# Can this paradox be resolved?

Asked by flutherother (34626) June 16th, 2013

There are two envelopes on a table, identical plain brown manila envelopes. One contains twice as much money as the other. You lift one of the envelopes from the table and you are then given a choice; to keep the contents of this envelope or to swap for the other envelope. What should you do?

Solution: If the envelope you chose contains £Y the other envelope will contain either £2Y or £0.5Y. You therefore stand to gain £Y by swapping but you can only lose half that or £0.5Y so clearly it is to your benefit to swap.

Paradox: What if we had chosen the other envelope. Wouldn’t the same logic apply? How can it always be beneficial to swap?

Observing members: 0 Composing members: 0

How is it a benefit to swap? You’re getting money either way, so I don’t think it’s fair to say that getting ½ of what you could have gotten is a benefit.

dxs (15160)

I would take neither and get a job.

talljasperman (21916)

Yea, “benefit” is a tricky term. It’s really just a measurement of risk.
It’s not that your outcome will be better or worse, it’s simply that your chances of the more desirable outcome are greater one way or the other.

ninjacolin (14246)

@dxs Well you are taking a chance when you swap. There is a 50% chance you will lose half your money but there is an equal 50% chance you will double it. Say you have £100 in your envelope, you might lose £50 but you have an equal chance of gaining £100. Doesn’t that sound like a good deal to you?

flutherother (34626)

You get to take one of the two envelopes, so you are guaranteed at least the amount of the smaller envelope.

filmfann (52295)

@filmfann But the math says it does, hence the paradox.

flutherother (34626)

GQ! I love this problem. I don’t have the answer, but I still love it.
It reminds me a bit of the Monty Hall problem .

LuckyGuy (43768)

This is a classical paradox, there’s even a wiki article on it. Link

PhiNotPi (12681)

I think it is the weight of the envelope that determining what envelope one picks. Heavier the better.

talljasperman (21916)

@talljasperman You are making an assumption about the denomination of the notes.

flutherother (34626)

@flutherother It sounds similar to Americas Deal or no Deal game show. Worst one can get is one penny and that is still considered a profit. So swapping you always get profit in the end.

talljasperman (21916)

@LuckyGuy I thought of that, too. Someone explained it to me once, but I forgot how it works.

dxs (15160)

Where can I find this opportunity? I could use a bit or a bit more, whichever.

Sunny2 (18842)

But if the envelope that you didn’t choose first contains £Y, then the envelope in your hand contains either 2£Y or 0.5£Y. Clearly, there is no benefit to switching.

glacial (12145)

To those who think that they have solved the puzzle, let me throw in an additional challenge.

There are two envelopes, one of which contains twice as much money as the other. You pick a random envelope, open it, and it contains \$50. You are given two options: keep the \$50, or switch to the other envelope. Which option do you chose?

The other envelope contains either \$100 or \$25, which averages out to \$62.5, which means that it is better to swap.

Now, notice that opening the envelope and finding \$50 doesn’t really mean anything. You could have opened it and found \$0.50, and the same logic will apply. (In fact, you don’t even need to open the envelope, because the amount of money doesn’t matter.) This leads to the conclusion that it is always better to switch. This, however, contradicts the fact that both envelopes have the same expected value.

The challenge: find the flaw in the reasoning above. I have my solution, and I’ll post it in a while.

PhiNotPi (12681)

I don’t see how you can use the money in the first envelope to predict the amount in the second envelope. Whether it is more or less is still a 50/50, regardless of if the winnings are more than the loss. In comparison to the Monty Hall problem that @LuckyGuy posted, the probability does not change.

dxs (15160)

Yes, the probability of gain/loss is 50–50, but how much gain and how much loss are very mathematically important. That’s how you calculate expected value.

In fact, the amount of win and loss has just as much influence (sometimes more) than the probabilities involved. There are things that can have a 95% chance of profit, but which turn out to be Taleb distributions with horrible long-term prospects.

PhiNotPi (12681)

Ugh. I wish I understood math better.

dxs (15160)

Once you’ve made the decision to swap you’re presented with the same chance all over again to make a more mathematical decision.

But.. the math only guarantees you a minimum amount of profit. It doesn’t compel you to choose either. It tells you what the math says about your “chances” but it doesn’t tell you when to stop asking what the math says. A buzzer in a game show, however, that’ll provide the last variable for time that resolves the loop.

Are you sure this qualifies as a paradox?

ninjacolin (14246)

Anyways, I’ve got a sleep cycle in the near future. I guess I’ll go ahead and tell my solution.

The flaw is simple. We have been saying all along that the probability of gain is equal to the probability of loss. Indeed, if that were true, then switching would actually be a good deal.

The problem, however, is that the probability of gain is not necessarily equal to the probability of loss. That is impossible to say. Here’s why:

First, let’s assign some things:
A = larger quantity of money
B = smaller quantity of money = A/2

Say that you open the envelope and find \$50, then one of two things must have happened:
There was \$100 and \$50, and you received \$50. (A = 100)
There was \$50 and \$25, and you received \$50. (A = 50)

If we are to say that the probability of gain is equal to the probability of loss, then that also implies that:
p(A = 50) = p(A = 100)
That is wild speculation. We don’t know anything about the underlying probability distribution.

PhiNotPi (12681)

@PhiNotPi As far as I can tell, you have just restated the problem, only with actual dollar amounts. What is your solution?

glacial (12145)

We can’t calculate the expected value of the swap without knowing the probability distribution. By averaging the two amounts, we are assuming something about the probability distribution that we don’t know.

PhiNotPi (12681)

How could this situation be any different from flipping a coin? What distribution would you think appropriate to this scenario?

glacial (12145)

Sorry, but it’s getting really late, I’ll have to hold off answering until tomorrow.

PhiNotPi (12681)

Here’s the intuitive side coming out of me, but to me you wouldn’t benefit nor gain here since you would have no way of determining which envelope you’ve picked up to begin with. You would have a 50% chance of attaining a loss or gain either way without knowing what you had picked up, right?

I think your intuition is right but the math says it is wrong. If you were to toss a coin and get £50 if a head comes up and lose £25 if it is a tail seems like a good deal.

If you keep playing, doubling your money with every head and halving it for every tail you will ultimately come out even but for just one toss of the coin it would seem a chance worth taking.

flutherother (34626)