Pascal’s Mugging and One-shot Problems

I’ve had some thoughts on Pas­cal’s Mug­ging which might be worth shar­ing. I’m as­sum­ing some fa­mil­iar­ity with Pas­cal’s Mug­ging in this post.


Be­fore we get re­ally started, let’s trans­form Pas­cal’s Mug­ging into a prob­lem that is eas­ier to rea­son about but still gets at the core idea.

First, for­get “util­ity”, for­get money, we’re max­imis­ing pa­per­clips. I know it’s kind of silly, but us­ing money or util­ity re­ally tends to muddy up think­ing. I was sur­prised how much eas­ier it was to rea­son about the prob­lem when I switched to pa­per­clips[0].

Now, here’s my ver­sion of the prob­lem. Sup­pose there’s a cas­ino, in which there is a googol­plex-sided die (10^(10^100) sides). You can pay 10 pa­per­clips to roll the die, which are de­stroyed. If you roll a 1 on the die, the cas­ino man­u­fac­tures 3^^^^3 pa­per­clips. Other­wise, noth­ing else hap­pens and you lost 10 pa­per­clips.

3^^^^3 is much, much, much larger than a googol­plex, so ex­pected util­ity max­imi­sa­tion over­whelm­ingly says you should play at this cas­ino if you’re try­ing to max­imise the num­ber of pa­per­clips.

(From this point on, I will re­fer to “ex­pected util­ity” as “EU”)


In some cases, EU is cor­rect. Here are 3 cases.

Case 1: If you ex­pect to live a googol­plex years, then pay­ing to roll the die a few times per year is a good idea, be­cause over that time frame the prob­a­bil­ity of win­ning at least once is very high.

Case 2: You have a nor­mal lifes­pan, but there are a googol­plex other pa­per­clip max­imisers on Earth. In this case, ev­ery­one plays, and the prob­a­bil­ity that at least one per­son wins is very high.

Case 3: You are the only pa­per­clip max­imiser in this uni­verse, but there are a googol­plex al­ter­nate uni­verses that con­tain al­ter­nate “you”s who are in the same situ­a­tion as you, and for whom the dice rolls are all in­de­pen­dent. In this case the prob­a­bil­ity that at least one “you” wins is high, so you should play.


Here is a differ­ent case.

Case 0: You are alone in ex­is­tence. There is no one else on earth, there are no al­ter­nate re­al­ities, you are liter­ally alone in the en­tirety of all ex­is­tence. This is the only de­ci­sion you will ever get the chance to make. Should you pay 10 pa­per­clips to roll the die?

I think it’s clear in Case 0 that you should not pay. If you pay, you lose 10 pa­per­clips[1] and, for all prac­ti­cal pur­poses, are cer­tain to lose. If you don’t pay, at least you get to keep your 10 pa­per­clips. Since we’re try­ing to max­imise the num­ber of pa­per­clips, the lat­ter wins.


The key differ­ence be­tween case 0 and the other 3 cases is that in case 0, you only get one chance to max­imise pa­per­clips. I’m call­ing this sort of sce­nario a one-shot prob­lem.

A one-shot prob­lem can ba­si­cally be de­scribed as try­ing to max­imise pa­per­clips by choos­ing from a finite set of choices, each choice be­ing mu­tu­ally ex­clu­sive and hav­ing a finite set of out­comes whose prob­a­bil­ity sums up to 1, and each out­come con­tain­ing a finite num­ber of pa­per­clips.

I think a key in­sight, which I don’t know enough to prove but which seems cor­rect, is that any finite se­quence of de­ci­sions can be trans­formed into a one-shot prob­lem, sim­ply by view­ing each pos­si­ble se­quence of de­ci­sions as a sin­gle choice.

As a sim­ple ex­am­ple, if choos­ing be­tween ei­ther flip­ping a coin or rol­ling a die is one de­ci­sion, then flip­ping a coin and then rol­ling a die is 2 de­ci­sions. But they can be com­bined to­gether as a sin­gle choice where you both flip a coin and then sub­se­quently roll a die. This com­bi­na­tion can be done even when the se­quence of de­ci­sions is more com­pli­cated, for ex­am­ple when choos­ing to flip a coin, and, if it’s heads, rol­ling a die, oth­er­wise flip­ping a coin again.

I ex­pect you can do some­thing similar to group the de­ci­sions of al­ter­nate “you”s into one, or even the de­ci­sions of other peo­ple on Earth, so long as their de­ci­sion-mak­ing pro­ce­dure is similar enough to yours.

If I’m right that this trans­for­ma­tion is pos­si­ble, that means one-shot prob­lems are iso­mor­phic to finite multi-shot prob­lems, and hence in­sights into one are ap­pli­ca­ble to the other. This means that a solu­tion to one-shot prob­lems should give a solu­tion to Pas­cal’s Mug­ging in gen­eral.


Solv­ing one-shot prob­lems means find­ing a de­ci­sion-mak­ing pro­ce­dure that max­imises pa­per­clips when you only have one de­ci­sion. One might ex­pect EU, which is all about max­imis­ing pa­per­clips, would at least provide some in­sight into this.

Sur­pris­ingly, EU doesn’t seem to help. The key prop­erty of an EU max­imiser is that as it makes more and more de­ci­sions, the prob­a­bil­ity that it will get more pa­per­clips ap­proaches 1.

For ex­am­ple, in my cas­ino ver­sion of Pas­cal’s Mug­ging, at 1 rep­e­ti­tion there is a very low prob­a­bil­ity of an EU max­imiser win­ning. But at 1 googol­plex rep­e­ti­tions, there is an ~63% chance of it win­ning at least once. At 2 googol­plex, that prob­a­bil­ity be­comes ~86%. In the long run, the prob­a­bil­ity that EU will come out on top ap­proaches 1.

This means that EU com­pletely sidesteps the prob­lem of how to make de­ci­sions un­der un­cer­tainty, by choos­ing the se­quence of de­ci­sions that has a vir­tu­ally 100% prob­a­bil­ity of win­ning out in the long run!


In sum­mary, Pas­cal’s Mug­ging oc­curs be­cause ex­pected util­ity de­pends on there be­ing a long time frame, and solv­ing what to do when there isn’t a long enough time frame is equiv­a­lent to solv­ing it for the sim­pler case where you only get to make a sin­gle de­ci­sion.

Thank you for read­ing this, and I hope to learn a lot from your replies!


[0] Switch­ing to pa­per­clip max­imis­ing also helps show why I think bounded util­ity func­tions are an in­com­plete solu­tion to Pas­cal’s Mug­ging. Which choice is op­ti­mal for max­imis­ing the num­ber of pa­per­clips in the world? This is a seem­ingly fac­tual ques­tion, and our best an­swer is ex­pected util­ity max­imi­sa­tion, which is vuln­er­a­ble to Pas­cal’s Mug­ging. This ques­tion is in­de­pen­dent of our util­ity func­tion, and can’t be re­solved by say­ing that we should use a bounded util­ity func­tion.

Us­ing pa­per­clip max­imi­sa­tion also helps re­move an­thropic prob­lems in Pas­cal’s Mug­ging. You can ar­gue that pro­duc­ing 3^^^^3 util­ity re­quires cre­at­ing 3^^^^3 peo­ple, which means the prob­a­bil­ity that you are one of those 3^^^^3 peo­ple coun­ter­bal­ances the re­ward from be­ing Pas­cal Mugged. But this rea­son­ing does not work if your “util­ity” is pa­per­clips.

[1] Note that the price be­ing 10 pa­per­clips is only a cour­tesy from the cas­ino. They could charge a billion pa­per­clips per roll and EU would still say that you should pay up.


Up­date: On fur­ther re­flec­tion, my crit­i­cism of bounded util­ity func­tions in the ze­roth foot­note is wrong. I’ve up­dated in this di­rec­tion due to Dagon’s sec­ond point in the com­ments (thank you!). Max­imis­ing the num­ber of pa­per­clips can be done us­ing a bounded util­ity func­tion as well, for ex­am­ple the func­tion 1-1/​2^p is bounded be­tween 0 and 1, where p is an non­nega­tive in­te­ger giv­ing the num­ber of pa­per­clips.

That this can be done is sur­pris­ing to me right now, and sug­gests that I need to think some more about all this.