PARADOX RESOLVED: The Kraitchik and 2-Envelopes Paradoxes
See also: Toward a Theory of Economic Relativity
© 2000 by Christopher Michael Langan
first consider the Paradox of
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:
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.
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 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.
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.
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:
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."
In this case the player reasons:
x1 = 2y, then gain1 = -y = -(x1)/2, but if x2 = y/2, then gain2 = y/2 =
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.
can just as easily formulate gain in terms of y; in the first subjective
frame, the player's expectation is
So the true expectation is (y1)/2 + -(y1)/2 = 0.