weaselman
weaselman
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August 13th, 2010 at 4:48:12 PM permalink
Quote: Wizard


I agree, but it is hard to explain why.



Why is it hard? It is not unusual that the probability changes after the outcome is partially revealed. In particular, if you knew that only one envelope contained money to begin with, the probability of finding it in one of the envelopes would be 50% before you start the experiment, and it would change to either 0 or 1 after an envelope is open.
The same exact thing happens here - while both envelopes are sealed, each has 50% probability to contain more money than the other, but once you open it, the probability changes, just like in the other case. The only difference is that the new value of probability is unknown, and we are psychologically inclined to assume that it is 50% because we don't know what it is.

Doc: You are right about 50% initial probability. All I meant when I said that 50% was special was that if you pick head or tails with a 50% probability, you will always have a 50/50 chance of guessing right, no matter how biased the coin is, and if you choose any other rule, than your chance of winning will no longer be 50% unless the coin is unbiased. It's a minor point, and not really related very much to the problem at hand. I regret that I ever brought it up ...
"When two people always agree one of them is unnecessary"
Ayecarumba
Ayecarumba
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August 13th, 2010 at 5:52:45 PM permalink
Quote: weaselman

while both envelopes are sealed, each has 50% probability to contain more money than the other, but once you open it, the probability changes, just like in the other case. The only difference is that the new value of probability is unknown, and we are psychologically inclined to assume that it is 50% because we don't know what it is.



I disagree that the probability of second envelope containing more than the opened one changed when the first envelope was opened. As the problem was laid out, one envelope contains more than the other. That fact has not changed with the unsealing of the first envelope. What you still don't know is whether you hold the higher or lower amount. The 1.25 EV is connected to the fact that half the time you will win $200, and half the time you will win $50. ($200+$50)/2 = $125 However, it costs $100 to play.

What if the opening of the envelopes were reversed, and the other envelope was opened to reveal $100. Is the value of trading an unknown envelope for one with a known amount different? Would you trade your sealed envelope (which you now know contains $50 or $200), for the sure $100?
Simplicity is the ultimate sophistication - Leonardo da Vinci
weaselman
weaselman
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August 13th, 2010 at 6:32:55 PM permalink
Quote: Ayecarumba

I disagree that the probability of second envelope containing more than the opened one changed when the first envelope was opened.


Do you agree it changes in the other case, when you know that only one envelope contains money? Before you open an envelope, the probability of finding $100 in either of them is 50/50, right? Once one envelope is opened, the probability changes to 0/100.
This demonstrates, that the event of opening an envelope changes the probabilities. The only difference is that in one case you know the new values, and in the other case they become unknown. That does not mean they have not changed, in some cases they will remain 50/50 (like in example I showed before with a uniform distribution, when the amount you found is below the middle of the interval), and in the others they won't (when the amount is above the middle). They may change or they may not, depending on the amount you found, and the distribution, since the latter is not known, it is impossible to tell what the new probabilities are, the 50/50 guess is no better than any other number.
"When two people always agree one of them is unnecessary"
Doc
Doc
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August 13th, 2010 at 7:06:21 PM permalink
Just curious, weaselman, what do you think of my suggestion that it is not that we don't know the probabilities but instead that it isn't even a probabilistic variable? I suggested that it is deterministic, determined by whoever put the money in the envelopes, and we don't know what they determined or have any information to predict what they determined. Does that make any sense? It leads to the same conclusion.
weaselman
weaselman
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August 13th, 2010 at 7:18:03 PM permalink
Well, I don't see the real difference. A deterministic variable is just a particular case of distribution, concentrated at a single point. When I say, there is a distribution, and we don't know what it is, one of the possibilities is that it is a delta-function (a peak at a single point), there are other possibilities as well.
"When two people always agree one of them is unnecessary"
Wizard
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Wizard
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August 13th, 2010 at 8:25:56 PM permalink
I think that since Dorothy posed the question she is obligated to weigh in. I'm waiting.
It's not whether you win or lose; it's whether or not you had a good bet.
Asswhoopermcdaddy
Asswhoopermcdaddy
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August 13th, 2010 at 8:32:46 PM permalink
Quote: Wizard

What bothers me about this problem too is that I have not had that "a-ha" moment either. At the core of this I think it is an abuse of the expected value formula to come up with an EV of the other envelope of $125. Still, I can't point to a specific reason why. It has something to do with the 50% and 100% being applied to two different amounts, not the same one.




This mathematical enigma is beautiful yet annoying. I have to agree with the Wizard here despite the popular evidence that suggests a 50/50 probability that creates a higher expected value. I think the EV formula is being abused.

1.) Once you choose the envelope, you have a finite event. Great you have $100. You can choose to lock this in or walk away.

2.) The second event is that you are presented with a choice. Choose the other envelope or keep the one you have.
----There's a real cost to this, which I believe the EV formula just doesn't factor in. Suppose the 2nd envelope was really an IOU in which it says you either get an extra $100 or you are required to pay $50. It's roughly the same logic and same final payoff for both scenarios, yet there is a negative expected value to this equation if you apply the math. EV = .5*100 + .5*(-50) = $25

3.) So far, most widely debated answers seem to suggest that once a finite event 1 occurs, you still have a 50/50 chance for event 2 to occur. I disagree with this interpretation, because these are not real random events. This to me is like abusing order of operation or independent vs. dependent variables.

Consider the following scenario:
A.) Suppose you have 3 envelopes with $50, $100, and $200. You choose 1 envelope and low and behold it contains $100. Now you are presented with a 2nd event. I take away your envelope and allow you to pick b/t the 2 leftover. 2 envelopes left, I can get 50 or 200. Ie, for the 2nd event you have a true path dependent variable with an EV of 50% of 50 and 50% of 200 = $125. It should make some sense that EV changes. Prior to the first event, the EV is 33% of 50, 33% of 100, and 33% of 200 = EV of 116.6(before 1st choice). Once you take away the envelope, then you have a true 50/50 probability. But if you have a choice to keep the original envelope or exchange it, then your EV has not truly changed in my opinion. Unfortunately, even this is hotly contested.

Perhaps this is all semantics as one other member described it.
Wizard
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Wizard
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August 13th, 2010 at 10:04:16 PM permalink
Quote: Asswhoopermcdaddy

...Once you take away the envelope, then you have a true 50/50 probability. But if you have a choice to keep the original envelope or exchange it, then your EV has not truly changed in my opinion. Unfortunately, even this is hotly contested.



I don't think I'm following the part quoted. Are you saying that if I have the choice of keeping my $100 envelope or choosing one of the other two, that my EV of choosing is not $125?
It's not whether you win or lose; it's whether or not you had a good bet.
Triplell
Triplell
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August 13th, 2010 at 11:39:01 PM permalink
It would seem that the person has a better chance by switching because he is going to earn $100 if he chooses correctly and only loses $50 if he chooses wrong. I think there is a psychological aspect to this problem that makes it seem like the person would rather switch do to the fact that the money received is given.

If there is in fact a 50/50 chance that switching will either double of halve his money, then it can be replaced by a coin. Taking out the aspect that the person received the money for free and you get the following example.

A person has $100 and is approached by a stranger. The stranger asks the man if he is willing to wager half the money he has on a coin flip. If the man chooses correctly, he doubles his money, however if the man is incorrect, he halves it. There is no difference whether he chooses heads or tails. If the man flips the coin and covers it up with his palm and asks the man heads or tails, the man could switch 1000 times between heads or tails, and the probability doesn't change.

I believe the error comes with the percentages. In the equation y = 0.5*(x/2) + 0.5*(2x), wouldn't the plus sign indicate logic OR, however the wording indicates logic AND? I believe the correctly equation would be y = 0.5*(x/2 * 2x), which would indicate 50% chance of halving AND a 50% chance of doubling.

Also I would like to note that by opening the envelope, no new information is added. You know you have an amount of money, however you still have no clue if you chose the higher or lower amount.

in the Monte problem, the host removes one of the losers. You now have a 50/50 chance of getting it right, and by taking into consideration that you had a greater chance of getting it wrong before, you switch.
mkl654321
mkl654321
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August 14th, 2010 at 1:04:16 AM permalink
Quote: Wizard

What bothers me about this problem too is that I have not had that "a-ha" moment either.



See what happens when all you mathy types try to explain a simple concept with equations? I am truly boggled by all the (metaphorically speaking) running around in confusion here--several dozen posts that seem to me to be getting more and more bewildered. Since this is actually a "DUH!" problem, where is mathematics failing here?

Perhaps it's because you can't express the following sentence as an equation:

No matter what envelope our hero eventually does open, that envelope will STILL be randomly chosen, no matter what extent or manner of paroxysms lead up to that event--therefore there will be no expected gain (or loss) by switching.

If I were indeed mathy (rather than wordy), I might try something that sums all the possible gains and losses from the different strategies, which might wind up as being the sum of an infitite series or something with that bizarre symbol that looks like a Z or something involving taking the derivative of something else, or maybe phlogiston, or...

So much simpler to just default to common sense, which says that switching CANNOT increase or decrease the expected value, and the relative differential between the amounts in the two envelopes is just a red herring.
The fact that a believer is happier than a skeptic is no more to the point than the fact that a drunken man is happier than a sober one. The happiness of credulity is a cheap and dangerous quality.---George Bernard Shaw

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