The Two Envelopes Problem: the most boring paradox in the world
iaindooley.com
iaindooley.com
Consider your position. I present you with two envelopes, and tell you that each contains a number, and one of those numbers is twice the other. You choose an envelope, and find that it contains the number 8.3467394756297. You can choose to swap, if you like, and your objective is to maximize the expected value of the number in the envelope you're holding.
Do you swap? Or not? Here's a line of reasoning.
Call the number I've chosen Z. The other envelope contains either Z/2 or 2Z, and it would appear to be 50:50 as to which it is. If you swap, my expected value is then (Z/2)0.5 + (2Z)0.5 = 5Z/4. That's bigger than Z, so you should swap.
But that means you should always swap, which doesn't seem to make sense.
Perhaps this version explains more clearly to you the paradox. Or not, maybe you still think it's all obvious. In that case, could you revise your explanation for this context, away from the $20 vs $10 or $40 issues?
You choose an envelope, and find that it contains the number 8.3467394756297
You're actually given the chance to switch without having first seen the value in the envelope.
If we forget the notion of switching and expected values for a second, I just want to focus on the very start of the statement where you say:
Call the number I've chosen Z. The other envelope contains either Z/2 or 2Z
The issue as I see it is that the probability exists that the other envelope contains Z/2 or 2Z, but not both - ie. the probability that one of those exists negates the other from existing, because in order for both to be possible you're introducing a 3rd value where only 2 existed at the start.
If you ignore all the envelopes and switching and whatnot you can boil it down to this:
I have two values A and B. I ask you to select one. What is the probability that the remaining value is C? Obviously it's zero.
So if A and B are envelopes, how does this change? If B is twice A how does it affect anything?
All the other information provided in this paradox just serves to distract you from the fact that in order for you to have selected a value, and for the probability to exist that either half that value or twice that value is in the remaining envelope, you would have had to have 3 values present at the start of the exercise.
Using your example, ignoring selection and envelopes and simplifying the goal of the answer: "Here's the letter B from a set of two consecutive letters. What's the other letter?"
The selection doesn't actually reveal anything. You're given the opportunity to switch prior to opening the envelope.
Using your example, ignoring selection and envelopes and simplifying the goal of the answer: "Here's the letter B from a set of two consecutive letters. What's the other letter?"
A and B are just labels for the 2 options present at the start of the exercise. You can't assume you've selected B and then say that the remaining option is either A or C because only A and B existed to begin with (or B and C, whatever the labels are is irrelevant).
To put it another way: if you select an envelope and attribute a dollar value to it (in the Wikipedia article this is $20) and then suppose it's either the higher or lower value, then compare the expected value of switching in each case, the things you're comparing have absolutely no relationship to each other.
If you assume the value you chose was $20, and it's the lower value, then your initial values were $20 and $40, and you should switch. If you assume the value you chose was $20, and it's the higher value, then your initial values are $20 and $10 and you shouldn't switch.
In either case the only relevant piece of information is whether or not you chose the higher or lower value and you can't include all 3 values ($10, $20 and $40) in an "expected value" calculation because only 2 values actually exist.
Actually it becomes even more starkly obvious if you change the dollar value between each statement, eg: assume you chose $20 and it's the higher value. If you switch, you lose $10. But what if you chose $5,000 and it's the lower value? If you switch you'd gain $5,000 so you should always switch because you're only risking a 50/50 chance of losing $10 for a 50/50 chance of gaining $5,000. This is clearly ridiculous!
This is a paradox only if you act like a computer. So what do I mean by that? This problem simply means, that if you selected an envelope, and you know that there is a chance that the other envelope will have higher value than then one you selected, then by probability, you could conclude that you are better off swapping to the other envelope. However, the paradox starts after you switch envelope because now you can start this all over again and think, hey, even though I just swapped, but it is still possible that the other envelope has value higher than the one I just selected!
To explain it in mathematical terms, when the value of the currently selected envelope is Z, the expected value of the other envelope will always be Z/2 * 0.5 + 2Z * 0.5 = 5/4Z higher than the one you selected.
The trick is to think procedurally, instead of trying to be realistic. In other words, it is always statistically possible that the other envelope has higher value than the one you selected, therefore you should always swap.
So there is 50% probability that the other one contains Z/2 and 50% that it has 2Z. Do you disagree with that statement?
I have two values A and B. I ask you to select one. What is the probability that the remaining value is C? Obviously it's zero.
No it doesn't. According to the problem, both 2Z and Z/2 are possible and you don't have the information to decide either way. Think of a simple coin toss, before I show you the result it is 50-50 to be heads or tails, even though it is just one of the two.
Yes! Well, in some circumstances. Unless you know how the values were placed into the envelope to begin with, how could you possibly assign a probability to it? I just found this great link today: http://lesswrong.com/lw/dy9/solving_the_two_envelopes_proble... which basically resolves the problem I'm having by taking into account how the values get into the envelopes in the first place.
If you have absolutely no knowledge of how the values got into the envelope in the first place then there's a 50/50 chance you got the higher value on your first pick, and a 50/50 chance you'll get the higher value on your second pick. You can't make a more informed decision unless you know how the values were initially chosen.
> The issue as I see it is that the probability exists that the other
> envelope contains Z/2 or 2Z, but not both - ie. the probability that
> one of those exists negates the other from existing, because in order
> for both to be possible you're introducing a 3rd value where only
> 2 existed at the start.
You could have flipped a fair coin to decide which envelope to choose in the first place. Therefore there is a 50% chance the non-selected envelope contains Z/2, and a 50% chance the non-selected envelope contains 2Z.
I don't understand where you think the impossible value C is being introduced. I don't understand where you think there is a third value.
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.
> 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.
That is the paradox, which you still appear not to have resolved.
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_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.