Timeline for Two envelope (sub)-problem / related problem
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| Mar 3, 2016 at 14:35 | comment | added | John | Finally, back to my OP, the sub-puzzle. In that variation I tell the player a lot more. I even tell them which envelope contains the amount that resulted from the coin flip. It's a no brainer which to go for. The point of the sub puzzle is to raise the question about the prior probabilities. What is the correct way to think about these? Is it even a sensible question? | |
| Mar 3, 2016 at 14:32 | comment | added | John | What is more, there can be a tendency to modify the puzzle in unwarranted ways to satisfy the maths. Above you argued for averaging by playing multiple times. In the final paragraph the focus is always on the concrete amount being the largest in the game. By symmetry I'm as likely to be looking at the smallest amount aren't I? I think we need to tackle the puzzle by discussing the simplest form. There are two envelopes on the table. One contains twice as much as the other. If pressed I will say that I just thought of a number and flipped a coin to put double or half in the other envelope. | |
| Mar 3, 2016 at 14:24 | comment | added | John | I'm happy with all of that. I'm even happy with the claim by a mathematician friend that when I claim to have just 'thought of a number', this notion has to be considered as analogous to some kind of pattern of selection from an unbounded set rather than pure inspiration. But, back to the beginning, is all this really useful or necessary? Is it an advance on thinking about a single pair in isolation, a pair that is to be thought of as coming from nowhere? I can argue that there is no expectation of gain through the X, 2X logic, which is sound even if very hard to keep hold of after opening. | |
| Mar 3, 2016 at 14:16 | comment | added | John | OK, so moving on to some of the mathematical arguments in a bit more detail. I have just discussed the idea of playing a fixed, finite, collection of pairs of envelopes. Beyond that we need to make some distinctions, in particular: between the concept of infinity and an unbounded set; and between an unbounded set and the problem / possibility of selecting an amount from an unbounded set. 'Infinity' is never the amount in an envelope, but the amounts can be selected from an unbounded set using a probability distribution that tails off towards zero (and sums to one). | |
| Mar 3, 2016 at 14:09 | comment | added | John | My second overview point is that some of the maths you have presented really just reduces to the above point. If we play a finite collection of envelopes and average over them, then the set will contain a largest amount and dropping from this as well as from other amounts with no higher pair balances out the gains leading to no gain overall. All fine, but no different from playing a finite set comprised of one pair. The logic is just the same as X, 2X in my previous comment. | |
| Mar 3, 2016 at 14:05 | comment | added | John | Hi, I'm afraid that I'm still not convinced! It'll take a few comments, so bear with me. Two overview points first: as I have argued before, the player can be told the following as a reason to expect no gain on swapping: if the smaller amount on the table is X, the larger is 2X, so swapping amounts to going up or down a fixed amount, X, with equal chance. It is quite easy to see this when the envelopes are sealed. It is the act of opening one and seeing an actual amount of money that throws the player. But really, what has changed? My first overview point re all the maths is: is it necessary? | |
| S Mar 3, 2016 at 9:40 | history | answered | delusionist | CC BY-SA 3.0 | |
| S Mar 3, 2016 at 9:40 | history | made wiki | Post Made Community Wiki by delusionist |