The Two Envelope Problem - a Mystifying Probability Paradox 

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

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

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.

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

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.

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.

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.

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.

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

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.

I like how this paradox assigns probabilities to only one of the unknown values. Edit: I thought of a way to represent it, which should have been obvious, but anyway -- If we assign the lower value x, then the higher value is 10x, and the average value is 5.5x, which they _both_ have, meaning that switching provides no additional value. Below are my original thoughts: The amount in your hand is also either 1/10th or 10 times the amount in the one you left behind. They both contain, on average, 5.05 times the amount in the other envelope. This is a strange way to represent it, but it shows that they're equal, at least. I'm sure there's a way to represent it that doesn't require saying they both have more than the other (on average). It can be done with example values easily enough. If you assume one is $10 and the other is $100, they both contain (100+10)/2= $55 on average. (I added that equation after the edit... which kinda makes the solution I have up at the top a lot more obvious. I think I didn't notice because I was doing the math in my head.)

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

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.

This is one of those lovely "doesn't actually apply in a real life version of the scenario it demonstrates" thought experiments in mathematics. Still very interesting.

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.

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.

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.

I never realized that the heart of the paradox was conditionally convergent sequences. Very well done. I thought it was one of those things that just completely mystified people, since the solution wasn't discussed in the math class in which it was brought up, most likely because it needs a result from a completely different area of mathematics that there's no guarantee students will have encountered.

You explained very well the fact that E(switch)>0 and E(switch)0 if X is the amount in our envelope, and E(switch|X)

i like how he starts with “there’s a random amount of money in the envelope and in the other one there is 10x the first one or 1/10 the first one” and then he proceeds to assign these amounts to each envelope with very specific probabilities for each amount

This was fascinating and I'm totally convinced by your explanation. One thing that confused me though is that your demonstration only applies to an infinite number of envelopes with rewards X that follow a specific probability distribution (arbitrary, aside that sums to 1). In my mind, I was mixing this scenario with a more "realistic" one, where the number of possible rewards is finite. For the records, in the finite case, assuming the rewards probabilities are uniformly distributed (and sum to 1), what matters is the existence of a cap, a highest reward. If the maximum is known, then it is indeed favorable to always switch to the non-opened envelope, unless the opened one shows the largest amount (in which case we choose that). Switching away from the maximum is the only source of negative return. If the maximum is not known, then switching doesn't matter because the expected value is 0 if we include the case that we may have opened the maximum reward that we can't exceed (it's a small probability of a high loss, that balances all the other potential gains).

"i will give you right now $10" vs "maybe when i feel like it i could give you either $100 or $1"

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.)

To me, "winning" is picking the envelope with the greater value. Using your defined rules, any value that is revealed (with the exception of $1) is twice as likely to be the greater of the two possibilities. So, the envelope that has is contents revealed is the one I would pick (unless it is $1, in which case I would take the unrevealed one).

I trust that most people realise that this is not a paradox; it is merely a mathematical sleight-of-hand con trick that makes it appear as if there is a paradox.

Great video! An interesting coda to this would be what happens if there are NOT infinite terms, and instead the maximum envelope holds something like 10^100 dollars. To total the probability to 1, there are a few options, but we could just say that the final envelopes are equal in value. Rigorously, the expected value is now defined, and intuitively, the in-paradox analysis of switching still holds in 99% of cases. Is that final miniscule probability of immense negative value enough to wipe out all the positive terms? I suspect it cancels them exactly.

what a cool variation of the monty hall problem. the only issue i have is that by introducing different probabilities for each possible pair of numbers, you changed the problem from the original problem. why should any pair of numbers be any more likely than any other pair?

I kept screaming at this video that the use of expected value made no sense in the argument. I was glad I persevered to the end to see the demonstration that it's an impossible criterion!

So many people here clearly coming from math puzzle videos. "Actually it's based on if the person who made the envelopes had enough money" "Actually you shouldn't switch because trying to get more money like that is greedy" "Actually it's free money so you shouldn't complain" "Actually it was always 50/50 so why bother?" No-one is arguing the morals or technicalities. It's a demonstration of a basic paradox using expected values, the typical method for determining optimal strategies. It isn't some riddle where you have to think outside the box. The answer is there is no strategy, because expected value breaks down with infinite values in both directions. That's all.

Its truly fascinating how unreasonable math can get. As long as your revealed envelope isnt 1$, then its a perfect 50-50 probability the 2nd one is better/worse, and the fact that math can prove it to be otherwise, is just disturbing.

If you were to do a simulation (whether completely in a computer program, or by acting out trials), you would implicitly address the issues that prevent a real answer from existing. In particular, you would not have an infinite number of possible values, and you would have some actual probability distribution of what values are in the envelopes. You'll get an answer. But the experimenter needs to understand that the answer is sensitive to the specific details chosen and doesn't apply to the original vague problem as a generalization.

I feel like there's a really key perspective on what's going on here that you didn't mention. The expected profit from switching is equal to the expected value in the other envelope minus the expected value in your envelope. If you calculate the sums, both expected values are _infinite_. And thus the expected profit is infinity minus infinity, which is indeed not well defined.

That was great! However, I got a little bamboozled when you put forth the table with amounts and their likelyhood. Suppose we can check OUR envelope and find 10$. That means we either picked the higher amount in the 1$-10$ pair or the lower amount in the 10$-100$ pair. And as you calculated, the former option is twice as likely. So an extra layer to the paradox seems that "average profit" tells you to switch, plain old probability simultaneously tells you not to. After all, you are twice as likely to already have picked the higher amount of money.

The whole thing ends after two pieces of information: 1. Any combination of amounts is possible, 2. You have a 50% chance of picking either. If the first wasn't true, then statistics could tell you when switching is EV+. But it is, so just pick your damn envelope and go about your day.

Okay now explain this paradox. When I ask my girlfriend what she wants to eat, she says she is okay with anything. But when i name all possible places to eat, she rejects every decision.

Hi, this was a really interesting paradox! One thing I don't understand, though: Why do you assume that as the money values of the pairs increase (1/10 to 10/100, etc), the probability decreases by half each time? Personally, I feel like the scenario of the paradox statement implied that both "The other envelope contains 1/10 the money" and "The other envelope contains 10x the money" should be assumed to be equal, and I feel like assigning the decreasing probabilities is an arbitrary way to make that untrue, rendering the paradox itself trivial without really addressing it. What are the experimental results (using a program) of that case?

Isn't the fundamental paradox here is that the expected value of the game infinite? Assume you always pick the smallest of the envelopes, your EV is 1/2*1+1/4*10+1/8*100+.... Each product is biggest than the one before, so the sum is infinite. So in the case none of the envelopes are opened, it does not matter if switching increases the EV by X or by X times (depending on what funny math you use), because infinity + X or infinity*X (X not zero) is just infinity.

There is no paradox, no one told you to start averaging things. There is no average. A or B. 10x or 0.1x. You will end up with either 90% more or 90% less than what you currently have by switching. The only factor worth considering is what number would be significant enough to see in your first envelope for you to not risk losing 90% of it.

Something interesting to consider is that E(X), the expected amount of money in the envelope you pick, is also undefined (blows up to infinity). I wonder if that necessarily implies that E(Y) is undefined too

Thank you so much, bro!!! I encountered this problem in the past and I thought that since the problem leads to contradiction, such a situation cannot exist. I thought that we cannot have two envelops with condition that one has 10 times more money than another. I could not imagine that this has something to do with conditionally convergent series. Honestly, this is one of most mind-blowing problems in math to me. Great video!!!!!

But you've resolved the infinite part of the paradox(switch without looking) but not the concrete part (look, then switch). Did I miss that part? If your are allowed to check your envelope, you should do so then switch. Likewise, if you can inspect the other envelope, do so and don't switch. The reasoning doesn't involve infinite series, you've stated that the expected values in both cases are correct but the conclusions are still paradoxical.

Conclusion: judge how likely it is that the guy showing you two envelopes actually _has_ ten times the value you see

If I learned something from ancient Greek mathematicians, is that if we reach a paradox, it means we made a flase asumption somewhere along the line. This one isn't an exception.

I would just swap any number below 10k since the money is so low the gamble is worth it but as soon as my envelope shows 10k or higher I'd stay

This was good and While the math of series wasn’t new to me it’s use here was novel to me and brilliant. I would have ended with a different set up where the expected value of an envelope is defined. You can do this if there a max payoff or a distribution. That decays sufficiently fast. Either way that will lead to a rule where if you observe a big enough number you don’t switch. And the math all falls in nicely Thanks.

This is one of those "paradoxes" that is actually an error. To say that the other envelope has 10x means that x is the smaller number. To say it has x/10 means x is the larger number. In combining the two in one expression the same variable x is being used to represent two different numbers.

I love how the conclusion is exactly what I would have expected from intuition alone based on the description of the problem.

So it’s a paradox for the same reason some series can converge at different values depending on the order of the terms. I got it now. Thanks.

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.

If you know the amount opened, switch if it's a fairly low amount, something you are fine with losing just in case you actually have the higher one. If it's a decent or high amount of money simply keep it as the chance of it being the lower isn't worth losing it. That amount would be different for everyone and I'd be curious about the average "cut off" numbers for when people feel better switching

I think that applying a utility theory based on risk aversion would be a good approach to make a decision for each value of the envelope.

Really interesting and clear demonstration of this concept, thanks! It's not really a "paradox" though, not anymore than dividing by 0 is a paradox. A simple way of avoiding the improper use of infinite sums is to ignore the monetary value and realize that no matter what envelope you open, there are only two possible scenarios - the other envelope has less money, or it has more money. In other words there is a "better" envelope and a "worse" envelope. You know for certain that the odds of having picked the better envelope were 0.5 because that's how the choice was defined. So switching is also a random 50/50 choice between a "better" and "worse" envelope (unless of course you hit the edge case where the envelope you opened contains $1 and you can switch for $10 every time).

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.

This isn't a paradox, it's sophistry. By the setup described, if you open an envelop with $100, 2/3rds of the time that envelop will be from the $10/$100 pair, and 1/3rd of the time from the $100/$1000 pair. Averaging the $10 and $1000 is pure sophistry that has absolutely nothing to do with the odds of which set of envelops you have.

This was a very good explanation of the mechanics of how the structure of the payout allows contradictory answers to be reached, but isn't that the whole paradox? When one envelope is observed the possibilities collapse to 4 which excludes an undefined value function of swapping because no infinite series is involved. Isn't the fact that which conclusion is logical depends on if you open your envelope, the other envelope, or leave both unopened arbitrary to any value you observe the paradox? edited for spelling

Man i have been getting so many these amazing math videos from literally unknown channels and I've been loving it! Thanks for the video formant and thank you youtube for these amazing recommendations! Love from India!

Ahh, the good old divergent double sum (or integral for that matter), where iterated sums taken in different orders give different results. Always be careful, Fubini's theorem says that coinvergence of double sum (integral) guarantees the convergence of both iterated sums (integrals), but not the other way around!

Probability is where my knowledge is constantly taken behind the woodshed by my lack of understanding.

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.

The thing is that the expectation or the expected value is a concept which is probabilistic in nature and not deterministic!! The expected value may give a closer estimate for a very large number of turns rather it is expected to give a closer estimate. However, with just 2 turns, the expectation is not reliable!! The law of large numbers!!

I had never heard of this paradox before, and for the first half of the video I was left wondering how you broke maths and if it's actually true that the grass IS always greener on the other side! Great explanation of both the puzzle and fallacy. Surely though the obvious answer is simply to steal both envelopes - switching be damned ;)

It's obvious from the beginning - there's no point in switching, cause no new information is given when opening the envelope.

Knowing the probability of each set of envelopes completely changes the nature of the problem.

Okay I am still not satisfied. Let's go back to the concrete cases. Here expected value makes sense. When I open the 1 Dollar envelope switching has an expected value of 9 Dollars. So I should switch. And switching from the 10 Dollar envelope still has 24 expected value. Nothing funny going on here. The fact that for any concrete envelope I should switch to maximize profit remains paradox to me. Let's say I open my envelope, observe the value inside, calculate the expected value from switching and decide to switch. But before doing so memory gets tampered with and I cannot remember the exact number in the envelope. Is switching now not advantageous for me anymore, because my situation is equivalent to not opening the envelope in the first place? I am confused.

Assuming you didn’t know anything about what possible values could be in each envelope and only knew that one was 10x the other, you could still reasonably infer that the expected value of switching was 0. Based on your explanation, it can’t be mathematically defined, because there is nothing to base it on, but when you have nothing to base it on doesn’t that mean it’s always 0?

The problem with this “paradox” is that you’re comparing envelope A against the average value of envelope B. So of course envelope B will always be the superior choice. However, when comparing envelope A against either a worse result or a better result in envelope B, then the logic remains 50/50…which is what common sense says it should be. TLDR - to compare your envelope versus the hypothetical average of the other is completely illogical.

I think a very important point is missing from the explanation here. The expected value of money in an envelope (left or right) is infinite. The rest follows from this. For example your expectation of profit is the difference between expectation values of the left and the right envelopes. Since both of the values are infinite, the expectation value for the profit is undefined (infinity minus infinity is undefined). That's it.

“Just like some animals have legs and others don’t, some random variables have expected value and some don’t.”

I'd like to point out that your examples of "reasons" that influence your choice, beyond the expected profit value being positive, didn't include any examples that were not symmetric. This gives the false impression that such external goals will always lead to symmetric results as well. My example, to add to the mix: You **need** $900 for some urgent expense or coveted purchase. This gives you a cut-off value rather than a smooth curve. If the participant opened the envelope and saw $1200, he should keep it and *not risk* switching to $120 ("bird-in-hand"). If, on the other hand, he found $100, then he should switch because keeping it will not help and switching gives the _possibility_ of getting $1000 which would be a successful outcome. For values outside of the 90-9000 range, it doesn't matter. We can bring in other psychological factors for these cases.

"Your envelope has $40. That means the other envelope has either $4 or $400, and on average, it has $202." Uhhh I don't think that's how it works. With this method, using the dollar amounts, you are essentially unfairly "weighting" the probability of getting more money. Because the absolute difference between any number _times_ 10 will always be larger than that number _divided_ by 10. All this is really saying is that you will always have _more dollars_ to potentially gain than to lose. But that has nothing to do with the probability that you _will_ gain vs lose. That is obviously 50/50. This isn't a paradox, this is plugging irrelevant numbers into a problem and then being confused why it doesn't make sense.

1:00 Well, if I open the envelope and the amount of money is *not* a multiple of 10 (of the smallest currency unit), I know it must be the smaller amount.

This is a very thorough explanation, but I think I have come up with a more intuitive/simpler one. Basically, in the real world, there aren't infinite possible values. There is some maximum finite value that the envelope could possibly contain. That might be $1,000. It might be $1,000,000,000,000,000, but whatever that value is, if you apply the rule "switch no matter what" there's always some non-infinitesimal chance that you are screwing yourself out of the largest possible payout. The negative expected value of a maximum value envelope will be exactly equal to the positive expected values of all the other envelopes combined despite the fact getting that envelope might be ridiculously unlikely.

Since we don't know anything about the distribution of x, the Maximum Likelihood principle says x is uniform over all values including negative ones. If x is negative, then 5.05x is LESS THAN x. If you calculate the return of switching envelopes now, you find that the expectation of switching is 0.

My goodness this is the answer to the question I was ruminating for years. I've known both the loose and rigorous definitions, and I do know that if the series is not absolutely convergent, the expectation is not defined. And yes, the reordering of not absolutely convergent series is something that was taught very early in my math classes. I have failed to connect the dots. Thank you so much. I've played with simulations under the rigorous definition and tried to prove that if one plays this game infinitely many times, the log expectation ratio of winnings is zero. The ratio of winnings is the amount you've got, vs the amount you could have gotten cumulatively. The log of this quantity is an interesting random variable with a finite distribution symmetric around zero. Ultimately, this was a wrong approach, but I learned so much from it.

As soon as you said the thing about adding up the conditional expected values multiplied by their probabilities I had one of the biggest "aha" moments I've had in a long time. Incredible reframing of the paradox as an old problem with infinite series

I still got one question though: from your explanation it looks like it is still paradoxically better to switch AND not switch in case you check the content before switching. If i'm not wrong, even if E(Y) does not exist, and therefore E(Y) is not equal to E(Y|X=x) for any x, it doesn't change the fact that the sum of E(Y|X=x) for any x is >0 and

Pick envelope. Other envelope has 5.05x as much. Switch. Now the other envelope has 5.05x. Switch again. Now the other envelope has 5.05x. ... Infinite Profit.

I didn't know this paradox existed and now I'm questioning everything.

Dude this was extremely helpful. I hope you make a sequel with more about this paradox.

When I think about it, this might actually be describing two different scenarios. It doesn't seem like it, since all that's changed between the two scenarios is knowledge of the money in one of the envelopes. But I contend that this knowledge, and the way the problem is framed, describes two different scenarios. And, accordingly, the different scenarios have different solutions. In the first scenario: The participant starts with nothing. They are given the option of two beneficial outcomes. Either they gain x dollars or gain 10x dollars. The participant will not lose any money. The participant will also not go home empty handed. The chance of each outcome: 50/50. In the second scenario, one of the envelopes have been open, and its contents are known. The second scenario starts off assuming that the contents of the known envelope are the participant's by default. So, in the second scenario, the participant is starting off with x dollars. The participant can wager those x dollars to go for an unknown amount. the unknown amount is either 10x dollars or x/10 dollars. There's a 50/50 chance of each outcome. Should the participant risk it? That's very different. My guess is this explains the discrepancy. It also explains why the option to switch or keep changes depending on which envelope is opened, as the answer changes depending on the dollar amount revealed, but does not change depending on whether the participant is holding the envelope with the known amount or not. Although, this made me come up with another scenario that combines the two: Let's assume a person is a player on a game show. The player is given an envelope that has money in it. The host holds another envelope. The participant is told that they can keep the money in their envelope, or they can exchange their envelope with the host's envelope and take the money in there instead. They are told that either the host's envelope has ten times the amount of money than player's envelope has, or it has ten times less money than the envelope than the player has. The host (correctly) assures the player that there's a 50 / 50 chance of either outcome. Should the player switch their envelope with the host? Does switching or holding make a difference? Although I'm no statistician, when I think about my hypothetical scenario more, one can look it is like this: The player will leave with x dollars. This is the lower amount between the two envelopes. This is certain. But, they have a 50 percent chance of winning an additional 9x dollars. Nothing will change this. It doesn't matter which envelope the player is holding. It doesn't matter if the contestant switches envelopes or not. It does not matter if the contents of one of the envelopes are revealed. The player is guaranteed to win x dollars and has a 50% chance of winning an additional 9x dollars. So switching does not matter in this case.

1. yes, summing a sequence is important in the order of it...HOWEVER, it does not change the overall equality. If you come up with a different number, then you have purposely negated another infinite series of values. In the harmonic for ln(2) it is specifically +-+- ordered and described....so if you take 1 subtraction followed by 2 additions, then you are pushing 1 subtraction behind for each set, and eventually you build an infinite summation pool that will immediately swing the answer to its correct state. MEANING, write a program that will automatically tag the missing fractions as you go along in a queue stack (FIFO), then plot the sum of that queue stack alongside the sum of the sequence you are doing. You will find that convergence to 3/2 ln(2) will result in the convergence of the queue stack sum will be -1/2 ln(2)....when you add those together, it will be ln(2) because that's the true harmonic series set. 2. you are correct that the expected does not exist, however there is the utilitarian expected which is a subjective evaluation that changes how you see the game. For example, offering the poorest person the opportunity, it may be more advantageous for them to swap any envelope less than $100 because their conscience ability to gain back any possible perceived loss at that small amount is easily washed away compared to what they could afford in their means on any profits, while anything above $100 is quite a bit of money for them not to pass up. But if you were to offer this to a billionaire, they may find it more advantageous to swap any envelope containing less than $100 million because it is literally chump change compared to the profits they could seek, while $100m still is enough "free money" to allow them investment opportunity to gain the rest without much effort. This means that the best thing for a person to do is to think of a number they are willing to stop at no matter what, and have the willpower to follow through ("deal or no deal" is a prime example). According to my past records of stock market returns, projecting an amount into a 10 year future based on those valuations, my number on a 10x game would be swapping anything less than $12,000.....but, as I said, it's all subjective.

I have been shown so many videos on the Monty Hall problem, and all concluded with "your intuition is wrong! you should always switch! it doesn't make sense! maths is crazy!" which is like ending a mystery novel with "the detective did it." on the last page, and no further explanations. Thanks for unfrustrating me. Much appreciated.

The simple solution is that before you pick an envelope, there’s a 50:50 chance that the larger amount is in the right envelope. After you have picked it, there is still a 50:50 chance that the larger amount is in the right envelope. After you open and observe the amount in your envelope, there is still a 50:50 chance that the larger amount is in the right envelope. The probability doesn’t change just because you pick and observe one envelope. The paradox only appears once you start using flawed methods to try to determine which strategy is the best.

Since the possible amount of money is continuous, you probably need to find a function that describes the density of probability and integrate it to get an even more rigorous proof

utility is not linear with $. so, if the amount is small, you switch (expected value, as now you have a probability distribution). If the amount is large (say, a billion and you're not Bazos), you hold, as the difference to you between a billion and zero is huge while the difference between a billion and ten billion is "whatever". That's why it matters if you see what's in the envelope.

This is beautifully argued. Thank you very much: you have done me a great favour.

A crystal clear explanation. Not simple, just very clear, and that is much more valuable. Thank you!

best - and most satisfying - breakdown of this paradox i've ever seen

I'll just pick whichever one is thicker.

Great video, it cleared up a lot of questions! It didn't answer the main original question though: should you switch when maximizing the expected money received? The answer "should" be "it doesn't matter" based on the symmetry. The "expected profit" (in whatever less rigorous sense, but "average" sense) from switching "should" be zero (based on the symmetry). I am still confused! One thing which I am surprised you didn't mention is that the expected money received by playing this game is +Infinity, because probabilities are halved each-next-right-column, but prizes are 10x-ed, so the infinite sum is +infinity, which partially explains why switching doesn't matter: you switch from one +infinity to another +infinity

20:09 This is incorrect. Since the regular harmonic series is at the border of convergence, both of these series converge to a limit. The second series is a geometric series and we can use the formula S=a1*1/(1/r) for r=1/2, a1=-1/2 to show that it converges to -1. The real requirement is absolute convergence: the sum of the absolute values of each term is finite. As previously mentioned, this does not hold for the harmonic series - it is infinite. Really, as another comment mentioned this whole problem is ill-defined, as there is an infinitesimal small chance of getting more money off to infinity. Since there is a lower bound to the amount of money in the envelope but no higher bound, we use tools without properly addressing this. If the series truncated as some high value, we would have an analogue to $1 where we always want to switch and we would find the expected profit to be zero.

2:00 After you switch, by this reasoning, you should switch back. And back again. And again. There's a reason we don't develop statistical models from one respondent.

I am sorry, but this video misses the point of the paradox - or rather, artificially replaces it with a different point. The original paradox makes no assumptions about the decision-making process of the envelope sender. This version completely and arbitrarily constructs the decision-making process of the envelope sender, while also, assuming that they have infinite money and infinite computing power (to choose from a truly unbounded 1/2^n). Which leads to a paradox, because, duh, playing games against someone with infinite resources and trying to analyze their actions can lead to paradoxes. Decision-making processes of letter-senders with finite amounts of money and finite computing power will *always* have a concrete preference of switching/not switching for any given amount which is trivial to compute if the decision-making process is known; there will be at least one case where you always switch (when the values in your envelopes is minimal) and one case where you never switch (when it's maximal) regardless of the probability distribution; the rest depend on a particular probability distribution for all the options. The numbers will also always add up in such a way that beforehand, the situation is symmetric. For example, if the sender has equal probabilities to send A: 1 and 10, B: 10 and 100, C: 100 and 1000, and D: 1000 and 10000 dollars, then you should always switch for 1, 10, 100, and 1000, but never for 10000. However, if the probability of B is 85% and of the rest A, C and D is 5% each, then you should not switch from 100, while switching from 10 is much more profitable. In each case, expected loss from switching in all scenarios is equal to the expected gain from switching in them; and beforehand, the situation is symmetric. However, not knowing the sender's decision-making process does not invalidate the problem or make it unsolvable. It's quite simple: the ONLY thing we know are values of the envelopes *relative to each other*, and *not* to the distribution the sender has. One envelope contains X dollars, and another contains 10X. This is true regardless of the sender's wealth and creativity which we will assume nothing about. This is true regardless of the range or distribution of possible X. You do not know whether you open an X or a 10X envelope. But in case you've opened the X envelope, you will lose 9X by switching, and if you've opened a 10X envelope, you will gain 9X. This is also true regardless of X. So, in total absence of knowledge of the decision-maker, the expected gain of switching is 0. (Note that, in a certain sense, it takes an infinite amount of knowledge to describe the decision-maker with an infinite table. Not according to Kolmogorov, but still. At the very least, it would requite infinite evidence to confirm their infinite resources and brain.) In practice, you could always do slightly better than "0 expected value" after the opening the envelope - by assuming some reasonable things about the decision-maker; extremely small or extremely large or perhaps "boring" amounts of money are less likely. If you open 1000 dollars, chance of 100 is a bit higher than of 10000, if you open 1 penny, chances are, the other envelope doesn't contain a cheque for 0.1 penny.

You're gonna end up popular, this is a great channel

In the "vague" case, I'm pretty sure the expected profit from switching/not switching is zero. Once you have selected an envelope, you are either in possession of x or 10x. Thus, there are two possibilities that arise from switching: You either gain 9x, or lose 9x, with probability 50% for each. In other words, your expected profit for switching is 0, since 0.5*9x+(0.5*-9x) =0.

This explanation is inadequate. You showed that the expected value diverges for a particular generative model for the values. But the original question does not assume any particular generative model. Let’s say you the generating distribution is uniform on the integers from 1 to 1000. That is, the probability that the smaller envelope contains any of the numbers is 1/1000. At the boundaries where you see a number less than 10 you of course know for sure you have the small envelope, and if you see a number greater than 1000 you know for sure you have the big envelope. Likewise if you observe a number not divisible by 10. But let’s say you pick and you get 30. By bayes rule, the probability that the other envelope has 300 is Pr(Other is 300, this is 30) / (Pr(this is 30) = 0.5 (1/1000) / (1/1000) = 0.5. So you still end up with the switching paradox. Before looking, both envelopes are equivalent. Now you’ve looked, you have an equal probability of being small or big, and yet the other envelope is more attractive by expected value. The core of the issue is not convergent series, or a deficiency with expectations, but rather a deficiency with linear utility functions. There’s no getting around the fact that this decision depends on subjective utilities. In the original presentation of the problem, before looking in an envelope, the problem is solely about choosing the bigger envelope. You could define the utility function = 1 if you end up with the big one, = 0 if you end up with the little one. Then, once you reveal a value, so long as you have the same utility function, there’s no paradox-the envelopes remain equivalent. The paradox comes from the object of the game changing. If you have a linear utility of money (which you don’t), then yes you should switch. If you have a logarithmic utility of money (much more realistic), then the envelopes are once again interchangeable in multiplicative problems like this. Any decision rule over some probability space following minimal axioms of coherence can be represented as an expected utility. As such, talking about decisions under uncertainty without utility is inadvisable. The loss probability rule you discuss here is fine, and can be represented as a simple binary utility, but I wouldn’t recommend making decisions based on win/loss probabilities in general since in a multivariate setting order statistics can become intransitive, resulting in completely incoherent decisions. Further, expectations are not just “nice sometimes, when they converge.” Expectations (not means or wrt any particular functions, but rather the linear operator itself) are the only single point summary of probability distributions that are well defined, that is to say don’t change with irrelevant reparameterizations.

An interesting alternate example to the Monty Hall problem, where in that case switching *is* the better strategy.

This is an example of misuse of statistics. You have a 50/50 chance of picking the better envelope. There is no "average winnings". You get one shot and you either win or lose. You are not a population of stats. You are one data point.

The expected amount of a random amount from the infinite set of all positive numbers is greater than any finite number. So any given number when the envelope is opened is always infinitely smaller than the amount in the other envelope. If the number is not simply drawn at random from all random numbers with equal probability, the other calculations change

This is maths that sounds so smart that it actually becomes stupid.