Note, yet again, that all the values that could be in the cases are known from the start. There is no speculation about possible outcomes — Srap Tasmaner
What’s the expected value of his box? — Michael
average total value — Srap Tasmaner
The resolution of the apparent paradox is that the probabilities are not 50:50 for most values of Y.there's a 50% chance that the other envelope contains £20 and a 50% chance that the other envelope contains £5. — Michael
"if Y = X then Z is 2X or X/2" only adds up to 3X in one instance, the rest results in false conclusions as it contradicts the premise that the total should always be 3X. — Benkei
In the above you have an inherent contradiction in your conditionals as X and 2X are both 10. — Benkei
1. If R = heads then ...
2. If R = tails then ...
1 and 2 have contradictory antecedents. But I'm not saying that both the antecedent of 1 and the antecedent of 2 are true. One is true and one is false, with a 50% chance of each being true. And the same with my example with the envelopes:
1. If X = 10 then ...
2. If 2X = 10 then ... — Michael
1. If X = 10 then ...
2. If X = 5 then ... — Michael
If X = 10 then the other is 20
If X = 5 then the other is 10
In both cases you have now assumed you're opening the smaller envelope and the other HAS to be the bigger envelope. — Benkei
Why do you need to include Y, Michael? — Jeremiah
It does not provide updated information for the original uncertainty of X or 2X
Now to revisit the idea of knowing the amount in the envelope: I think using amounts like £5/10/20 is misleading because £5 intuitively feels like a throwaway amount that anyone would be happy to lose. Instead, what if your chosen envelope contained a cheque for £10 million? Would you throw away £5m chasing an additional £10m that may not even exist?
And here it gets interesting for me because... given a £10 envelope, I really would switch because a £5 loss is nothing. Given a £10m envelope, I'd stay.
Hi Benkei. Nice to see you join this discussion.Let's name the envelopes Y and Z (note, they do not denote amounts). The expression "if Y = X then Z is 2X or X/2" only adds up to 3X in one instance, the rest results in false conclusions as it contradicts the premise that the total should always be 3X. Knowing that Y is either X or 2X, we get four possibilities:
If Y = X then Z = 2X for a total of 3X is true.
If Y = X then Z = X/2 for a total of 1.5X is false.
If Y = 2X then Z = 2X for a total of 4X is false.
If Y = 2X then Z = X/2 for a total of 2.5X is false.
This suggests that replacing the variable of one envelope with a fixed amount or a fixed placeholder messes up things. I'm not sure why. Maybe andrewk can tell me. — Benkei
Imagine three variations of this game:You are playing a game for money. There are two envelopes on a table.
You know that one contains $X and the other $2X, [but you do not
know which envelope is which or what the number X is]. Initially you
are allowed to pick one of the envelopes, to open it, and see that it
contains $Y . You then have a choice: walk away with the $Y or return
the envelope to the table and walk away with whatever is in the other
envelope. What should you do?
ou can't just enumerate a set of cases, and claim each is equally likely. If you could, there would be a 50% chance of winning, or losing, the lottery. — JeffJo
2. Treat Y as known and model X using a Bayesian prior. This leads to a rule under which the player can calculate a value c such that her expected switch gain is positive if Y<c and negative if Y>c. — andrewk
c is not an observer-independent item that can be known or not. It is a feature of the Bayesian prior distribution the player adopts to model her uncertainty.I think the issue is that even if you know Y from opening the initial envelope, the expected gain from switching is still zero if you don't also know c. — Andrew M
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