The Doubling Box

Let’s say you have a box that has a to­ken in it that can be re­deemed for 1 utilon. Every day, its con­tents dou­ble. There is no limit on how many utilons you can buy with these to­kens. You are im­mor­tal. It is sealed, and if you open it, it be­comes an or­di­nary box. You get the to­kens it has cre­ated, but the box does not dou­ble its con­tents any­more. There are no other ways to get utilons.

How long do you wait be­fore open­ing it? If you never open it, you get noth­ing (you lose! Good day, sir or madam!) and when­ever you take it, tak­ing it one day later would have been twice as good.

I hope this doesn’t sound like a re­duc­tio ad ab­sur­dum against un­bounded util­ity func­tions or not dis­count­ing the fu­ture, be­cause if it does you are in dan­ger of am­pu­tat­ing the wrong limb to save your­self from para­dox-gan­grene.

What if in­stead of grow­ing ex­po­nen­tially with­out bound, it de­cays ex­po­nen­tially to the bound of your util­ity func­tion? If your util­ity func­tion is bounded at 10, what if the first day it is 5, the sec­ond 7.5, the third 8.75, etc. As­sume all the lit­tle de­tails, like re­mem­ber­ing about the box, trad­ing in the to­kens, etc, are free.

If you dis­count the fu­ture us­ing any func­tion that doesn’t ever hit 0, then the growth rate of the to­kens can be cho­sen to more than make up for your dis­count­ing.

If it does hit 0 at time T, what if in­stead of dou­bling, it just in­creases by how­ever many utilons will be ad­justed to 1 by your dis­count­ing at that point ev­ery time of growth, but the in­ter­vals of growth shrink to noth­ing? You get an ad­justed 1 utilon at time T − 1s, and an­other ad­justed 1 utilon at T − 0.5s, and an­other at T − 0.25s, etc? Sup­pose you can think as fast as you want, and open the box at ar­bi­trary speed. Also, that what­ever solu­tion your pre­sent self pre­com­mits to will be fol­lowed by the fu­ture self. (Their de­ci­sion won’t be changed by any change in what times they care about)

EDIT: Peo­ple in the com­ments have sug­gested us­ing a util­ity func­tion that is both bounded and dis­count­ing. If your util­ity func­tion isn’t so strongly dis­count­ing that it drops to 0 right af­ter the pre­sent, then you can find some time in­ter­val very close to the pre­sent where the dis­count­ing is all nonzero. And if it’s nonzero, you can have a box that dis­ap­pears, tak­ing all pos­si­ble util­ity with it at the end of that in­ter­val, and that, lead­ing up to that in­ter­val, grows the util­ity in in­ter­vals that shrink to noth­ing as you ap­proach the end of the in­ter­val, and in­creas­ing the util­ity-worth of to­kens in the box such that it com­pen­sates for what­ever your dis­count­ing func­tion is ex­actly enough to asymp­tot­i­cally ap­proach your bound.

Here is my solu­tion. You can’t as­sume that your fu­ture self will make the op­ti­mal de­ci­sion, or even a good de­ci­sion. You have to treat your fu­ture self as a phys­i­cal ob­ject that your choices af­fect, and take the prob­a­bil­ity dis­tri­bu­tion of what de­ci­sions your fu­ture self will make, and how much util­ity they will net you into ac­count.

Think if your­self as a Tur­ing ma­chine. If you do not halt and open the box, you lose and get noth­ing. No mat­ter how com­pli­cated your brain, you have a finite num­ber of states. You want to be a busy beaver and take the most pos­si­ble time to halt, but still halt.

If, at the end, you say to your­self “I just counted to the high­est num­ber I could, count­ing once per day, and then made a small mark on my skin, and re­peated, and when my skin was full of marks, that I was con­stantly re­fresh­ing to make sure they didn’t go away...

...but I could let it dou­ble one more time, for more util­ity!”

If you re­turn to a state you have already been at, you know you are go­ing to be wait­ing for­ever and lose and get noth­ing. So it is in your best in­ter­est to open the box.

So there is not a uni­ver­sal op­ti­mal solu­tion to this prob­lem, but there is an op­ti­mal solu­tion for a finite mind.

I re­mem­ber read­ing a while ago about a para­dox where you start with $1, and can trade that for a 50% chance of $2.01, which you can trade for a 25% chance of $4.03, which you can trade for a 12.5% chance of $8.07, etc (can’t re­mem­ber where I read it).

This is the same para­dox with one of the traps for wannabe Cap­tain Kirks (us­ing dol­lars in­stead of utilons) re­moved and one of the un­nec­es­sary vari­ables (un­cer­tainty) cut out.

My solu­tion also works on that. Every trade is analo­gous to a day waited to open the box.