Therefore, she should retain her $10, as her prior for Joe having included $20 is sufficiently low. Regardless, before she inspects the second envelope, both outcomes ($5 or $20) remain possible. — Pierre-Normand
Your assertion that 'only two values are possible' for the contents of the envelopes in the two-envelope paradox deserves further exploration. If we consider that the potential amounts are $(5, 10, 20), we might postulate some prior probabilities as follows: — Pierre-Normand
You can conclude either strategy is optimal if you can vary the odds (Bayes or nonconstant probability) or the loss function (not expected value). Like if you don't care about amounts under 20 pounds, the optimal strategy is switching. Thus, I'm only really interested in the version where "all results are equally likely", since that seems essential to the ambiguity to me. — fdrake
As I wrote, the prior probabilities wouldn't be assigned to the numbers (5,10,20), they'd be assigned to the pairs (5,10) and (10,20). If your prior probability that the gameshow host would award someone a tiny amount like 5 is much lower than the gigantic amount 20, you'd switch if you observed 10. But if there's no difference in prior probabilities between (5,10) and (10,20), you gain nothing from seeing the event ("my envelope is 10"), because that's equivalent to the disjunctive event ( the pair is (5,10) or (10,20) ) and each constituent event is equally likely — fdrake
Edit: then you've got to calculate the expectation of switching within the case (5,10) or (10,20). If you specify your envelope is 10 within case... that makes the other envelope nonrandom. If you specify it as 10 here and think that specification impacts which case you're in - (informing whether you're in (5,10) or (10,20), that's close to a category error. Specifically, that error tells you the other envelope could have been assigned 5 or 20, even though you're conditioning upon 10 within an already fixed sub-case; (5,10) or (10,20).
The conflation in the edit, I believe, is where the paradox arises from. Natural language phrasing doesn't distinguish between conditioning "at the start" (your conditioning influencing the assignment of the pair (5,10) or (10,20) - no influence) or "at the end" (your conditioning influencing which of (5,10) you have, or which of (10,20) you have, which is totally deterministic given you've determined the case you're in).
envelope_pair(X, [X,Y]) :- A is 2 * X, B is div(X, 2), ( maybe -> member(Y, [A,B]) ; member(Y, [B,A]) ). pick_envelope([A,B], X) :- ( maybe -> member(X, [A,B]) ; member(X, [B,A]) ).
envelope_pair(10, Pair), pick_envelope(Pair, Mine)
?- envelope_pair(10,Pair), pick_envelope(Pair,Mine). Pair = [10, 20], Mine = 10 ; Pair = [10, 20], Mine = 20 ; Pair = [10, 5], Mine = 5 ; Pair = [10, 5], Mine = 10.
?- envelope_pair(10,Pair), !, pick_envelope(Pair,Mine). Pair = [10, 5], Mine = 10 ; Pair = [10, 5], Mine = 5.
?- envelope_pair(4,[A,B]), !, pick_envelope([A,B],Mine). A = 4, B = Mine, Mine = 8 ; A = Mine, Mine = 4, B = 8.
If we assume that all results are equally likely, the EV of switching given that the chosen envelope was seen to contain n is (2n + n/2)/2 - n = 1.5n. Hence whatever value n might be seen in the initially chosen envelope, it is irrational not to switch (assuming only our goal is to maximize EV). This gives rise to the paradox since if, after the initial dealing, the other envelope had been chosen and its content seen, switching would still be +EV. — Pierre-Normand
What does the cut correspond to? — Srap Tasmaner
However, it seems that you've misunderstood my use of Bayesian updating. I am not arguing that observing the $10 allows me to switch between cases. Rather, I'm saying that, given an observation of $10, I can update my beliefs about the probability of being in the (5,10) case or the (10,20) case. — Pierre-Normand
Seeing that there is $10 in one envelope still leaves it open that there might be $5 or $20 dollars in the other one. — Pierre-Normand
Since you're an R user, you might find it interesting to define a model in RStan, using different choices for the prior for the smallest amount S put into a envelope. Provided the chosen prior P(S) is proper, a sample from the posterior distribution P( S | X) , where X is the observed quantity of one of the envelopes, will not be uniform, resulting in consistent and intuitive conditional expectations for E [ Y | X] (where Y refers to the quantity in the other envelope) — sime
There's the question of whether the "Bivariate Distribution Specification" reflects the envelope problem. It doesn't reflect the one on Wiki. The reason being the one on the wiki generates the deviate (A,A/2) OR (A,2A) exclusively when allocating the envelope, which isn't reflected in the agent's state of uncertainty surrounding the "other envelope" being (A/2, 2A).
It only resembles the one on the Wiki if you introduce the following extra deviate, another "flip" coinciding to the subject's state of uncertainty when pondering "the other envelope": — fdrake
What does the cut correspond to?
— Srap Tasmaner
Answering that gives you the origin of the paradox, right? — fdrake
This doesn't require an additional coin flip. She either is in the (5, 10) case or the (10, 20) case, with equal prior (and equal posterior) probabilities in this scenario. However, this is just one hypothetical situation. — Pierre-Normand
Taking into account all three possibilities regarding the contents of her initially chosen envelope, her EV for switching is the weighted sum of the updated (i.e. conditioned) EVs for each case, where the weights are the prior probabilities for the three potential contents of the initial envelope. Regardless of the initial bivariate distribution, this calculation invariably results in an overall EV of zero for switching. — Pierre-Normand
This approach also underlines the flaw the popular argument that, if sound, would generate the paradox. If we consider an initial bivariate distribution where the potential contents of the larger envelope range from $2 to $(2^m) (with m being very large) and are evenly distributed, it appears that the Expected Value (EV) of switching, conditioned on the content of the envelope being n, is positive in all cases except for the special case where n=2^m. — Pierre-Normand
The fallacy in the 'always switch' strategy is somewhat akin to the flaw in Martingale roulette strategies. — Pierre-Normand
Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch?
You’re not given any more information, so I really don’t follow the rest of the calculations. — Mikie
Given that the expected value of the unchosen envelope is greater than the value of the chosen envelope, it is rational to switch. — Michael
I see this has been gone over quite a bit — Mikie
There's not even agreement among analysts about whether this is a probability problem. (I don't think it is.) — Srap Tasmaner
So what we’re left with is the claim that going through the motion of picking one envelope and then — with absolutely nothing else changing— switching and picking up the other envelope is somehow rational. That’s simply wrong. — Mikie
So if Bob is mad enough to reason with subjective probability distributions (which IMO should never be used in science, and which can be avoided even when discussing credences by using imprecise probabilities — sime
So a more accurate formulation is:
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