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The Backgammon Solution

Posted By: Casper Van der Tak
Date: Sunday, 16 October 2016, at 9:21 a.m.

In Response To: The Backgammon Solution (Rick Janowski)

I am not sure the logical process goes wrong anywhere. The prize information is used to define the possible scenarios (2000-4000 and 1000-2000), and then you can assess the relative likelihood of the two possible scenarios and calculate the expected gains or losses from the switch.

I thought Timothy's two envelopes link also interesting, also because that is different (you are invited to switch before opening). In that case, the argument goes that the amount in the selected envelope is "A", meaning that in the other envelope there is either 2A or 0.5A, and you also could consider to switch. However, after you switch, you could call the amount you select B, use the same logic, and keep on switching forever. In that case, it is more clear that something goes wrong.

My personal take (and apologies if I make points already made in the link, I did not read through all) is that you could, having selected the envelop, call the amount in the other one A, and you either hold 2A or 0.5A, so you lose from switching. That shows there must be a flaw in the logic. Now if you believe a total of 3X is divided between the two envelopes, one contains 2X and the other X. You can then see that whichever way you chose to define "A", depending on the scenario it will be either 2X or X. Because A is not constant but twice as large in the one scenario than in the other, you cannot use it to calculate the gain from switching as it is normally done (0.5*(2A-A) - 0.5*(A-0.5A) or with signs reversed), because on the right hand side the amount A represents is twice as large than on the left side.

The upshot is that in absence of further information (so number of participants, entry fee, etc, say if I'd win a tv quiz) I would pay a 100 or so in the Lamford formulation, and I would not pay to switch in the two envelop problem. Call me crazy!

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