the flaw here is assuming that the value being higher on average in the envelope is a sufficient criteria to make swapping the best choice. this is not how probability works, if i have a million empty envelopes and one with a billion dollars in it, the average value of each envelope being £1000 doesn't mean anything in the face of the 1 in a million chance of actually picking the right one.
Two envelopes problem
21–30 of 96 posts
Re: Two envelopes problem
#22You can even explicitly define the distribution as P($2^i) = 1/(2^i) and the 'paradox' remains. The problem is that you are reasoning about the expected gain of swapping but that expectation is a sum over a series which is not absolutely convergent (http://en.wikipedia.org/wiki/Absolute_convergence) so it's value depends on the order in which you sum over the different cases.
In fact, if you repeatedly simulate the problem (make sure you use bignums) you will find that the mean gain from swapping is not well behaved at all and refuses to converge. The law of large numbers only applies if the expectation is well-defined.
The lesson here is: when in doubt, explicitly write out the probability space over which you are working. Problems like this and the Monty Hall problem are trivially solvable on paper. There is a reason that mathematicians get all hot and bothered about formalism - it gives you a solid base from which to build correct intuitions.
EDIT Let's write this down properly.
i | gain from swapping if I have the smallest | gain from swapping if I have the largest
1 $2 -$2
2 $4 -$4
3 $8 -$8
etc
The first is that the situation is symmetric so you can't possibly gain. That is:
E(Gain) = (1/4 * $2 + 1/4 * -$2) + (1/8 * $4 + 1/8 * -$4) ...
= $0 + $0 + $0 + $0 ...
= $0
E(Gain) = (1/4 * $2) + (1/4 * -$2 + 1/8 * $4) + (1/8 * -$4 + 1/16 * $8) ...
= $0.5 + $0 + $0 + $0 ...
= $0.5
Re: Two envelopes problem
#23That 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…
Yes, it's not a paradox it's just seductive flawed reasoning. Yes, at any point EV of picking an envelope at random is 3/4n (n being higher amount of money out of the two). It is all there is to it. The "paradox" is introduced by silent assumption that distribution of amounts put in envelopes is uniform which is impossible (because you can't pick numbers from infite set uniformly even if there was infinite amount of…
You can explicitly state the distribution and still run into the same problem: https://news.ycombinator.com/item?id=6387344 .
The underlying problem is basically that probability theory in non-finite spaces has some gotchas - one of which is that the expectation of a random variable does not always exist.
Re: Two envelopes problem
#24the flaw here is assuming that the value being higher on average in the envelope is a sufficient criteria to make swapping the best choice. this is not how probability works, if i have a million empty envelopes and one with a billion dollars in it, the average value of each envelope being £1000 doesn't mean anything in the face of the 1 in a million chance of actually picking the right one.
Actually, that's exactly how expected value works.
Re: Two envelopes problem
#25The 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.
but you are implicitly turning the scenario into one with three values, 1/2X, X, and 2X, so something went wrong with what you're doing. There was only ever a universe of two values.
Re: Two envelopes problem
#26You can even explicitly define the distribution as P($2^i) = 1/(2^i) and the 'paradox' remains. The problem is that you are reasoning about the expected gain of swapping but that expectation is a sum over a series which is not absolutely convergent ( http://en.wikipedia.org/wiki/Absolute_convergence ) so it's value depends on the order in which you sum over the different cases. In fact, if you repeatedly simulate the…
This is either way over my head or not at all clear. More explanation would be appreciated.
Re: Two envelopes problem
#27That 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…
Suppose you restate it. You own a $100 stock. It has a 50% chance of doubling, and a 50% chance of going down 50%.
Should you sell it, or hold onto it? Expected value of holding is ($200 + $50)/2 = $125 . So it seems like you should hold on!
If you repeat that wager indefinitely, the standard deviation of the net wins (number of ups-downs) goes up according to a square root law.
And the value of your stock goes up exponentially as the number of net wins goes up.
So over time the EV goes to infinity like (2 ^ (n/2)). The EV of say the -1SD outcome goes to zero. The EV of the +1SD outcome goes to 1/that. So the average EV overall goes to infinity.
And yet the expected value of the growth rate is 0. For every 16x win there's a loss down to 1/16th. But the average of those 2 outcomes diverges to infinity.
All this to say, when you're looking at exponential returns (or other processes), you need to measure growth rates, not average outcomes. And it has nothing to do with log utility.
Re: Two envelopes problem
#28You can even explicitly define the distribution as P($2^i) = 1/(2^i) and the 'paradox' remains. The problem is that you are reasoning about the expected gain of swapping but that expectation is a sum over a series which is not absolutely convergent ( http://en.wikipedia.org/wiki/Absolute_convergence ) so it's value depends on the order in which you sum over the different cases. In fact, if you repeatedly simulate the…
This is either way over my head or not at all clear. More explanation would be appreciated.
Sorry, I rushed it. Did the edit make it clearer?
Re: Two envelopes problem
#29The 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…
> If you run the math without adding assumptions, it works out that swapping makes no difference, statistically.
I beg to differ :)
Re: Two envelopes problem
#30This is one of my favorite problems. Here is one variation can help clarify thinking about it:
"There's two envelopes. One contains twice as much money as the other. No envelope contains more than $N."
This changes the problem dramatically. If your envelope contains more than $N/2, of course you should not switch: the other envelope necessarily contains less. If it contains $N/2 or less, perhaps you should switch: the other envelope may or may not contain more.
Say your envelope contains $x, and you don't know what $N is. There's two possibilities:
1. $x 2. $x > $N/2. If you switch, you will get $x/2, since no envelope contains more than $N.
If we assume the distribution is uniform on the range [0, N], then these possibilities are equally likely. Therefore the total expectation value is the average of the EVs of the two possibilities (3x/2 and x/2), which is $x. So we recovered the naive expectation of "it doesn't matter" from this variation. Now we can take the limit as $N goes to infinity, and while the EV of $x approaches infinity, the fact that switching does not matter does not change.
As others have said, the underlying problem is the assumption that a uniform probability on an infinite set makes sense, which it does not. However, we can instead take the limit for finite sets, in which case we recover the intuitive result that switching does not matter.