On Tue, 11 May 1999 GSLevy.domain.name.hidden wrote:
> I was refering to a problem posed by ids.domain.name.hidden (Iain Stewart):
> > Let box one contain m pounds and box two 2m pounds, where m is
> > any real positive number. You may choose a box look at the
> > amount of money inside and decide either to keep this amount or
> > the other box's amount. (For this exercise you may think of
> > cheques with real numbers on it and a special bank account where
> > you can actually bank any real number of pounds.)
Yes.
> This is a well know problem in probabilities. I am not sure if the solution
> which I propose here is also known: The expected value for the money in the
> box can be calculated as the "average" between (1/2)m and 2m. If the average
> is taken as an arithmetic average we get 1.25m which implies that we should
> switch our choice. This obviously does not make sense. However, if the
> "average" is taken as a geometrical average (equivalently, an arithmetic
> average along a logarithmic scale) then we get sqrt(2m x 0.5m) = m which
> implies that switching results in exactly the same expected value as not
> switching, which makes sense.
No, it doesn't. The other box contains either m/2 or 2m. It does
not contain log(m/2) or log(2m). Apparently I was being too Socratic by
merely asking for clarification; you didn't get the hint. The only way
your equation could even have meant anything is if you were averaging
values of something else, e.g. a utility function, not directly
proportional to m.
> Another way to state the same thing is that the distribution is such that
> there are as many number greater than m as smaller than m. (m is smack in the
> middle of the scale, and a sufficient reason for this to occur is that the
> scale must be logarithmic).
If you want to assume that m was initially drawn from a
distribution that is uniform for log m on (-00,00), so far so good. But
that's completely different from what you did above with the geometric
average.
Suppose you make that assumption. Then the value of m will most
likely be either extremely small or extremely large. That's the source of
the apparent paradox: 1.25 x = x is OK if x = 0 or x=00.
If you assume that it came from a more reasonable distribution,
you will still conclude that switching doesn't matter. It is interesting
to use the Bayesian proceedure. Assume that M_min < m < M_max. Given m,
estimate M_min and M_max with a prior as above.
Now answer these three questions:
- What # should you treat as the probability that the other box
contains 2m?
- If your utility function is directly proportional to what you get,
should you switch?
- How do you explain the above 2?
I'll tell you what I think if you want.
> As to the relationship with MWI I would like to recall the proposal I made a
> few months ago about extending the Cosmological Principle to the MW. I'll
> provide some background first by recalling that the conventional Cosmological
> Principle, assumes that in the large scale, the properties of the Universe
> are identical everywhere, and that no matter "where" we are, the universe
> looks approximately the same in every direction. [...]
> So I propose to make the size of the field of view as large as the MW. The
> Cosmological Principle thus extended, then state that the MW looks identical
> from any point in the MW. Rephrasing this in terms of probabilities of life
> and death: From the PERSPECTIVE OF THE OBSERVER The probability of continued
> life for a 25 year old OBSERVER is identical to the probability of continued
> life for of a 100 year old very sick OBSERVER.
Do you mean that it is some kind of additional postulate that you
make, not derivable from standard MWI physics?
In any case it is easily proved that it's false, since if you were
immortal you would expect to be very old.
- - - - - - -
Jacques Mallah (jqm1584.domain.name.hidden)
Graduate Student / Many Worlder / Devil's Advocate
"I know what no one else knows" - 'Runaway Train', Soul Asylum
My URL:
http://pages.nyu.edu/~jqm1584/
Received on Wed May 12 1999 - 14:07:45 PDT