Interesting Mathematical Paradox?

[jason1990]

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The indifference principle assumes that there is no effect.

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Is that correct?

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Let E be the event that we have the smaller envelope. Unconditioned, P(E) = 1/2. If we apply the indifference principle after looking at the value in the envelope, A, then we would conclude that P(E|A) = 1/2. Notice that this implies

P(A = x and E) = P(A = x)P(E | A = x) = P(A = x)P(E).

That is, under the indifference principle, A and E are independent. This is what I mean by "no effect."

[PairTheBoard]

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I was a little dissatisfied with my Random Switchers, since I was supposed to be replying to your idea about calling off the fixed $100 bet. So here is an idea about Random Call-Offers.

There is a sequence of Groups. Group n decides whether or not to call off the bet in this way. They generate a random number U, uniform on (0,1), and call off their bet if

A > sqrt{n|ln(U)|}.

Their EV is

50(2e^{-y^2/n} - e^{-4y^2/n}),

where the values in the envelopes are y and 2y. For small n, Group n will not do very well, since they will be calling off their bet too often. But the EV will increase monotonically with n. At about n = 1.64y^2, the EV of Group n will be about the same as the Never Call-Offers, $50. But the EV will continue to increase, reaching a maximum at about n = (3y^2)/ln(2). This Group will have an EV of about $59. After that, the EV will decrease monotonically, with the limit being $50. In other words, all Groups with n sufficiently large will outperform the Never Call-Offers.

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That's remarkable and I think an interesting new addition to the archive of 2+2 Two Envelope Facts, Results, and Implications.

I'm a little suprised because I thought the 2-1 odds were so great as to put any improvement out of reach. I guess it should be noted that for any fixed envelope amounts y,2y picking n large enough improves on the Never-Call-Offers. However, no matter how large n is chosen, if the envelope amounts y,2y are large enough the Never-Call-Offers still do better.

But wait a minute. This is actually amazing isn't it? Why can't you just wait until you see the envelope amount before you choose which n to use for your random choice? You won't know whether you're looking at y or 2y. But just choose n large enough to work for either. That way you can always insure an improved decision over just Not Calling-Off. Are you sure this is right? If so I think its a significant advance.

Wouldn't this varying n method also work for asssumed prior distributions for Y (other than the point mass)?

PairTheBoard

[jason1990]

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But wait a minute. This is actually amazing isn't it? Why can't you just wait until you see the envelope amount before you choose which n to use for your random choice? You won't know whether you're looking at y or 2y. But just choose n large enough to work for either.

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Suppose you see A. You know that either y = A or y = A/2. The worst case for you is y = A. So you could choose n = CA^2 for some suitable large C. The problem is this. The smaller n is, the more likely you are to call off your bet. With this strategy, in the case that A = y, you will be more likely to call off your bet than in the case A = 2y. This is exactly the opposite of what you want to be doing. You will screw yourself with this strategy and actually end up doing worse than the Never Call-Offers no matter how large C is.

[PairTheBoard]

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But wait a minute. This is actually amazing isn't it? Why can't you just wait until you see the envelope amount before you choose which n to use for your random choice? You won't know whether you're looking at y or 2y. But just choose n large enough to work for either.

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Suppose you see A. You know that either y = A or y = A/2. The worst case for you is y = A. So you could choose n = CA^2 for some suitable large C. The problem is this. The smaller n is, the more likely you are to call off your bet. With this strategy, in the case that A = y, you will be more likely to call off your bet than in the case A = 2y. This is exactly the opposite of what you want to be doing. You will screw yourself with this strategy and actually end up doing worse than the Never Call-Offers no matter how large C is.

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I thought my idea sounded too good to be true. But I guess I'm not understanding how this is working. You said,

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For small n, Group n will not do very well, since they will be calling off their bet too often. But the EV will increase monotonically with n. At about n = 1.64y^2, the EV of Group n will be about the same as the Never Call-Offers, $50. But the EV will continue to increase, reaching a maximum at about n = (3y^2)/ln(2). This Group will have an EV of about $59. After that, the EV will decrease monotonically, with the limit being $50. In other words, all Groups with n sufficiently large will outperform the Never Call-Offers.



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I understand you get screwed with small n. But in my idea you'd be choosing n large. I understand you can't insure it's the optimum n. But if you choose n larger than 1.64A^2 and A=y then you've chosen n greater than 1.64y^2 and you improve over Never-Calling-Off. And if A=2y, then you've still chosen n greater than 1.64y^2 and you still improve over Never-Calling-Off. In either case you've improved over Never-Calling-Off. What am I missing here?

PairTheBoard

[tisthefire]

yea, this confused me for a bit, if you realize neutral ev isn't 0 it works, the wiki thing makes it seem impossible which is far from accurate

[jason1990]

I think what you may be missing here is that my EV calculations are based on the fact that A and n are independent. If they are dependent, then that EV formula and subsequent analysis goes out the window. If you look back at the Call-Off criterion, you will see that if n is of the form CA^2, then you will just be calling off your bet with some fixed probability that does not depend on A. Your EV will then simply be $50*(prob. of not calling off).

Let me throw something else out there. Suppose you have a function f taking values between 0 and 1. You can use this function to generate a strategy for calling off. Namely, if you see A, then you call off your bet with probability 1 - f(A). Conversely, whatever strategy you might employ will correspond to some f. For example, the Never Call-Off strategy corresponds to f(A) = 1 for all A. The Group n strategy corresponds to f(A) = e^{-A^2/n}.

If your strategy corresponds to f, then your EV will be 50(2f(y) - f(2y)). In order to outperform the Never Call-Offers, you need this to be greater than 50. Let us call a strategy "uniformly dominating" if

2f(y) - f(2y) >= 1 for all y.

I claim that the only uniformly dominating strategy is the Never Call-Off strategy. I will leave it as a puzzle to prove this result.

[PairTheBoard]

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I think what you may be missing here is that my EV calculations are based on the fact that A and n are independent.

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ok. I think I just fell into another version of the Two Envelope Paradox flawed thinking. n is not a random variable. But if I choose n based on A I turn it into a random variable. In fact, I do so in a way that actually worsen my results, as you point out here:

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Suppose you see A. You know that either y = A or y = A/2. The worst case for you is y = A. So you could choose n = CA^2 for some suitable large C. The problem is this. The smaller n is, the more likely you are to call off your bet. With this strategy, in the case that A = y, you will be more likely to call off your bet than in the case A = 2y. This is exactly the opposite of what you want to be doing. You will screw yourself with this strategy and actually end up doing worse than the Never Call-Offers no matter how large C is.


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Another way to look at it is in an example. Say y=10. Say I look in the envelope and see 10. Say I then choose the random decision based on n=200. Will my results be better than Never-Calling-Off? Well, no. If I repeat the experiment for this fixed y and look at all the times I see 10 in the envelope, I will be calling off the bet some of the time which is poor because the Never-Calling-Off strategy wins every time I see 10 in the envelope. This even though the n=200 Decision strategy works better than the Never-Calling-Off strategy for y=10. That's because when you repeat the experiment for that EV you get to see A=20 half the time.

Introducing the function f is a nifty way to express the decision and generalize.

It looks like

2f(y) - f(2y) >= 1 for all y

would imply

f(2^n *y) <= 2^n(f(y)-1) +1

with right hand side going negative for large n, unless
f(y)=1

I'm going to have to think about all this a little bit. I think it has shed a little more light on the Two Envelope Paradox.

PairTheBoard

[dukemagic]

Huh I kind of assumed this thread had died off so I didn't check on it. Good to see all of the responses.

I was satisfied with the explanation of my friend and of what I think Midge was saying above:

Prior to opening either envelope, assign one envelope the value X and the other 2X. When you draw one it doesn't matter whether or not you switch, since you are 50/50 of having either one and so you stand to either gain X or lose X by switching. That is, if you have X and you switch, you gain X to now have 2X; the opposite is true if you have the envelope with 2X.

Why can't that be true after you open the envelope? You still have either X or 2X, with equal probability. The problem I was making before was then assigning new values to the second envelope of .5x and 2x, which you obviously can't do correctly, because that assigns a fourth possible value to the 2nd envelope (now you have 4 values - .5X, X, 2X, and 4X, whereas you started with only X and 2X).

So it seems to me that a simple explanation is that switching after opening is neutral EV. You stand to gain either X (if you had X and switch to 2X) or lose X (if you had 2X and switch to X). Since we are keeping the values in the envelopes consistent throughout the problem and you are indeed initially 50/50 to have either envelope, does this not make switching neutral EV?

Sorry if I'm rehashing a point we already disproved or something - I don't have enough practice with probability notation and math to understand the later responses.

[arahant]

I don't have the patience, either, since we've done this one a few a times. The optimal strategy is picking a number beforehand such that you switch if the first envelope contains less than that number. Probably, somebody already said that.

[PairTheBoard]

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The problem I was making before was then assigning new values to the second envelope of .5x and 2x, which you obviously can't do correctly, because that assigns a fourth possible value to the 2nd envelope (now you have 4 values - .5X, X, 2X, and 4X, whereas you started with only X and 2X).


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If you leave this part out, what you're saying is correct from the viewpoint that the envelope amounts are fixed and probabilities should be assigned based on repeating the experiment based on the fixed envelope amounts.

However you are wrong to indicate that after seeing amount A in the envelope you can't talk about the other envelope possibly having .5A or 2A. Even though only one is actually possible, either is possible according to what we know. So we can talk about them. We just have to be careful what we say. The mistake comes in assuming they have probabilites of 50%.

PairTheBoard