I haven't contributed to the list recently but probability is a topic 
that interests me and which I discussed several years ago. I have a 
"relativist" interpretation of the MW.

To apply Probabilities to the MW _every probability should be stated as 
a conditional probability, that is conditional on the existence of the 

For example:
P{event X} is meaningless
P{ event X / observer A} is the probability that observer A sees event X.

Obviously we have:
P{ Observer A / Observer A } = 1

Things become interesting when we have two observers A and B observing 
the same event X. (Recall Einstein thought experiment on simultaneity).

_Case 1: Classical case: Event X totally decoupled from the existence of 
observer A and B_

When the existences of A and B are not contingent on X we have

P{X/A} = P{X/B}

and both A and B agree on the "objectivity" of their observation. They 
call this probability P{X} even though strictly speaking P{ X} is 

This case represents the classical case: all observers see an objective 
reality in which all events have the same probabilities.

_Case 2: Tegmark case: Existence of A is 100% contingent on X._

In this case, the observed probabilities are different:

P{X/A} <> P{X/B}

For example let's consider Tegmark Quantum Mechanics suicide thought 
experiment. Let us say that A is the observer playing the lottery event 
X and B is passive.

B may observe the probability of A winning the lottery as

P{ X/B } = 0.000001
Since A is contingent on X:
P{ A/B} = 0.000001

Note that if B attempts to use Bayes rule to compute "P{X}" (or "P{A}")  
he'll use

P{X} = P{X/B} P{B}; However B has no access to P{B}. He actually uses 
P{B/B}. So for B Bayes rule becomes:
P{X} = P{X/B} P{B/B} = 0.000001 x 1 = 0.000001  ; B is a "third person." 
Most of the time he sees A dying.

Since A is 100% contingent on X and vice versa, A observes

P{X/A} = 1

If A attempts to compute "P{X}" using Bayes rule he'll get:

P{X} = P{X/A} P{A}; However P{A} does not make sense. A must use P{A/A}. 
So for A Bayes rule becomes:
P{X} = P{X/A} P{A/A} = 1 x 1 = 1; A is the first person. He always sees 
himself alive.

_Case 3. Both A and B are contingent on X in different degrees._
Assume that A is test pilot flying a very dangerous plane. B is in the 
control tower. C is far away.
X is a successful flight;  X1 is a plane crash on the ground killing A; 
X2 is the plane crashing in the control tower killing A and B.
Let P{X/C} = 0.7; P{X1/C} = 0.2, P{X2/C} = 0.1
P{X} as seen by C = 0.7.

Calculating P{X} according to B is more tricky. The events that B sees 
are the successful flight and the crash in the ground. He does not see 
the crash in the control tower.

To get P{X} as seen by B we need to normalize the probability to cover 
only the events seen by B:
According to B:   P{X} + P{X1} = 1
Therefore: P{X} = 0.7/(0.7+0.2) = 0.77   and P{X1} = 0.23.
So according to B P{X} = 0.77.

A does not see any of the crashes. So:
P{X} as seen by A = 1.0

This last example illustrates how three different observers can see 
three different probabilities.

George Levy

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