Hi 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 observer:_ 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 meaningless. 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 --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to [EMAIL PROTECTED] To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~----------~----~----~----~------~----~------~--~---