No, I can get $1001,000. If I randomly choose to take one box the first time, then both boxes will contain money the second time, where I might randomly choose to take both.
(Unless randomising devices are all somehow forced to come up with the same result both times)
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.
The first time you enter the room, the boxes are both empty, so you can’t ever get more than $1,000,000. But you’re otherwise correct.
No, I can get $1001,000. If I randomly choose to take one box the first time, then both boxes will contain money the second time, where I might randomly choose to take both.
(Unless randomising devices are all somehow forced to come up with the same result both times)
Sorry, my mistake. I misread the OP.
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.