From: "Noel Vaillant" <vaillant@netcomuk.co.uk>
Subject: Re: Two Envelopes: One Has Twice as much Money as the other
Date: Tue, 4 Jan 2000 21:19:34 -0000
Newsgroups: sci.math
Summary: [missing]

I was asked this quiz before: here is the answer I eventually came up with:

10-Oct-99:

This little quizz is a lot harder than it looks, I think.
If the amounts in my two enveloppes are represented
by two random variables X and Y, where X is the amount
initially chosen, I have seen a lot of my collegues argue:

E[Y|X] = (1/2)*(X/2 + 2*X)  = 5X/4 > X

and therefore that it should be optimal to switch enveloppes.
However, although E[Y|X]>X is a sensible criteria for switching
(One may consider the variance in relation to excess return, for
more sophisticated criteria), the above algebra is wrong in general
(I think), as it assumes that given a realisation of  $100 for X, the
likelyhood of having chosen the 'big' enveloppe is unaffected and still
equal to 1/2...

I will outline a possible formal answer to your quizz:

Let X1 be an arbitrary random variable with values in [0,+oo[.
(X1 representing the 'low' amount of money)
I assume that the distribution of X1 is diffuse with a strictly positive
density q(x). i.e:

P(X1 <a) = \int_{0}^{a}q(x)dx (X1 has values in R+)

Let E be a random variable with values in {0,1}, assumed to
be independent of X1. (E represents 'which' enveloppe I choose)

We define X= X1    if E=0 and X = 2X1 if E=1.
We define Y = 2X1 if E=0 and Y = X1    if E=1.

X represents the amount in the enveloppe I have
initially chosen, whereas Y is the amount in the
other enveloppe.

The crunch of the arguement is to calculate
the conditional expectation: E[Y|X]

Using E[Y|X] = (2X)P(E=0|X) + (X/2)P(E=1|X)

we are reduced to computing the conditional
probabilities P(E=0|X) and P(E=1|X):
With a bit of work, I got:

P(E=0|X) = q(X)/(q(X)+q*(X))
P(E=1|X)= q*(X)/(q(X)+q*(X))

where q is the above density in R+, and q*(x)=q(x/2)/2.

It follows that:
E[Y|X] = X(2q + q*/2)(q+q*)

and it is optiomal to switch: (at least as far as expected return is
concerned) if:

(2q(100) + q*(100)/2)/(q(100)+q*(100)) > 1 (for $100 in the enveloppe)

In practice, whoever is playing the game may not know the
actual distribution q of X1 (driving X and Y)...


Let me know if you think I got it wrong somewhere.

Regards. Noel.

-------------------------------------------
Dr Noel Vaillant
http://www.probability.net
vaillant@probability.net
==============================================================================

From: israel@math.ubc.ca (Robert Israel)
Subject: Re: Two Envelopes: One Has Twice as much Money as the other
Date: 4 Jan 2000 21:32:48 GMT
Newsgroups: sci.math

In article <20000104145824.20792.00000139@ng-cg1.aol.com>,
 doyle60@aol.com (DOYLE60) writes:
> There is a famous (I think) question that goes:
> I have two envelopes and one has twice as much money as the other. I give you
> one and you peak inside.  I ask "Do you want to switch?"  Should you switch?
> 
> 1) If you have heard of this riddle before, is it necessary for me to mention
> that I know which envelope I am handing you -- the smaller or larger?  Or
> should it be stated that I am doing it blindly?  
> 
> 2) Also, do I have to mention that we are going to play this game repeatedly?
> 
> 3) And, do I have to mention that I will always ask if you want to switch?

The most relevant question is (1).  If you (the giver) know the contents 
of the envelopes, can choose which one to give or whether to give me the 
opportunity to switch, and are interested in minimizing the amount of 
money I get, then this becomes a two-person game rather than just a 
probability question.

But much more relevant is the question of how the two amounts of money are 
chosen.  Without knowing this, there is no "optimal" decision procedure.
On the other hand, there is a well-known method by which (assuming you
don't know which envelope you gave me) I can ensure that I have probability
greater than 1/2 of getting the larger amount.

Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2
==============================================================================

From: briggs@eisner.decus.org
Subject: Re: Probability puzzle
Date: 8 Jun 2000 12:15:44 -0500
Newsgroups: sci.math

In article <393F0E38.E912A135@math.missouri.edu>, Stephen Montgomery-Smith <stephen@math.missouri.edu> writes:
> I was recently told this probability puzzle - I hope this
> hasn't been hashed out here before.
> 
> I have two numbers that I picked in a way unknown to you.
> I give one of them to you, so you know the number I gave
> you, but you don't know the number I have.  (I use an
> unbiased coin to decide which of the two I give you.)
> 
> You have to decide whether the number you have is smaller 
> or bigger than mine.  One way is that you toss a coin and
> use that to decide, but that strategy has only a 50% chance
> of working.
> 
> Can you think of a strategy for deciding this that has a 
> greater than 50% chance of working?

Yes, this puzzle has been hashed out.  (It's a variant of the "two
wallets" puzzle.  A quite nicely specified variant)

Choose any monotone increasing function whose range is the reals and
whose domain is [0,1].  Call this function f.

Call the number you have been handed x.

Answer "bigger" with probability f(x) and "smaller" with probability 1-f(x).

Unless your number selection technique is a constant function, this
strategy will yield better than 50/50 results.

	John Briggs			briggs@eisner.decus.org
==============================================================================

From: Stephen Montgomery-Smith <stephen@math.missouri.edu>
Subject: Re: Probability puzzle
Date: Thu, 15 Jun 2000 05:38:15 GMT
Newsgroups: sci.math

Here is how I understand it.

So I have two distinct numbers x and y, and give one of them to
you with even odds.  You know the number I gave you.

You pick a random number using a continuous distribution 
whose support is the whole real line (e.g. normal).
If the number you pick is bigger than
the number I gave you, you decide that the number I have is
biggest.  Otherwise you decide that the number you have is
bigger.

Why does this work?  Let us suppose that x<y.  You have one
of these numbers.

Now for the number you picked at random, let us write
A = probability it is less than x
B = probability it is between x and y
C = probability it is bigger than y.

So A+B+C=1, and A, B and C > 0.

So what is the probability you will be correct?  If you
happen to have the number x, then you will be correct
with probability B+C.  If you have the number y, the
probability you will be correct is A+B.

So the probability you will be correct is

0.5(A+B) + 0.5(B+C) = 0.5(1+B) > 0.5.

-- 
Stephen Montgomery-Smith
Department of Mathematics, University of Missouri, Columbia, MO 65211
Phone 573-882-4540, fax 573-882-1869
http://www.math.missouri.edu/~stephen  stephen@math.missouri.edu
==============================================================================

Here is a Maple simulation: we run 10000 iterations using the cutoff function
f(x) = 1/2 + arctan(x)/pi  and various methods of selecting two points  x,y.
Note that it is critical that after  x,y  are selected, we are given one
of them (x) with a 50-50 chance. We show our total payoff (where we collect
each time the net surplus of our envelope over the other), as well as the
number of times we guessed correctly.

f:=proc(x)  evalf( arctan(x)/Pi+1/2 )  end:

doit:=proc() local T,S,i,p,x,y:
S:=0:T:=0:
for i to 10000 do p:=sel():x:=p[1]:y:=p[2]:
  if rand()/10^12 > 1/2 then p:=x:x:=y:y:=p: fi:
  p:=rand()/10^12: 
  if p + f(x) > 1 then S:=S+(x-y); if x>y then T:=T+1: fi:
                  else S:=S+(y-x); if y>x then T:=T+1: fi:
  fi:
od:
[S,T];end:

sel:=proc() evalf([rand()/10^12,rand()/10^12]): end:  doit();  
#                              [296.2027680, 5254]
#i.e. a net profit of 2.96 cents per game if two amounts in [$0, $1] are
#chosen at random. We end up with the larger amount 52% of the time.
#Of course this is random. Here's another run:
#                              [408.3810400, 5403]
#The true expected value in this case is (Pi-4+2*ln(2))/(4*Pi) = .0420079...
#and the true win rate is  (3*Pi-4+2*ln(2))/(4*Pi) = .5420079...

sel:=proc() [2.0,3.0]: end:  doit();  
#                                 [444.0, 5222]
#i.e. a net profit of 4.44 cents per game if we are repeatedly given
#either a $2 or a $3 envelope (and somehow don't notice the pattern...)
#We manage to end up with the larger one 52.22% of the time.

sel:=proc() local p: p:=evalf(-log(rand()/10^12)): [p, 2*p]: end: doit();
#                              [685.7756637, 5310]
#A classic variant, in which it is known that one amount is double
#the other (but how the amount is chosen is unknown). IN this formulation,
#if the envelopes contain  x  and  2x  my expected return is
# x(f(2x)-f(x)) >0  and I will make the right choice  (1/2) + (f(2x)-f(x))/2
#i.e. more than half of the time. (With my  f  and with this  sel()  I find
#an expected win rate of  53.805%  and an expected payoff of .0739 .)

sel:=proc() evalf([exp(-rand()/10^11), sin(rand())*rand()/10^11*(2*rand()/10^12-1) ]): end:  doit();  
#                              [6250.071528, 6226]
#Something goofy this time to select the amounts in the envelopes.
#And yet we still win handily.

ct:=0:
sel:=proc() global ct: ct:=ct+1: if ct<998 then [1,10] else [10,1000] fi:end:
doit();
#                                 [19917, 6161]
#So you see the selection algorithm can even change in time! (Need not be
#a probability distribution on the plane).