The paradox depends on the possibility of always being able to double the amount. This isn't true.
Swapping once will either double or half the amount; swapping back will just do the opposite.
Two envelopes problem
11–20 of 96 posts
Re: Two envelopes problem
#12I don't see the paradox... If you have 2x and swap you lose x. If you have x and swap, you gain x. 0.5 (-x) + 0.5 (x) = 0, So you should be indifferent to swapping. A I missing something? Not to say I don't make mistakes, but I have a BS in mathematics so maybe this is only obvious for people with a background in math? EDIT: No need for dollar values.
I don't have a BS in maths, but the important bit, as I read it, is that you don't know the values involved. I.e. if you have 20, there is either 10 in the other envelope, or 40. You have no way of knowing which is the case, so it's in your interest to swap since the benefits outweigh the risks. This is totally counterintuitive, though, so I'm fully willing to accept I'm missing something! And I don't buy the 'indefi…
Once you have the second envelope in hand, since you don't actually know what's in it, the exact same expected-value argument applies to switching back: there might be $10 or $40 in the other (first) envelope, "so it's in your interest to swap since the benefits outweigh the risk".
While the second swap would reverse any advantage possibly gained, it would also reverse any possible harm sustained You dont know. The situation is perfectly symmetric; picking and switching is the same as just picking the second one at first. So just pick one.
Re: Two envelopes problem
#13All the probability math therein, for a problem whose solution is highly intuitive (if you swap, you'd be just as inclined to swap envelopes indefinitely, is all you need to realize), reminds me of this quote: "The intuitive mind is a sacred gift and the rational mind is a faithful servant. We have created a society that honors the servant and has forgotten the gift." - Albert Einstein Shameless plug for a blog I lik…
It wasn't Einstein
http://ianchadwick.com/blog/it-wasnt-einstein-who-said-it/
Re: Two envelopes problem
#14The way I see it the flaw is in the first sentence of the "example":
> Assume the amount in my selected envelope is $20.
You can't just pull that assumption out of your ass. You can only assume what the problem states, which is that the values in the envelopes are X and 2X and you had a 50% chance of choosing either.
If you run the math without adding assumptions, it works out that swapping makes no difference, statistically.
Though it is a very clever trick :).
Re: Two envelopes problem
#15Just take both envelopes and run.
I agree. Plus, both have money. Win win.
Re: Two envelopes problem
#16The way I see it the flaw is in the first sentence of the "example": > Assume the amount in my selected envelope is $20. You can't just pull that assumption out of your ass. You can only assume what the problem states, which is that the values in the envelopes are X and 2X and you had a 50% chance of choosing either. If you run the math without adding assumptions, it works out that swapping makes no difference, stati…
I think the key assumption is there is a 50% chance of getting double and a 50% chance of getting half in the swap scenario. Once you pick an envelope initially the chance disappears; you can't parlay the chance into the new context.
Re: Two envelopes problem
#17That is a very long article based on flawed argument. Given no other information, assuming someone gave you 2 envelopes and told you one has $40 vs $20, common sense dictates choose 1 randomly and walk away - with no other information it is illogical to reason any other way. The chance you choose the lower value is 1/2. Now, if you are allowed to look inside the envelope (which gets introduced further down) then it b…
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Re: Two envelopes problem
#18The way I see it the flaw is in the first sentence of the "example": > Assume the amount in my selected envelope is $20. You can't just pull that assumption out of your ass. You can only assume what the problem states, which is that the values in the envelopes are X and 2X and you had a 50% chance of choosing either. If you run the math without adding assumptions, it works out that swapping makes no difference, stati…
Taking out the specific dollar amount doesn't change the math at all. If X designates the amount in the envelope I've selected, than 50% chance the other envelope contains .5 * X and a 50% chance it contains 2 * X, so the expected value of the other envelope is .5 * .5 * X + .5 * 2 * X = 1.25 * X which is greater than X.
Re: Two envelopes problem
#19That is a very long article based on flawed argument. Given no other information, assuming someone gave you 2 envelopes and told you one has $40 vs $20, common sense dictates choose 1 randomly and walk away - with no other information it is illogical to reason any other way. The chance you choose the lower value is 1/2. Now, if you are allowed to look inside the envelope (which gets introduced further down) then it b…
Common sense in the Monty Hall problem says just pick a door and stick with it. The chance you chose the goat is 1/2, right?
No. Common sense is often wrong.
Re: Two envelopes problem
#20Since you don't know what is in either envelope surely both theoretically contain both x and 2x (al a Schrödinger's cat). Only upon opening an envelope will we know the amount; which one you take or how many times you swap make no difference.
Also swapping, in my understanding, doesn't increase your probability. In the first choice you 1 in 2 chance of gettng the higher amount. In the second choice (the ability to swap, which is fundamentally the same as choosing between the two envelopes) you have a 1 in 2 chance of getting the higher amount.
Correct me if I'm wrong.