Envelope Problem

You have two indistinguishable envelopes, each of which contains a positive sum of money, but one envelope contains twice as much as the other. You may pick one envelope and keep whatever amount it contains. You pick one envelope at random but before you open it you are given the chance to take the other envelope instead

Now suppose you reason as follows:

I denote by A the amount in my selected envelope.

The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.

The other envelope may contain either 2A or A/2.

If A is the smaller amount, then the other envelope contains 2A.

If A is the larger amount, then the other envelope contains A/2.

Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.

So the expected value of the money in the other envelope is (½)(2A)+(½)(A/2)=(5A/4)

This is greater than A, so I gain on average by swapping.

After the switch, I can denote that content by B and reason in exactly the same manner as above.

I will conclude that the most rational thing to do is to swap back again.

To be rational, I will thus end up swapping envelopes indefinitely.

As it seems more rational to open just any envelope than to swap indefinitely, we have a contradiction.

Fun problem, try it out! Good luck!

Note by Tim Ye
5 years, 7 months ago

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Sort by:

- 5 years, 5 months ago

why don't you just see which one is heavier?

- 5 years, 6 months ago

- 5 years, 6 months ago

The symbol A in the expression (½)(2A)+(½)(A/2) pretends to represent at the same time 2 numbers, in the first occurrence "the smaller sum" and in the second one "the larger sum", which cannot be.

- 5 years, 6 months ago

Please read the set up (reason's first line). The symbol $$A$$ is chosen to represent the amount of money in the envelop that you picked. Then we considered the possible outcomes of the second envelope, and shown that it has an expected value of $$\frac {1}{2} (2A) + \frac {1}{2} (A/2)$$.

Staff - 5 years, 6 months ago

I think you refer to "I denote by A the amount in my selected envelope". Then let's go now up down, after having gone down up, again I say that $$A$$ is not a number but a random variable $$A(X|x=Amount, x=2\times Amount)$$, whom you can't treat as a number.

- 5 years, 6 months ago

$$A$$ is the amount of money in the envelop that you picked. It is not set up as a random variable. If you want to define a probability distribution on it, it would be better to use a another symbol, and not let your notation do double duty.

Consider your statements very carefully, and see if they make sense. You could be close to understanding what is wrong with Tim's presented argument.

Staff - 5 years, 6 months ago

Hi Tim,

you suble cheater tried to confuse us..

Then, if A is the smoller emount, the expected value of the other envelope is not (½)(2A)+(½)(A/2) but (½)2A+(½)A=(3/2)A. Which is the same expected value of the one you chose. There are, then, no rational reason to swap.

- 5 years, 7 months ago

Please review the concept of Expected Value, and how to calculate it. Read the setup to ensure that you understand it properly.

Staff - 5 years, 7 months ago

The expected value (or expectation, mathematical expectation) of a random variable is defined by: $$\sum_{i=1}^k x_{i}p_{i}$$ where $$x_{1}, x_{2},... x_{k}$$ are the values the variable can take and $$p_{1}, p_{2},... p_{k}$$ the corresponding probabilities. In our case: (½)2A+(½)A. Where was I wrong? It may be I was wrong...elsewhere and I beg pardon. Anyway I intended to be joking.

- 5 years, 7 months ago

Read Tim's reply. You have the wrong setup of the problem. The other envelope either contains twice as much, or half as much, as the original envelope that you're holding on to.

Also, the expected value of the envelop that you chose is just $$A$$, and not $$\frac{3}{2} A$$. This is because with probability 1, it contains $$A$$.

Staff - 5 years, 7 months ago

A is not the smaller amount, A is the number of $in the first envelope you chose. So, the other envelope has a 50% chance of being half and a 50% chance of being double, hence (½)(2A)+(½)(A/2). - 5 years, 7 months ago Log in to reply $$A + B = 3x$$ $$A + \frac {5A}{4} = 3x$$ $$\frac {9A}{4} = 3x$$ $$A = \frac {4x}{3}$$ Now, we know that B must be either twice as much as A and add up with A to 3x, or half of A and add up with A to 3x. However, both contradict this: $$\frac {4x}{3} + \frac {1}{2}(\frac {4x}{3}) = 3x$$ $$2x \neq 3x$$ $$\frac {4x}{3} + 2(\frac {4x}{3}) = 3x$$ $$\frac {12x}{3} = 3x$$ $$4x \neq 3x$$ - 5 years, 7 months ago Log in to reply When writing up a solution, it's helpful to state what you're trying to show. It is unclear to me what you hope to achieve with these statements, apart from contradicting the procedure given in the problem. Also, when writing up a proof, please ensure that your statements make sense, and that you explain what you're doing. For example, in the first line, when you tried to replace $$B$$ with $$\frac {5x}{4}$$, such a statement is not true. All that has been shown, is that the expected value of $$B$$ (denoted as $$E[B]$$ is equal to $$\frac {5x}{4}$$. Staff - 5 years, 7 months ago Log in to reply I agree with Nathan M. in that my intuition is that the probability of picking either the large or small sum should be equal each time you switch. However in either case it seems weird to me that it would be (5/4)A with A being whatever is in my hand. If I think of it in terms of the scenario before I have picked up an envelope it makes more sense to me. I would rephrase the situation let A=the smaller sum. The expected amount I would pick up would be (1/2)(A)+(1/2)(2A)=(3/2)A. So I would think the expected outcome is between the small and large sum when I first choose an envelope. I think that should be the same each time I switch as well. I do not like how in the question A is either the large or the small amount, but do not know how to argue against it. - 5 years, 7 months ago Log in to reply Because A is the number of$ in the first envelope you chose, you don't know if it's the small or large sum.

- 5 years, 7 months ago

Yeah, the way I would change it makes it into a choosing problem not a swapping problem, which is not the point of the problem, so I'm wrong.

- 5 years, 7 months ago

But since the total amount of money in both envelopes is constant regardless of what you have in hand, the expected value of the money in the envelope in your hand is 5A/4. Hence the probability will still be 1/2 and if you keep changing envelopes the probability of getting 2A or A/2 is the same.

- 5 years, 7 months ago

Can you be more explicit about what is the error, and make sure that you're not committing it?

I'm quite certain that if the selected envelope in my hand has $$A$$, then it has expected value of $$A$$.

Staff - 5 years, 7 months ago