Re: [R] Two envelopes problem

From: S Ellison <>
Date: Tue, 26 Aug 2008 18:40:04 +0100

>>> Duncan Murdoch <> 26/08/2008 16:17:34 >>>
>>If this is indeed the case, switch; the expected gain is
>>positive because _you already have the information that you hold the
>> median value of the three possibilities_. The tendency when
> >with the problem is to reason as if this is the case.

>No, you don't know that A is the median. That doesn't even make
>based on the context of the question: there is no 3-valued random
>variable here.

This is my point; apologies if I put it badly. The fact is that you _don't_ hold the median value and that this is indeed a silly way of looking at it. My assertion was that this is the way many folk DO look at it, and that this results in an apparent paradox.

In fact, you inadvertently gave an example when you said
>The unopened envelope can hold only two values, given
>that yours contains A.

True - for a rather restricted meaning of 'true'. As written, it implicitly allows three values; A for the envelope you hold, and two more (2A and 0.5A) for the alternatives you permit. The usual (and incorrect) expected gain calculation uses all three; 2A-A for one choice and 0.5-A for the other. To do that at all, we must be playing with three possible values for the contents of an envelope.

This clearly cannot be the case if there are only two possible values, as we are told in posing the problem. The situation is that you hold _either_ A _or_ 2A and the other envelope holds (respectively) 2A or A. We just don't know what A is until we open both envelopes. So for a given pair of envelopes, it is the choice of coefficient (1 or 2) that is random.

If I were to describe this in terms of a random variable I would have to assume an unknown but - for this run - fixed value A multiplied by a two-valued random variable with possible values 1 and 2, would I not? We surely can't have both 0.5 and 2 in our distribution at the same time, because the original proposition said there were only two possibilities and they differ by a factor of 2, not 4.

>You have no basis for putting a probability distribution on those
>values, because you don't know the distribution of X.
But we do know; or at least we know the distribution of the coefficient. We have two envelopes of which one is selected at random, and the ratio of values is 2. On that basis, assigning a 50:50 probability of ending up with A or 2A on first selection seems uncontroversial.

But I'm more than willing to have my intuition corrected - possibly off-line, of course, since this stopped being R a while back!

Steve E

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