The Exchange Paradox

Photograph by Jake Slagle @ flickr

 

“A swami puts m dollars in one envelope and 2m dollars in another. You and your opponent each get one of the envelopes (at random). You open your envelope and find x dollars, and then the swami asks you if you want to trade envelopes. You reason that if you switch, you will get either x/2 or 2x dollars, each with probability 1/2. This makes the expected value of a switch equal to (1/2)(x/2) + (1/2)(2x)=5x/4, which is greater than the x dollars that you hold in your hand. So you offer to trade.

 

The paradox is that your opponent has done the same calculation. How can the trade be advantageous for both of you?”

 

———-

source:

Casella, G.  and Berger, R. (2002). Statistical Inference. Duxbury Press.

 

 

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8 thoughts on “The Exchange Paradox

  1. Antoine says:

    The amounts in the envelopes are m and 2m. By swapping, you may gain m or lose m. The trade isn’t advantageous for anyone, because the potential gain is equal to the potential loss for both.

    • epanechnikov says:

      Hello Antoine. Say your envelope contains $10. Your potential excess gain is $10 if your envelope contains the least amount of dollars. But if your envelope contains the most, you will exchange the envelope for another which contains $5 (and not $0!) – hence your potential loss is $5 and not $10. Hence (subject to a few assumptions) your expected gain would be

      1/2*$10-1/2*$5=$2.5

      while generally if your envelope contains $x your expected gain will be

      1/2*($2x-$x)+1/2*($1/2x-$x) = $1/4x >0

      • Antoine says:

        I didn’t use x (which is different for you and your opponent), but only used m (which is the same for you and your opponent, although both don’t know what m is. There is only a 50% chance that m=x) as you defined in the text.
        So there is 50% chance for you to win or loose an amount m, and the same for your opponent. You can also see the symmetry of the situation for you and your opponent, so It shouldn’t matter. The percentages of profit or loss compared to x are different, but the absolute value in dollars of the profit or loss is the same: m (even though you don’t know for sure what m is)

      • epanechnikov says:

        That is however a slightly different game. In your solution you are not utilizing the fact that the player has observed an amount $x. Say that instead of m vs 2m we had m vs 100m. You open the envelope and you see 1 dollar. Would you exchange it or not? I guess most people would do so. Additionally, you are right to highlight the symmetry of the game but that does not explain the paradox. Taking conditional expectations on what has been observed may lead to a trade which seems to be advantageous for both (but it can not be).

        But lets forget for one moment the second player. Would you be willing to participate in a game which pays double the participation fee if you win and half the fee if you loose (when each event has a 50% chance of happening)? This game seems identical to the one described on this post. But is it?

  2. Antoine says:

    My analysis also holds when you don’t open your envelope and don’t know x, but if you understood that non-fractional amounts of dollars were put in the envelopes, then if you see 1 dollar, you know the other must contain $100, so I would switch.
    If I’d see 1 billion dollars I would not switch, because I don’t think the swami would have 101 billion to put in the envelopes (no man on earth has that much right now).
    The second game you described is not identical, and that one I would like to play as much as possible, because your gain would be 25% of the participation fee on average.

    • epanechnikov says:

      I don’t think you have suggested any solution to this apparent paradox. We need to know why the expected values estimated here are misleading. Here is an approach I suggest to attack this problem. As you imply in your last message the posterior (i.e. post observation) probabilities are functions of the observed amount. We have no good reason to assume those probabilities should always be identical to 50%. The initial (frequentist) probabilities are relevant only to the unobserved amounts m and 2m and not to the posterior probabilities of 1/2x vs 2x. Furthermore we have no reason to assume linear utility functions. Hence the marginal marginal of gaining $99 could be much higher (in absolute terms) than the marginal utility of loosing 99 cents. Hence if I see $1 I might be more tempted to switch rather than not (subject to my subjective posterior probabilities with which I weigh those outcomes of course).

  3. Markus says:

    Please excuse my poor English …

    So for me there seems not to be a paradox with this calculation. I mean the expected value 5/4 $x (respectively 1/4 $x as a long-term profit) as a mean value is the correct value for a random experiment as described below. The crux of the matter: both persons mean to have exactly this random experiment (by the way i suppose they know about calculation of probabilities and random experiments, don’t they?!) but actually it is a different scenario for obvious reasons explained below.

    The following experiment – which might be considered by the two persons in a simliar way – may lead to the 5/4 $x (respectively 1/4 $x as a long-term profit) calculation:

    E1.) There are two persons A and B trading envelopes n times (n is quite a large number …).
    E2.) B is the person who gets envelope with $x and A with $y in the envelope. Both x and y are fixed numbers for the whole series (n draws).
    E3.) Person A and B agree to trade each draw.

    (So this experiment is likely to make person A and B get equal amounts of money without risk at the end … so the very amount is n*5/4 $x or 5/4 $x per “game”)

    Now the procedure is:

    first trade: B gets the envelop sees $x and knows A has $y=1/2*$x or $y=2*$x each the same chance. A and B do exchange their envelopes and don’t look inside and put them aside.

    second trade to n-th trade … the same game …

    So n trades later B has as an average of n/2 * 1/2*$x + n/2 * 2*$x = n*5/4 $x. That means 1/4 $x profit each game. And this is the amount that is wrongly assumed by A and B as the gain even only for a one step experiment.

    So what’s the wrong assumption by A and B? 1.) As mentioned there only one trade is done, not a whole series of trades, this means the averaging which finally leads to the value 1/4 $x is quite stupid. 2.) with the n-th time draw there will be the problem that persons know about the amount of money in their envelope from previous draws, so additionally to E1.) E2.) and E3.) you must demand that persons decide the same way independently of what they found in their own envelopes last time or with other words A and B simply must forget previous draws.

    • epanechnikov says:

      Hello Marcus. I’ve just seen your message. Apologies for my late reply.

      “B gets the envelop sees $x and knows A has $y=1/2*$x or $y=2*$x each the same chance”

      Could you please explain why you believe we should allocate the same chance on those two events?

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