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From: <GSLevy.domain.name.hidden>

Date: Thu, 13 May 1999 00:00:26 EDT

In a message dated 99-05-12 07:45:08 EDT, Gilles Henri wrote:

<<

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. >>

and Jacques Mallah wrote

<< 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. >>

Gilles and Jacques, we are going in opposite direction in our reasoning and

this is the reason we don't meet. You are using conventional logic, and I am

using anthropic logic.

Assumption 1)

I start out by assuming that m is what it is, however improbable it is From

an outside observer, the probability that m is what it is approaches zero. In

fact the range of choice for m is the whole set of real numbers.However,

this is the m that we observe. (ANTHROPY) and, given this m, the probability

of me seeing this particular m is one.

Assumption 2)

It is common sense that it does not make any sense to switch. (I am willing

to play some money gambling games with you if you disagree on this one). The

expected value of the money in the other box is m. Because of the PERFECT

SYMMETRY of the situation, there is NO REASON for the expected value in the

other box to be other than m. Lack of causes for the boxes to be different

translates into perfect symmetry and the equivalence in the expected value

between the two boxes. The laws of physics are the same in all reference

systems (after choosing box A or after choosing box B)

Conclusion: The distribution of m depends on the problem and must be such

that the expected value for the other box is equal to m. For the problem

0.5m and 2m the distribution MAY BE logarithmic (sufficient condition) but

there may be other distributions. For the problem sqrt(m) and m**2, the

distribution is such that

fn( sqrt(m)) + fn( m**2) = m. I am too lazy to work out all the possible

functions that fn could be.

In other words the problem itself defines the distribution.

<< In any case it is easily proved that it's false, since if you were

immortal you would expect to be very old.>>

It all depends how you measure my age. In fact, my lifeline extends

uninterrupted probably four billions years (or possibly more), since it first

appeared on earth, and we may even be cousins. And by the way, your lifeline

also extends that much. Joyeux Anniversaire! Happy Birthday! :-)

George

Received on Wed May 12 1999 - 21:02:48 PDT

Date: Thu, 13 May 1999 00:00:26 EDT

In a message dated 99-05-12 07:45:08 EDT, Gilles Henri wrote:

<<

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. >>

and Jacques Mallah wrote

<< 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. >>

Gilles and Jacques, we are going in opposite direction in our reasoning and

this is the reason we don't meet. You are using conventional logic, and I am

using anthropic logic.

Assumption 1)

I start out by assuming that m is what it is, however improbable it is From

an outside observer, the probability that m is what it is approaches zero. In

fact the range of choice for m is the whole set of real numbers.However,

this is the m that we observe. (ANTHROPY) and, given this m, the probability

of me seeing this particular m is one.

Assumption 2)

It is common sense that it does not make any sense to switch. (I am willing

to play some money gambling games with you if you disagree on this one). The

expected value of the money in the other box is m. Because of the PERFECT

SYMMETRY of the situation, there is NO REASON for the expected value in the

other box to be other than m. Lack of causes for the boxes to be different

translates into perfect symmetry and the equivalence in the expected value

between the two boxes. The laws of physics are the same in all reference

systems (after choosing box A or after choosing box B)

Conclusion: The distribution of m depends on the problem and must be such

that the expected value for the other box is equal to m. For the problem

0.5m and 2m the distribution MAY BE logarithmic (sufficient condition) but

there may be other distributions. For the problem sqrt(m) and m**2, the

distribution is such that

fn( sqrt(m)) + fn( m**2) = m. I am too lazy to work out all the possible

functions that fn could be.

In other words the problem itself defines the distribution.

<< In any case it is easily proved that it's false, since if you were

immortal you would expect to be very old.>>

It all depends how you measure my age. In fact, my lifeline extends

uninterrupted probably four billions years (or possibly more), since it first

appeared on earth, and we may even be cousins. And by the way, your lifeline

also extends that much. Joyeux Anniversaire! Happy Birthday! :-)

George

Received on Wed May 12 1999 - 21:02:48 PDT

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