Alime.209
net.math
utzoo!decvax!harpo!ihnss!houxi!houxe!lime!glenn
Thu Apr 8 23:31:46 1982
Newcombe again... (L O N G !)
Well, I got two replies on my question about Newcombe's Paradox... One
said he would sign his articles if I sent him a description of the paradox,
so he wouldn't have to look through back-issues of Sci Am; I did, and got
no response, so I forgot about the whole issue. Then I got a another reply
from a latecomer who said to post it and start all kinds of arguments. So,
with apologies for its length, here it is: (~~75 lines)
----------- Newcombe's Paradox ----------------
You are led into a room which contains two boxes, Box A and Box B. Box A
contains $1000. [This is known, no gimmicks.] Box B contains either:
(1) $1,000,000 (a million bucks).
OR
(2) Nothing. (no bucks)
Your mission is to maximize the amount of money you leave the room with,
BUT your choices as to which box(es) you take are limited to the following
(i) You can take Box B alone
OR
(ii) You can take both Box A and Box B.
Now, the contents of Box B has been determined in a rather unusual way:
One week previous to the present (assuming "the present" is when you're
led into the room with the boxes) you were subjected (with your knowledge)
to scrutiny by a super-psychologist-type computer [obviously not "Eliza"]
which attempts to predict your actions one week in the future, based on the
state of your neurons [or whatever, the method is irrelevant] at the present
time. (It doesn't tell you what its prediction is.) It then determines the
contents of Box B as follows: If it predicts that one week from then, you
will take Box B alone (choice (i) above) it puts the $1000000 into Box B.
If it predicts that you will be greedy and take both boxes (choice (ii) above)
then it puts nothing in Box B. At that time, the money is put into the boxes,
and left untouched until you walk in a week later and make your choice. The
computer's probability of correct prediction is 99.9%.
Think about it for a few minutes, to get the feel of the thing.
OK, now that you've decided that (ii) is the obvious choice, consider
the following scenario:
The week has elapsed. You are standing in a long line leading to the
door where the boxes are. Each person in front of you has gone through the
same computerized scrutiny procedure the week before, and the computer has
made a prediction for each of them, and placed money accordingly in the
Box B marked for each person. [Assume there are lots of Box A's around, all
with $1000 in them.] When its your turn, the computer's secretary (huh?) looks
up your name on a list, gets the Box B that was pre-packed for you a week ago,
puts it next to a $1000-filled Box A and lets you into the room to make your
choice. Now, suppose there were 2000 people before you that day, and you've
been keeping statistics on how the computer's predictions are holding up.
The people in front of you were evenly divided in the way they chose: half
of them (1000) chose both Box A and Box B, half chose only Box B. True to the
probability quoted (99.9%), of the 1000 that chose both boxes, 999 got only
$1000 (nothing in Box B as predicted) and one lucky soul got $1001000.
Of the 1000 who chose Box B alone, 999 left very cheery with $1000000 in their
little paws, and one poor slob got skunked - nothing in Box B - but again,
the probabilities were as advertised. [Of course, even if P = .999 as
advertised you wouldn't expect to get *exactly* one miss in 1000 tries, but
if that bothers you, change the scenario to make the number of people in
front of you arbitrarily large, and assume you've been observing for years
before your turn comes up.]
Hmmm. Now (ii) doesn't look so obvious... After all, from
the statistical point of view in the above scenario, you'd be a *fool*
to go in like those 1000 before you who chose both boxes, 999 of whom got
nothing... right? Hmmm. Well, on the other hand, you *know* the money has
been placed in the boxes a week ago. Nothing can alter that. So you *know*
you can maximaze your money by taking both boxes, 'cause no matter
what is in Box B, Box A + Box B is always greater than or equal to Box B
alone. So you can't lose by taking both. But on the other hand, *all those
people* who just tried that (all except one that is) got only $1000! Is it
worth it just to gamble for $1000 when you have a 99.9% chance of getting
a cool million? Sheesh, you could get a dual-processor VAX, 5 RP06's,
couple a meg of memory... But on the other hand, whatever money is in there,
is in there *now* and you can walk out with *all of it* by taking both boxes...
But on the other hand, thats probably what all those other greedy A&B'ers
were thinking and look what happened to them. Well, except that one guy
who really hit the jackpot. But on the other hand...
Confused? Puzzled? Probably not. You're probably damned sure
that what you've decided is absolutely right, and anybody that thinks
the other way is just being silly. That is one of the most interesting
things about this paradox - it polarizes people instantly, as Gardner
found out when he published his version of it in Sci. Am. He got
tons and tons and tons (I think it holds the record for his column)
of mail about it. He devoted an issue's column to responses, then
another half a column some time later to more responses. It was
really incredible the arguments that ensued! It is a elegant paradox.
Comments?
- Glenn Golden (vax135!lime!glenn or ihnss!houxg!glenn) (BTL)
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