Thank you for the input. I think point 4 is the key problem.
> The overall average payoff is not the simple combination of probability at X times payoff at X plus probability at Y times payoff at Y.
This is hard to justify by usual probability calculations. Putting it into the context of the sleeping problem we get:
“The overall probability for Heads is not the simple combination of probability “today is Monday” times p(Heads|Monday) plus the probability “today is Tuesday” times p(Heads|Tuesday).”
This is typically denied by most. And the Achilles’ heel of double-halving.
What I mean is that you can just calculate each of...
A: Fraction of decisions made at X
B: Average payoff of decisions made at X
C: Fraction of decisions made at Y
D: Average payoff of decisions made at Y
E: Average payoff of all runs (some of which have multiple decisions)
...and observe that AB + CD = E does not hold. The reason for this is that some payoffs are the result of multiple decisions.
The overall probability for Heads is not the simple combination of probability “today is Monday” times p(Heads|Monday) plus the probability “today is Tuesday” times p(Heads|Tuesday).
I’m not a double-halfer, so I don’t have a problem denying this.
I think the interesting thing is what does AB+CD actually means? If we treat the fraction of decisions at X as the probability here is X, same for Y, then AB+CD should be the expected payoff of this decision. Typically the best decision should be derived by maximizing it. But clearly, that leads to wrong results such as 4⁄9. So what’s wrong?
My position is that AB+CD is meaningless. It is a fallacious payoff because self-locating probabilities are invalid. This also resolves double-halfers problems. But let’s leave that aside for the time being.
If I understand correctly, your position is maximizing AB+CD is not correct. Because when deciding we should be maximizing the payoff of runs instead of this decision. Here I just want to point out the payoff for runs (planning utility function) does not use self-locating probabilites. You didn’t say if AB+CD is something meaningful or not.
Aumann thinks AB+CD is meaningful. However, maximizing it is wrong. He pointed out that the decision at X and at Y are causally disconnected, yet the two decisions ought to be the same. A symmetrical Nash equilibrium. So the correct decision is a stable point of AB+CD. The problem is when there are multiple stable points, which one is the optimal decision, the one with the highest AB+CD? Aumann says no. It should be the point that maximizes the planning payoff function (the function not using self-locating probabilities).
I am not convinced by this. First of all, it lacks compelling reason. The explanation is ad-hoc “due to the absentmindedness”. Second, by his reasoning, AB+CD effectively plays no part in the decision-making process. The decision maximizing planning utility function is always going to be a stable point of AB+CD, and it is always going to be chosen regardless of what value it gives for AB+CD. So the whole argument about AB+CD being meaningful lacks substantial support.
Each of A,B,C,D are measurable, and so AB+CD is measurable as well, but since that combination isn’t the expected value we want to be maximizing, I would certainly say it’s not useful and at that point I wouldn’t really care whether it’s meaningful (if it does “mean” anything, it would be something convoluted and not really relevant to the decision).
When I actually run the experiment many times, I find the following when using p=4/9:
The fraction of decisions made at X is indeed 1/(p+1), in this case 9⁄13.
Of all the times a choice is made at X, the average resulting payoff is 96⁄81 (as expected).
Of all the times a choice is made at Y, the average resulting payoff is 24⁄9 (as expected).
The overall average payoff is not the simple combination of probability at X times payoff at X plus probability at Y times payoff at Y.
The discrepancy is cause by the driver-at-Y having to “share” a portion of the expected payoff with the previous driver-at-X.
Thank you for the input. I think point 4 is the key problem.
> The overall average payoff is not the simple combination of probability at X times payoff at X plus probability at Y times payoff at Y.
This is hard to justify by usual probability calculations. Putting it into the context of the sleeping problem we get:
“The overall probability for Heads is not the simple combination of probability “today is Monday” times p(Heads|Monday) plus the probability “today is Tuesday” times p(Heads|Tuesday).”
This is typically denied by most. And the Achilles’ heel of double-halving.
What I mean is that you can just calculate each of...
A: Fraction of decisions made at X
B: Average payoff of decisions made at X
C: Fraction of decisions made at Y
D: Average payoff of decisions made at Y
E: Average payoff of all runs (some of which have multiple decisions)
...and observe that AB + CD = E does not hold. The reason for this is that some payoffs are the result of multiple decisions.
I’m not a double-halfer, so I don’t have a problem denying this.
I think the interesting thing is what does AB+CD actually means? If we treat the fraction of decisions at X as the probability here is X, same for Y, then AB+CD should be the expected payoff of this decision. Typically the best decision should be derived by maximizing it. But clearly, that leads to wrong results such as 4⁄9. So what’s wrong?
My position is that AB+CD is meaningless. It is a fallacious payoff because self-locating probabilities are invalid. This also resolves double-halfers problems. But let’s leave that aside for the time being.
If I understand correctly, your position is maximizing AB+CD is not correct. Because when deciding we should be maximizing the payoff of runs instead of this decision. Here I just want to point out the payoff for runs (planning utility function) does not use self-locating probabilites. You didn’t say if AB+CD is something meaningful or not.
Aumann thinks AB+CD is meaningful. However, maximizing it is wrong. He pointed out that the decision at X and at Y are causally disconnected, yet the two decisions ought to be the same. A symmetrical Nash equilibrium. So the correct decision is a stable point of AB+CD. The problem is when there are multiple stable points, which one is the optimal decision, the one with the highest AB+CD? Aumann says no. It should be the point that maximizes the planning payoff function (the function not using self-locating probabilities).
I am not convinced by this. First of all, it lacks compelling reason. The explanation is ad-hoc “due to the absentmindedness”. Second, by his reasoning, AB+CD effectively plays no part in the decision-making process. The decision maximizing planning utility function is always going to be a stable point of AB+CD, and it is always going to be chosen regardless of what value it gives for AB+CD. So the whole argument about AB+CD being meaningful lacks substantial support.
Each of A,B,C,D are measurable, and so AB+CD is measurable as well, but since that combination isn’t the expected value we want to be maximizing, I would certainly say it’s not useful and at that point I wouldn’t really care whether it’s meaningful (if it does “mean” anything, it would be something convoluted and not really relevant to the decision).