4 ms·
It depends on how the envelopes are prepared. If the two envelopes have x and 2x in them and one is randomly handed to you, there's a symmetry between them that
by pontus 4y ago
It depends on how the envelopes are prepared. If the two envelopes have x and 2x in them and one is randomly handed to you, there's a symmetry between them that tells you there is no benefit in switching.
However, suppose x is placed in an envelope and handed to you. Then a fair coin is flipped. If it comes up heads, 2x is placed in another envelope and if it comes up tail, x/2 is placed in that envelope. Then you should switch.
It's a subtle difference in how the envelopes are prepared, but it makes all the difference.
- zmgsabst 4y agoOh, weird. I’m not sure I’d have naively realized that the fixed amount in your envelope makes so much difference. For people confused like me, think about the outcomes: When the two are prepared together, say $2 and $4, then when you exchange your options are +2 and -2 with 50:50 odds. When the second envelope is prepared based on $2 in your envelope, then when you exchange your options are +2 and -1 with 50:50 odds. (I had to diagram this all out; unintuitive to me!)
- pontus 4y agoI did a pretty deep analysis of this problem a while back here in case you're interested: https://mindbowling.wordpress.com/2020/09/14/two-envelope-paradox/ https://mindbowling.wordpress.com/2020/09/14/two-envelope-pa...
- Groxx 4y agoIn some ways it's similar to the Monty Hall problem: https://en.wikipedia.org/wiki/Monty_Hall_problem https://en.wikipedia.org/wiki/Monty_Hall_problem The introduction of a second "stage" based on your action changes things in a non-obvious way. Your first action is less impactful than your second one, regardless of what you did.
- bottled_poe 4y agoThis is new information which changes the definition of the problem though.
- pontus 4y agoYeah, I agree. This setup though is the one to which the original "naive" argument applies which is why I think it's interesting to ponder.