Wizard
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Wizard 
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August 14th, 2010 at 8:10:36 AM permalink
Quote: mkl654321


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 question at hand is not should you switch, but what is the flaw in the expected value argument that says the value of the other envelope is 0.5*(50+200)=125? It is a worthy question for debate, and famous paradox in mathematics.
It's not whether you win or lose; it's whether or not you had a good bet.
Doc
Doc
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August 14th, 2010 at 9:17:35 AM permalink
Quote: Triplell

I believe the correctly equation would be y = 0.5*(x/2 * 2x)

Well, that would suggest that y(the expected value) is equal to half of x squared. If x is in dollars, is y in dollars squared? I think you need to re-consider your math; there are some errors in it.


I'm starting to enjoy this thread, now that I have begun to believe that I understand the problem. I would like to offer two lines of discussion:

#1. For those people who firmly believe that the fact the mystery envelope must contain either $50 or $200 means it has an expected value of $125, I offer you this opportunity. I have an envelope that I absolutely guarantee you contains either exactly $50 or exactly $200. What do you think is the expected value of the contents of the envelope? What would you offer for the contents of the envelope? Would you offer me $100? If you bargain real hard, I might be willing to part with it for $80, or even $75. How about it?

No, I don't think this forum is likely to have many suckers who would fall for that. I presented that a bit too blatantly, and most people who started off thinking they believed the $125 figure would have retrenched. After they rethink it, I expect the folks who believe in the $125 figure for the original problem would probably claim that it has something to do with the 50% chance of selecting on the first try the envelope with $100 vs. the mystery envelope. For those folk, I offer a different line of thinking.

#2. Suppose this game is going to be played numerous times, similar to the way Monte Hall played his game. Let's assume the game is offered on TV by a host named Bob. Bob cannot always offer the same two amounts in the envelopes, or people would get wise and know whether to swap envelopes or not. So Bob needs some plan for deciding how much to put in the envelopes for the next game. Let's assume he has a rule that he always follows but which he doesn't reveal to the audience.

You get to be a participant in Bob's game. You choose an envelope, open it, and find $XX. This could be $100 as in the original problem, but just on a whim, I'm going to suggest that your envelope contains $120. You realize that your mystery envelope contains either $60 or $240. You calculate the expected value of this mystery envelope as 0.5*($60+$240)=$150. (This is, of course, 1.25 times the amount you found in the envelope you opened.) Do you switch envelopes? Do you feel that it is obvious that, on average, the mystery envelope must be worth more than the $120 in your first envelope?

Before Bob asks, "Is that your final answer?", perhaps you should consider this. Suppose Bob's secret rule for deciding how much money to place in the envelopes is this: "The lesser-valued envelope should contain a random amount of money between $40 and $79, and the other envelope should contain twice as much." In that case, if we actually knew about this secret rule, we would always swap if our first envelope contained $79 or less and never swap if it contained $80 or more.

Now we don't know whether that is really Bob's rule, but it could well be. This rule, or most any such rule, establishes conditions where it is obvious that 0.5*(2X)+0.5*(X/2) is not a good method for estimating the value of the mystery envelope. It shouldn't take too much thought to realize that we don't have any basis for the assumption that we multiply both 2X and X/2 by 0.5 -- yes, there are only two possibilities for what might be in the mystery envelope (just as there are only two possibilities for the contents of the mystery envelope in the offer I presented in #1 above), but there is no reason to think they are equally likely. (Are you sure you don't want to offer $80 for my envelope?)

The truth is, we have no basis for knowing whether the factors should be 0.5 and 0.5, 0.9 and 0.1, 1.0 and 0.0, or anything else. If we don't know Bob's rule, we have no basis for deciding whether to swap or not. That is why I earlier said, "Take an envelope and run. You don't even have to open it until later."

How many people, if any, found #1 and #2 to offer a convincing presentation? I'm trying to work on my atrophied instructional skills.
DorothyGale
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August 14th, 2010 at 10:31:33 AM permalink
Quote: Doc

I'm starting to enjoy this thread, now that I have begun to believe that I understand the problem. I would like to offer two lines of discussion:


This problem is not related to the "Monte Hall" problem. Instead, it is most closely related to the "Deal or No Deal" where the player is given the option of switching on the very last two suitcases. He knows there are two values left, so should he switch? Also, in "Deal or No Deal" the banker offers less than the expected value until the last two suitcases, then he always offers *more* than the expected value for the final offer. What's with that???

--Dorothy
"Who would have thought a good little girl like you could destroy my beautiful wickedness!"
Doc
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August 14th, 2010 at 10:43:28 AM permalink
Thank you, Dorothy. I only referred to Monte Hall because you had put his name in the title of this thread. My suggestion #2 about Bob's game is indeed more like "Deal or No Deal", which is why I put in the quip "Is that your final answer?" Except in Bob's game, in contrast to "Deal", you don't have any basis for calculating an expected value at all.
weaselman
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August 14th, 2010 at 10:44:47 AM permalink
Quote: DorothyGale

Instead, it is most closely related to the "Deal or No Deal" where the player is given the option of switching on the very last two suitcases. He knows there are two values left, so should he switch?



But there he also knows which values are left, and the probabilities of finding each in each of the suitcases (50%), if I am not mistaken about the rules of that game, so, mathematically, the decision is a "nobrainer", he knows the expected value, and should switch if it is more than what he already has.
At this point, the psychology takes over math - if I have a $1000, I will, probably, give it up for a 50% shot at a million, but if it is, say $300,000 ... I know, I should still go for it mathematically, but, probably, would not anyway.
"When two people always agree one of them is unnecessary"
Asswhoopermcdaddy
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August 14th, 2010 at 10:46:03 AM permalink
Quote: Wizard

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?



I think Doc might have better explained my line of reasoning. Without knowing the true basis behind what's in those other envelopes, a 50/50 probability may represent 2 paths, but not the true EV. In which case, wouldn't the true EV revert back to 116.6? The fact that you are allowed to keep a known value can be inferred as an alternative probability.

What is the probability of you choosing the $100 or the 2nd envelope which contains either 50 or 200. Would it be subsequently fair under the same logic to say EV = .5*100 + .5{[.5*50+.5*200]} = 112.5? Math seems to work out, but we'd be confusing the definitions of expected value with the most probable value. Paths maybe binary, but the weights are not.

Gotta love the Monte Hall problem. There is a variation on wikepedia, that may be helpful:

http://en.wikipedia.org/wiki/Monty_Hall_problem
scotty81
scotty81
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August 14th, 2010 at 10:46:54 AM permalink
I'd like to get past the 50% ambiguity. The Wizard is the only one who can make this decision.

Wizard: Can we all assume from this point on that the original problem assumes a probability of 50% for the contents of the second envelope?

As near as I can tell, the problem remains the same, and the Wizard's paradox is still intact.
Prediction is very difficult, especially about the future. - Niels Bohr
weaselman
weaselman
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August 14th, 2010 at 10:50:40 AM permalink
Quote: scotty81

I'd like to get past the 50% ambiguity. The Wizard is the only one who can make this decision.

Wizard: Can we all assume from this point on that the original problem assumes a probability of 50% for the contents of the second envelope?

As near as I can tell, the problem remains the same, and the Wizard's paradox is still intact.



If you know the probability to be 50%, then the correct strategy is to switch, and there is no paradox. The paradox is rooted in assuming it is 50% when in fact it isn't.
"When two people always agree one of them is unnecessary"
Doc
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August 14th, 2010 at 10:54:37 AM permalink
And yes, Scotty, the original statement of the problem does make that mistaken assumption.
Wizard
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Wizard 
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August 14th, 2010 at 11:00:19 AM permalink
Quote: scotty81

I'd like to get past the 50% ambiguity. The Wizard is the only one who can make this decision.

Wizard: Can we all assume from this point on that the original problem assumes a probability of 50% for the contents of the second envelope?

As near as I can tell, the problem remains the same, and the Wizard's paradox is still intact.



I'm not sure I understand what you're asking. Personally, I don't think the key to this paradox is in analyzing whether the other envelope has a 0%, 50%, 100%, x% chance of having the smaller/larger amount. Rather, I think the flaw is in looking at the ratio of the two envelopes, rather than the difference. I've been thinking all morning how to put it in words, but still have not reached the "a-ha" moment, but think I'm getting closer.
It's not whether you win or lose; it's whether or not you had a good bet.

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