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.
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.