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Saying that I chose $20, and there's a 50% chance of there being $10 in the remaining envelope, and a 50% chance of there being $40, takes into account 3 amount

by dools 13y ago

Saying that I chose $20, and there's a 50% chance of there being $10 in the remaining envelope, and a 50% chance of there being $40, takes into account 3 amounts: $10, $20 and $40. However if you're presented with 2 envelopes, there are only 2 amounts to begin with. The only way you could assign a probability to what the second envelope contained would be if you started with 3 amounts then chose 2. For example I have a 5c coin, a 10c and a 20c coin. I pick the 10c coin and flip it: heads I pair it with the 5c coin and tails I pair it with the 20c coin, then I put each under a cup and invite you to choose. Now if you lift the cup and see 10c you know there is a 50% chance the other cup has 5c and a 50% chance it has 20c. You can take into account 3 amounts (ie. the one you chose and 2 possible remaining values) because the question starts off with 3 amounts. If your knowledge of the system commences with "here are 2 cups" you have no extra information -- you only have 2 amounts. The higher and the lower. You will pick the higher one 50% of the time on the first go, and 50% of the time if you switch.

ColinWright 13y ago
> Saying that I chose $20, and there's a 50% chance of there > being $10 in the remaining envelope, and a 50% chance of there > being $40, takes into account 3 amounts: $10, $20 and $40. Right, but that's after you have opened the envelope. > However if you're presented with 2 envelopes, there are only > 2 amounts to begin with. I don't see why you think this is an issue, or relevant. > The only way you could assign a probability to what the second > envelope contained would be if you started with 3 amounts then > chose 2. That doesn't make sense. Suppose I flip a coin, and choose an envelope at random. Without opening it, suppose I designate the value it holds as Z. Now I know with probability 50% that the other envelope contains Z/2, and with 50% the other envelope contains 2Z. > For example I have a 5c coin, a 10c and a 20c coin. I pick the > 10c coin and flip it: heads I pair it with the 5c coin and tails > I pair it with the 20c coin, then I put each under a cup and > invite you to choose. > Now if you lift the cup and see 10c you know there is a 50% chance > the other cup has 5c and a 50% chance it has 20c. You can take into > account 3 amounts (ie. the one you chose and 2 possible remaining > values) because the question starts off with 3 amounts. This seems (a) completely different, and (b) unenlightening. It's completely different because I seem to know what all the options were for what you did and the amounts involved. > If your knowledge of the system commences with "here are 2 cups" > you have no extra information -- you only have 2 amounts. The > higher and the lower. You will pick the higher one 50% of the time > on the first go, and 50% of the time if you switch. Yes. So now, without lifting the cup, let's consider some statistics. Let's call the unknown amount under the cup I've chosen Z. The other cup must contain Z/2 (with probably 50%, as you say) or 2Z (also with probability 50%). If I switch, my expected return is then 5Z/4, which is larger than Z. Hence, if given the option, I should switch, even though I don't know what the amount is that I've chosen. That is the paradox, which you still appear not to have resolved.
- dools 13y ago
  The other cup must contain Z/2 (with probably 50%, as you say) or 2Z (also with probability 50%). Right that's not true. If you're presented with 2 cups/envelopes, whatever, and you're told that one contains double the amount in the other, then you have 2 amounts: Z and Z/2. If you choose a cup, then the other one either contains Z or Z/2. Whether or not you refer to those initial amounts as Z and 2Z, or Z and Z/2 is up to you, but there aren't 3 options, there are only 2. This seems (a) completely different, and (b) unenlightening. It's completely different because I seem to know what all the options were for what you did and the amounts involved. Yep that's my point: in order to attribute an expected value with probabilities for both Z/2 and 2Z assuming the amount you chose was Z, you have to know a bunch of additional information. This explains it much better than I can: http://lesswrong.com/lw/dy9/solving_the_two_envelopes_problem/ http://lesswrong.com/lw/dy9/solving_the_two_envelopes_proble... But if all you get are two envelopes, and the information that one is double the other, the situation is much simpler: you're choosing between 2 amounts and there is absolutely no difference if you switch.

by dools 13y ago

ColinWright 13y ago

> Saying that I chose $20, and there's a 50% chance of there > being $10 in the remaining envelope, and a 50% chance of there > being $40, takes into account 3 amounts: $10, $20 and $40. Right, but that's after you have opened the envelope. > However if you're presented with 2 envelopes, there are only > 2 amounts to begin with. I don't see why you think this is an issue, or relevant. > The only way you could assign a probability to what the second > envelope contained would be if you started with 3 amounts then > chose 2. That doesn't make sense. Suppose I flip a coin, and choose an envelope at random. Without opening it, suppose I designate the value it holds as Z. Now I know with probability 50% that the other envelope contains Z/2, and with 50% the other envelope contains 2Z. > For example I have a 5c coin, a 10c and a 20c coin. I pick the > 10c coin and flip it: heads I pair it with the 5c coin and tails > I pair it with the 20c coin, then I put each under a cup and > invite you to choose. > Now if you lift the cup and see 10c you know there is a 50% chance > the other cup has 5c and a 50% chance it has 20c. You can take into > account 3 amounts (ie. the one you chose and 2 possible remaining > values) because the question starts off with 3 amounts. This seems (a) completely different, and (b) unenlightening. It's completely different because I seem to know what all the options were for what you did and the amounts involved. > If your knowledge of the system commences with "here are 2 cups" > you have no extra information -- you only have 2 amounts. The > higher and the lower. You will pick the higher one 50% of the time > on the first go, and 50% of the time if you switch. Yes. So now, without lifting the cup, let's consider some statistics. Let's call the unknown amount under the cup I've chosen Z. The other cup must contain Z/2 (with probably 50%, as you say) or 2Z (also with probability 50%). If I switch, my expected return is then 5Z/4, which is larger than Z. Hence, if given the option, I should switch, even though I don't know what the amount is that I've chosen. That is the paradox, which you still appear not to have resolved.

dools 13y ago
The other cup must contain Z/2 (with probably 50%, as you say) or 2Z (also with probability 50%). Right that's not true. If you're presented with 2 cups/envelopes, whatever, and you're told that one contains double the amount in the other, then you have 2 amounts: Z and Z/2. If you choose a cup, then the other one either contains Z or Z/2. Whether or not you refer to those initial amounts as Z and 2Z, or Z and Z/2 is up to you, but there aren't 3 options, there are only 2. This seems (a) completely different, and (b) unenlightening. It's completely different because I seem to know what all the options were for what you did and the amounts involved. Yep that's my point: in order to attribute an expected value with probabilities for both Z/2 and 2Z assuming the amount you chose was Z, you have to know a bunch of additional information. This explains it much better than I can: http://lesswrong.com/lw/dy9/solving_the_two_envelopes_problem/ http://lesswrong.com/lw/dy9/solving_the_two_envelopes_proble... But if all you get are two envelopes, and the information that one is double the other, the situation is much simpler: you're choosing between 2 amounts and there is absolutely no difference if you switch.