kkwan…what are you talking about? My point was that step 7 was calculating expected value using impossible values, not that expected value is not a reliable principle in decision theory. And since it is using impossible values, it is an invalid expected value calculation. Why was it posited using impossible values in the first place? Because of steps 4-6, and I’ve already explained what’s wrong with them.
The expected value was not calculated using impossible values. In fact, it was calculated using possible values as:
Thus, step 7 is impeccably valid and your objection is not justified.
“In the two envelopes problem (TEP), there is no such guide to switch at all because the probabilities of selecting the envelope with the higher or lower amount are the same”
If one is given one choice, there’s no way of discerning or ‘strategizing’ how to pick the one with the most $.
Beyond that, i’m not really getting what’s so ‘intriguing’ about this ‘quandary’.
What I wrote must be taken in the context of post 130 where I was comparing the TEP to the Monty Hall problem i.e in comparing initial selection probabilities. It should not be taken to mean that the solution is simple.
The probabilities are the same in the initial selection process where we select one envelope randomly as the two envelopes are identical.
However, we don’t know whether the envelope we have selected has the higher or the lower amount. Hence, we are in a quandary. This is intriguing because we have selected an envelope with an unknown amount and comparative value wrt what is in the other envelope.
After the selection process comes the switching process (whereby the offer to switch is considered) and there is no such guide to switch or not to switch as in the Monty Hall problem.
This is the second quandary. It is also intriguing because the expected value (as a guide) leads to the logical absurdity to switch ad infinitum.
the problem is in the incorrect use of “expected value.” Another way of looking at is that the expected value is not the value that one would expect is in the other envelope, it’s simply a fancy statistical term for the average of the amounts in the two envelopes.
Another try. The text as kkwan quoted (bold by me):
Let us say you are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You may pick one envelope and keep whatever amount it contains. You pick one envelope at random but before you open it you are offered the possibility to take the other envelope instead.
Assume one envelope has A dollars in it, the other 2A.
Now let’s change the procedure a little: you pick an envelope, and you show the contents to me: it is A dollars. Now I take the other envelope, retract me from your view, and toss a coin. Dependent on the result I put A/2 dollars into it, or 2A dollars. Now you can loose A/2 to take the other envelope, or gain A. Now your formula applies. But the difference with the original formulation is clear: the total amount is fixed, it is 3A. In the second it is variable: it is minimal 1.5A (you took the envelope with A, and I tossed for 0.5A), maximal 6A (you took the envelope with 2A and then I tossed for doubling it, so 6A).
It shows the same fact again and again: in your example the A’s are not the same, and you cannot combine them in the way you do.