1. If the player does not know the amount in the chosen envelope then the expected gain from switching is zero. — Andrew M
Random selection, which means equal probability, mitigates observational bias by treating each n in a population the same. — Jeremiah
The equiprobability that I am talking about is the posterior equiprobability between the two possible contents of the second envelope: either X/2 or 2*X. — Pierre-Normand
So far so good. But we cannot do this:
— Srap Tasmaner
If you right-click on a TeX formula and select 'Show Math As...' then 'TeX commands', then you can copy and paste the code for that in between ... — andrewk
... Isn't this the same as: ... — Michael
Sorry, I'm a bit confused by your response. Did you read me as saying "this isn't the same as"? — Michael
Of course you won't. You don't like facts that disagree with your beliefs.I will not further debate such specifics. — Jeremiah
As an aside, I think we're saying the same thing from different angles. — fdrake
I see the situation as this:
Assume there's £10 in my envelope and that one envelope contains twice as much as the other. The other envelope must contain either £5 or £20. Each is equally likely and so the expected value of the other envelope is £12.50. I can then take advantage of this fact by employing this switching strategy to achieve this .25 gain. — Michael
Even in that more general case, the Bayesian approach can give a switching strategy with a positive expected net gain. — andrewk
The error is "Each is equally likely," and it is wrong because we need to work with F, not S. My very first post gave an example. Suppose I fill 9 pairs of envelopes with ($5,$10) and one pair with ($10,$20). I choose a pair at random, and present it to you saying "one contains $X and the other $2X, [but you do not know which envelope is which or what the number X is]." — JeffJo
The point is that, in the correct version for your calculation E=($V/2)*P1 + ($2V)*P2, the probability P1 is not the probability of picking the larger value. It is the probability of picking the larger value, given that the larger value is $10. In my example, that 90%. In the OP, you do not have information that will allow you to say what it is. — JeffJo
Jeremiah oversimplifies.If you use a distribution you are making assumptions not included in the OP. I pointed this out before. — Jeremiah
And that unknown value in his pocket has a distribution. We don't need to "check it," as long as the symbolic probability space we use satisfies the requirements of being a probability space.I agree in the strictness sense of the definition there is an unknown distribution, but as far was we know it was whatever was in his pocket when he filled the envelopes. We can't select a distribution to use, as we have no way to check it. — Jeremiah
The simple truth of this is, if I walked up to you on the street and handed you one envelope and said one of these has twice as much as the other, you'd have no clue as to range, distributions or anything of that sort, even after seeing Y. You could make speculations, but that is all they would be and they would carry a high degree of uncertainty you could never account for in your calculations, as they would just be wild guesses.
You could spin all types of models, but you're still just shooting in the dark. The truth of it is, in that moment, switching or don't, you stand to gain x and you stand to lose x. That is the one thing we know is constant. — Jeremiah
You don't "guess" a prior. Priors have to be justified. If you don't know you use an uninformative prior. — Jeremiah
And if I see £10 then I stand to gain £10 and I stand to lose £5. — Michael
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