statistics question:
a man and his wife won a price. the chairman had 2 envelops. one
envelop with  X%, and in the other  0.5X or  2X.
 He gave them one of the envelops, they open it and found  100$.
After that he asked them if they agree to exchenge the envalop with
second envelop in which there are  50$ (50% ) or 200$ (50%).
the man agreed, because there are 50% to win 100$ and 50% to lost 50$.
the expectation is  100*1/2 +50*1/2 = 1.15 > 1 and therefor I want to
exchange the envelops.
The wife said that it is silly to exchange the envelops, since the
same expectation we will have even if we didnt open the envelop, so we
could exchange the envelop just after we got it, and this is a silly,
since we dont know with of them has the the bigger amount.

I know that the wife is right.
The question is why the man is wrong. The expectation is realy  1.25,
so why should they exchange the envelops? ( they can exchange it many
times, every time they will have expection=1.25.... )
                                         Thanks

Hello gadtrab!
Searching through Google I've found out that several people are
puzzled by this problem. See, for example,

http://people.cs.uchicago.edu/~gmkrishn/snippets/Puzzle/Puzzle

for some twists that just make this problem become weirder, especially
because it would seem that even before looking at the envelope you
got, you want to switch it (as you say, the expectation appears to be
1.25X).

Fortunately, I found, in another page, the solution to this problem.
It explains quite clearly what's the mistake in that "1.25" expected
value. The answer to why there are no gains from switching BEFORE you
see the contents of your envelope is quite straightforward; while the
understanding of why there are no gains from switching once you see
the $100 requires some basic knowledge of probability. Check the
answer at

http://www.faqs.org/faqs/puzzles/archive/decision/

This page has many puzzles (and their solutions). To find the one
you're looking for use Edit - Find (On This Page) - "envelope.p".

If you need further assistance with this question, please request a
claeification.


Hope this helps!

elmarto-ga


search terms used:
0.5X 2X envelope puzzle switch

The man's calculation of expectation that yields 1.15 is wrong. 
Consider it this way: he is being given the chance to buy for $100 one
of two envelopes, one containing $50, and one containing $200, but he
cannot distinguish them.  His expectation is 0.5 x $50 + 0.5 x $200 =
$125.  This is better than his cost of $100, and so he should take the
risk.

The wife's logic - as I understand it - is wrong.  Without opening the
envelope they had been given, their expectation is about 0.33 x $50 +
0.33 x $100 + 0.33 x $200 = $116.67.  Once they have opened the
envelope and know it contains $100, their expectation for the next
envelope rises to $125.

From the link given:
http://www.faqs.org/faqs/puzzles/archive/decision/
<quote>
OK, so let's consider the case in which I looked into the envelope and
found that it contained $100.  This pins down what X is: a constant.

Now the argument is that the odds of $50 is .5 and the odds of $200
is .5, so the expected value of switching is $125, so we should
switch.
However, the only way the odds of $50 could be .5 and the odds of $200
could be .5 is if all integer values are equally likely.
<unquote>
This is erroneous.  There are two possible outcomes - each equally
likely.  The probability of each is therefore  0.5.  The notion that
"all integer values" be equally likely has no bearing whatsoever.

This question is at

http://rec-puzzles.org/decision.html

under the title "envelope".  The solution is discussed there. 
Basically the issue is that you're making an implicit assumption that
*any* amount of money is equally likely to be in the envelope, but
that is not possible.  If that were true, the probablity of finding a
finite amount of money in the evelope would be zero.

A minor oops - I misunderstood the scenario a bit - initially I
thought there were envelopes with $0.5X, $X, and $2X.

However, I remain unconvinced by the arguments presented thus far.
The situation confronting the subject is the possibility of exchanging
$100 for an envelope that may contain either $50 or $200, with equal
probability.  Have I managed to grasp this correctly?

If so, I invite anyone interested to undertake the following thought
experiment:
In this game, it costs $100 to play a round.  Each round, I take the
$100 and toss a fair coin.  If it is "heads", I give you $50, if it is
"tails", I give you $200.  Do you want to play?

To me, this seems a logical equivalent to the "envelope" puzzle.
Lets play 1000 rounds.  In total, you give me $100,000.  Slight
randomness aside, 500 times the toss will be "heads" and I will give
you $50, and 500 times, it will be "tails" and I will give you $200. 
In total, I will give you 500x$50+500x$200=$125,000.  You make $25k! 
Your expectation was 1.25 as calculated.

Bring on your envelopes - I'll play!

The point is that a $50,$100 pair is more probable than a $100,$200 pair.
Or at least that a 0.5X,X pair is more probable than a X,2X pair, once X gets
high enough.  Because the sum of all probabilities over all X has to be one.

wondering-ga:
"The point is that a $50,$100 pair is more probable than a $100,$200
pair.
Or at least that a 0.5X,X pair is more probable than a X,2X pair, once
X gets
high enough.  Because the sum of all probabilities over all X has to
be one."

This analysis too is erroneous.
Any consideration of probabilities for all X is irrelevant, since we
know X to be = $100.  The envelope contains either $50 or $200.  There
are just these two possibilities - either could be true, and there is
no basis to suggest that either is more likely.

And, BTW, if, as is suggested, there is some value of X at which the
(0.5X, X) pair becomes more probable, what is this "magic" value?

"Any consideration of probabilities for all X is irrelevant, since we
know X to be = $100.  The envelope contains either $50 or $200."

I disagree.  Suppose you have the *extreme* probability distribution
that it's *always* $50,$100.  In that case when you open the envelope
and it contains $100: would you switch?  Your statement that "the
envelope contains either $50 or $200" then still would be true (in
fact it would contain $50 :-)), but that doesn't help.

Your fallacy is to think that because you chose a random envelope,
the $50 and $200 have to have the same probability.  But the fact
that you found out that it's $100 gave you extra information, and
therefore that is no true anymore.

"if, as is suggested, there is some value of X at which the
(0.5X, X) pair becomes more probable, what is this "magic" value?"

It depends on the probability distribution :-)

But we were GIVEN the probability distribution!
"After that he asked them if they agree to exchenge the envalop with 
second envelop in which there are  50$ (50% ) or 200$ (50%). "

Half the time, the envelope would contain $50, and half the time $200.
 The scenario of an *extreme* distribution does not apply.

The reason I "think that because you chose a random envelope, 
the $50 and $200 have to have the same probability" is because that is
exactly what was specified in the scenario given.  It is an axiom, not
a fallacy.

xarqi--

Whether or not the problem states that all values are equally likely
to be found in the evelope or not does not change the fact that that
cannot be true.  Think of it this way:  if any amount of money between
zero and infinity were equally likely, what would be the mean
(expectation value) of the amount of money in the envelope if you
played the game many times?
The problem implies that the man *assumes* there is a 50% chance of
each of the two possible outcomes, not that that is actually the case.
 In reality, the amount of money in the envelopes is chosen from some
pdf.  If $100, the amount found in the first envelope, is less than
the expectation value based on that pdf, the man should switch, but if
it's more, he should not switch.  Since he has no knowledge of the
underlying pdf, he has no statistical reason to switch.

Sorry - still no sale
"Whether or not the problem states that all values are equally likely
to be found in the evelope or not does not change the fact that that
cannot be true."
Nonsense.  The problem DOES NOT state that ALL values are equally
likely, it says quite clearly that it is either $50 or $200 with a 50%
chance of each.  I repeat, this is an axiom, a matter of fact, a
definite element of the described situation.

"The problem implies that the man *assumes* there is a 50% chance of
each of the two possible outcomes, not that that is actually the
case."
It beats me how you infer this from the description of the situation:
"...in which there are  50$ (50% ) or 200$ (50%)."  The man has not
merely assumed the 50% chance, it is a fact.  It IS actually the case.

Of course if you *know* about the probability distribution
that for $100 the chances are 50/50, you should switch.

But whatever the prob. distr., the chance that you lose
money on the switch will be bigger the higher the amount
that you find in the envelope is, exactly compensating for
the 1.25.

wondering-ga
Half right - yup given that it is 50/50 it makes sense to switch - but
this is REGARDLESS of the amount at stake - so the decision DOES
depend on the probability distribution.  Fortunately, we were given
this.