This really works to make CDT decisions! Try thinking through what the market would do in various decision-theoretic problems.
I thought through whether it works in Newcomb’s problem and it was unexpectedly complicated and confusing (see below) and I now doubt that it always recovers CDT in Newcomblike problems. I may have done something wrong.
Newcomblike problems are obviously not the reason we want casual decision markets. But more generally I’m realising now that the relationship between evidential and causal decision markets is quite different from the relationship between EDT and CDT as normally conceived. EDT and CDT agree with each other in everyday problems and only disagree in Newcomblike problems, whereas evidential and causal decision markets disagree in everyday problems.
So perhaps ‘EDT’ and ‘CDT’ are not good terms to use when talking about decision markets.
Does the proposed scheme get us CDT in Newcomblike problems?
Consider Newcomb’s problem. Say I make markets for “If I twobox, will I get the million?” and “If I onebox, will I get the million?” and follow the randomisation scheme. In the nonrandomised cases, I should twobox if 1000000p1+1000>1000000p2, where p1 is the first market probability and p2 is the second. That is, if p1+1/1000>p2.
Let’s assume omega can’t predict when I’ll randomise or what the outcome will be, and so always fills the boxes according to what I do in the nonrandomised cases.
If p1+1/1000>p2, then I’ll twobox in the nonrandomised cases, so the box will be empty even in the randomised cases, so both the twobox and the onebox market will resolve NO and so both p1 and p2 should be bet down.
If p1+1/1000≤p2, then I’ll onebox in the nonrandomised cases, so the box will always be full, and both the twobox and onebox markets will resolve YES and so both p1 and p2 should be bet up.
I think the only equilibrium here is if p1 and p2 are both 0, in which case I’ll twobox and in the randomised cases both markets will correctly resolve NO. That does agree with CDT in the end but it’s kind of a weird way to get there. Not sure if it generalises.
Alternatively we could assume that omega can predict the randomisation. But then the twobox market will be at 0% and the onebox at 100%, so I would onebox.
I thought through whether it works in Newcomb’s problem and it was unexpectedly complicated and confusing (see below) and I now doubt that it always recovers CDT in Newcomblike problems. I may have done something wrong.
Newcomblike problems are obviously not the reason we want casual decision markets. But more generally I’m realising now that the relationship between evidential and causal decision markets is quite different from the relationship between EDT and CDT as normally conceived. EDT and CDT agree with each other in everyday problems and only disagree in Newcomblike problems, whereas evidential and causal decision markets disagree in everyday problems.
So perhaps ‘EDT’ and ‘CDT’ are not good terms to use when talking about decision markets.
Does the proposed scheme get us CDT in Newcomblike problems?
Consider Newcomb’s problem. Say I make markets for “If I twobox, will I get the million?” and “If I onebox, will I get the million?” and follow the randomisation scheme. In the nonrandomised cases, I should twobox if 1000000p1+1000>1000000p2, where p1 is the first market probability and p2 is the second. That is, if p1+1/1000>p2.
Let’s assume omega can’t predict when I’ll randomise or what the outcome will be, and so always fills the boxes according to what I do in the nonrandomised cases.
If p1+1/1000>p2, then I’ll twobox in the nonrandomised cases, so the box will be empty even in the randomised cases, so both the twobox and the onebox market will resolve NO and so both p1 and p2 should be bet down.
If p1+1/1000≤p2, then I’ll onebox in the nonrandomised cases, so the box will always be full, and both the twobox and onebox markets will resolve YES and so both p1 and p2 should be bet up.
I think the only equilibrium here is if p1 and p2 are both 0, in which case I’ll twobox and in the randomised cases both markets will correctly resolve NO. That does agree with CDT in the end but it’s kind of a weird way to get there. Not sure if it generalises.
Alternatively we could assume that omega can predict the randomisation. But then the twobox market will be at 0% and the onebox at 100%, so I would onebox.