A gambling game that creates money!

[Fiksdal]

Going through probability, I was having a talk with my math teacher about gambling and Expected Value.

He explained to me a following scenario:

You are playing against an opponent. The casino gives both you, and you opponent an envelope. The casino guarantees that both envelopes contain money, and that one of them contains double the amount of the other one. You do not know which contains the most.

Anyway: The casino offer, that if you both agree, you can swap envelopes. By common sense, this should have a neutral EV to both of you. But my math teacher explained it to me like this:

The current EV of the money in your envelope is X.

You know that your opponents envelope contains either double that, or half that. So either:

-2X (50% chance)
or
- 0.5X (50% chance)

So on average you will by swapping envelope win

(2x + 0.5X)/2 = 1.25X!

You actually end up with 0.25X more than you would have gotten by keeping your original envelope! wtf.

Example: You take a peek inside your envelope, and see that it contains $10000. You now realize, that your opponents envelope contains either $5000, or $20000. There is an equal chance of both, so if you choose to swap, your new EV is ($20000+$5000)/2 = $12500. And your original EV was $10000! You just discovered that you can increase your EV by $2500 if you choose to swap.

Obviously, your opponent will also find swapping to be favorable. So swapping envelopes is +EV for both you and your opponents. And after you swapped (assume no one ever opened their envelope), you would like to swap again! Because it is still +EV.

Of course, this goes against all rules of EV. You can't just create EV. And the value of the two envelopes added up remains the same no matter how many times you swap. Obviously.

So I assume something in this reasoning is wrong. But: what is it that is wrong?

[AlienBoy]

Each time you swap you do not change the EV.

For an envelopes with X and another with 2X:

When you are first given the envelopes, your Expected Return is 1.25X

The first time you swap the EV is 1.25

The second time you swap the EV is still 1.25

It does not matter how many times you swap, unless you LOOK in the envelope first.

IF:

You are handed the envelope, and you see that inside it has $10,000, then you know the other envelope has either 5000 or 20000.

NOW, if you switch, you have an expected value of 12,500.

BUT If you switch again, you have an expected value of 10,000.

Why? because on the second switch you KNOW that the envelope has 10000 in it. When you switch the 1,000 envelope with the unknown envelopt, you are now betting 10,000 that the other envelope has either 5000 or 20000.


If you don't know the contents of either envelope then it's random, and you're either winning X or 2X, for a total EV of 1.25X, regardless of how many times you swap.

AB


(Note that this is different from the Monty Hall scenario, where you can choose 3 boxes, and after choosing ONE, Monty Hall then shows you an empty box you didn't pick - here if you *now* switch boxes, you improve your chances of picking the prize box 50%)

[Evenkeal]

The half of the time that you are dealt the smaller envelope, you switch and gain a net profit of half (NOT the whole) of the largest envelope. The other half of the time that you are dealt the larger envelope, you switch and lose a net amount of half the largest envelope.

[ChrisV]

This is actually a pretty hard paradox and none of the replies so far have grasped the nature of the problem. The answer to the problem in this form is that x is not a constant value through the argument. There are however more complicated variants of the paradox. See here.

[kordothebear]

its pretty simple, its because when u change envelopes u either gain 10k or lose 5k. If it was set up where u either gain 10k or lose 10k there wouldnt be any EV in the trade. This really is very simple.

[ill rich]

actually it's not quite gambling.

if you and a friend are both given envelopes, you're EVEN at $100. it's like saying when you lose a pot in a session of poker it's impossible to "get even" because you already are.

so essentially you will either gain or lose $100.

you have a 50% chance of gaining $100, and a 50% chance of losing $100.

we let $100 equal X, so... (X-X)/2 = 0. You're true EV is zero.

[MelloAceCV]

If you are given an envelope, how could youe EV be neutral? You are guaranteed X, or 2X. Regardless, can you really go wrong, whether you open your envelope or switch with the other player? The OP never mentions having to wager to gain the money, so how can you lose here? If the envelopes are $100 and $200, you either gain $100 or $200, right? Am I missing something?

[ill rich]

if the envelopes are given to you, you're guarnteed at least the lower of the prize. if the envelopes are $100 and $200, switching the envelopes is a gamble of $100.

50% chance you gain $100, and 50% chance of losing $100 = EV of 0.

the problem with the question here is that to get the initial stake IS NOT A GAMBLE. the only gamble here is involved in switching the envelopes. what are you wagering to win an envelope? nothing. the casino is offering it to you. of course getting a FREE GIFT is +EV.

when you consider the actual gambling, switching the envelopes, the EV will be zero.

[SheetWise]

OP- Why would you create a hypothetical X to represent the money in your envelope, and then state that the money in your envelope was either .5X or 2X? You've already defined X as your expectation.

Simply define X as one of the outcome values, and the expectation becomes obvious.

[efiusa]

Simplify things

To begin with, there are 2 envelopes. One contains X and the other contains 2X

Since you have a 50/50 chance of being handed either envelope, your initial expected value is 1/2 (X + 2X) or 1.5X.

The original premise is wrong that you begin with EV of X. Because the original premise is wrong, the remaining argument is moot.

When you go through the calculations that you have a 50% chance of receiving the envelope with X and 50% of receiving 2X, you will see that whether you keep the envelope or switch, your expected value is always 1.5X