Srap Tasmaner

That we're driven to use such a phrase is apparently the whole problem. Can this sort of thing be done rigorously? What would we have if we did?

You have an average total value of 22.5 and an average envelope value of 11.25. Both of those values always turn out to be wrong.

When I calculate the total value of our envelopes to be U + 10, and the average to be U/2 + 5, I'm right. Whatever U turns out to be, these calculatio...

Unknown. In DOND, after each case is opened I can tell you the total value and the average value of all the remaining cases. To the penny. With no gue...

For DOND you accept any offer in the neighborhood of the expected payout, because the banker usually low-balls you. (There was extensive discussion am...

How did you choose an envelope in the first place? Suppose you have chosen, perhaps by flipping a coin, if the facilitator then offers to tell you the...

And it doesn't bother you that if you know the value of A you want B, but if you know the value of B you want A?

Here's a proof (which you won't accept) that opening the envelope is irrelevant, and that your reasoning should be symmetrical. Suppose you choose an ...

If only the amount in the first envelope, the envelope you chose and perhaps are even allowed to open, is fixed, and the second envelope is then loade...

Agreed. But it would be nice, knowing that the argument leads to absurdity and is therefore false, to pinpoint the step we should disallow. Like figur...

Because you're ignoring the de dicto/de re distinction that Sapientia isn't: That has a nice ring to it, but only because it is idiomatically suppress...

There are two natural and apparently sound approaches, one of which, the one you mention, produces the correct result. The puzzle is figuring out what...

Have a look at the SEP article on Formal Epistemology Here's the first paragraph: This might be trouble though. I think it turns out that to go this w...

The rest of the point being that envelopes worth less than yours, yours being worth Y, have an average value of Y/2 0, as a matter of fact. The envelo...

Which is to say that mean > mean. I've been thinking some about how this works. If you tried, as the player, to broaden your view of the situation, it...

The other big picture issue that has gotten short shrift in this thread, by focusing on the open envelope, is the paradox of trading an unopened envel...

There are two ways to look at this: (1) Knowing the rules of the game, when you get the $10 envelope, you use your amazing Math Powers to deduce that ...

The coins and colored balls thing is different because there are four possible outcomes, you're just choosing in two steps, maybe because you don't ha...

Even using terms like "expectation" or "expected value" is too fancy here. We're just talking about a mean of two values. What's the mean of A and B? ...

You cannot calculate an expectation for X if you do not know what the sample space for X is. It really should be called the "Grass is Always Greener" ...

Take a step back from all the math and the modeling. Try to see the forest here. (1) There are two envelopes. (2) You end up with one of them. Maybe y...

See how red blue and green are all mentioned by name as possibilities.

It is suggestive that the only probabilities ever in play are 50% and the only expected value calculation I have any faith in tells us absolutely noth...

The strangest thing about the puzzle to me is that you need only designate an envelope to get into trouble. Wikipedia says Smullyan thought it was a l...

Just a typo

The more I look at it, the more I miss seeing U=X and U=2X. Maybe it's best to leave them alone. Seeing P(U=5...) when 5 might not even be a possible ...

I'm with you. I just hadn't thought of doing it this way before. It gives you a way to acknowledge what you've learned by learning the value of one of...

Right, good point. In which case, defining the sample space that way is the mistake -- turns our problem into the Ali Baba problem. Didn't mean to do ...

For bogarting your L.

Sorry, man. I think a more complete, realistic answer is around here, yes. Still, for a single trial, your guess would have to be awfully lucky to be ...

Here's another slightly different version with a little twist. As above, we'll have an envelope loading event, this time . I'm just going to do the ca...

I'm going to meet you half-way, Michael. In essence what I'm saying is that P(U = 2X | Y = 10) = P(U = 2X), which entails that (U = 2X ? Y = 10) = ?. ...

No problemo.

You are taking a weighted average of one value which is not only possible but actual, and one value that is not possible. You cannot calculate an expe...

The sample space for U has two values in it. One of them is 10, and that's taken by Y. That leaves exactly one slot open in the sample space for eithe...

But this can't be it. You'd accept something short of a Vulcan mind-meld as communication, yes? Here's a couple thoughts. Suppose you're a lifelong fa...

There are facts that represent other facts. They do this, roughly, by their elements being arranged the same way the elements of the facts represented...

It's really not. There are two different sorts of things here. One is the conditional probability of an event occurring, one is the value in our sampl...

Okay, so pictures and reality. A picture represents a way things might stand in logical space. I've been calling that sort of thing a possible partiti...

And yet my use of X and 2X leads to the correct conclusion, while your substituting Y and Y/2 leads to the wrong conclusion. Since that is the only di...

That's not an argument, it's just a calculation, and it didn't. It weights the possible values known to be in the sample space for \small U, namely \s...

I know it seems that way to you. That is your subjective view of the situation. But we are told no such thing. We are not told X is selected by some r...

Here we go. Thinking I'll post bits as I get them done, then you can read and comment while I'm doing the next bit. (Here and there you'll notice me f...

1. Y = 10 2. Y = X ? U = 20 3. Y = 2X ? U = 5 These statements are all true. *4. Y = X *5. ? U = 20 (1,2,4) *6. Y = 2X *7. ? U = 5 (1,3,6) One of (4) ...

I suppose I could also have thrown in this: \small \begin{align} E(U \mid Y=10) &= P(U=2X \mid Y=10)(2X) + P(U=X \mid Y=10)(X) \\ &=\frac12 2X + \frac...

Usually a good idea to use different variable names when you add or remove quantifiers, so you remember they're not the same variables. ((And use the ...

@"Michael", @"andrewk" Nothing here that @"Jeremiah" and @"Snakes Alive" haven't already said, I think, just arranged a little differently. Check my m...

Honestly this take on the problem is far more interesting than the "paradox" and of real-world use. I'd rather be thinking about that. But it's also c...

It is absolutely true that in iterated play you can learn stuff about the sample space and its distribution, and that you can develop strategies that ...

And in every single case, whatever the value of Y, the player will choose to switch. I model the results of switching, which are quite clearly the sam...