The Two Envelope Paradox: An Experiment in EV

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

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 0.5x20000+0.5x80000-40000, 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?

[BluffTHIS!]

Read King Yao's Weighing the Odds in Hold'em Poker for the chapter entitled "The Monty Hall Problem". Same thing.

[JoshuaD]

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Read King Yao's Weighing the Odds in Hold'em Poker for the chapter entitled "The Monty Hall Problem". Same thing.

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I understand that problem, but I don't see the connection.

edit: O I C.

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Read King Yao's Weighing the Odds in Hold'em Poker for the chapter entitled "The Monty Hall Problem". Same thing.

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I dont think its the same thing. It is always correct to rechoose in the "Monty Hall problem" as you increase your odds of success from 1/3 to 1/2. The OP has somehow applied this reasoning to a dilema which I dont think you can do. Your odds stay 1/2 in the OP. I dont have the math knowledge to show why the OP is wrong but I think it is. Your average return on rechoosing envelopes should be zero. Maybe not opening the first envelope is the problem, I dont know.

[BluffTHIS!]

You are right. It isn't the same, as I just reread the OP. You have a guaranteed win here which isn't the case in the MH problem. If you are always offered the choice of the second envelope, then it doesn't matter which one you pick or whether you reselect.

[PLOlover]

http://mathproblems.info/prob6s.htm
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Suppose the envolopes contained $1 and $2. By switching you would either gain $1 or lose $1. Regardless of the actual amounts in the envelopes the point remains the same, that the average gain of the two possibilities is zero.


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+1 or -1

EV .5(.5)+.5(2) -1 = 1/2 or 2 , give up 1
EV .5(1)+.5(4) -2 = 1 or 4 , give up 2

So it seems that +1 or -1 is represented as double or half.

What we have is two variables, but turning into three. For example, we get envelope with 2 dollars, the possible amounts are 1,2,4.

But they should really be represented as -1,0,1 or 1,2,3. Note how in -1,0,1 you can gain or lose one.

somehow we have to come up with a metric where half or double is +1 -1.

The answer here is a bit tricky

If you switch no matter what amount you see in the first envelope, your EV overall will remain unchanged

The key here is that the game-maker cannot make it so that there will always be a 50/50 that the other envolope will be the larger amount, given the amount that you see in the first envelope

Here's what I mean: Suppose he includes $10 as a possible amount. That means he has to include $5 as a possible amount, or else when you have $10 you will know the other evelope is $20. This also means he has to include $2.5, and $1.25, and $.625 etc.

Same thing for the high side, he will have to include $20, and therefor $40, and therefor $80 etc.

And they will all have to have the same probability, this is the only way to ensure that given one amount in one envelope there is a 50/50 that the other evelope is double (Actually, you could alternate probabilities, so that only the amounts 4x and (1/4)x have to be the same probability, but the same problem remains).

The problem here is if some one created a game to ensure this 50/50 rule, there is no upper bound on the amount of money that could be in the envelope, and in fact the average amount of money in an envelope would be infinity

So, in real life, you may be able to guess what the game-maker's upper and lower limits are about, and therefore make a EV-maximizing decision which may be either switching (probably because you think the amount is on the low side of his range) or not switching (probably because you think there's a reasonalbe chance you have hit the upper limit already)

I suck bad at explaining things, so I apologize if I confused anyone

[PairTheBoard]

The explanation for why the EV calculation is wrong is simple but uncomfortable for a lot of people.

Once the First Envelope is opened there is no longer a 50% probability the Second Envelope contains double and a 50% probability the Second Envelope contains half.

After the First Envelope is opened there is Now Either a 100% probabilty the Second Envelope contains Double, OR a 100% probabilty the Second Envelope contains Half, we just don't know which.

Most people refuse to accept this but that's the way it is. If you apply this principle in less clearly necessary ways you are likely to actually get Flamed for it, as I was when I did so in this Probabilty Forum Thread.

Which Twin Has the Tony?

PairTheBoard

[Jman28]

Interesting side-problem

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Interesting side-problem

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Agreed, that is interesting