PARADOX RESOLVED:   The Kraitchik and 2-Envelopes Paradoxes

 See also: Toward a Theory of Economic Relativity

 © 2000 by Christopher Michael Langan


Letís first consider the Paradox of Kraitchik

Two people are offered a chance to bet on whose wallet contains the lesser amount of money; the one with less gets all of the money in both wallets.  Both players reason as follows:

"If I lose, I lose just what I have, but if I win, I win more than I have.  Therefore, given that the win and loss probabilities are indifferent, my expectation is positive and I should bet."

The paradox resides in the application of this rationale to both players, implying a positive expectation for each in a 0-sum game in which the total amount of money is fixed.  The resolution is effected by showing what's wrong with the rationale, and more precisely with its quantitative interpretation.  

The problematic interpretation is based on the tacit assumption that the amount of money in the player's own wallet is "context-free" or fixed no matter how the game ends, but that the amount in his opponent's wallet is "context-sensitive" and varies according to the game's result.  Obviously, the opposite assumption is just as valid from a subjective viewpoint; one may consider the opponent's amount fixed and one's own amount as varying with the result of the game.  When symmetry is restored by separately interpreting the rationale in light of both assumptions - i.e., with respect to the two subjective "cases" associated with these assumptions - the mathematical expectation is seen to be 0.

Specifically, let the player's stake = x, his opponent's stake = y, "if I lose" = condition 1 (x1 > y1), and "if I win" = condition 2 (x2 < y2).

Then the problem is simply this: each player lopsidedly treats x as fixed and independent (and therefore unindexed), and y as mutable and dependent (and therefore in need of an index 1 or 2). That is, each player thinks "If I lose, then I lose just what I have, but if I win, I win more than what I have", which becomes "If y1 < x, then gain1 = -x, but if y2 > x, then gain2 = y2 = x+n, n>0." 

In fact, y could just as well be the fixed quantity and x the conditionally dependent one.  In this case, the player would reason "If x1 > y, then gain1 = -x1, but if x2 < y, then gain2 = y (< x1) = x1-n, n>0."

In the first subjective frame, the player's expectation is
(gain1 + gain2)/2 = [-x + (x+n)]/2 = n/2, n>0. 

In the second subjective frame, it is [-x1 + (x1-n)]/2 = -n/2, n>0.  

Since n is arbitrary, the true expectation according to this essentially correct rationale is n/2 + -n/2 = 0.   

Paradox resolved.

Now consider the 2-envelopes game with a slight modification: each envelope is in the possession of a person, and the options are now to switch or not to switch.  

In this modified version of the 2-envelopes game, the same rationale applies.  Where x is the amount in the player's own envelope, the amount to be won by switching (x) is larger than that to be lost (x/2), the player again reasons as follows: "If I lose, I lose half what I now have, but if I win, I win an amount equal to what I now have.  Therefore, given that the win and loss probabilities are indifferent, my expectation is positive and I should switch."  And again, the paradox arises from the applicability of this rationale to both players in a 0-sum game.  Since the only difference is a minor additional constraint - namely, that the winning envelope contains twice what the other one does - the paradox is resolved exactly as was that of Kraitchik. 

Specifically, let the player's stake = x, his opponent's stake = y, "if I lose" = condition 1 [x1 = 2(y1)], and "if I win" = condition 2 [x2 = (y2)/2].  Then the problem is again that each player lopsidedly treats x as fixed and independent (and therefore unindexed) and y as mutable and dependent (and therefore in need of an index 1 or 2).  

Given that all variables must be expressed in terms of the objective maximum total value T of the game, the half-or-double condition requires that the fixed variable in terms of which the value of the conditionally dependent variable is expressed has the value T/3; double this amount equals 2T/3 and half equals T/6, and both variables take these conditional values in turn.

In other words, since the player doesn't begin with a knowledge of the game's objective total value, he has to deal with a hypothetical maximum value, and where the opponent's wallet may contain twice his own money, the maximum total value of the game is three times as much. Alternatively, we could keep the original fractions and stipulate that T is the maximum amount in either wallet.

That is, each player thinks:

"If I lose, then I lose half what I have, but if I win, I win an amount equal to what I have", which becomes "If 2(y1) = x, then gain1 = -y1 = -x/2, but if (y2)/2 = x, then gain2 = (y2)/2 = 4(y1)/2 = 2(y1) = x."

But in fact, y could just as well be the fixed quantity and x the conditionally dependent one.  

In this case the player reasons:

"If x1 = 2y, then gain1 = -y = -(x1)/2, but if x2 = y/2, then gain2 = y/2 = (x1)/4."

In the first subjective frame, the player's expectation is
(gain1 + gain2)/2 = (-x/2 + x)/2 = x/4.  

In the second subjective frame, it is [-(x1)/2 + (x1)/4]/2 = [-(x1)/4]/2 = -(x1)/8 = -x/4. 

So the true expectation according to this (essentially correct) rationale is x/4 + -x/4 = 0. 

We can just as easily formulate gain in terms of y; in the first subjective frame, the player's expectation is
(gain1 + gain2)/2 = [-y1 + 2(y1)]/2 = (y1)/2, while in the second, it is [-y + y/2]/2 = (-y/2)/2 = -y/4 = -2(y1)/4 = -(y1)/2. 

So the true expectation is (y1)/2 + -(y1)/2 = 0. 

Paradox resolved.

At no point did either player have to consider the viewpoint of the other; all each had to do was interpret his own straightforward rationale in terms of all arithmetical possibilities.