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  #91  
Old 07-02-2006, 06:50 PM
NaobisDad NaobisDad is offline
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Default Re: The envelope problem, and a possible solution

I have read this thread so far with great pleasure. I am not in a position to add anything useful. But Jason, PTB and Aaron, thank you very much.
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  #92  
Old 07-02-2006, 07:23 PM
jason1990 jason1990 is offline
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Default Re: The envelope problem, and a possible solution

The system I had in mind was to generate a random variable U which is uniform on (0,1). I would decline the bet if the amount in my envelope was greater than (1/U)-1. I believe this is +EV for me, regardless of how you choose T, unless of course you choose T=0. I will leave it to you to check my math, if you're so inclined.
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  #93  
Old 07-02-2006, 07:35 PM
jason1990 jason1990 is offline
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Default Re: The envelope problem, and a possible solution

Here's the dubious argument from the original post that we've been debating.

[ QUOTE ]
Argument 1: It's +EV to switch. You had a 50/50 chance of picking the high or low envelope so there's a 50% chance that the other envelope is the high and a 50% chance it's the low. Therefore, EV of switch = 0.5*(+100) + 0.5*(-50) = +25.

[/ QUOTE ]
Aaron, here's some of your comments about it.

[ QUOTE ]
Aaron-
I claim the envelope paradox is a true paradox, two strong arguments that lead to opposite conclusions. Yes, people invent rigorous ways to resolve it, but only by torturing commonsense ideas of uncertainty.

[/ QUOTE ]

[ QUOTE ]
Aaron-
Before we open our envelope, call the amount inside X. We know that the other envelope has a 50% chance of holding 0.5*X and a 50% chance of holding 2*X. Its expected holding is 1.25*X ... When there's no reason to think A is better or worse than B, we assume there's a 50% chance that A is better. Then if told we have a 50% chance of losing 0.5*$X and a 50% chance of gaining $X, we calculate our expected gain is 0.25*$X. It's hard to come up with general principles that prevent us from making errors in real problems.

[/ QUOTE ]

[ QUOTE ]
Aaron-
[The Paradox] asks a common sense question that appears to have two opposing common sense answers. People layer the mathematics on to it. That's fine, but making technical assumptions to make one argument right and one wrong doesn't answer the fundamental question of the paradox.

[/ QUOTE ]

[ QUOTE ]
Aaron-
The envelope paradox is simplified to make the basic contradictions clear, which makes it easy to "resolve" with ad hoc techniques. But those techniques are not much practical help in realistic problems.

[/ QUOTE ]
So you claimed that in order to counter Argument 1, we need to layer on extraneous mathematical assumptions which are designed solely to avoid this Paradox, which are typically ignored by practicing statisticians, and which are of little help in real life problems.

I then claimed that the only assumptions we need in order to counter Argument 1 are the elementary principles of probability.

[ QUOTE ]
Jason-
<ul type="square">
[1] if A and B are events, then P(A | B) = P(A and B)/P(B),
[2] if A and B are independent events, then P(A | B) = P(A),
[3] if A and B are mutually exclusive events, then P(A or B) = P(A) + P(B),
[4] if A is an event, then 0 &lt;= P(A) &lt;= 1.[/list]As far as I can tell, if we assume these four things, then it cannot be the case that

P(Y = 2X | X = k) = 0.5

for all k.

[/ QUOTE ]
(For those who haven't been following the notation of this thread, in the quote above, the statement "P(Y=2X|X=k)=0.5 for all k" means "the probability you chose the smaller envelope, given that you found it to contain $k, is 50%, regardless of the value of k".)


[ QUOTE ]
Jason-
the only techniques we need to use in order to arrive at the "sensible" conclusions are the techniques of elementary probability, such as assumptions [1]-[4] of my previous post. These are consistent, they are certainly not ad hoc, and they are extremely helpful in a wide array of realistic problems. Are you denying this?

[/ QUOTE ]
Here are some of your replies.

[ QUOTE ]
Aaron-
I accept all four of your assumptions, and don't know anyone who does not. But how do they lead to the conclusion?

The only problem I know of with P(Y = 2X | X = k) = 0.5 for all k is that the expected value of X must be infinite. I understand there are technical complexities with dealing with distributions that allow this statement to be true, but that's not the same as proving it false. Even if you outlaw them entirely, that's only a formal resolution of the paradox, not a refutation of the force of its argument.

[/ QUOTE ]

[ QUOTE ]
Aaron-
I maintain the ways people deny the force of the always switch argument result in restrictions that are commonly ignored, and if not ignored would make statisticians useless.

[/ QUOTE ]
I then showed you, using the probability space that you yourself suggested, how those assumptions lead to the conclusion. In other words, I showed that Argument 1 violates the very foundations of probability. No ad hoc assumptions are needed. No layers of ignorable mathematics are necessary. Only the axioms of probability which are given above. I apparently showed you something you had never seen before, since you said the only counter to Argument 1 that you had ever seen involved infinite expectations. Here is part of your reply.

[ QUOTE ]
Aaron-
Jason1990 does a thorough job of showing that no consistent prior distribution on T justifies the idea that the probability of switching is always 50%.

[/ QUOTE ]
In other words, I did a thorough job of showing that the logic in Argument 1, the same logic that you repeatedly claimed could only be refuted with ad hoc assumptions, violates the four assumptions given above, which are the fundamental principles of probability theory.

[ QUOTE ]
Aaron-
However, there are consistent prior distributions that justify always switching, but they have infinite expectation for T.

[/ QUOTE ]
What??? Suddenly, this is what you've been talking about all along? Who cares if there are other ways to justify always switching besides Argument 1? We've been discussing Argument 1, and you've been making claims about Argument 1 this whole time. And all of a sudden, that's not what the Paradox is about? And then ...

[ QUOTE ]
Aaron-
I think both Jason1990 and PairThe Board have redefined the original paradox

[/ QUOTE ]
Priceless. Incidentally, I am in complete agreement with PTB, and his perspective is covered as a special case of the proof I gave you by taking T to be a constant random variable. The fact that you think he and I are talking about two different things shows me that you really don't understand what either of us are saying.

I originally thought I misunderstood you. Then I thought you were just confused. Now I think you may be intentionally trying to confuse the readers of this forum. Why else would you suddenly change your tune 85 posts into the thread?

Argument 1 logically contradicts the axioms of probability theory. That's a fact. This may violate your intuition. But so will a lot of other things in mathematics. In fact, as you probably know, so will a lot of things in "real life". If you want to maintain that Argument 1 is valid in any sense, then you must either reject the axioms of probability or be logically inconsistent. I only hope that my participation in this thread has made this clear to at least one person who would have otherwise been confused. If you wish to go beyond this simple fact and discuss additional topics such as infinite expectations, Bayesian philosophy, or baseball statistics, then have fun.
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  #94  
Old 07-03-2006, 09:13 AM
AaronBrown AaronBrown is offline
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Posts: 2,260
Default Re: The envelope problem, and a possible solution

[ QUOTE ]
Fair enough. I would say the lesson of the Two Envelope Paradox is this:

Beware of Proposition Bets where the Outcome Determines the Size of your Bet.

[/ QUOTE ]
A fair summary for a poker forum, even in the Probability section. But we could also learn:

Beware of People Who Claim Statistical Certainty where the Experiment's Outcome was Determined before the Calculation was Done.

Gamblers understand this. The only civilized way to settle a dispute is to make a bet about a future event. Nongamblers prefer to fight wars over differing interpretations of known past events. Once the evidence is in, interpretation will always be a matter of opinion (unless we get a real solution to the envelope paradox). But a fair bet made before the dice are rolled puts the contest above silly arguers.
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  #95  
Old 07-03-2006, 01:41 PM
PairTheBoard PairTheBoard is offline
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Join Date: Dec 2003
Posts: 3,460
Default Re: The envelope problem, and a possible solution

[ QUOTE ]
Beware of People Who Claim Statistical Certainty where the Experiment's Outcome was Determined before the Calculation was Done.

Gamblers understand this. The only civilized way to settle a dispute is to make a bet about a future event. Nongamblers prefer to fight wars over differing interpretations of known past events. Once the evidence is in, interpretation will always be a matter of opinion (unless we get a real solution to the envelope paradox). But a fair bet made before the dice are rolled puts the contest above silly arguers.

[/ QUOTE ]

I'm afraid this is not making much sense to me. Perhaps you could give some examples to illustrate what you're saying. Also, in the Two Envelope problem, what is the statistical certainty being claimed that you should beware of?

In my lesson, "Beware of Proposition Bets where the Outcome Determines the Size of your Bet", I think the principle is simple and clear. After having chosen the envelope, even though the outcome has been determined there's still no problem betting that the amount you see in your envelope is the smaller amount. If you can get better than 1-1 odds on your bet it's a good bet. If you can get 2-1 odds on your bet it's a very good bet. However, it's not a good bet if the size of your bet is being forced by the outcome that has already been determined. If you are forced to bet small when that outcome makes you a winner and forced to bet big when that outcome makes you a loser, it's clearly no longer such a good bet.

That's exactly what's happening when you are offered the chance to switch envelopes. Switching envelopes is equivalent to betting half your envelope amount. However the already determined (but unknown) outcome of the bet has also determined your envelope amount - ie. the size of your bet. Whether you are a winner or loser has already been determined and you are being forced to bet small when you are winning and big when you are losing. That fouls your EV calculation.

Consider this Proposition Bet. I have a standard deck of 52 cards. I will randomly deal one of these cards face down and allow you to place a bet that it's a Red Card. If it is a Red Card I will pay you 2-1 odds. What's your EV on every dollar you bet? It should be +$.50 right? Well I have a wrinkle in the Proposition. On the back of every card I've written a dollar amount. When the card is dealt face down you are only allowed to bet the amount written on the back of the card? Now, the cards are shuffled, a card is dealt, it happens to have $1 written on its back. If you make the forced $1 bet that the card is Red do you still have +EV of $.50? Do you still like the Proposition?


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