There are two reasonable ways to assign variables to the set from which you will select your envelope: {x, 2x} and {x, x/2}. Either works so long as you stick with it, but if you backtrack over your variable assignment, you have to completely switch to the other assignment scheme. — Srap Tasmaner
I don't see any disagreement. — Srap Tasmaner
Besides an opportunity to play with our respective toys, do you get anything out of this? — Srap Tasmaner
I could imagine using it for teaching probability modelling. Get students to analyse the problem. Then do it IRL with both sampling mechanisms. Should be a cool demonstration of "physical" differences between what's seen as a merely "epistemic" probability assignment! — fdrake
Eh, probability modelling also includes assigning random variables. It has a lot to do with what random variables you put in play. — fdrake
Given a set {10, 20}, the expected value of a number selected from that set is 15. There's nothing wrong with your first set of equations, and it gives the right answer. You don't have to go through all that; you just need the average. — Srap Tasmaner
(1) What are the chances that y = x and the chances that y = 2x if y is chosen randomly from a set {x, 2x}? (You may, if you like, write it backwards as x = y and x = y/2.)
(2) What are the chances that a y chosen randomly from a set {x, 2x} was chosen from a set {y, 2y} and the chances it was chosen from a set {y/2, y}? — Srap Tasmaner
We can only consider this from the perspective of the participant, who only knows that one envelope contains twice as much as the other and that he picked one at random. His assessment of and can only use that information.
Is it correct that, given what he knows, ?
Is it correct that, given what he knows, ?
If so then, given what he knows, .
Perhaps this is clearer if we understand that means "a rational person's credence that his envelope contains the smaller amount given that he knows that his envelope contains £10".
But given the set {10,20}, E(z)=15=(5/4)12, and 12 isn't the value of the chosen envelope. — Michael
Where have I gone wrong? — Srap Tasmaner
It's contradictory expectation values don't appeal to faulty reasoning given acceptance of the premises. — sime
Rather the switching argument is unsound, for among it's premises is an improper prior distribution over x, the smallest amount of money in an envelope. And this premise isn't possible in a finite universe.
Intuitively, it's contradictory conclusions makes sense; if the smallest amount of money in an envelope could be any amount of money, and if the prior distribution over the smallest amount of money is sufficiently uniform, then whatever value is revealed in your envelope, the value of the other envelope is likelier to be higher. — sime
I believe it does, as I showed above. It covertly redefines y
such that when it concludes E(z)=54y is no longer the value of the chosen envelope. — Michael
No covert redefinitions of y are happening — sime
Notice that in E(z) the variable y stands for 3 different values. In one case it stands for the value of the smaller envelope (10), in another case it stands for the value of the larger envelope (20), and in the final case it stands for a different value entirely (12). — Michael
Your two definitions of E[z] aren't equivalent. — sime
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