# Gary_Drescher comments on A problem with Timeless Decision Theory (TDT)

• I already saw the \$1M, so, by two-box­ing, aren’t I just choos­ing to be one of those who see their E mod­ule out­put True?

Not if a coun­ter­fac­tual con­se­quence of two-box­ing is that the large box (prob­a­bly) would be empty (even though in fact it is not empty, as you can already see).

That’s the same ques­tion that comes up in the origi­nal trans­par­ent-boxes prob­lem, of course. We prob­a­bly shouldn’t try to re­cap that whole de­bate in the mid­dle of this thread. :)

• That’s the same ques­tion that comes up in the origi­nal trans­par­ent-boxes prob­lem, of course. We prob­a­bly shouldn’t try to re­cap that whole de­bate in the mid­dle of this thread. :)

Don’t worry; I don’t want to do that :). If I re­call the origi­nal trans­par­ent-boxes prob­lem cor­rectly, I agree with you on what to do in that case.

Just to check my mem­ory, in the origi­nal prob­lem, there are two trans­par­ent boxes, A and B. You see that A con­tains \$1M and B con­tains \$1000. You know that B nec­es­sar­ily con­tains \$1000, but A would have con­tained \$1M iff it were the case that you will de­cide to take only A. Other­wise, A would have been empty. The con­clu­sion (with which I agree) is that you should take only A. Is that right? (If I’m mis­re­mem­ber­ing some­thing cru­cial, is there a link to the full de­scrip­tion on­line?) [ETA: I see that you added a de­scrip­tion to your post. My rec­ol­lec­tion above seems to be con­sis­tent with your de­scrip­tion.]

In the origi­nal prob­lem, if we use the “many choosers” heuris­tic, there are no choosers who two-box and yet who get the \$1M. There­fore, you can­not “choose to be” one of them. This is why two-box­ing should have no ap­peal to you.

In con­trast, in your new prob­lem, there are two-box­ers who get the \$1M and who get their E mod­ule to out­put True. So you can “choose to be” one of them, no? And since they’re the biggest win­ners, that’s what you should do, isn’t it?