re: The Two Envelopes Problem (A Monty Hall Problem Variant)

First, the Monty Hall Problem is very different. Both depend on how poor most people's understanding of probability is. But how each problem uses this ignorance is very different.

The flaw in the Wikipedia solution is in this statement:

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Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.

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To see this, imagine that your benefactor started with two identical, empty envelopes, one $20 bill, one $10 bill, and nine $5 bills.

He puts the $10 bill in one envelope, and seals it. Then he randomly selects one of the other bills, and seals it in the second envelope.

It is true that the chances are 1/2 that you pick the larger envelope, and 1/2 that you pick the smaller. But if you try to associate the random variable A with your envelope, it depends on what A is.

What the misunderstanding is, is that a random variable, like this A, represents every possible value at the same time. But each value has a weight, called the probability.

There is a 45% chance that A=$5, a 50% chance that A=$10, and a 5% chance that A=$20. This makes the expected value

Exp(A) = $5*0.45+$10*0.50+$20*0.20 = $8.25.

But what about the other envelope (call it B)? If A=$5, there is a 100% chance that B=2A=$10. If A=$20, there is a 100% chance that B=A/2=$10. And if A=$10, there is a 90% chance that B=A/2=$5, and a 10% chance that B=2A=$20. Note that the total chance that B=A/2 is 50%, but it is never 50% for any particular value.


And the expected value of B is

Exp(B) = $5*0.9/2 + $10*0.50 + $20*0.10/2 = $8.25.

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You can't calculate the expected value of the other envelope without knowing the distribution of values going into th envelopes. But you can say each has the same distribution now. If you open an envelope, the answer still depends on the distribution for A/2 and 2A, and you don't know that. (If you do, and they have equal probability? Then it is advantageous to switch.)

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depends on the distribution for A/2 and 2A, and you don't know that. (If you do, and they have equal probability? Then it is advantageous to switch.)

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Would work the other way too, leading u to switch back, right? This can’t be the flaw

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The problem is that they defined A as being dependent on the value in the other envelope - so you can't then directly use those definitions in the expected value problem.

If you set up an independent variable to define the value in the envelopes it shows no advantage in switching via expected value.

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A can be thought of as dependent on the other, but really it’s the other way around. How would you set up an EV equation with independent variables?

It's because you can't have assumption where your envelope is the good and bad envelope at the same time.