Monty Hall-esque question

[Masquerade]

Of course as a general strategy it makes no difference whether you switch or not in terms of EV.

To see this imagine an omniscient observer who puts X and 2X into the envelopes and runs the scenario many times.

Of those who never switch half will choose X and half will choose 2X initally so their EV is 3/2 X. Clearly those who always switch have the exactly the same EV.

The fallacy is in thinking that once you've chosen an envelope there is a 50-50 chance of doubling or halving. There is a 100% chance of doubling and 0% of halving - or vice versa. The outcome is completely determined.

[AaronBrown]

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His statement that all of his points have been proven rigorously is dead wrong?

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No, my objection is to what followed immediately afterwards, that:

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but people still argue because the only way these are intuitive is if you have studied this before. Math is hard.

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People do not argue only because they can't follow the math. That's the dangerous belief.

Let's take another paradox, where perhaps positions are not so rigid. Zeno told the story of a race between Achilles and the Tortoise. The Tortoise has a head start. Achilles cannot win, because by the time he gets to where the Tortoise started, the Tortoise has moved forward some amount. By the time Achilles gets to that point, the Tortoise has moved forward again. However many times you iterate, Achilles is still behind the Tortoise.

If Achilles moves at constant speed A, the Tortoise moves at constant speed T and the headstart is distance H, it takes time H/A for Achilles to get to the original start, by which time the Tortoise has moved forward T*H/A. It takes Achilles T*H/A^2 to get there, then T^2*H/A^3 and so on. The sum from i = 0 to infinity of T^i*H/A^(i-1) = H/(A - T), so it will take Achilles that long to catch up with the Tortoise. The key is the sum of an infinite number of terms can have a finite sum.

That's all true, and can be made rigorous. But it just verifies a calculation that anyone can do without infinite sums. Achilles gains on the Tortoise with speed A - T, and has to make up H, so it will take H/(A - T). Everyone knows that is what will really happen.

We have not addressed the other side of the paradox. The math requires assumptions about space and time that are not physically true. A deeper consideration of the paradox can lead to insights into important issues in physics and mathematics.

If you say everyone who continues to argue about Zeno's paradox doesn't know enough math to sum an infinite series, you go down a dangerous road. The same thing is true of the envelope paradox.

[PairTheBoard]

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The fallacy is in thinking that once you've chosen an envelope there is a 50-50 chance of doubling or halving. There is a 100% chance of doubling and 0% of halving - or vice versa. The outcome is completely determined.

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I used to make this argument, and it is true from a certain perspective. But I don't think it's satisfying. Certainly, prior to opening the envelope you would be willing to bet it's the smaller envelope and you'd be happy to take 2-1 odds on the bet. So in a reasonable sense most gamblers would say there's a 50-50 chance it's the smaller envelope, even though the outcome has really been determined by your choice. It's just like being dealt a card face down. Either it's a red or black card and the outcome has been determined already. But until you know the outcome you still say the probabilty is .5 that it's red and you'd bet on it at 2-1 odds, happy to take the +EV.

Once the envelope is opened, the outcome has been no more determined than it was before. And unless you have some inside information on the prior distribution for envelope amounts you'd probably still be happy to bet it's the smaller envelope given 2-1 odds. Here's where the problem comes in though. It's not that the unknown outcome has been determined that fouls the EV. It wouldn't foul it if you could make an arbitrary bet. The problem is that you are required to bet an amount that has been dictated by that determined outcome. Switching is equivalent to betting half the envelope amount at 2-1 that it's the smaller envelope.

I tried to promote the idea you are talking about in a thread called "Which Twin has the Tony?" Twin girls are presented wearing hats. One has a Tony hairdo the other doesn't. What is the probabilty that the one on the left has the Tony? I argued that it wasn't 50-50 because the outcome was already determined. Either she had the Tony or she didn't. It's not like you got to flip a coin to decide. While a couple of experts agreed with my technical point it clearly wasn't satisying to the crowd - or to me for that matter. I've since revised my thinking on this kind of thing.

PairTheBoard

[mykey1961]

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This is the best explanation you'll ever see for what's going on here. Credit for it goes to me,

PairTheBoard

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I sure hope not. All you've done is restate ev = (50+200)/2 as getting 2:1 on 1/2 of your envelope.

Both make assumptions that aren't stated in the "paradox".

Both compare unknown apples to imaginary oranges.

2:1 is true under certain conditions.

(50+200)/2 is true under certain conditions.

Both ignore other conditions.

Neither compares the EV of the 1st decision with the EV of the 2nd decision.

[PairTheBoard]

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This is the best explanation you'll ever see for what's going on here. Credit for it goes to me,

PairTheBoard

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I sure hope not. All you've done is restate ev = (50+200)/2 as getting 2:1 on 1/2 of your envelope.



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That's not what I said. When you open the envelope, see $10, and are offered a switch, you are being offered a gamble. The $10 is now yours. It's your money. If you switch and you have the smaller envelope you will win $10. If you switch and don't have the smaller envelope you will lose $5. That's the gamble. What I said was, that gamble is equivalent to betting half the amount in your envelope at 2-1 odds on the proposition that you have the smaller envelope.

But that's only the first step. To look at it in those terms. Given those terms it appears to be a +EV bet. The question remains, why isn't it?

Is it because seeing the $10 means there is no longer a 50% chance that it's the smaller envelope? Well, yes and no. pzhon's first point says yes, and of course he's correct from the perspective of possible prior distributions on the amounts put in the envelopes. In order for all possbile envelope amounts to have equal chance of being the smaller, the envelope amounts would have to be drawn from a uniform distribution on the positive real numbers. There is no such distribution, as elindauer points out.

Masquerade's point also says Yes from the perspective that the envelope amounts are fixed and any repetition of the experiment will use the same envelope amounts. In that case, seeing the $10 determines the outcome. There's no longer a 50% chance that it's the smaller. Either it is or it isn't. Masquerade's point is also correct and agrees with pzhon's points if you consider the prior distribution for the smaller envelope amount to be a point mass distribution.

How can the answer be "Yes and No" if the above points are correct. Well, it depends on what you mean by probabilty, and it's exactly on this "Yes and No" that Aaron basis his opinion that the problem really is not so well understood.

The sense in which the answer is No, seeing the $10 does not change the probalility of 50% that it's the smaller envelope is from a gambler's perspective. Prior to opening the envelope, anybody here would be delighted to bet $100 that it's the smaller envelope at 2-1 odds. We would see it as 50-50 proposition with favorable odds despite Masquerade's perspective that the envelope has already been chosen and so the 50% probabilty has been nulified.

Furthermore, once the envelope has been opened and we see it has $10 in it, I think most people here would still be willing to bet the $100 at 2-1 odds that it's the smaller envelope. We might look at the $10 and consider elindauer's objection and pzhon's first point. But those points don't really effect the EV for our $100 bet as long as we would have been allowed the same $100 bet if we had chosen the other envelope instead. As far as our $100 bet is concerned the chances are still 50% that it's the smaller envelope. And our $100 bet has +EV of $50.

So why doesn't our $5 bet - equivalent to switching - also have +EV of $2.50? Is it really because of elindauer, pzhon, and Masquerade's points? Well, ok, if you try to use a probabilty model to calculate the EV you will run into their points. But if you simply compare the $100 bet with the $5 bet I think the reason is transparent. The $100 bet has +EV because you are betting an amount of your choosing. The $5 bet is not +EV because it is a betting amount that has been determined by the outcome of the bet. A gambler doesn't need to know any fancy probabilty theory to know he's getting suckered there. Flip a coin and bet on heads, I'll pay you 2-1. Great says the gambler. Except you have to bet $10 if it's heads and $50 if it's tails. Not going for that says the gambler. How about $10 for heads and $20 for tails? Why should I says the gambler. Nothing in that for me. What if I hide the same proposition in two envelopes? The gambler is still indifferent.

If you think about it mykey, I'm in complete agreement with your point.

PairTheBoard

[PairTheBoard]

.

[southerndog]

Why not just actually play the game?

[jason1990]

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Why not just actually play the game?

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Are you suggesting we build a computer simulation to repeatedly play the game and actually compute (i.e. estimate) these probabilities? Well then, you have to decide on the details. For example, which of the following rules do you incorporate in your game:

(1) With each play, the computer uses the same dollar amounts for the envelopes.
(2) With each play, the computer randomly selects new dollar amounts for the envelopes.

You make a choice of either Rule (1) or Rule (2) and then run the simulation. You then want to estimate the conditional probability that the higher envelope was chosen, given that chosen envelope contains some given number X. So you let N be the number of times the chosen envelope contains X and you let M be the number of those times that this is the higher envelope. If you had chosen Rule (1), then you will find M/N is either 0 or 1, depending on the value of X. If you had chosen Rule (2), then you will find M/N is converging to some number (not necessarily 0.5) which depends on the method by which the computer is randomly filling the envelopes. So you would find that the simulation confirms the resolutions to the paradox which are given in this thread, and there is nothing left to argue about...

Unless someone says, "Hey, the original Two Envelope Paradox is a one-time shot. You are not allowed to repeat it." Then you can throw your simulation in the trash, because it won't convince them. And now you're back to arguing.

[mykey1961]

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[ QUOTE ]
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This is the best explanation you'll ever see for what's going on here. Credit for it goes to me,

PairTheBoard

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I sure hope not. All you've done is restate ev = (50+200)/2 as getting 2:1 on 1/2 of your envelope.



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That's not what I said. When you open the envelope, see $100(ed), and are offered a switch, you are being offered a gamble. The $100(ed) is now yours. It's your money. If you switch and you have the smaller envelope you will win $100(ed). If you switch and don't have the smaller envelope you will lose $5. That's the gamble. What I said was, that gamble is equivalent to betting half the amount in your envelope at 2-1 odds on the proposition that you have the smaller envelope.

But that's only the first step. To look at it in those terms. Given those terms it appears to be a +EV bet. The question remains, why isn't it?

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It could be +EV to switch, it could be -EV to switch, and it could be EV = 0 to switch. We don't have enough information to determine whether it is, or isn't +EV in this case.


What exactly do we know from the "Paradox"?

We know that one envelope containes $100. We know that the other envelope contains either $50 or $200.

What we don't know is how likely the other envelope is to contain $50, or $200.

If we assume both amounts are equally likely:

Yes it is +EV to switch.

The EV of switching would be 50*(1/2)+200*(1/2) = $125


If the other envelope has a >= 2/3 chance of $50:

EV of switching <= 50*(2/3) + 200*(1/3)

EV <= 100

Bottom line is we don't have enough information to make a solid decision.

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Is it because seeing the $100(ed) means there is no longer a 50% chance that it's the smaller envelope? Well, yes and no. pzhon's first point says yes, and of course he's correct from the perspective of possible prior distributions on the amounts put in the envelopes. In order for all possbile envelope amounts to have equal chance of being the smaller, the envelope amounts would have to be drawn from a uniform distribution on the positive real numbers. There is no such distribution, as elindauer points out.


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We have no need for "all possbile envelope amounts to have equal chance of being the smaller". We wouldn't always have picked the $100 envelope. All we know is that we did this time.

[mykey1961]

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[ QUOTE ]

Why not just actually play the game?

[/ QUOTE ]
Are you suggesting we build a computer simulation to repeatedly play the game and actually compute (i.e. estimate) these probabilities? Well then, you have to decide on the details. For example, which of the following rules do you incorporate in your game:

(1) With each play, the computer uses the same dollar amounts for the envelopes.
(2) With each play, the computer randomly selects new dollar amounts for the envelopes.

You make a choice of either Rule (1) or Rule (2) and then run the simulation. You then want to estimate the conditional probability that the higher envelope was chosen, given that chosen envelope contains some given number X. So you let N be the number of times the chosen envelope contains X and you let M be the number of those times that this is the higher envelope. If you had chosen Rule (1), then you will find M/N is either 0 or 1, depending on the value of X. If you had chosen Rule (2), then you will find M/N is converging to some number (not necessarily 0.5) which depends on the method by which the computer is randomly filling the envelopes. So you would find that the simulation confirms the resolutions to the paradox which are given in this thread, and there is nothing left to argue about...

Unless someone says, "Hey, the original Two Envelope Paradox is a one-time shot. You are not allowed to repeat it." Then you can throw your simulation in the trash, because it won't convince them. And now you're back to arguing.

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You can randomly pick however you want.

Just remember you have to ignore any trial where the envelope selected isn't $100.