What if we just say that, having observed the value of our envelope to be a, then the expected value of the other is 3X - a for some unknown X? That formula, unlike the expected value formula, doesn't require any probabilities to be filled in. It's uninformative, but so what? — Srap Tasmaner
To what purpose? It doesn't help you to answer any of the questions. — JeffJo
And what you seem to be avoiding with that attitude, is that the expectation formula (v/2)/2 + (2v)/2 is already assuming: — JeffJo
I am not an advocate for that expectation formula, so I don't see why you'd think I am avoiding those objections to it. - andrewk
Maybe I was mixing Andrews up. I apologize. — JeffJo
So what am I supposed to do if I don't know how the values are selected? As I said here, if I don't have any reason to believe that X = 5 is more likely than X = 10 then why wouldn't I switch? I think the principle of indifference is entirely appropriate in this circumstance. There's more to gain than there is to lose, and a loss of £5 is an acceptable risk. — Michael
And that's because of what I explained here. — Michael
The final criterion for deciding whether a model is "good" is whether it yields good and useful information. The motivation for using mathematical models lies primarily in their utility.
And that's because of what I explained here. — Michael
We've already established that the expected gain is
E(B∣A=a)=P(X=a∣A=a)2a+P(2X=a∣A=a)a/2 — Michael
The objective probabilities of X=a∣A=a and 2X=a∣A=a depend on how the host selects the value of X.
If he selects it at random from a distribution that includes a2 and a then the objective probability of X=a∣A=a is 0.5 …
So there would be an objective expected gain.
Choosing any explicit distribution for the OP is indeed misguided, which is why your simulations were misguided. That, and the fact that your conclusions could be proven without such modeling.I think in terms of modeling the actual OP the inclusion of distributions or switching strategies is misguided. — Jeremiah
However, these models do provide a useful platform to investigate properties that may be applied to other probabilistic aspects.
My simulations were there to display the inherit ambiguity in defining a prior distribution. X is an unknown, treat it like an unknown. — Jeremiah
And my response to these sentiments has always been that you can't define/calculate the prior distribution, and that it was a misguided effort to even try (as you did).I have also already shown that trying to calculate expected returns is a misguided effort. — Jeremiah
you can't define/calculate the prior distribution, and that it was a misguided effort to even try (as you did) — JeffJo
Personally, I think in terms of modeling the actual OP the inclusion of distributions or switching strategies is misguided. If our goal is to assess the situation before us then they move outside that scope. However, these models do provide a useful platform to investigate properties that may be applied to other probabilistic aspects. — Jeremiah
I’m going to assume that A means your envelope, B means the “other” envelope, and X means the lower value. But A, B, and X mean the abstract concept of the random variable; it is a, b, and x that mean values. So “2X” is meaningless. What I think you mean here is:
E(B∣A=a) = Pr(X=a∣A=a)*(2a) + Pr(X=a/2∣A=a)*(a/2) — JeffJo
Do we know that both P(X = a) and P(X = a/2) are non-zero? We know that at least one of P(X = a | Y = a) and P(X = a/2 | Y = a) is non-zero, but we do not know that both are. Without Y = a, we wouldn't even know that at least one of P(X = a) and P(X = a/2) are non-zero. Without knowing that both are non-zero, we can't even safely talk about the odds P(X = a):P(X = a/2). — Srap Tasmaner
Yes. You learn something about the distribution when you open an envelope (namely, that it had an envelope with that seen amount). But not enough to calculate anything useful. It's like getting a bicycle with one wheel. You might wonder whether you could get somewhere with it, but you probably can't. — Andrew M
Interestingly, if the game only allows whole dollars - or even whole cents, and the player knows that, they can use it as the basis for another strategy: if the number is odd, switch, otherwise don't. That's because if the number is odd it cannot be the doubled value, so the other one must be.(I am assuming that the game only allows for amounts in whole dollars, for simplicity — Pierre-Normand
(...) To remove that option, I recast the problem with the envelopes containing IOUs rather than cash, for an amount that is a real number of cents, with an arbitrary but large number of decimal places shown. The amount is only rounded to the nearest cent (or dollar) when the IOU is cashed in. — andrewk
Do we need to assume that X is not continuous? If it is, all these probabilities are just 0, aren't they?
*** Urk. Forgetting that at least one of them has to be non-zero. — Srap Tasmaner
Wouldn't that information be useful? — Pierre-Normand
To remove that option, I recast the problem with the envelopes containing IOUs rather than cash, for an amount that is a real number of cents, with an arbitrary but large number of decimal places shown. The amount is only rounded to the nearest cent (or dollar) when the IOU is cashed in. — andrewk
Yes. But I think the OP is asking for a general solution for one run with no special assumptions about the context (such as whole dollar amounts or million dollar limits). — Andrew M
the equiprobability condition that alone grounds the derivation of the unconditional 1.25X expectation from switching. — Pierre-Normand
I'm still confused. This makes it sound like the switching argument isn't fallacious -- it just makes an unwarranted assumption. So if every value of X were equally probable, then it would be true that you can expect a .25 gain from switching? I see how the math works, but if that's true, isn't it true whether you know the value of your envelope or not? And if that's true, isn't it also true for the other envelope as well? — Srap Tasmaner
I think I'm just reluctant to see the simple situation of choosing between two envelopes of different values in terms of the strange behavior of infinity. — Srap Tasmaner
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