OT -- Monty Hall-type problem

[Ganchrow]

Well, I’m not going to pretend to be able to fully follow all of that, especially the equations, but I think I’m at least closer to the Eureka moment I’m looking for. I think your discussion of “terminal” values is what is mostly opening my eyes.
-snip-

Yeah, you have it here. You certainly seem to get it. I would just mention one issue. If you noticed I chose only to consider possible that were integer powers of 2. Doing so simply makes the problem more tractable and easier to directly apply to the infite continuous value case (where the boxes could contain any real values unbounded from above).

The difference is that by only considering powers of 2 (or really any sequence of powers of 2 where the exponents increase by 1 with each go), we don't need to worry about simply being in the neighborhood of a terminal value as there will no longer be any values which can solely be approached from above or below. And again this is exactly what we'd see were the distribution of possible values continuous -- if the player knew the maximum possible value ex-ente then switching would always be correct were he to draw anything other than the maximum value.

Mind you, there's absolutely nothing wrong with your approach and in fact could be considered the more general solution. Nevertheless, there is an additional elegance when all non-terminal values are equally attainable.

So in other words, when you write:

Let’s think of it in terms of my original $500-$20,000 prize range. Note how few values there are for your box where you could infer that it is certain or probable the other box contains double that. It would have to be less than $2,000 down to the $500 minimum. Compare that to all the values there are for your box where you could infer that it is certain or probable the other box contains half that. Anything over $5,000 up to the $20,000 cap.

And this will be true (I think) for all possible ranges. There will always be a much wider area above the 50-50 area than below it.

while this may be true in general, these considerations may be relaxed and the problem will still hold. For example, if you were dealing solely with integer powers of 2, then there would be exactly 1 value where it would be certain that the other box contained half and exactly 1 value where it would be certain that the other box contained double.

Anyway, that's pretty academic on my part ... I just found it interesting. I'll see if I can't think of some intuitive explanation of why the paradox also breaks down in the infinitely-many-possible-values case.

[Arilou]

It is funny, but I would switch if the price was cheap in this spot, because the number was 9000. I find it unlikely that one of the envelopes contains 4500, since that is less round and looks worse on a game show. If you had given me 10000, I would have kept it.

Not that this solves the paradox, of course. But this actually seems pretty simple to me. I´d explain it like this. Assume some set of potential evelopes, E, and seperate them into the set of smaller ones, S, and the 2x ones that correspond to them, L. No matter what distribution you select for the envelopes, whether it is chosen at random or not, the amount of money in L averages twice the size of S for obvious reasons. So no matter what the distribution of values is, the more money you see the more likely it is you have the large value. That takes care of the problem well enough for me to stop worrying about it.

[so im zach]

Good old thread.

[reno cool]

that is an interesting question. Man Ganchrow used to catch a lot of shit. He did make everything seem complex.
I can't read all those formulas.
If you were given a box and told there is 50% chance switching would double your prize and 50% chance it would lower it, then you would want to switch. But that presumes 3 boxes. What exactly makes 2 boxes different I can't quite comprehend. Somehow there must be less than a 50% chance the other box contains double.

ok, once you get the higher amount there's 0 chance that it can be doubled. That has more negative value than the 100% chance of doubling the lower amount.

[MrX]

Sure, it does get you to the right answer ... but in a manner somewhat akin to explaining away Zeno's paradox with the contention that Achilles never actually lived.

Zeno burn. I miss Ganchrow.

[NeedProtection]

what happened to ganch?