I already did. — Jeremiah
But if we try to model this from the participant's point of view — Srap Tasmaner
we use Y. Then the expected gain loss sample spaces are [-Y/2, 0] and [2Y, 0], aren't they? — Srap Tasmaner
I'm tempted to think andrewk's point is relevant here: — Srap Tasmaner
Is the objection to Michael's approach not the assignment of the prior but simply that he is using sample spaces he cannot justify using? — Srap Tasmaner
You said this in reply to my pointing out that arguments based on expected values prior to opening an envelope are wrong because we don't even know if there is an expected value. To rebut that, you need to show that there is an expected value, eg that we can be confident the lower amount X is not drawn from a Cauchy distribution.Not true. — Jeremiah
It was a good choice. In the version in wikipedia it is easy, and uncontroversial, to conclude that there is no reason to switch. Not so in this case.I picked this version to generate more discussion. I think this version of the problem forces more conflict of ideas, which in turn generates more discussion. — Jeremiah
This is not well-defined. It needs re-stating to make it unambiguous. Here are some options.There are two possible amounts for the contents of the two envelopes either X or 2X. You don't know which is which — Jeremiah
That is not consistent with your use of the term 'expected value' in the following:This has nothing at all to do with distributions. — Jeremiah
'Expected value' only has meaning in the context of a probability distribution.In both cases our expected value then is:
1/2(X) + 1/2(2X) = 1/2X+X.
Which is the mid point between X and 2X. — Jeremiah
That distribution is (1) unknown, as X is unknown and (2) not appropriate for estimating expected gains from switching, since it does not use all the available information after the envelope has been opened. The expected gains from switching, in the absence of any knowledge of how X was selected, is as @Michael calculated it earlier. But the question in the OP does not explicitly ask for expected gains from switching. It just asks 'should you switch?'. What is supposed to be the metric used to determine whether to switch?Well technically this is a probability distribution: 1/2(X) + 1/2(2X). 50% is distributed to X and 50% is distributed to 2X.
That distribution is (1) unknown, — andrewk
The conclusion of symmetry is incorrect, because X is £5 in the first case and £10 in the second case. So we can't say that there's no difference because we lose X in one case and gain X in the other.[Jeremiah's] reasoning is:
I have £10. If X is 5 then I lose £X by switching. If X is 10 then I gain £X by switching.
So it's either -£X or +£X. This is symmetrical. — Michael
Interesting. The problem is different. Unlike in the OP, in the Wiki case, the envelope has not been opened before the option to swap. So the player has no new information in the wiki case. I think their analysis in the Simple Case is wrong, because it assumes the existence of an expected value that may not exist, but I think the conclusion that there is no reason to switch may in spite of that error be correct in that case, but not in this one. — andrewk
Why not? I've opened the envelope and seen that I have $10. That's in the rules as specified in the OP. And knowing the rules of the game I know there's a 50% chance that the other envelope contains $5 and a 50% chance that the other envelope contains $20. — Michael
Is there some rule of statistics that says that one or the other is the proper way to assess the best strategy for a single instance of the game (where you know that there's $10 in your envelope)? — Michael
I would have thought that if we want to know the best strategy given the information we have then the repeated games we consider require us to have that same information, and the variations are in the possible unknowns – which is what my example does. — Michael
The problem is making the randomizing element dependent on the first chosen envelope amount instead of simply as the choice of the envelope. — Andrew M
That claim is not consistent with the following formal calculations from my earlier post:On choosing an envelope and learning the amount in it, there is still no reason to prefer one envelope to the other. So there is no reason to switch. — Andrew M
which say that, after opening the first envelope and finding 10 pounds in it, the expected gain from switching is 2 pounds fifty.we use conditional probabilities to calculate the expected value of the gain G from switching [after seeing 10 pounds in the first envelope] as follows:
E[G] = 0.5 * E[G | X= 5] + 0.5 * E[G | X= 10] ] . . . . . . . . . . . (0)
= 0.5 * E[-X | X= 5] + 0.5 * E[+X | X= 10] . . . . . . . . . . . (1)
= 0.5 * E[-5 | X= 5] + 0.5 * E[+10 | X= 10] . . . . . . . . . . . (2)
= 0.5 * (-5) + 0.5 * (+10) . . . . . . . . . . . (3)
= £2.50 . . . . . . . . . . . . (4) — andrewk
4. From 2 and 3, there's a 50% chance that my £10 envelope is the X envelope and a 50% chance that my £10 envelope is the 2X envelope. — Michael
This is false. Whether the amount in your envelope is the smaller or the larger of the pair offered you is not a matter of chance. Whether you choose that envelope is. — Srap Tasmaner
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