6. A simpler version

I will discuss a version of Newcomb’s problem where you must choose between taking box A and box B, rather than between both boxes and box B.

This is slightly more natural and it will also help me explain my position.

So here’s my simplified version of Newcomb’s problem. Two boxes A and B stand before you and you know that one of these two situations obtains, though not which one:

A contains $1,000 while B contains nothing.
A contains $1,001,000 while B contains $1,000,000.

In short, B contains either nothing or a million dollars and A always contains a thousand dollars more.

You may take exactly one box and keep its contents. However, if the man predicted you would take A, he has (already) put you in the first situation, whereas if he predicted you would take B, he has (already) put you in the second.

Everything else is as before. Should you take A or B?

Since you believe the man has correctly predicted your choice, you must accept that if you take B, the second situation will prove to obtain, earning you a million dollars, whereas if you take A, the first situation will prove to obtain, earning you only a thousand. So it seems that you should take B. (Same as one-boxing in the original problem.)

On the other hand, whatever this man’s predictive powers, the contents of the boxes are already fixed, and since you know that A always contains a thousand dollars more than B, it seems that you should definitely take A. (Two-boxing.)

In what follows, I won’t discuss the case for taking B, which I believe is essentially correct. My concern is with the case for taking A, which appears as convincing as ever. What’s wrong with it?