3 ms·

No. If I understand correctly, your last two outcomes are wrong. There are 6 possible outcomes, not four. This is because when you choose the 2X correctly, you

No. If I understand correctly, your last two outcomes are wrong. There are 6 possible outcomes, not four. This is because when you choose the 2X correctly, you may receive 2(2X) or X. The 6 outcomes are: You select X, then you choose to stay, and wind up with X. You select 2X, then you choose to stay, and wind up with 2X. You select X, then choose to swap and wind up with 1/2X You select X, them choose to swap and wind up with 2x You select 2X, then choose to swap and wind up with X You select 2X, then choose to swap and wind up with 4X

sillysaurus2 13y ago
Actually, you always wind up with X or 2X, never 0.5X and never 4X. Remember, the envelopes contain either X or 2X, and you always end up with either one or the other.
- iamshs 13y ago
  I myself understood the problem with Afforess's reply. You are explicitly pinning the value in the envelopes, but remember the other envelope has either half or double of the value than the one in your hand, and this is the crux of this paradox. Hopefully, I am thinking in correct terms.
  - delinka 13y ago
    But he has assigned the identifiers "X" and "2X" to the envelopes. He could have called them "A" and "B" and his logic still holds. The problem I see with the wiki article explanation is: while using actual dollar amounts as examples, it gives the impression that you get to open the envelope that you picked first to see the amount, but that you don't know if it's the smaller value or larger value. So I choose "A" and I don't know if it contains $X or $2X. I don't even get to open it to know what amount is in the envelope. Whatever it contains, if I choose to switch, I get envelope "B" - I always have a 50% chance of choosing the larger amount or switching to the envelope with the larger amount because no information is revealed after the first choice. for A->$X, B->$2X: Choose A, stay with A, receive $X Choose A, switch to B, receive $2X Choose B, switch to A, receive $X Choose A, stay with B, receive $2X for A->$2X, B->$X: Choose A, stay with A, receive $2X Choose A, switch to B, receive $X Choose B, switch to A, receive $2X Choose A, stay with B, receive $X 50% chance of getting either amount, regardless.
deleted 13y ago
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sillysaurus2 13y ago

Actually, you always wind up with X or 2X, never 0.5X and never 4X. Remember, the envelopes contain either X or 2X, and you always end up with either one or the other.

iamshs 13y ago
I myself understood the problem with Afforess's reply. You are explicitly pinning the value in the envelopes, but remember the other envelope has either half or double of the value than the one in your hand, and this is the crux of this paradox. Hopefully, I am thinking in correct terms.
- delinka 13y ago
  But he has assigned the identifiers "X" and "2X" to the envelopes. He could have called them "A" and "B" and his logic still holds. The problem I see with the wiki article explanation is: while using actual dollar amounts as examples, it gives the impression that you get to open the envelope that you picked first to see the amount, but that you don't know if it's the smaller value or larger value. So I choose "A" and I don't know if it contains $X or $2X. I don't even get to open it to know what amount is in the envelope. Whatever it contains, if I choose to switch, I get envelope "B" - I always have a 50% chance of choosing the larger amount or switching to the envelope with the larger amount because no information is revealed after the first choice. for A->$X, B->$2X: Choose A, stay with A, receive $X Choose A, switch to B, receive $2X Choose B, switch to A, receive $X Choose A, stay with B, receive $2X for A->$2X, B->$X: Choose A, stay with A, receive $2X Choose A, switch to B, receive $X Choose B, switch to A, receive $2X Choose A, stay with B, receive $X 50% chance of getting either amount, regardless.

deleted 13y ago

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