The Envelope Paradox

22 September 2006 at 5:02 am 6 comments

| Lasse Lien | 

Here’s something to annoy you over the weekend. If you already know the envelope paradox, don’t read on. If you do not, and you are a bit of a nerd, I guarantee you’ll be facinated. The following version of the paradox is cynically stolen from  Amos Storkey’s homepage.  

You are taking part in a game show. The host introduces you to two envelopes. He explains carefully that you will get to choose one of the envelopes, and keep the money that it contains. He makes sure you understand that each envelope contains a cheque for a different sum of money, and that in fact, one contains twice as much as the other. The only problem is that you don’t know which is which.

The host offers both envelopes to you, and you may choose which one you want. There is no way of knowing which has the larger sum in, and so you pick an envelope at random (equiprobably). The host asks you to open the envelope. Nervously you reveal the contents to contain a cheque for 40,000 pounds.

The host then says you have a chance to change your mind. You may choose the other envelope if you would rather. You are an astute person, and so do a quick sum. There are two envelopes, and either could contain the larger amount. As you chose the envelope entirely at random, there is a probability of 0.5 that the larger check is the one you opened. Hence there is a probability 0.5 that the other is larger. Aha, you say. You need to calculate the expected gain due to swapping. Well the other envelope contains either 20,000 pounds or 80,000 pounds equiprobably. Hence the expected gain is 0.5×20000+0.5×80000-40000, ie the expected amount in the other envelope minus what you already have. The expected gain is therefore 10,000 pounds. So you swap.

Does that seem reasonable? Well maybe it does. If so consider this. It doesn’t matter what the money is, the outcome is the same if you follow the same line of reasoning. Suppose you opened the envelope and found N pounds in the envelope, then you would calculate your expected gain from swapping to be 0.5(N/2)+0.5(2N)-N = N/4, and as this is greater than zero, you would swap.

But if it doesn’t matter what N actually is, then you don’t actually need to open the envelope at all. Whatever is in the envelope you would choose to swap. But if you don’t open the envelope then it is no different from choosing the other envelope in the first place. Having swapped envelopes you can do the same calculation again and again, swapping envelopes back and forward ad-infinitum. And that is absurd.

That is the paradox. A simple mathematical puzzle. The question is: What is wrong? Where does the fallacy lie, and what is the problem?

Entry filed under: - Lien -, Ephemera, Former Guest Bloggers.

Evidence That Demands A Verdict The Pareto Criterion and Ethics

6 Comments Add your own

  • 1. Peter Klein  |  22 September 2006 at 9:22 am

    Isn’t this simply a scaling issue, rather than a behavioral anomaly or a flaw in how expected values are calculated? Calculating percentage changes raises the same problem. If I go from $1 to $2, my wealth has increased by 100%, but if I go from $2 to $1, my wealth has fallen by only 50%. That’s why we typically use arc elasticities to calculate percentage changes.

    If you calculate the expected gains and losses in percentage terms, using an arc elasticity formula, does the paradox disappear?

  • 2. pj  |  22 September 2006 at 10:45 am

    Well, first, it’s wrong to say “Having swapped envelopes you can do the same calculation again and again, swapping envelopes back and forward ad-infinitum.” This is not the same situation as an infinite series of envelopes, the (n+1)st envelope always having either half or double the amount of the nth envelope. In that series you end up with a binomial distribution, and you have some probability of gaining a very large amount, N * 2^n. Whereas swapping two envelopes back and forth the number of halves and doublings can never vary by more than one, and you always have an expectation of either N (even number of swaps) or 0.5 * (N * 2 + N / 2) (odd number of swaps).

    A side observation is that your preference will depend on your utility function for money. If utility goes as the log of the amount of money, you will be indifferent between a doubling and halving of N. For most people, the amount in the envelopes will be a marginal contribution to their total wealth, and so the utility function will be more linear. Therefore, they should happily switch envelopes when promised a doubling or halving with equal probability.

    Also, of course, you have to trust the host. If he would be more likely to offer you the swap when you would lose money, then it becomes a bad idea to swap.

  • 3. Jung-Chin Shen  |  23 September 2006 at 1:52 pm

    http://www.anc.ed.ac.uk/~amos/doubleswapsoln.html

  • 4. Lasse  |  24 September 2006 at 4:58 pm

    Well, according to Wikipedia, there is no universal agreement among probability theorists about the solution to the paradox, which is kind of scary in my mind.
    However, by following the link supplied by Jung-Chin, you’ll find Amos Storkeys solution – which does make sense to me. But I can’t say for sure whether it is flawless…

  • 5. Siyaah  |  27 September 2006 at 9:11 pm

    Very interesting.

    The analysis shown only considers ‘expected gain’. How about ‘expected losses’. How would that change the scenario?

    Kind of reminds me of Kahneman & Tversky.

  • 6. Lasse  |  29 September 2006 at 8:21 am

    Siyaah, this is a very natural association to get, but the pardox is a pure mathematical one about the calculation of expected values. It abstracts away from any and all behavioral assumptions on the part of the players. Certainly, such behavioral issues would be very important in a real situation

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