The mistake could also be found in the assumption that the envelope I hold has a determinate amount X of which the values of the other envelope is derived. — Benkei
One venue I'm thinking about is that this falsely suggests there are three possible values for the envelopes: X, 2X and X/2. But we know there are only two; X and 2X.
If the envelope I'm holding is X then switching gives me 2X but if it's 2X then switching gives me X. Profit and loss are equal.
The mistake could also be found in the assumption that the envelope I hold has a determinate amount X of which the values of the other envelope is derived. — Benkei
If only the amount in the first envelope, the envelope you chose and perhaps are even allowed to open, is fixed, and the second envelope is then loaded with either half or twice the amount in yours, then switching is the correct strategy. This is the variant Barry Nalebuff calls the Ali Baba problem. — Srap Tasmaner
I recall an earlier discussion where another poster (a frequentist) said that once a coin had been tossed it would be wrong to say that it's equally likely to be heads as tails (even if we haven't looked); instead if it's actually heads then it's not possible that it's tails and if it's actually tails then it's not possible that it's heads. I believe Jeremiah and Srap (and perhaps you?) would take this same reasoning and say that if it's actually £5 in the other envelope then it's not possible that it's £20 and if it's actually £20 in the other envelope then it's not possible that it's £5. — Michael
We're dealing with a situation where we know that there's £10 in our envelope. What's the value of the other envelope? It's possible that it's £5, as the envelope set could be £5 and £10, and it's possible that it's £20, as the envelope set could be £10 and £20. It's not possible that it's £1 as the envelope set can't be £1 and £10. — Michael
Your expression however allows for both and therefore has to be wrong by necessity. — Benkei
The total sum possible for both envelopes in the above assuming one envelope contains 10 GBP is either 3x = 30 or 3x = 15 but we know it's either one of the two, it cannot be both.
I also refer to my earlier comment that the above expression, if we allow for unlimited switching of envelopes, would entail having to switch indefinitely wich is absurd. — Benkei
We wouldn't, because we've opened an envelope in this example. I know that there's £10 in my envelope. If from this we can deduce an expected value of £12.50 in the other envelope then once we switch we have no reason to switch back. — Michael
Would you agree that your expression allows the envelopes to carry values of either X, 2X or X/2? — Benkei
We wouldn't, because we've opened an envelope in this example. I know that there's £10 in my envelope. If from this we can deduce an expected value of £12.50 in the other envelope then once we switch we have no reason to switch back. Instead we have a reason to stick — Michael
Suppose you choose an envelope and then the facilitator tells you the other envelope has $10 in it. Then you would choose not to switch because yours has an expected value of $12.50. — Srap Tasmaner
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