Andrew M

The OP asks, "What should you do?" Think of the problem as being in the same family as Pascal's Wager, involving decision theory and epistemology.

You seem to be saying that with an unknown distribution, there is an expected gain from switching for one game even though over repeated games (with a...

Yes, I messed up the math. Have edited...

You are referring to two different probabilities. The first is P(lower) which is 50%, the second is P(lower|amount) which depends on the specifics of ...

Yes, I think it's reasonable to assume discreteness (with a finite but unspecified precision and range). Yes. But I think the OP is asking for a gener...

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 c...

Isn't 2X just a transformation of X that doubles the possible values in X? So Pr(2X=a) would be equivalent to Pr(X=a/2). Here's an academic example us...

I agree. Regarding the OP, the math provides no reason to be anything other than indifferent to sticking or switching. (Of course, people may wish to ...

Yes your account of objective probabilities explains it. But that is just probability simpliciter where the player's expected gain calculations are ba...

You can make a similar argument for keeping. Suppose you choose an envelope but, in this game, instead of opening your envelope the host opens the oth...

I don't advocate it either. I advocate E = ((v/2) * Pr(v/2,v) + (v*2) * Pr(v,2v)) / (Pr(v/2,v) + Pr(v,2v)).

Yes. That is because you know what the distribution is. Without knowing the distribution you don't know whether both the envelope pairs were possible ...

There is an expected gain of X from strategic switching if the algorithm generates a random number between the X and 2X envelope amounts. In this case...

It seems to me that there are some claims about the Two Envelopes problem that everyone might agree on. 1. If the player does not know the amount in t...

Yes, assuming that the player does accurately estimate the maximum possible payout (and the procedure for generating the envelope amounts). If she doe...

Can you give a concrete example where such a value would be used?

I think the issue is that even if you know Y from opening the initial envelope, the expected gain from switching is still zero if you don't also know ...

I disagree that the 50% chance can be assumed. If the sample space is finite, then there is a non-zero chance that the $10 amount is the maximum amoun...

The problem is that you use the $10 starting amount to generate the random half ($5) or double ($20) amount for the second envelope. That is equivalen...

Before choosing, there is no reason to prefer one envelope to the other. On choosing an envelope and learning the amount in it, there is still no reas...

The Wikipedia entry uses the expected value being the same for both envelopes as their simple resolution. Do they have that wrong, in your view?

No, but I can make the same argument using concrete amounts (see below). You can't assume you have a starting envelope of, for example, $10, and that ...

As Jeremiah points out, your code doesn't reflect the problem in the OP. Before an envelope is picked, the expected value of each envelope is the same...

Yes it does. But considering myself, my utility function would be roughly linear for the amounts being talked about here (around $20). Which is to say...

The expected value of each envelope is $(X + 2X)/2 = $3X/2 = $1.5X. You should be indifferent between switching or not. However consider this variatio...

So the argument is that is necessary to index the quantum state to a participant (broadly conceived). As an analogy, there is nothing mystical about o...

More recent philosophical discussion has moved on from the reality versus mysticism debates between the founders of quantum mechanics. "The war over r...

Agreed. Yes. Do you mean there can be different standards for measurement, depending on the context?

It hasn't been proven - it's an interpretational issue. Per the quantum interpretations table on Wikipedia, roughly half are local interpretations, in...

What Linde is saying is that, taken as a whole, the universe is predicted to be static and unchanging (per the Wheeler-DeWitt equation). In order to p...

Welcome back! I regard them as two distinct epistemic perspectives. One is Beauty's on Sunday (or Wednesday) who doesn't condition on being interviewe...

Yes, in the sense that one should condition on being interviewed. He's simply saying that if you interview different people for each permutation inste...

Normally you would just give 1 M&M. But it is ultimately a question about what sample space and rules are appropriate in the circumstances.

Yes that's the nature of the experiment. There are two ways of looking at it. In my view, probability is a measure of the state that the agent is in. ...

It happens! Here's how I see it. Lewis' halfer view fails because it gives an absurd result of 2/3 when conditioning on Monday. But the double-halfer ...

1/2.

Beauty knows beforehand that she will be awakened and interviewed. But P(Heads|Awake) does not become relevant until she is actually awakened in the e...

I don't think the halfer view ultimately flies. I think it conflates the question of the nature of fair coins (which we all agree come up heads half t...

When Beauty is being interviewed, what probability should she assign to Monday and Tails? If 1/2 then Tuesday and Tails would be 0 which doesn't seem ...

That is of no use to her. When awakened, she doesn't know whether she is in an awake state that she should assign a probability of 1/2 to or 1/4 to. S...

I think you are both looking at the experiment from an independent observer's perspective (or Beauty's Sunday perspective) and not from Beauty's persp...

OK, that's the double-halfer view. When Beauty is told it is Monday, all the Tuesday and Tails probability is reallocated to Monday and Tails which vi...

I see those equations as a formalization of the rules that we learn in familiar settings. In unfamiliar settings, the question would be whether it is ...

I'd like to analyze the halfer's P(Heads|Monday) = 2/3 consequence further because I think it is key to how we see the Sleeping Beauty scenario. Just ...

Agreed. But that scenario is equivalent to randomly waking Beauty on either Monday or Tuesday if tails, but not both days. To be analogous to the Slee...

That's all we're asking Beauty about. That is, the probability that the next marble to be drawn (or the interview that is being conducted) will be ass...

I was referring to the background sample space in my earlier post that included all combinations of day outcomes and coin toss outcomes and that are a...

I was assuming the two possible interview days in the experiment. But since the probabilities are all equal, it doesn't matter how large the backgroun...

Yes, it would be good to hear some halfer reasoning for the P(Heads|Monday) = 2/3 consequence. I agree. That's why I think we should seek to derive th...

I'm suggesting that we should start with a background sample space that includes all possible combinations of days and coin toss outcomes and then ass...