get $1000 (2-box on the second day). Omega could just use a randomizer if it thinks you will, in which case people would say “Omega always guesses right, unless you use a randomizer. But it’s stupid to use one anyway.”
Where p is the probability of 1 boxing, $E = p^2 $1,000,000 + p(1-p $1,001,000 + (1-p)^2 * $1,000 = $999,000 p + $1000.
So the smart thing to do is clearly always one-box, unless showing up Omega who thinks he’s so big is worth $499,500 to you.
I completely agree that to maximise your expected gain you should one-box every time. I was thinking of the specific case where you really, really need $1001,000 and are willing to reduce your expected gain to maximise the chance of getting it.
Hang on a minute though
1 box then 2 box = $1,001,000 1 box then 1 box = $1,000,000 2 box then 2 box = $1,000 2 box then 1 box = $0
$2,002,000 divided by 4 is $500,500. Effectively you’re betting a million dollars on two coinflips, the first to get your money back (1-box on the first day) and the second to get $1000 (2-box on the second day). Omega could just use a randomizer if it thinks you will, in which case people would say “Omega always guesses right, unless you use a randomizer. But it’s stupid to use one anyway.”
Where p is the probability of 1 boxing, $E = p^2 $1,000,000 + p(1-p $1,001,000 + (1-p)^2 * $1,000 = $999,000 p + $1000. So the smart thing to do is clearly always one-box, unless showing up Omega who thinks he’s so big is worth $499,500 to you.
I completely agree that to maximise your expected gain you should one-box every time. I was thinking of the specific case where you really, really need $1001,000 and are willing to reduce your expected gain to maximise the chance of getting it.