Monty Hall-esque question

[elindauer]

I knew I'd be glad I asked. Thanks for your well considered response.

FYI, I'd be really interested to read more about this:

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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.

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For both the Envelope problem and the Achilles problem, your thoughts on the deeper insights available if you study the other side of the paradox sound pretty intiguing.

-eric

[PairTheBoard]

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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.

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The point is that people incorrectly conclude that the probabilty that it's the smaller envelope is 50% because all envelope amounts are equally likely and so $50 and $200 must be equally likely for the other envelope. elindauer's point that there is no probabilty distribution for envelope amounts such that all envelope amounts are equally likely speaks to this flawed assumption by people who base their incorrect EV calculation on it.

PairTheBoard

[mykey1961]

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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.

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The point is that people incorrectly conclude that the probabilty that it's the smaller envelope is 50% because all envelope amounts are equally likely and so $50 and $200 must be equally likely for the other envelope. elindauer's point that there is no probabilty distribution for envelope amounts such that all envelope amounts are equally likely speaks to this flawed assumption by people who base their incorrect EV calculation on it.

PairTheBoard

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Examples can be shown with "probabilty distribution for envelope amounts such that all envelope amounts are equally likely" and where they are not equally likely, and still the "paradox" applies.

The bottom line is, sometimes the EV of switching will be greater than the first EV, sometimes it will be less, and sometimes the same. Opening one envelope doesn't give us enough information to deduce which is the case.

[PairTheBoard]

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Examples can be shown with "probabilty distribution for envelope amounts such that all envelope amounts are equally likely"

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By "all envelope amounts" I mean all amounts in the positive real numbers. There is no such example. If you think there is feel free to exhibit it.



PairTheBoard

[mykey1961]

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Examples can be shown with "probabilty distribution for envelope amounts such that all envelope amounts are equally likely"

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By "all envelope amounts" I mean all amounts in the positive real numbers. There is no such example. If you think there is feel free to exhibit it.



PairTheBoard

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Ok I don't understand why you feel $8 Billion has to be as likely as $2.95.

[PairTheBoard]

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Examples can be shown with "probabilty distribution for envelope amounts such that all envelope amounts are equally likely"

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By "all envelope amounts" I mean all amounts in the positive real numbers. There is no such example. If you think there is feel free to exhibit it.



PairTheBoard

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Ok I don't understand why you feel $8 Billion has to be as likely as $2.95.

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Try rereading this:

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The point is that people incorrectly conclude that the probabilty that it's the smaller envelope is 50% because all envelope amounts are equally likely and so $50 and $200 must be equally likely for the other envelope. elindauer's point that there is no probabilty distribution for envelope amounts such that all envelope amounts are equally likely speaks to this flawed assumption by people who base their incorrect EV calculation on it.


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Let me try saying this another way. The point is that many people assume that all envelope amounts Are equally likely. It's from this hidden incorrect assumption that they draw the incorrect conclusion that there Must be a 50% chance that they are looking at the smaller envelope. They then use that incorrect conclusion to make the incorrect EV calculation. The hidden assumption that all envelope amounts are equally likely is not only arbitrary, as you point out, but in fact impossible.

And you are of course right in pointing out that since the prior distribution for the envelope amounts is unknown we don't really know the true probabilty for the unseen envelope. So therefore the EV calculation can't be made.

PairTheBoard

[mykey1961]

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Examples can be shown with "probabilty distribution for envelope amounts such that all envelope amounts are equally likely"

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By "all envelope amounts" I mean all amounts in the positive real numbers. There is no such example. If you think there is feel free to exhibit it.



PairTheBoard

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Ok I don't understand why you feel $8 Billion has to be as likely as $2.95.

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Try rereading this:

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The point is that people incorrectly conclude that the probabilty that it's the smaller envelope is 50% because all envelope amounts are equally likely and so $50 and $200 must be equally likely for the other envelope. elindauer's point that there is no probabilty distribution for envelope amounts such that all envelope amounts are equally likely speaks to this flawed assumption by people who base their incorrect EV calculation on it.


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Let me try saying this another way. The point is that many people assume that all envelope amounts Are equally likely. It's from this hidden incorrect assumption that they draw the incorrect conclusion that there Must be a 50% chance that they are looking at the smaller envelope. They then use that incorrect conclusion to make the incorrect EV calculation. The hidden assumption that all envelope amounts are equally likely is not only arbitrary, as you point out, but in fact impossible.

And you are of course right in pointing out that since the prior distribution for the envelope amounts is unknown we don't really know the true probabilty for the unseen envelope. So therefore the EV calculation can't be made.

PairTheBoard

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Ok, so lets rewind.

How is getting 2:1 on half your envelope the best explanation?

Remember in your "best explanation" post, you addressed the amount you bet when you win, vs the amount you bet when you lose. Nowhere in the post did you mention that you have no idea how often that you win.

Also, since we aren't offered this 2:1 wager until after we have opened an envelope, we are wagering the same amount win or lose.

So I'm at a loss as to how that is a "best explanation".

[PairTheBoard]

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Ok, so lets rewind.

How is getting 2:1 on half your envelope the best explanation?

Remember in your "best explanation" post, you addressed the amount you bet when you win, vs the amount you bet when you lose. Nowhere in the post did you mention that you have no idea how often that you win.

Also, since we aren't offered this 2:1 wager until after we have opened an envelope, we are wagering the same amount win or lose.

So I'm at a loss as to how that is a "best explanation".

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I think it's the insight that best satisfies this psychological condundrum. Prior to opening the envelope I think everyone here would say it's 50-50 that it's the smaller envelope. We'd all be willing to gamble on that given favorable odds. We'd certainly bet on it given 2-1 odds. After opening the envelope I doubt many here would want to call off the bet. As far as our bet goes, opening the envelope should not change the 50% chance that it's the smaller envelope. If we played this game repeatedly and every time just ignored the envelope amount we would on average win $25 for every $50 bet.

I think this is the psychology in the back of people's minds when they look at switching. Switching after seeing $100 in the envelope amounts to betting $50 at 2-1 odds. It's got +EV of $25 in the above scenario. Why doesn't it have +EV of $25 now? The mathematical reasons given by elindauer, pzhon, Masquerade, and yourself all explain it and they are incontrovertable. But imo there remains a nagging psychological condundrum that something's just not right. The wager explained above has +EV of $25 for every $50 bet. Why doesn't switching the $100 envelope as well?

Put in this light I think the answer satisfies this psychological condudrum. The reason switching the $100 envelope (ie. betting $50 at 2-1) doesn't have the $25 +EV that the first wager does is because it is a betting amount that is being dictated by the already determined outcome.

The example I gave with the cards illustrates this point further. You don't need to know any math to see you're getting suckered in the Red-Black Cards example. If you look at it the right way you can see that switching envelopes amounts to the same kind of bet.

PairTheBoard

[mykey1961]

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Ok, so lets rewind.

How is getting 2:1 on half your envelope the best explanation?

Remember in your "best explanation" post, you addressed the amount you bet when you win, vs the amount you bet when you lose. Nowhere in the post did you mention that you have no idea how often that you win.

Also, since we aren't offered this 2:1 wager until after we have opened an envelope, we are wagering the same amount win or lose.

So I'm at a loss as to how that is a "best explanation".

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I think it's the insight that best satisfies this psychological condundrum. Prior to opening the envelope I think everyone here would say it's 50-50 that it's the smaller envelope. We'd all be willing to gamble on that given favorable odds. We'd certainly bet on it given 2-1 odds.

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And they would be correct because they know the distribution of thier choices. Not the amounts, just fact that they have a 50% chance of selecting the lower envelope.

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After opening the envelope I doubt many here would want to call off the bet. As far as our bet goes, opening the envelope should not change the 50% chance that it's the smaller envelope.

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Once we open the envelope, we know the other envelope contains either $50 or $200. Our choice had nothing to do with this distribution. Whoever filled the envelopes determined the distribution.

Lets assume p is the probability that the other envelope is $50, and q is the probability that the other envelope is $200.

That would mean the probability of selecting the $100 envelope would be:

p/2 + q/2 = (p+q)/2 = 1/2

This means once we know the amount of the envelope, we have eliminated 1/2 of all random occurances. We can not use the $100 amount in any computation of the EV without also including the unknown p, and q.

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If we played this game repeatedly and every time just ignored the envelope amount we would on average win $25 for every $50 bet.

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This is both wrong, and a very bad example.

Do you mean we would on average win 1/2 the amount wagered? Or do you mean for just those times we bet exactly $50, we would average a $25 win.

Both are wrong.

Lets assume the amounts possible are A, B, and C, where B = A*2 and C = B*2

Let p be the probability that the envelopes contain A and B.
Let q be the probability that the envelopes contain B and C.

The probability we select envelope A is p/2.
The probability we select envelope B is p/2 + q/2 or 1/2.
The probability we select envelope C is q/2.

We will select envelope A, switch and win A with probability p/2 and lose A/2 with probability 0.
We will select envelope B, switch and win B with probability q/2 and lose B/2 with probability p/2.
We will select envelope C, switch and win C with probability 0 and lose C/2 with probability q/2.


When we open an envelope.

EV represents the gain/loss from switching, not the average amount we will end up with.

If we assume we are opening envelope A.
A = $100, B = $200, C = $400.
We will select envelope A, switch and win A with probability p/2 and lose A/2 with probability 0.
EV = $100*p/2
Since we don't know p, EV is anywhere from 0 to $50

If we assume we are opening envelope B.
A = $50, B = $100, C = $200
We will select envelope B, switch and win B with probability q/2 and lose B/2 with probability p/2.
EV = 100*q/2 - 100/2*p/2
Substituting 1-p for q
EV = 100 - 150*p
Since we don't know p, EV is anywhere from -$50 to $100

If we assume we are opening envelope C.
A = $25, B = $50, C = $100
We will select envelope C, switch and win C with probability 0 and lose C/2 with probability q/2.
EV = - C/2*q/2
Substituting 1-p for q
EV = -$25 + $25*p
Since we don't know p, EV is anywhere from -$25 to $0

If we make no assumptions about which envelope we opened:
Our EV is A*p/2 + B*q/2 - B/2*p/2 - C/2*q/2
Substituting 1-p for q, A*2 for B, and A*4 for C we get
EV = A*p/2 + A*2*(1-p)/2 - A*2/2*p/2 - A*4/2*(1-p)/2
EV = 0

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I think this is the psychology in the back of people's minds when they look at switching. Switching after seeing $100 in the envelope amounts to betting $50 at 2-1 odds. It's got +EV of $25 in the above scenario. Why doesn't it have +EV of $25 now? The mathematical reasons given by elindauer, pzhon, Masquerade, and yourself all explain it and they are incontrovertable. But imo there remains a nagging psychological condundrum that something's just not right. The wager explained above has +EV of $25 for every $50 bet. Why doesn't switching the $100 envelope as well?


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We never knew the EV was specifically $25.

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Put in this light I think the answer satisfies this psychological condudrum. The reason switching the $100 envelope (ie. betting $50 at 2-1) doesn't have the $25 +EV that the first wager does is because it is a betting amount that is being dictated by the already determined outcome.

The example I gave with the cards illustrates this point further. You don't need to know any math to see you're getting suckered in the Red-Black Cards example. If you look at it the right way you can see that switching envelopes amounts to the same kind of bet.

PairTheBoard

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Knowing the pot odds is of no value if you have no clue of the probability that you will win the pot.

[PairTheBoard]

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If we played this game repeatedly and every time just ignored the envelope amount we would on average win $25 for every $50 bet.


[/ QUOTE ] This is both wrong, and a very bad example.

Do you mean we would on average win 1/2 the amount wagered? Or do you mean for just those times we bet exactly $50, we would average a $25 win.

Both are wrong.

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You took this quote out of context. In its original context it read,

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Prior to opening the envelope I think everyone here would say it's 50-50 that it's the smaller envelope. We'd all be willing to gamble on that given favorable odds. We'd certainly bet on it given 2-1 odds. After opening the envelope I doubt many here would want to call off the bet. As far as our bet goes, opening the envelope should not change the 50% chance that it's the smaller envelope. If we played this game repeatedly and every time just ignored the envelope amount we would on average win $25 for every $50 bet.


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Read in context it should be clear that the "bet" I'm talking about in what you quoted is the wager being discussed in that paragraph. The one that is made after the envelope is picked but before it is opened. The one that is not called off after the envelope is opened. It doesn't matter whether new envelope amounts are chosen each time, or the same ones are used each time, the statement remains true that for THIS wager, with the chance to call it off after seeing the envelope amount, If we played this game repeatedly and every time just ignored the envelope amount we would on average win $25 for every $50 bet.

For example, for THIS wager, if we decide to bet $1000 our EV will be:

(.5)2000 - (.5)1000 = 500

ie. +$25 for every $50 wagered.

Try reading my post again.

PairTheBoard