[extropy-chat] what is probability?

The Avantguardian avantguardian2020 at yahoo.com
Thu Jan 18 01:14:55 UTC 2007

--- gts <gts_2000 at yahoo.com> wrote:

> Another pesky Principle of Indifference problem...
> A train leaves at noon to travel a distance of
> 300km. It travels at a  
> constant speed of between 100km/h and 300km/h. What
> is the probability  
> that it arrives before 2pm?
> At least two solutions with different answers:
> 1. For the train to arrive before 2pm, its velocity
> must be greater than  
> or equal to 150km/h. By the principle of
> indifference, all velocities  
> between 100 and 300 km/h are equally likely. The
> probability of it being  
> greater than or equal to 150km/h such that the train
> arrives before 2pm is  
> therefore 3/4.
> 2. The train must arrive between 1pm and 3pm. 2pm is
> half way between  
> these two. By the principle of indifference, the
> train is as likely to  
> arrive before 2pm as after 2pm. So the probability
> must be 1/2.
> What would Scooby Do?

I don't know what Scooby would do but I have a third
solution. Unlike some problems involving the PoI, this
one is actually a physics problem. Whereas I am
unaware of any experimental verification of PoI in
terms of time or space, I do know that there is a
verifiable principle in statistical thermodynamics
that seems somewhat similar to PoI only it is defined
objectively instead of epistemologically. This
principle is called Boltzman's equipartition theorem.
This theorem is only defined for classical systems but
as a train is a classical system, it should apply.

Briefly the theorem states that the energy of a system
will divide itself equally amongst all possible
degrees of freedom of the system. Since one can
describe the velocity of a train as the kinetic energy
of the train, one can say that the range of allowed
velocities of the train (100km/hr to 300km/hr) can be
described as degrees of freedom of the kinetic energy
of the system. Thus we could theoretically say that
all possible kinetic energies of the train between 1/2
mass * (lowest velocity)^2 and 1/2 mass * (upper
velocity) should be equally probable.

We further make the assumption that the train's mass
is constant throughout the trip although realistically
it would be burning fuel and getting lighter
throughout the trip. Since it is constant, we can
eliminate mass from the kinetic energy expression
leaving us with units of (km/hr)^2. When we do this we
find out that given the boundary conditions of the
problem, the lowest possible (mass reduced) energy the
train can have is (100 km/hr)^2 and the highest energy
is (300 km/hr)^2. This gives us a range of energies
between 10,000 (km/hr)^2 and 90,000 (km/hr)^2, where
any energy in the range is equally likely.

In order to arrive before 2pm the least amount of
kinetic energy the train can have is 22,500 (km/hr)^2.
To determine a probability one has merely to look at
what proportion of possible kinetic energies allow the
train to arrive before 2pm versus the total range.
Thus the probability that the train will arrive at or
before 2pm is P(<=pm) = (90,000 - 22,500)/(90,000 -
10,000) = 0.84375 which is different from either one
of the answers given above.

So does this help? *wicked grin*

Stuart LaForge
alt email: stuart"AT"ucla.edu

"The probability that we may fail in the struggle ought not to deter us from the support of a cause we believe to be just." -Abraham Lincoln

Any questions? Get answers on any topic at www.Answers.yahoo.com.  Try it now.

More information about the extropy-chat mailing list