RobinZ comments on Newcomb's Problem and Regret of Rationality - Less Wrong

64 Post author: Eliezer_Yudkowsky 31 January 2008 07:36PM

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Comment author: nhamann 26 July 2010 05:54:58PM *  7 points [-]

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.

Comment author: RobinZ 26 July 2010 09:00:46PM *  3 points [-]

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.