Let C be identically distributed to but statistically independent from A| you'll know A (not any specific value of A). C and A| you'll know A can each have infinite expected utility, by assumption, using an extended definition of “you” in which you get to expand arbitrarily, in extremely remote possibilities.C−$100 is also strictly stochastically dominated by A| you'll know A, so C−$100≺A| you'll know A.
Now, consider the following prospect:
With probability p=P[ you'll know A], it’s C−$100. With the other probability 1−p, it’sA| you'll know A. We can abuse notation to write this in short-hand as
A strictly stochastically dominates D, so D≺A. Then the rest of the money pump argument follows, replacing B−$100 with D, and assuming “Find out A” only works sometimes, but enough of the time that A| you'll know A still has infinite expected utility.[1] You don’t know ahead of time when “Find out A” will work, but when it does, you’ll switch to D, which would then be C−$100, and when “Find out A” doesn’t work, it makes no difference. So, your options become:
you (sometimes) pay $50 ahead of time and switch to A−$100 to avoid switching to the dominated D, which is a sure loss relative to sticking through with A when you do it and irrational.
you stick through with A (or the conditionally stochastically equivalent prospect X) sometimes when “Find out A” works, despite C−$100 beating the outcome of A you find out, which is irrational.
you always switch to C−$100 when “Find out A” works, which is a dominated strategy ahead of time, and so irrational.
Let C be identically distributed to but statistically independent from A| you'll know A (not any specific value of A). C and A| you'll know A can each have infinite expected utility, by assumption, using an extended definition of “you” in which you get to expand arbitrarily, in extremely remote possibilities.C−$100 is also strictly stochastically dominated by A| you'll know A, so C−$100≺A| you'll know A.
Now, consider the following prospect:
With probability p=P[ you'll know A], it’s C−$100. With the other probability 1−p, it’sA| you'll know A. We can abuse notation to write this in short-hand as
D=p(C−$100)+(1−p)(A|you won't know A)=p(C−$100)+(1−p)XThen, letting X=A| you'll know A, we can compare D to
A=p(A| you'll know A)+(1−p)(A|you won't know A)=p(A| you'll know A)+(1−p)XA strictly stochastically dominates D, so D≺A. Then the rest of the money pump argument follows, replacing B−$100 with D, and assuming “Find out A” only works sometimes, but enough of the time that A| you'll know A still has infinite expected utility.[1] You don’t know ahead of time when “Find out A” will work, but when it does, you’ll switch to D, which would then be C−$100, and when “Find out A” doesn’t work, it makes no difference. So, your options become:
you (sometimes) pay $50 ahead of time and switch to A−$100 to avoid switching to the dominated D, which is a sure loss relative to sticking through with A when you do it and irrational.
you stick through with A (or the conditionally stochastically equivalent prospect X) sometimes when “Find out A” works, despite C−$100 beating the outcome of A you find out, which is irrational.
you always switch to C−$100 when “Find out A” works, which is a dominated strategy ahead of time, and so irrational.
Or otherwise beats each of its actual possible outcomes.