The Two Envelope Problem - a Mystifying Probability Paradox

There are two envelopes in front of you, and you know that one of them has ten times more money than the other. You pick randomly one envelope, but before taking it home, you are given the option to switch, and actually take the other envelope. Should you switch?

That was (one version of) the infamous two-envelope paradox. It's a paradox, because there is a seemingly convincing argument for why switching is pointless (the two envelopes are symmetrical), but also a seemingly convincing mathematical argument for why switching increases your expected winning. This paradox has baffled some of the greatest minds, including the king of recreational mathematics, Martin Gardner.

Unlike most other famous probability paradoxes (e.g., the Monty Hall problem, the "boy or girl" paradox, and Simpson's paradox), the two-envelope problem has a relatively intricate solution. At its heart, the solution is related to the limitations of the notion of expected value, and to an often overlooked caveat regarding conditioning and the law of total expectation (aka the tower property). The paradox solution is also related to the mathematics behind conditional convergence of series, and to Riemann's rearrangement theorem (aka Riemann series theorem).

Created by Yuval Nov for the 2021 "Summer of Math Exposition" (SoME1) competition, hosted by the one and only 3Blue1Brown (Grant Sanderson).

#paradox #math #probability #mathematics #envelopes #puzzle #3b1b #SoME1

[d.e.p.-j.]

This is the best explanation of the paradox I've ever seen. Thanks!

[aleksandersabak]

I heard the paradox multiple times, including from teachers at schools, and this is the first time I've had it explained at all, not to mention the quality. Thank you, good person.

[Yitzhk]

I feel like the best strategy to avoid the "sense of remorse" is to not switch, and to be sure to never learn what was in the other envelope :P

[Reddles]

In any realistic situation with a finite amount of money involved, you would look at your envelope and decide to switch or not based on whether the amount in the envelope was above or below the expected payout. The problem is only counterintuitive because you've set it up so that the expected payout is infinite, so when you see any finite value in the envelope it's still below average and you should switch.

[andrewweirny]

This is a pretty rare situation: we know our intuition is right but we have to torture the math to get to it. Usually we’re having to use math to show our intuition is wrong.

[templetonbob]

I think your choice on switching depends completely on the amount. If the values are really infinite, then being handed an envelope with $45, 382, 139 will certainly be worth not switching because the money is life changing. Being handed an envelope with $63 is worth switching because the chance of a few hundred is worth losing $50.

[rogerkearns]

This experiment can be carried out at a cost of a mere minus one-twelfth of a dollar.

[Drotdog]

So in the end, it’s exactly as non statisticians would see it, a 50:50 chance with no impact on chances whether or not you choose to switch.

[eklektikTubb]

The right strategy is: Pick the envelope, save some time by not switching and save some more time by not looking what into another envelope. You will end up with random amount of money, but a little more time.

[Ulkomaalainen]

Just wanted to mention: if you do not use a "larger amount is 10 times as large" as here or "2 times" as in the more common variant, but settle for some multiplier between 1 and 2, ay 1.5 (which of course would quickly give ugly numbers) while keeping the introduced "probability of pairs" (1/2 for the lowest pair - say 1 and 1.5 -, 1/4 for the next - 1.5 and 2.25 -, then 1/8, and so on), then both the series of the positive and negative sums would converge and we could actually calculate the expected value of switching, which turns out to be 0.

[harmonicarchipelgo]

I mean, if hypothetically switching was better than not switching (without opening the envelope) then after you switch you are in the exact same scenario and should switch back. This continues infinitely.

Thus the solution cannot be that switching is better than not switching, since not switching is just switching twice.

So I am glad we were able to resolve the oddity of the expected value being weird.

[AdmiralJota]

I think the best strategy depends on the utility of a given amount, which is something you have to decide personally. For example, I could afford to retire immediately if I won $10M, but not if I only won $1M. On the other hand, while $100M would be nice, it wouldn't make *that* big of a change to my lifestyle over $10M, because there's nothing I need that costs that much. So if I see $1M, I'll switch, because the hope of getting $10M is worth more to me than the risk of just getting $100k. But if I see $10M, I'll keep it, because I don't want to risk losing it.

[cannot-handle-handles]

Great video!
And a great idea to essentially mix the Two Envelope Problem with the St. Petersburg lottery paradox. (The fact that the expected value of the money in each envelope is infinite also illustrates nicely why this paradox cannot really arise in the real world.)

[lubomirkurcak]

This is an excellent video!

This paradox is similar to the Cauchy distribution, which is a symmetric PDF which surprisingly has no mean (i.e. expected value) or variance.

This was really hard for me to swallow until I realized the fact that averaging samples does not change the distribution, namely:
if X ~ Cauchy(0, 1) and Y ~ Cauchy(0, 1) then (X + Y)/2 ~ Cauchy(0, 1).

In other words, no matter how many samples of Cauchy(0, 1) you average out, you can still expect the same value as if you took just a single sample.
This goes contrary to the Central limit theorem (CLT), which assumes that averaging samples of any distribution converges to a normal distribution.

This fact was enough for me to accept that Cauchy(0, 1) has no mean or variance. By averaging any number of samples, the next sample can be so extreme it totally throws off your previous average. You never settle on any value no matter how many samples you take.

[iczyg]

The problem with this paradox is that it exists in a vacuum and in any scenario youd have to pick, thered be other factors and other values at play. Such as setting. Is it part of a game show? How much are prize amounts usually on this game show? Is it a single benefactor giving you this choice? There are so many other factors at play that can be used to define which envelope probably holds the higher amount, that the math here is trivial. This isnt a mental excercise to help understand some higher mathematical concept.

So the answer is that it will always depend, but it will never realisticislly just depend on the factors given here.

[bastiaan]

If only the host knows the amounts, and offering a choice is forced, and there are two envelopes, then switching is 50/50. If there are more than two, and the host knows the amount, and the offering of a choice is forced, and the other amount is forced to be the highest left, then switch. If there are more than two, host knows the amount, and is forced to be the lowest left, then don't switch. If the offering of a switch isn't forced, host wants to give you as little as possible, but does give you a choice, then the choice is always less so don't switch. If the host doesn't know anything and neither do you, it doesn't matter.

[glennmelzer]

Yuval, I know that many smart people have analyzed this paradox as you initially described it, but it seems like there is a simple solution. What do you think:
The expected value equation used to determine the value of the other envelope was the arithmetic mean of the two equally possible outcomes (2X or ½X). However, the original problem is described as two equally possible factors (either 2 or ½) that is to be applied to the selected envelope’s value to determine the other envelope’s value. Therefore, determining the expected value of the other envelope requires knowing what is the expected factor. Calculating the expected factor requires using not the arithmetic mean, but the geometric mean formula:
Expected Factor = squareroot ((2)(1/2)) = 1
Expected Value of other envelope = 1X
This shows that the expected value of the other envelope is A, the same value as the selected envelope and there is no advantage to switching.
The arithmetic mean formula would have been appropriate if the value of the other envelope had been defined in terms of an addend amount (as opposed to a factor) that could either be added to or subtracted from the selected envelope’s amount with equal probability.

[lilyofluck]

I just spent 20 minutes learning about a scenario that will never realistically happen, and I'm not disappointed by that fact.

[yeahuh]

1. Look at the thickness
2. Pick the thicker one?

[PlusdB]

If your envelope has an odd number of dollars in it, then you can deduce that you have the envelope with the small amount in it.
ALWAYS switch if your envelope contains an odd number of dollars.
The probably that your envelope has an odd number of $ is 1/2, if the amount of $ in the envelope is random.
If a random amount of $ is a property of the problem, then you always have at least a 50% chance of getting the big amount.