Sure. So what? If Omega is just 99.99999999999% of the time correct, how does that change in practice whether you should one-box or two-box?
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; otherwise I two-box. With a prior of 13 nines, getting down to three would require ten decades of evidence; if I shared any feature with one person who Omega wrongly identified as a one-boxer but not with the first 10^10 people who Omega correctly identified as a two-boxer, I think that would be strong enough evidence.
As a programmer, I’ll tell you it’s equally determistic whether you multiply 3 5 every time, or if you only multiply it once, store it in a variable and then return it when asked about the product of 35...
Unless you are doing the math on a Pentium processor...
“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.
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; otherwise I two-box. With a prior of 13 nines, getting down to three would require ten decades of evidence; if I shared any feature with one person who Omega wrongly identified as a one-boxer but not with the first 10^10 people who Omega correctly identified as a two-boxer, I think that would be strong enough evidence.
Unless you are doing the math on a Pentium processor...
??? 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.