“If I estimate a probability of less than ~99.9001% ($1,000,000/$1,001,000) that Omega will be correct in this specific instance, I one-box; ”
??? Your calculation seems to be trying to divide the wrong things. One boxing gives you $1,000,000 if Omega is right, gives you $0 if Omega is wrong. Two boxing gives you $1,001,000 if Omega is wrong, gives you $1,000 if Omega is right.
So with Omega being X likely to be right: -the estimated payoff for one-boxing is X $1,000,000 -the estimated payoff for two-boxing is (100-X) ($1,001,000) + X * $1000.
One-boxing is therefore superior (assuming linear utility of money) when X $1,000,000 > (100-X) ($1,001,000) + X * $1000.
One-boxing is therefore superior (always assuming linear utility of money) when Omega has a higher than X> 50.05% likelihood of being right.
??? Your calculation seems to be trying to divide the wrong things.
One boxing gives you $1,000,000 if Omega is right, gives you $0 if Omega is wrong.
Two boxing gives you $1,001,000 if Omega is wrong, gives you $1,000 if Omega is right.
So with Omega being X likely to be right:
-the estimated payoff for one-boxing is X $1,000,000
-the estimated payoff for two-boxing is (100-X) ($1,001,000) + X * $1000.
One-boxing is therefore superior (assuming linear utility of money) when X $1,000,000 > (100-X) ($1,001,000) + X * $1000.
One-boxing is therefore superior (always assuming linear utility of money) when Omega has a higher than X> 50.05% likelihood of being right.
Yeah, looking at it as a $500,000 bet on almost even money, odds of about 50% are right.