If Omega is fallible, then the value of one-boxing falls drastically, and even adjusting the amount of money doesn’t help in the end;
Assume Omega has a probability X of correctly predicting your decision:
If you choose to two-box:
X chance of getting $1000
(1-X) chance of getting $1,001,000
If you choose to take box B only:
X chance of getting $1,000,000
(1-X) chance of getting $0
Your expected utilities for two-boxing and one-boxing are (respectively):
E2 = 1000X + (1-X)1001000 E1 = 1000000X
For E2 > E1, we must have 1000X + 1,001,000 − 1,001,000X − 1,000,000X > 0, or 1,001,000 > 2,000,000X, or
X < 0.5005
So as long as Omega can maintain a greater than 50% accuracy, you should expect to earn more money by one-boxing. Since the solution seems so simple, and since I’m a total novice at decision theory, it’s possible I’m missing something here, so please let me know.
So as long as Omega can maintain a greater than 50% accuracy, you should expect to earn more money by one-boxing. Since the solution seems so simple, and since I’m a total novice at decision theory, it’s possible I’m missing something here, so please let me know.
Your caclulation is fine. What you’re missing is that Omega has a record of 70% accuracy because Omega always predicts that a person will one-box and 70% of people are one-boxers. So Omega always puts the million dollars in Box B, and I will always get $1,001,000$ if I’m one of the 30% of people who two-box.
At least, that is a possibility, which your calculation doesn’t take into account. I need evidence of a correlation between Omega’s predictions and the participants’ actual behaviour, not just evidence of correct predictions. My prior probability distribution for how often people one-box isn’t even concentrated very tightly around 70% (which is just a number that I remember reading once as the result of one survey), so anything short of a long run of predictions with very high proportion of correct ones will make me suspect that Omega is pulling a trick like this.
So the problem is much cleaner as Eliezer states it, with a perfect record. (But if even that record is short, I won’t buy it.)
Oops, I see that RobinZ already replied, and with calculations. This shows that I should still remove the word ‘drastically’ from the bit that nhamann quoted.
Wait—we can’t assume that the probability of being correct is the same for two-boxing and one-boxing. Suppose Omega has a probability X of predicting one when you choose one and Y of predicting one when you choose two.
E1 = E($1 000 000) * X
E2 = E($1 000) + E($1 000 000) * Y
The special case you list corresponds to Y = 1 - X, but in the general case, we can derive that E1 > E2 implies
X > Y + E($1 000) / E($1 000 000)
If we assume linear utility in wealth, this corresponds to a difference of 0.001. If, alternately, we choose a median net wealth of $93 100 (the U.S. figure) and use log-wealth as the measure of utility, the required difference increases to 0.004 or so. Either way, unless you’re dead broke (e.g. net wealth $1), you had better be extremely confident that you can fool the interrogator before you two-box.
Assume Omega has a probability X of correctly predicting your decision:
If you choose to two-box:
X chance of getting $1000
(1-X) chance of getting $1,001,000
If you choose to take box B only:
X chance of getting $1,000,000
(1-X) chance of getting $0
Your expected utilities for two-boxing and one-boxing are (respectively):
E2 = 1000X + (1-X)1001000
E1 = 1000000X
For E2 > E1, we must have 1000X + 1,001,000 − 1,001,000X − 1,000,000X > 0, or 1,001,000 > 2,000,000X, or
X < 0.5005
So as long as Omega can maintain a greater than 50% accuracy, you should expect to earn more money by one-boxing. Since the solution seems so simple, and since I’m a total novice at decision theory, it’s possible I’m missing something here, so please let me know.
Your caclulation is fine. What you’re missing is that Omega has a record of 70% accuracy because Omega always predicts that a person will one-box and 70% of people are one-boxers. So Omega always puts the million dollars in Box B, and I will always get $1,001,000$ if I’m one of the 30% of people who two-box.
At least, that is a possibility, which your calculation doesn’t take into account. I need evidence of a correlation between Omega’s predictions and the participants’ actual behaviour, not just evidence of correct predictions. My prior probability distribution for how often people one-box isn’t even concentrated very tightly around 70% (which is just a number that I remember reading once as the result of one survey), so anything short of a long run of predictions with very high proportion of correct ones will make me suspect that Omega is pulling a trick like this.
So the problem is much cleaner as Eliezer states it, with a perfect record. (But if even that record is short, I won’t buy it.)
Oops, I see that RobinZ already replied, and with calculations. This shows that I should still remove the word ‘drastically’ from the bit that nhamann quoted.
Wait—we can’t assume that the probability of being correct is the same for two-boxing and one-boxing. Suppose Omega has a probability X of predicting one when you choose one and Y of predicting one when you choose two.
The special case you list corresponds to Y = 1 - X, but in the general case, we can derive that E1 > E2 implies
If we assume linear utility in wealth, this corresponds to a difference of 0.001. If, alternately, we choose a median net wealth of $93 100 (the U.S. figure) and use log-wealth as the measure of utility, the required difference increases to 0.004 or so. Either way, unless you’re dead broke (e.g. net wealth $1), you had better be extremely confident that you can fool the interrogator before you two-box.