X-Mailer: ELM [version 2.3 PL11]
Newsgroups: bit.listserv.edstat-l
Date:         Wed, 7 Oct 1992 14:52:15 EDT
Reply-To:     Mary Ann Metzger <metzger@UMBC3.UMBC.EDU>
From:         Mary Ann Metzger <metzger@UMBC3.UMBC.EDU>
Subject:      Envelopes: a general solution

 This is for all who, like me, are enjoying the envelope discussion but
are not yet completely happy with the treatments given so far. I offer the
following argument, which I think is a solution of the general envelope
problem.  By 'general' I mean that it is specified that of the pair of
envelopes, one contains amount of money, X (a positive integer) and the
other contains 2X.  Having drawn an envelope and inspected the contents (Y),
the player is given the option to keep the contents (stay) or to give them
up and keep the contents of the other envelope (switch).
 Part 1: If Y is odd, then switch.  An odd integer must always be the smaller
amount.  It is not elegant, but it is true.
 Part 2: We do not have to consider any pair for which X is odd.  Any other
pair can be represented as [w*2**(k-1),w*2**k],  for some even integer w
and some integer k>0. For 'a*b' read 'a times b' and for '2**k' read ' 2 to
the kth power'.  The integer w can be factored out of all the pairs to be
considered in the following argument, so without any loss of generality, we
can consider only pairs of the form [2**(k-1),2**k].  Every such pair can
be imbedded in a sequence of N pairs where N is any positive integer.  The
general sequence of pairs is
[2**0,2**1] ... [2**(k-2),2**(k-1)] [2**(k-1),2**k] [2**k,2**(k+1)] ...
                                              ... [2**(N-1),2**N]
I apologize for the non-compact notation.  The handwritten version is more
comprehensible. For each Y, the proportion of occurrences of Y in the sequence
of pairs is P(Y=2**0)=P(Y=2**N)=1/(2N) and for 0<k<N, P(Y=2**k)=1/N.
 For each strategy, the gain G for each value of Y in the sequence
of pairs is given in #1, #2, and #3.
1. For Y=2**0: G(stay)=2**0;  G(switch)=2**1
2. For Y=2**N: G(stay)=2**N;  G(switch)=2**(N-1)
3. For Y=2**k: G(stay)=2**k;  G(switch)=(2**(k-1)+2**(k+1))/2
 Note that both of the ideas that have been so widely discussed on EDSTAT-L
are used here.  No pair is more likely to occur than any other and, for
middle pairs, the gain from switching is 1.25Y.
 From the definition of P and #1, #2, and #3, after some simplification,
the expected value (EV) for each strategy over all values of Y may be
calculated.
EV(stay)=(1/(2N))[2**0+2**N]+(1/N)[sum(2**k)]
EV(switch)=(1/(2N))[2**1+2**(N-1)]+(1/N)[sum((2**(k-1)+2**(k+1))/2)]
where 'sum' means 'summation from k=1 to N-1'.
 Algebraic operations will show that EV(stay)=EV(switch) for any N.  The
algebra is easier when the identity "sum(2**k)=(2**N)-(2**1)".
 The result is true for any Y and for any N.  Therefore, unless it
is allowed that some pairs in a given sequence may be more likely
to occur than others, it is the solution to the general problem.
 I believe that the paradox disappears when reasoning is explicitly
limited to finite quantities.  It seems that because there are end
pairs, the advantage correctly attributed to switching when Y<2N is
completely wiped out by the reduced gain that results from switching
when Y=2**N.
Agree?  Disagree? -Mary Ann