From: phsieh@lambert.waterloo.edu (Paul Hsieh)
Subject: Re: Dying young
Organization: University of Waterloo
Date: Wed, 8 Apr 1992 07:35:36 GMT

In article <farrell.702613794@coral.cs.jcu.edu.au> farrell@coral.cs.jcu.edu.au (John Farrell) writes:
>In <7452@uqcspe.cs.uq.oz.au> warwick@cs.uq.oz.au (Warwick Allison) writes:
>>tolman%asylum.utah.edu@cs.utah.edu (Kenneth Tolman) writes:
>>>You are put in a room by evil Nazi's.
>
>>1. Assume everyone else does whatever you do.
>>2. Pick a number between 1 and 100 (using dice).  If the number is 1 or
>>two, push the button.  
>
>>I'm not very good at probability (I have two sisters, one of which is
>>younger than the other), but there must be some assignment of probabilities
>>that is optimal, since 100% and 0% are both fatal (all people die).
>
>  Well, are we talking about optimal for you or optimal for everyone? The
>simple fact is that you cannot lose if you don't press the button. You can
>make a massive gain for everyone else, but so what if you're dead? Before you
>can talk about an optimal strategy you have to define what optimal means.
>
>  I am not stupid, ...
            ^^^^^^^
Debatable.
>               ... I am not expendable, ...
                             ^^^^^^^^^^^
Once again, debatable.
>                                    ... and I am not pressing the button.
>
>Friendless
 ^^^^^^^^^^
Not too surprising.

I cannot believe you jumped on this posting with little else to add but "Hey
I'm a shallow moron whose going to waste bandwith with pitiful comments with
no value or insight whatsoever".  I mean seriously.  Why tell us this?  Its
clear you have not thought about the problem in any meaningful way.   Anyways,
on to answering it:

This game is interesting because it is not degenerate and it is not zero sum.
(I.e. You are not tring to win by making your opponent lose).  Now consider the
approach given above:  If you take "Friendless"'s (aka putty brain) approach
what's to say you don't have 99 other putzs doing the same thing?  The
probability of living in such a scenario?  0.  Of course this seems to be a
worse case scenario but on the other hand if you intend to live with such an
approach, you are DEPENDING on the fact that at least someone is not going to
follow this approach.  But then using whatever reasoning you did for chosing
that approach that someone should reason the same way and hence has made a
mistake (assuming that this approach is an good) and should not do press the
button.  Contradiction (The kind of contradiction that would ensure your death).

So what to do?  We've seen from above that symmetry will cause this (and any
other) fixed strategy to fail.  The probability arguement sounded like a good
idea, so why don't we pursue it?  Let us suppose that we adopt the following
strategy.  With the aid of a uniform random number generator we commit
ourselves to press the button with probability p.  Then we apply symmetry; i.e.
assuming this is the best strategy, we assume that everyone else will adopt it.
So, using elementary probability theory, the probability that you will die
using this strategy, P, is p+(1-p)^(100).  Using calculus this is minimized
when p = 1-(1/100)^(1/99).  This gives P = 1-(99/10000)*(100)^(98/99) which is
about 11/200 or 5.5% chance of dying,  i.e. a 94.5% chance of living.

You might argue that the first "attempt" would fair pretty well (even better!)
if the other 99 people were to use this probabilistic strategy.  But then
again think of it from that persons perspective.  There is no way that s/he can
ever know for sure what strategy other people are employing and hence cannot
conclude that people will use a strategy that is different from him/her.  So it
still remains unreasonable to adopt that strategy.

But since this strategy almost seems to work, given that other are inclined to
use the probabilistic method, why not use a meta-probabilistic approach which
is a hybrid of the two approaches?  I.e., chose another probability q, and with
probability q you will employ the p probabilistic approach, otherwise you will
simply not press the button.  But when I worked this out, it ended up just
being the p probabilistic method in disguise.  This convinced me that there
was no better approach.

Of course I have not really given a proof of the fact that my approach is
"best" and in fact I am not entirely sure that it is.  I am quite interested in
what other intellegent (take notice of this adjective Mr. Friendless) comments
or idea people have about this problem, since non-zero sum games have
fascinated me for a couple of years now.

Anyways, enough of me shooting my mouth off, let's get back to the regular
dribble ...

Paul Hsieh