Re: Bayesian boxes and Independence of Scales

From: Gilles HENRI <Gilles.Henri.domain.name.hidden>
Date: Wed, 12 May 1999 13:47:22 +0200

À (At) 20:01 -0400 11/05/99, GSLevy.domain.name.hidden écrivait (wrote) :
>Hi Jacques,
>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.)
>
>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.
>Another way to state the same thing is in terms of logarithms:
>
> >Expected Value = (1/2) (Log(m/2) + Log(2m)) =
> >(1/2) (Log(m) - Log2 + Log(m) + Log2) = Log(m)

May be but how do you justify to take a geometrical average? And what if
you assume m and m^2 in the boxes instead of m and 2m?

I think the problem is that m is not restricted and can take any value from
zero to infinity (even with discrete values). So actually the probability
of finding any particular value of m is strictly zero, which does not make
sense from the beginning.

If m is restricted to be lower than m_max, the paradox disappears, because
following the value you find in the box, the probability of having opened
the "m" box and the "2m" box are no more equal. If the value you find is
between m_max and 2*m_max (p= 1/4), you are 100% sure that you must keep
it. If it is between 0 and m_max (p=3/4), the a posteriori (bayesian)
probability of having opened the "2m" box is 1/3 and that of having opened
the "m" box is 2/3, and you must choose the other box. The overall expected
value is = 1/4 * 2m + 3/4 * (1/3*m+2/3*2m)=1.75m, larger than 1.5 m because
you use the information brought by the opening of the first box AND the
knowledge of m_max. Of course letting m_max go to infinity causes much
trouble...

Gilles
Received on Wed May 12 1999 - 04:44:58 PDT