A Simple Version of Our Explanation

Eric Schwitzgebel and Josh Dever

The Paradox.  You are presented with a choice between two envelopes.  You know one envelope contains twice as much money as the other, but you dont know which contains more.  You arbitrarily choose one envelope -- call it Envelope A -- but do not open it.  Call the amount of money in that envelope X.  Since your choice was arbitrary, Envelope B is 50% likely to be the envelope with more and 50% likely to be the envelope with less.  Applying the standard expectation formula, you then calculate the expected value of switching to Envelope B as .50 * 1/2 X + .50 * 2X = 1.25X.  This is essentially the same calculation as you would perform for a coin flip that gave you a 50% chance of doubling your money and a 50% chance of halving it.  Since double or nothing is a fair bet on a coin flip, double or half is more than fair.  You ought to switch to Envelope B.  But obviously that is absurd: A symmetrical calculation could persuade you to switch back to Envelope A.  Hence the paradox.

But why is it absurd?  Where has your calculation gone wrong?

An Analogously Absurd Case.  Our solution to this paradox essentially analogizes the reasoning above to the following reasoning, where the source of the problem is more obvious: You are presented with an envelope containing either \$1, \$2, \$10, or \$20 with equal probability.  You are given the choice between two wagers.  On the first, you receive twice the amount of money in the envelope, if the amount in the envelope is \$1 or \$2, or just the amount of money in the envelope if the amount in the envelope is \$10 or \$20.  The second wager is the reverse: You receive twice the amount of money in the envelope if the envelope contains \$10 or \$20 and just the amount of money in the envelope if it contains \$1 or \$2.  Assigning X to the amount in the envelope, you reason that on either bet there is a 50% chance you will receive X and a 50% chance you will receive 2X, so you are indifferent between the two bets.

Wager 1                                Wager 2

\$1 * 2                                            \$1

\$2 * 2                                            \$2

\$10                                           \$10*2

\$20                                           \$20*2

Clearly, however, the second wager is preferable.  It's much better to have the chance to double \$10 or \$20 than to have the chance to double \$1 or \$2.  The proposed fallacious calculation is fallacious because it does not take that into account.  (The actual expectation, which can be calculated on a case-by-case basis, is \$9 for the first wager and \$15.75 for the second.)  In Wager 1, the expected value of X in the "2X" part of the formula is much lower than the expected value of X in the "X" part of the formula; in Wager 2, the reverse is the case.

The Solution.  Analogously, in the Two Envelope Paradox, the expected value of X in the "2X" part of the formula (where Envelope A is the envelope with less) is less than the expected value in the "1/2 X" part of the formula (where Envelope A is the envelope with more).  You would expect less in Envelope A if you knew that it was the envelope with less than you would if you knew it was the envelope with more.  Allowing X to have different expectations in different parts of the formula in this way is like comparing apples and oranges.  The "X" in the "2X" just isn't the same as the "X" in the "1/2 X" part.

The better calculation in the Two Envelope Paradox involves setting X to the amount in the envelope with less and calculating the expected value of Envelope B as (.5)X + (.5)2X = 3/2 X -- and the expected value Envelope A likewise as (.5)X + (.5)2X = 3/2 X.  In these calculations the expectation of X in the first term of each equation is identical to its expectation in the second term: The expected amount of money in the envelope with less does not change depending on whether Envelope A is the envelope with more or Envelope B is.

In general, we propose as a constraint on the use of variables within the expectation formula that their expected value be the same at each occurrence in the formula.  For a more technical paper on this matter see our "The Two Envelope Paradox and Using Variables Within the Expectation Formula".

Email me with comments, if you like, at eschwitz at domain- ucr.edu.