The Two Envelope Problem - a Mystifying Probability Paradox
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- Опубліковано 28 вер 2024
- 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
This is the best explanation of the paradox I've ever seen. Thanks!
You are welcome, and thanks for the compliment!
At the beginning I thought "Grab the two envelops and run away"
I have seen many youtubers explain this, but I have never seen them really point out a key important fact: under the rules of the game expressed by @formant the expected value of the envelope is infinite. In other words he assumes on average that the envelopes contain infinite dollars. If you don't make that assumption, your mathematics don't go off the rails like this and everything behaves nicely.
It not a paradox it makes sense you just don’t understand it
Very good explanation! Thanks!
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.
You are most welcome. Glad you enjoyed the video.
Particles interact independent of time and space. Wrap your mind around that. Two linked particles, one on earth, one on Mars... interact faster than the speed of light... and we all know that nothing can move faster than light speed... we think.
@@mmercier0921 What happens in entanglement is that a measurement on one entangled particle yields a random result, then a later measurement on another particle in the same entangled (shared) quantum state must always yield a value correlated with the first measurement. Since no force, work, or information is communicated (the first measurement is random), the speed of light limit does not apply (see Quantum entanglement and Bell test experiments). In the standard Copenhagen interpretation, entanglement demonstrates a genuine nonlocal effect of quantum mechanics, but does not communicate information, either quantum or classical.
@@error.418 I understand nothing but I'm glad there is a counter argument lol
There's a reason it's "the first time you had it explained". You hadn't. What this dude explained is a similar problem he just thought of himself.
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.
What is it, to know? What is knowledge?
@@Zhengrui0 och get over yourself. You know exactly what knowledge is.
facts, information, and skills acquired through experience or education; the theoretical or practical understanding of a subject.
awareness or familiarity gained by experience of a fact or situation.
Your question is one of those philosophical fallacies that seems like it is "digging into what really matters" whereas really you're just poking around linguistics
@@aceman0000099 please write some more paragraphs in the comments about how *I* need to get over myself
In this case it's our intuition about math that fails, and our intuition about money that success.
What really happened is that we forgot about some variables and oversimplified the computation.
This is very similar to the money change scam, where you trick someone into giving you an extra $10 when making change from a $20.
I would say yes and no, because my intuition tells me conflicting things. If I look at the problem one way, my intuition says that it can't matter if I switch or not, because the problem is symmetrical, and because knowing the contents of one envelope shouldn't change anything. But looking at the problem another way, my intuition tells me that if I picked the envelope with $100, I should switch, since the potential gain ($900) is much greater than the potential loss ($90). But that's why it's a paradox.🙂
I feel like the best strategy to avoid the "sense of remorse" is to not switch, and to be sure to never learn what was in the other envelope :P
Yeah, just be happy you got some free cash.
Or to switch, and be sure to never learn what was in your original envelope. 😉
Just like the train and switch problem. If I never know, I won’t be bothered. xd
The best strategy is to check first, and then not switch, unless it’s 1$, in which case switch.
the real solution is to find who is in charge in envelope land and show up with a bribe
but when you show up with your bribe to the ceo of envelopes, be sure to bring TWO envelopes, one with a much larger bribe...
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 would argue, even more, that it's not just puzzling, that is exactly that Paradox. we don't need to calculate E(switch) to reach it. so, in my understanding, the video doesn't really solve the Paradox.
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.
Paradoxes can't exist. Every paradox is not really a paradox. No surprise.
Yeah, very interesting. Personally I would switch if the amount of money did not matter - getting $0.10 or $1 does not matter, but $100k or $1M does matter. I mean to say that one can select the criteria "does this money have an impact", where amounts of money over x do have an impact, and amounts of money under x do not. Someone with debts might decide $10k should be kept, whereas someone with a mortgage could switch in hopes of changing that with $100k
@@Jordan-Ramses I feels like sometime people come up with mathematical paradoxes as an excuse to not properly solve the problem.
@@Jordan-Ramses Tons of paradoxes exist and have existed for millennia - you are just making a huge mechanical and causal assumptions about the world - neither of which is supported nor can they be proved
I would definitely disagree that this is a simple 50-50. For example, if I saw $10 in the envelope, I would swap every time. Intuitively, it's like a double-or-nothing, except it's 10x or 0.1x. If you want a practical answer to the question, you should use a finite set of possible values for the money in the envelope, even if you aren't sure of the upper bound.
Suppose you have a hunch that the maximum amount of money that could be in the envelope is 10 million. You can model this scenario the same way that is shown in the video, except using a truncated geometric sequence for the possibilities. If you do this, you will find that your expected profit from switching is always positive unless the envelope you open contains 10 million. Of course, this is assuming you were right with your hunch. If instead the maximum value was 1 million , then swapping from 1 million will always give you 100,000. If you take a weighted average of these two possibilities, you an get a better picture for the optimal strategy. The bigger the value you see in the envelope, the lower the average expected value of swapping, until eventually it goes negative.
Also, this is assuming that your marginal value of money is constant, which for most people is not the case. You might decide that the value of 10 million dollars is less than 10 times the value of 1 million dollars. By removing the infinite series from series, you can calculate whatever expected value you like without issue.
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 imagine it’s just got to do with their monetary value. It’s more likely someone hands you an envelope of $10 than an envelope of $10million
I don't think this is a variation of Monty Hall problem, that's a completely different mathematical problem.
@@imbaby5499 it’s certainly very similar, I would say it’s more like a cousin to the Monty Hall problem, I wouldn’t say they’re “completely different”.
- They both consider how probability changes when new information arises.
- They both have you stick with or switch a choice.
- They both have confusing reasoning behind their probabilities.
@@Astrussy it's mathematically a different problem. Monty Hall is about the change in sample space upon discovering that one of the envelopes is empty.
@@imbaby5499 I’m not disputing that they’re mathematically different: I’m just saying they aren’t “completely” different, they are in the same family of thought experiments, not even as brothers or sisters, but as cousins.
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.
I think you could create the same paradox even if you tapered things such that the expected value of an envelope was finite. The traditional way of doing it with multiples of two instead of 10 gives rise very easily to a finite expected value where the paradox still holds.
@@stevenglowacki8576 multiples of 2 is still infinite EV. For example 1/2*1 + 1/4*2 + 1/8*4 + ….
@@stevenglowacki8576 And multiples/exponent of 2 is the lowest number where you get an infinite EV and that’s also the first number where you can blindly switch. It actually doesn’t matter if you switch when you see anything but 1 in the envelope. For example if you see an 8, then 2/3*4+16/3=8. Any lower exponent means that the only case when you want to switch is when there is no downside, so you can’t say that switching is beneficial even without opening the envelope, which was the paradox.
I’m going to explain the paradox better than the video. Your math is wrong because you are making 2 variables out of this questions when really there should be one. Because both envelopes contain the same potential outcomes they are in fact the same fuckin variable! Stop making the problem in X & Y format, you make yourself look like a novice mathematician when I can obviously tell you have some advanced college mathematics education.
If your going to be stupid and solve the question using a 2 variable equation, then why don’t you first define what X is and how it’s separate from Y. If your going to say something stupid like X is the 1st envelope & Y is the second then you obviously need your head examined because this isn’t a permutations questions but a combinations question (since order doesn’t effect the outcomes.) So please try and define what the 1st or 2nd envelope is in this question since both envelopes share identical potential outcomes. Once you realize both envelopes are the same variable there is no paradox as to why they share identical probabilities.
Btw, I’m an expert in combinatorics and have even developed my own theorem that bridges the gaps between combinatorics and statistics; so don’t feel about I’m educating you on this subject.
P.S. I’m aware that you set X = to the value of the 1st envelope and Y = to the expected value. When I’m talking about X & Y I’m strictly speaking about how you immediately defined the envelopes as separate variables without even doing any explaination as to why, as if you had X & Y variable equations in your mind as the solution from the very beginning like a college student who gets stuck solving X & Y equations to pass an exam that there brains are on autopilot whenever they encounter a problem that looks like an X & Y variable problem.
Also you are wrong about the “order issue” as you like to call it in regards to dealing with continuous variables that approach infinite. The order does not make, what caused the change in its value as it approaches the limit of infinite and negative infinity is that you changed the sequence in which the infinite series was constructed. By making it so that you 2 positive fractions for every 1 negative fraction you have increased the the value of the sums infinite series only because their is now a delay in the amount of negative fractions to positive fractions. At any point, if the series was stopped their would always be twice as many positive fractions and negative fractions so NO SHIT the value sum has changed! Just because a series goes on forever does not mean it will eventually contain the same number of a negative fractions as positive. The only way the two series could equal each other is if somehow the series runs out of positive fractions and just starts using the other half of the negative fractions which haven’t caught up to the positive fraction values, but then by definition it’s not an infinite series!!!!!
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!
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.
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.
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.
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.
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 Monty Hall is a classic 'amateur math youtuber' video, because they think it sounds cool to make a video where "woah the probabilities don't add up to 1! your brain is wrong!" when really, they add up just fine, and intuition works just fine, they're just misrepresenting the odds and fuzzying the details.
The very basic version is 66% of the time, you pick a bad door. Therefore Monty MUST open the other bad door. Therefore the last door has the prize. They like to fuzz the math by acting like the final 2 doors are a 50/50 and not influenced by Monty's knowledge of the prize door, but that's just not true.
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.
BINGO! The expected value of an envelope is infinite. That means on average you get infinite money in an envelope. When you play games where you average infinite amounts of money, don't expect the usual rules of probability to work. Of course since infinite money doesn't exist, this game can't exist. Garbage in, garbage out.
@@MrPerfs what? Just because both are infinity doesn't mean they can't be compared. In this case their ratio happens to be undefined, but this is not a general case.
Also nothing about a hypothetical scenario not being possible irl makes it "garbage in, garbage out". That's just not how anything works.
@@randomnobody660 but they are actually identical infinities. Even if they weren't, it doesn't matter. Applied over the entire game, an envelope contains infinite amount of money. The "paradox" here is only rooted from a pedantic insistence of not continuing to apply the EV to the infinte series.
EV switch = EV dont switch
This is exactly a "garbage in garbage out" situation. The base assumptions and expectations are loaded to the question in a way that sets it up for fail before it even begins.
As long as there are infinite options for the amount of money an envelop can have in it, the answer to the question "Should you switch" will be undefined.
However, of course if you constraint yourself to a set of money amounts and their respective conditional probabilities, you can solve the game.
Garbage in garbage out is an actual sentence used a lot in the field and it holds here. Garbage assumptions = garbage results.
This is as much of a "paradox" as zeno's "paradox".
@@GameNationRDF You are missing the point.
Unless I miscomprehended, the OP here more or less asserted any comparison between infinities is automatically invalid ("...infinity minus infinity, which is indeed not well defined.")
The the gent I replied to on the other hand seems to be claiming any scenario that cannot happen irl is automatic "garbage in". ("since infinite money doesn't exist, this game can't exist. Garbage in, garbage out.")
I should also mention that your claim ("As long as there are infinite options for the amount of money an envelop can have in it, the answer to the question "Should you switch" will be undefined.") is also incorrect. In the scenario in the video, the expected value is only undefined because the series diverge (easiest tell is that each value is larger than the last, those can't ever converge). Further, as this video explained, the rearranging of terms messes up the sum of infinite series only when both all positive and negative terms sum to infinity.
With both in mind, it's very easy to see that we can have the expected value be the sum of a well behaving, in fact absolutely converging, series simply by reducing the ratio of the money in the two envelopes. In those scenarios, there will be a well defined, non infinite, expected value for the amount of money you can get out of an envelope, and the "should you switch" question is easily answered with "if you got less than average", all this despite there still being infinitely many, and arbitrarily large, options being available.
To be sure, I don't have a problem with the conclusion. I understand when the expected value is infinite/undefined all sorts of things we take for granted (here mostly addition being commutative) are broken. However even towards a correct conclusion there are correct reasonings and incorrect ones.
@@randomnobody660 Indeed, with a convergence things are resolved. The 5^n series here explodes however with a proper and quickly declining proobability distribution the game can be solved as well.
Thanks for pointing this out, I missed it myself.
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.
Exactly channels like these are promoting gambling which is absolutely stupid. The creator here should play this game with mob loan sharks. He will wind up only 34% dead...
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.
Math is not unreasonable and does not prove this paradox.
This paradox is a mere misunderstanding of averages.
Knowing the probability of each set of envelopes completely changes the nature of the problem.
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
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?
The probability not decreasing seems pretty absurd.
So 100€ has the same p as 10€ and the same p as 1€.
But then 0.1€ has the same p as that. And 0.01€. 0.001€, 0.0001€ and so on.
In fact the probability that a reasonable amount (let's say between 1Cent and the sum of all money that ever existed) of money is in the envelopes is basically 0.
That our mathematical Intuition breaks down in such a scenario doesn't seem weird to me at all.
I think he did he good job by simplefying the problem so that the apparent positive expected value was preserved.
If the probability of each pairing were equal, and there were an infinite number of possible pairs (1+10, 10+100, 100+1000, etc.), what would be the probability of getting a particular pair?
@@Ockerlord Obviously there would be some lower and upper limit to the money (say, 0.01 for the lower and 1000000 for the upper). However, that alone doesn't necessarily imply that the probability is decreasing, especially not given the statement of the original question.
I think that it could be equally valid for all values between those extremes to be equally likely. Obviously the strategies of "Keep if it's 1000000 and trade if it's 0.01" are trivial, but otherwise the paradox seems to remain unresolved with this scenario.
@@AdmiralJota See my other comment. "Probability shouldn't decrease" and "Highest possible value is unbounded" aren't necessarily synonymous.
This is another problem of the paradox. I actually struggled with this problem for a long time. The problem lies on that there isn't a uniform distribution of unbounded values. It is simply not possible to pick between say numbers (1,2,3,... to infinity) with an equal probability. This is very counter-intuitive but once this is understood I think you will understand why there needs to be a distribution like described in the video. The distribution that the video picked is not necessarily everyone's pick (and there will indeed be different results based on the distribution one picked, but again, a uniform distribution is not pickable as it does not exists), but by picking this specific distribution this video shows a scenario that actually the two-envelop paradox exists even if the "uniform distribution of unbounded value" thing is taken into consideration so the paradox is much deeper than I thought.
Very good video. But I have a remark: You have shown that the paradox arises from the fact, that the expected value for 'profit from switching' is not defined. But the reason that this expected value is not defined is that you have chosen a probability distribution for the amounts in the envelopes which also has no expected value. (It would calculate to be infinity and in some sense this is the heart of the paradox: If you expect an infinte amount, then no observed amount is ever enough...)
If you chose any realistic distribution with a defined expected value, the whole paradox vanishes. You can choose a finite number of envelopes or even an infinte number like in your example, just with different probabilities/amounts. One example: For amounts of (1,2) with p=3/4, (2,4) with p=3/16, (4,8) with p=3/64, ... the expected value is 9/4. Given that you are allowed to open the first envelope, the expected profit for switching is 1 for an observed amount of 1, -2/5 for an amount of 2, -4/5 for an amount of 4 and so on. So you should switch if you observe 1 and you shouldn't for every other amount. The 'overall' expected value for 'profit from switching' is defined, its value is 0, as we would expect.
And the same is true for every realistic probability distribution (i.e. one with a defined expected value) for the amount in the envelopes. Depending on the distribution, there is a defined strategy when to switch given you know the exact probabilities and are allowed to open the first envelope. The 'overall' expected profit from switching will be exactly 0, so if the player isn't allowed to open the first envelope the situation is symmetrical as it should be.
I have one question: Say there’s only a finite amount of possible letters (which in reality, will be the case) and I get to see the letter distribution beforehand. Then we can calculate the expected value, and it will still be positive and negative. The problem doesn’t depend on infinite series. So there’s another catch to find.
I agree. The ev is positive for switching after you see 1 envelope, but you will lose money 2/3 of the time based on the distribution he had with a finite amount of envelopes with a known maximum cap. So the best strategy if only given one chance to play is to switch at all smaller amounts and dont switch when the amount becomes too much for you personally to risk losing.
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.
the assumption is that the amount of money available is infinite. You can do all sorts of mathematical absurdities when you treat infinity like a number.
No. Math is a paradox in itself. The sets of all sets does not contain itself hence it's not a set of all sets.
If you look at it in the finite case for finite series. You are better switching except of the last value where you will lose a lot. In the infinite case, you are better switching except for the "limit" value which occurs with infinity low probability but with infinite loss.
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?
Or if we get to the opening the envelope point. You can assume the other envelope averages 202. that's fine to assume. but in the sense of it valuing more than your envelope is still just a 50/50. it doesn't suddenly disconnect itself from the rest of the information. So I don't know how its a paradox. its always a 50/50 on getting the "larger number". and if we add the human element, you just win for playing. either envelope is free money, it just comes to if x satisfies you.. if it does, great keep your envelope. if it doesn't, gamble and switch, you weren't happy with it to begin with.
@@SeriosSkies92 The paradox is that it is simultaneously always better to switch, and not.
@@Winasaurus except it isn't. "the average is a bigger number" is flawed to begin with.
@@SeriosSkies92 Except, it isn't. Something with a 2/3 chance of losing some money, but a 1/3 chance of gaining a lot of money has an average expected payout. The envelopes just represent the average expected payouts of the potential options.
@@Winasaurus you never lose money. You always gain money.
It's always a 50% chance the envelope could have more or less money than your envelope. There is no shift in the percentage like the three door problem.
Dude this was extremely helpful. I hope you make a sequel with more about this paradox.
Look at my explanation to the paradox, I also have an explanation for the different values of the infinite ordered sets. This video actually has several false explanations which hurt the math community.
One envelope contains x. The other envelope contains 10x. On average, they contain 5.5x.
As there is no way to tell the difference between the envelopes, you can only use the average. So, you pick one up. It contains an average of 5.5x.
The other envelope also contains an average of 5.5x.
So, what value in switching?
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
Exactly. This was left unanswered and confuses me a lot.
here are a few easier to think about situations. suppose we have two envelopes, each containing an independent uniform integer between $1 and $10. our strategy would be quite simple in such a case, if we open our envelope and see 1,2,3,4,5 we will switch, otherwise we won't. similarly in the opposite direction if we open the other envelope.
now let's imagine we have in our envelopes 2 independent identically distributed random variables with infinite mean (for example drawn from a half cauchy or pareto(1) distribution). we can imagine that whenever we open an envelope and see a finite value x, we are ALWAYS in some sense disappointed. we were expecting infinity and any finite value must necessarily fail to satisfy us. even if we see $10000000000, theoretically, we can still expect more from the other envelope. therefore if we opened our envelope we switch, if we opened the other envelope we don't switch.
similarly, in the problem described in the video, both the envelopes have infinite expected money in them. so the paradox of us always wanting the unknown envelope over the known finite one has some sense to it (although the dependency makes it a bit trickier to understand).
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
To explain this without math or infinities: Opening the envelope matters in a way that just pretending to open the envelope does not. They aren't the same problem anymore once you open an envelope. And just pretending to open it and doing the thought experiment doesn't change anything. Because both envelopes are closed, it's still 50/50 if you don't open them.Opening the envelope reveals information about itself, which locks it in place. It is now no longer possible for you to treat the two envelopes as identical, because one of them is open, and the other one is different. If yours is the one that's open, switch, because the uncertainty and chance of winning was placed into the other.
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
If you are allowed to look then the symmetry is broken all the other terms in the series disappear and you now only have finite number of terms and being finite means the expected value now actually exist. Looking what is inside means it is no longer an infinite series because you know exactly what is inside of one of the envelop and the possibilities of what inside the other envelop becomes finite.
And yet, if you try to run a simulation, you will see that switching and not switching will net you the same amount of money, no matter what you know...
@@Beregorn88 A naive simulation won't actually. What the expected value not existing means is that a finite number of trials doesnt converge predictably as the number of trials increases. If you compare switching and keeping in some large number of trials one will outperform the other because the low probability outcomes have an outsized effect on the result. Doing several sets of trials will show that which outperforms the other is a pure 50/50 chance, but thats not the same thing as the strategies converging on the same value.
I think you are correct and that's why the video didn't really solve the Paradox. if you analyze the problem by looking at your envelope, you will always switch (because E(Y|X=x) > 0) and if you analyze it by looking at the other envelope, you will always stay.
No need to conclude that E(Y) > 0
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.
I don't understand why the first pair ($1 and $10) has probability of 1/4 each, and second pair has probability of 1/8... Aren't they all equally probable?
I didn't know this paradox existed and now I'm questioning everything.
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.
What you just described is actually described in behavioral economics. Marginal utility of money is decreasing so when ppl are in the profit meaning they gain to win they act in a risk avoidant manner and try to avoid losses. However, when ppl are in the loss they become risk seeking and are willing to take bets to recoup the losses they have incurred.
It seems like there is a definitional problem with the use of the term "choose". If you pick envelope 1 and keep it then you chose envelope 1. If you pick envelope 2 and then switch then again you chose envelope 1. In each case, you chose envelope 1.
Suppose you are allowed to switch as often as you want. In that case, you might go back and forth 100 times but in the end, you will choose either 1 or 2. The preliminary choices are not real choices because they mean nothing.
There's a flaw in the reasoning. If you're shown one envelope, the expectation given is not for the other envelope, but it's the expectation for each of the envelopes. They have the same expected value, so knowing what's in one does not give you better information about knowing what's in the other in the sense of which is more valuable.
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.
x is the same number. If I look in my envelope, x, and see 10, then the other envelope either has 100 or 1. x hasn't changed, and isn't different. It's still 10.
@@Winasaurus Suppose the two envelopes contain $10 and $100. Either you have $10 and the other envelope contains $100, or you have $100 and the other envelope contains $10. Either way the expected value is the same irregardless of whether you swap. What you don't have is an envelope that contains $1. You're taking two separate problems and trying to combine them, and the result is nonsense.
@@jamessmith2522 That only in the case where there's 2 envelopes with only 2 values. Which doesn't apply to this particular paradox. You need the potential to be in a different 'pair' worth more. With only 2 possible values, of course it doesn't work, and it's pretty ridiculous to imply it's even related. We already have an example of a game with 2 possible outcomes and no amount of switching changes the odds. The paradox is specifically referring to a game where there's far more than 2 outcomes.
@@Winasaurus Didn't watch the whole video. Have now. With the series of pairs of envelopes there's one strategy: choose the envelope that you haven't seen. Your knowledge of the contents of one of the envelopes alters the probabilities, giving you the advantage. "Always switch" and "never switch" is nonsense; there's no paradox there. If you don't know the contents of either envelope then it doesn't matter if you switch or not; the odds are the same. You choose one of a pair of envelopes at random; if you happen to choose the top one you make a profit by switching; if the bottom you make a loss. But the probabilities and the numbers all cancel each other out and it comes to zero. I agree with his comments about harmonic sequences; I've seen a few good "paradoxes" built on them; but he has no reason to be producing an infinite harmonic sequence, because you can do the problem with simple probability.
@@jamessmith2522 The paradox is that any unopened envelope is, on average, worth more than any other envelope. Which means in the case of 2 unopened envelopes, they're both worth whatever the other is, times some factor. Which is the paradox because obviously a = b*? and b=a*? must be mutually exclusive aside from a=b=0. So even though it makes sense an unopened one is worth more, 2 unopened ones are actually not worth more.
the big issue with the paradox is that the expected gain of picking any envelope is infinite so when you try to figure out if you are better off switching you are comparing infinites. In reality the amount of money is bound to the ability of the gamemaker to pay you, so the distribution is finite and falls down quickly, also you get diminishing returns when the value is very large (is it really 10 time better to be offered 10^101 dollars than having 10^100? In both cases you won't be able to spend it in a lifetime, and you won't be able to have it in physical form)
Another example of the mathematical world and the real world diverging.
IRL, the answer is to look at the money in the opened envelope, consider that money to be yours, and then rephrase the problem. Now that you have $100 (or whatever the amount is) do you gamble YOUR $100 for a 2/3 chance of losing it versus a 1/3 chance of gaining $900 more? Your decision shouldn't be based on math, but on factors such as whether you enjoy gambling, whether you can afford to lose the money, or if you're in some unusual situation that turns normal reasoning on its head (maybe rent is due tomorrow, and $100 is useless to you but $1000 would save you from eviction).
I really loved your video. It really made me think. But in part I disagree, this is more of a gambling question than a switching question. The way I see it the first envelope you open is a gift to you; say $100. Then the question is do you want to spend $100 for a chance to win $1,000 on a coin flip? I would always choose to give up $100 to win a $1,000 for a 50/50 chance. If you do this over and over again you will end up being rich.
The general envelope problem: For each trial t there is an amount a(t) > 0 and a multiplier m(t) > 1 and one envelope has a(t) in it and the other has m(t)*a(t) in it. You have a 50% chance of having initially selected a(t) and a 50% chance of having selected m(t)*a(t). With no switching this is your final selection. With switching you have a 50% chance of ending up with m(t)*a(t) and a 50% chance of ending up with a(t). These are the same odds. You can switch or not, switching is not relevant. (Equivalently instead of saying m(t)*a(t) we could just as easily have referred to an amount b(t) > a(t).) End of story. Why complicate it with nonsensical analysis and integer series? Why even call it a paradox?
Honestly I think people just like how it sounds, paradox.
Thank you, finally someone has posted the correct answer. The entire video is utter garbage because at @3:19 he states, with zero justification at all, that the $1/$10 pair has a 50% chance of being selected. Which comes from no where, and is completely wrong. The remaining 25 minutes of the video are wrong.
I’m going to explain the paradox better than the video. Your math is wrong because you are making 2 variables out of this questions when really there should be one. Because both envelopes contain the same potential outcomes they are in fact the same fuckin variable! Stop making the problem in X & Y format, you make yourself look like a novice mathematician when I can obviously tell you have some advanced college mathematics education.
If your going to be stupid and solve the question using a 2 variable equation, then why don’t you first define what X is and how it’s separate from Y. If your going to say something stupid like X is the 1st envelope & Y is the second then you obviously need your head examined because this isn’t a permutations questions but a combinations question (since order doesn’t effect the outcomes.) So please try and define what the 1st or 2nd envelope is in this question since both envelopes share identical potential outcomes. Once you realize both envelopes are the same variable there is no paradox as to why they share identical probabilities.
Btw, I’m an expert in combinatorics and have even developed my own theorem that bridges the gaps between combinatorics and statistics; so don’t feel about I’m educating you on this subject.
P.S. I’m aware that you set X = to the value of the 1st envelope and Y = to the expected value. When I’m talking about X & Y I’m strictly speaking about how you immediately defined the envelopes as separate variables without even doing any explaination as to why, as if you had X & Y variable equations in your mind as the solution from the very beginning like a college student who gets stuck solving X & Y equations to pass an exam that there brains are on autopilot whenever they encounter a problem that looks like an X & Y variable problem.
Also you are wrong about the “order issue” as you like to call it in regards to dealing with continuous variables that approach infinite. The order does not make, what caused the change in its value as it approaches the limit of infinite and negative infinity is that you changed the sequence in which the infinite series was constructed. By making it so that you 2 positive fractions for every 1 negative fraction you have increased the the value of the sums infinite series only because their is now a delay in the amount of negative fractions to positive fractions. At any point, if the series was stopped their would always be twice as many positive fractions and negative fractions so NO SHIT the value sum has changed! Just because a series goes on forever does not mean it will eventually contain the same number of a negative fractions as positive. The only way the two series could equal each other is if somehow the series runs out of positive fractions and just starts using the other half of the negative fractions which haven’t caught up to the positive fraction values, but then by definition it’s not an infinite series!!!!!
They're purposely complicating the question early on to then go back and explain why it's wrong. I suppose it gets clicks
@@kylekristian27 I’m going to explain the paradox better than the video. Your math is wrong because you are making 2 variables out of this questions when really there should be one. Because both envelopes contain the same potential outcomes they are in fact the same fuckin variable! Stop making the problem in X & Y format, you make yourself look like a novice mathematician when I can obviously tell you have some advanced college mathematics education.
If your going to be stupid and solve the question using a 2 variable equation, then why don’t you first define what X is and how it’s separate from Y. If your going to say something stupid like X is the 1st envelope & Y is the second then you obviously need your head examined because this isn’t a permutations questions but a combinations question (since order doesn’t effect the outcomes.) So please try and define what the 1st or 2nd envelope is in this question since both envelopes share identical potential outcomes. Once you realize both envelopes are the same variable there is no paradox as to why they share identical probabilities.
Btw, I’m an expert in combinatorics and have even developed my own theorem that bridges the gaps between combinatorics and statistics; so don’t feel about I’m educating you on this subject.
P.S. I’m aware that you set X = to the value of the 1st envelope and Y = to the expected value. When I’m talking about X & Y I’m strictly speaking about how you immediately defined the envelopes as separate variables without even doing any explaination as to why, as if you had X & Y variable equations in your mind as the solution from the very beginning like a college student who gets stuck solving X & Y equations to pass an exam that there brains are on autopilot whenever they encounter a problem that looks like an X & Y variable problem.
Also you are wrong about the “order issue” as you like to call it in regards to dealing with continuous variables that approach infinite. The order does not make, what caused the change in its value as it approaches the limit of infinite and negative infinity is that you changed the sequence in which the infinite series was constructed. By making it so that you 2 positive fractions for every 1 negative fraction you have increased the the value of the sums infinite series only because their is now a delay in the amount of negative fractions to positive fractions. At any point, if the series was stopped their would always be twice as many positive fractions and negative fractions so NO SHIT the value sum has changed! Just because a series goes on forever does not mean it will eventually contain the same number of a negative fractions as positive. The only way the two series could equal each other is if somehow the series runs out of positive fractions and just starts using the other half of the negative fractions which haven’t caught up to the positive fraction values, but then by definition it’s not an infinite series!!!!!
"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.
I guess the paradox would resolve itself if you just use the median instead of arithmetic mean. In that case, for £1 you would need to switch and for all other values you would need to hold, and it would simply invert if the other envelope is opened, i.e. if other envelope has £1 you hold and switch for all other values.
Exactly.
It is how it works, because this is using expected profit, the typical method of determining optimal playstyle for many games. Moves that result in high returns and low losses are preferred. This is how chess robots make their moves, by analysing expected values. If a move puts them at a significantly higher advantage, they take that move.
In this case, the 'advantage' is the amount of money. Profit is payout*chance of payout. This is the strange case where the playstyle is undetermined, as expected profit doesn't work properly with the infinite values in question.
@@Winasaurus But an AI playing chess is doing so in a real world scenario. It's method allows it to win games.
In a real world scenario, switching envelopes will not result in an increase of gains regardless of the expected value. The entire "paradox" is using a framing device to manipulate you into thinking there is expected value at all. That framing device being that you already have one of the envelopes in your hand. Switching envelopes is exactly the same as choosing one or the other; the fact that you hold one in your hands is meaningless. It's smoke and mirrors and nothing more.
@@jaypee9575 Consider this as a real-world scenario, and attempt to find the optimal strategy.
The issue stems from the fact that the odds are technically impossibly good, and break common sense. Any unopened envelope has, on average, more than any other unopened envelope. That's the root of the paradox. Even if you have none in your hands, the odds issue still exists.
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.
Ohhhh. A paradox! I had to rewind three times. I understood the opening line up be: "Here is a pair of ducks"! 😂
😂😂
NOT A PARADOX
@@DanteHaroun Ah, must be ducks then. They're probably hiding in the envelopes.
Best video seen all YEAR ! Subbed.
So the best approach is to switch the given pair of envelopes to another pair (repeat few times depending on available pairs) and then select one of the two
You can conclude that the value of switching is undefined, because if switching was positive, and you were allowed any amount of switches, it would create a loop. No particular switch in such a loop is better than any other, so you might as well do none.
Interestingly enough, if you are allowed to look at the envelope, then for any *finite* group of values it is worth it to switch.
For example, if I decide to switch only if I see between 1,000-100,000,000 in the first envelope, then on average I will end up gaining a positive expected outcome
What a deeply unsatisfying resolution. The original problem was not an infinite series.
The answer being a measure of remorse is such a mood
I just discovered your wonderful video analysing precisely that paradox. I have some thoughts about it that I would like to share.
Firstly, the symmetry argument seem to demonstrate that in fact, the expected value actually exists and is zero. And you may - should - attain this value by a particular ordering of the terms in the infinite series.
Secondly - or, more precisely, moreover, there is a similar problem in quantum electrodynamics (and also in quantum chromodynamics, though much more acute in the latter case due to the high value of the coefficient replacing the fine structure constant value 1/137.06 of electrodynamics). All the essential computations in this theory rely on an infinite series, which is always absolutely divergent, and Richard Feynman and all the theoricists devised a way (the "renormalisation") to set the order of the terms so that the series actually converges. But as in the case of the two envelopes - or of any infinite series absolutely divergent - if you choose a different order, you may obtain a different result. And there are disputes on the ways to renormalise - or at the least to justify these.
And there is in physics another occurence of the same problem - though settled much more easily - the Madelung constant calculation. There the way to set the order of terms seems undisputable, which the observations and measurements confirm.
I suspect that this phenomenon is very common in the nature - for example, in the chaos theory - and explains its complexity and the seemingly endless hole in which every part of physic is trapped when trying to lift the veal on the basic mechanisms. Even the standard model - one of the rock solid dogmas of the third century physic - is put on question by now.
When you encounter a paradox, it is just an assumption error.
So going with 5.05x > x is not the right call to make. Where is the issue in this approach? It's simple: we make the mistake to calculate the gains AFTER the pick, not before. So we compare 5.05x to x but that's wrong because we did NOT pick x. The sum of both envelopes has to be 11x. But 5.05x + x is 6.05x... so we are missing something, obviously. If we pick something it's either x or 10x, so what we pick is 5.5x to begin with, regardless the envelope, cause each envelope does not only have 50% chance to be picked but also a 50% chance to be filled with either amount. So we end up having 5.5x (11x/2) in both envelopes (which finally sums up to our 11x, hooray). So we have to compare 5.5x to 5.5x in the end... and we find out: it does not care at all. Which was infact the obvious result to begin with.
Case closed.
PS: I really wonder that so many people did not realize that case A had 11x (10x+1x) and B had 1.1x (1/10x+1x) but we filled the enveloped with 11x in total, so the case of 1.1x in total was obviously not possible from the origin. And that A-B and B-A were always right calls... that is impossible. And the end up calling it a paradox instead of being just clumsy in math... :D What should make it super obvious that this is totally wrong: chain your assumption: if a switch is 5.05x times the value... switch again... and again... and again... you will switch to infinity. So OBVIOUSLY the 5.05x are wrong in the first place... if you do A-B-A and your result is not A... well... something went wrong. And why are people sure they picked x but not 10x, but at the same time, they say oh both cases are the same, let's skip one... :D So many nonononos. If you would do a math program - and you do it properly - pick 10^6 times A and 10^6 times B and the avarage will be 5.5x what you get. And half of it you switch. Still 5.5x. Running code but not actually trying to see if the paradoxon is visible by coding... so funny. :D
Elegantly done. I hope you do more in the future!
Please filter out some of the lowest frequencies from the audio next time for a more pleasant listening experience.
The problem with the initial example is that x represents 2 different numbers. the options aren't x/10 or 10x, it's simply x or 10x. One option is to lose 9x, the other option is to gain 9x. This supports the neutral outcome that our brain intuitively tells us is right.
Excellent! Please keep explaining how we are so easily confused, or any other thing that comes to your mind. Thanks!.
Instead of P(losing > 500) you could instead try to maximise P(choose the better envelope) and again get that the two strategies are equivalent.
The probability is just as likely to go the opposite way. With one being a positive number and the other a negative number.
Lost at 8:22. There is a probability of 1/2 that it is the 1:10 case and a probability of 1/4 that it is the 10:100 case, so it is more likely you are in the 1:10 case so you should not switch.
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.
No you're missing the key point... In your example you're not in general better off when you switch given a known amount in the second envelope. Because in 50% of the cases you already know you have the bigger envelope and you're actually worse of (that's the part that gets lost when you're dealing with infinite sums)
Why do you still have the switching paradox? Doesn't that require that it seems you should switch regardless of the value you would see in your envelope? In your example you can just calculate the exepcted value of switching given what was in your envelope and sometimes it is positive, sometimes negative.
I’m going to explain the paradox better than the video. Your math is wrong because you are making 2 variables out of this questions when really there should be one. Because both envelopes contain the same potential outcomes they are in fact the same fuckin variable! Stop making the problem in X & Y format, you make yourself look like a novice mathematician when I can obviously tell you have some advanced college mathematics education.
If your going to be stupid and solve the question using a 2 variable equation, then why don’t you first define what X is and how it’s separate from Y. If your going to say something stupid like X is the 1st envelope & Y is the second then you obviously need your head examined because this isn’t a permutations questions but a combinations question (since order doesn’t effect the outcomes.) So please try and define what the 1st or 2nd envelope is in this question since both envelopes share identical potential outcomes. Once you realize both envelopes are the same variable there is no paradox as to why they share identical probabilities.
Btw, I’m an expert in combinatorics and have even developed my own theorem that bridges the gaps between combinatorics and statistics; so don’t feel about I’m educating you on this subject.
P.S. I’m aware that you set X = to the value of the 1st envelope and Y = to the expected value. When I’m talking about X & Y I’m strictly speaking about how you immediately defined the envelopes as separate variables without even doing any explaination as to why, as if you had X & Y variable equations in your mind as the solution from the very beginning like a college student who gets stuck solving X & Y equations to pass an exam that there brains are on autopilot whenever they encounter a problem that looks like an X & Y variable problem.
Also you are wrong about the “order issue” as you like to call it in regards to dealing with continuous variables that approach infinite. The order does not make, what caused the change in its value as it approaches the limit of infinite and negative infinity is that you changed the sequence in which the infinite series was constructed. By making it so that you 2 positive fractions for every 1 negative fraction you have increased the the value of the sums infinite series only because their is now a delay in the amount of negative fractions to positive fractions. At any point, if the series was stopped their would always be twice as many positive fractions and negative fractions so NO SHIT the value sum has changed! Just because a series goes on forever does not mean it will eventually contain the same number of a negative fractions as positive. The only way the two series could equal each other is if somehow the series runs out of positive fractions and just starts using the other half of the negative fractions which haven’t caught up to the positive fraction values, but then by definition it’s not an infinite series!!!!!
I'd switch either way, just to satisfy my curiosity. Idc if i gain less money
I can't believe I've watched the whole thing. Very well done.
Not bad. The problem with the original paradox is that you have too little information about the distribution to apply valid probability theory. There's a limit to how little information you can have and still apply valid probabilistic calculations. Although, the Bayesians might have something to say about that.
In the second example, you have a distribution with an undefined (infinite) mean. So, again any calculations of expectation are not necessarily valid.
Another thing you should consider is whether the person is going to shrug their shoulders at $40 and go for $400 reasoning that $40 is 'nothing' so they may as well go for $400.
If you can look one the envelopes, decide whether it is close to the expected budget for the experiment. If it is well below your estimate, choose the other.
If you cannot, grab an envelope and run before the experimenter changes their mind.
Short explanation: The use of simple expected value is incorrect because the envelope is locked in. If you look at from gain perspective - let envelopes have X and 10X, so if you switch, 1/2 time you will gain 9X; 1/2 you will lose 9X, so E(Gain) = 0.
Long explanation: E(other envelope) is NOT 1/2(X + 10X). E(other envelope) =
1/2 * E(other envelope has 10X, given your envelope has X) +
1/2 * E(other envelope has X, given your envelope has 10X). For example,
E(other envelope has X, given your envelope has 10X) is not X because you actually lose 10X and gain X if you switch, so you need to look at the totality of gains from both envelopes.
Shorter explanation:
Intuitive way to solve - well, it's 50/50
Mathematical wat to solve - well, we make it this way around, now we walk this way, but actually we can't, but if we look this way... Yeah, it's 50/50
The expected profit is always > 0 as long as you don't have to pay for the envelope and if the content of the envelope is > 0. Now let's imagine that the choice is "You open one of the envelopes, I open the other. If mine has a larger amount, you have to pay me the difference. If mine has a smaller amount, I pay you the difference."
That's what the profit odds are for. Swapping from 100 to 10 is 'losing' 90, swapping from 100 to 1000 is 'gaining' 900. Hence the profit calculations leading to strange outcomes.
Just an idea... You are using the average to compare the possible "profit" by switching.
I propose that the problem is an exponential one. And that is why we can't wrap our head around it.
Let me explain: The money amounts differ in the exponent. 10^0, 10^1, 10^2, and so on.
Let's say you picked $10, that is 10^1. You are told that this is the case. Switching you can get either 10^0 or 10^2 dollars. And now take the number in the exponent and average it:
(0+2)/2=1
There you have it. You are neither making sure profit nor loss switching, but the chances are equal.
Unless you pick the $10^0 envelope, of course. But does having a lowest amount not also mean you should have a highest amount? Why isn't there a $10^-1 envelope? If the amount is infinite in its amount >$1, shouldn't it also infinite
The expected amount of money in an unopened envelope is infinity. So as soon as you open one envelope, the expected value of both envelopes becomes infinitely smaller.
Conclusion: don't ever open them and just pick one at random.
So -- the solution to this problem depends upon your personal definition of the "marginal utility of money". I can buy that entirely. The solution varies with the individual. Great video!
before you even start explaining past 2:11, I have a hypothesis where the symmetry comes in: it's weighted by the probability that the person giving you the envelopes had access to a given amount of money, right?
The assumption that probabilities diminish with increasing money values is arbitrary and unjustified. How does the math work without that assumption, or if a different assumption is used?
Well that was surprising
I thought not switching was the best strategy
My thinking behind this was
If I'm switching then it means I'm thinking the other envelope has more money(as in it was part of higher probability set, by switching now we are saying we think we are in the lower probability set)
but as per the probability distribution we have seen, let's say you see that your envelope has 100 dollars. Then by choosing to switch we are saying that, that 100 dollar belong to 100 and 1000 dollar set which has lower probability than 10 and 100 dollar set
I see the faulty reasoning that makes the Paradox seem like switching or not switching provides a positive expected value. But I disagree that the expected value is undefined. The original problem is that one envelope has x and the other has 10x and you don't know x. If you pick one that is a 50-50 chance, so no matter what you see you can only say you have a 50-50 chance of that number being x or 10x. The other envelope will have the converse or a 50-50 chance of that number being 10x or x. So the expected profit is 9x and -9x in equal amounts or 0. And if you simulate this that is exactly what is observed. You cannot make an infinite series out of it since the original problem is that one envelope has x and the other has 10x, just because x could be anything.
For example if x is 10, and you don't know it is 10, then you pick an envelope that has 10 and switch then you get 100, or if you picked 100 and switched then you would get 10. Do the switch or don't switch forever with infinite people who never know beforehand what x is and you will see both strategies will fluctuate around 0. So the expected value is clearly 0, not undefined.
The choice only matters if the total amount is appreciable pick the thicker envelope.
Wow! Brilliant explanation. More videos, please! 😃
I love this video, but it's hilarious that the 'resolution' to the paradox is by saying that you can't define its answer
As soon as i saw the thumbnail i had my answer and after watching the video i still hold it.
Unless there is a limit (like 1) your choice doesnt change anything at all. Its always a 50/50 if you dont know the answer. Even if there is 1/2, 1/4 possibilities of numbers showing, that probability only matters for the one you got, you could consider yourself lucky for getting 1.000.000 and 10.000.000 as options, but losing value is out of your control, and thats actually the point of the whole switching or not.
I never understood this, symmetrical structure of problem immediately shows that expected value of profit from switching must be zero. Even if we cannot define it mathematically!
This is brilliant!
And it’s crazy what we can do with random variables without expected value damnn
I am not able to understand this simple point.when we went about with the problem with average money strategy,why do we switch because the average is more than x?
Sometimes your voice gets lower and murkier, like if you'd hit a vape or sigarette every other sentence. That's really hits me in the ear every time.
But other than that - great video!
You have to go by risk of ruin. If you pick one that has 5 million, you should keep. If you have one with 5 dollars, you should switch.
If someone really needs money they should not switch because that would be gambling when not in a financial state to do it. If someone is comfortable switching is the funner option.
Always switch. 10-1=9 so -9 or 100-10=90 or -90. So it is always better to switch because if you are wrong you lose 9, but if you dont switch and your wrong, you lose 90.
If i was ever actually in a situation like this where i gotta pick I'd only switch if i feel the amount isn't life changing money. For example 10K is nice money to just be handed to you but i don't think in the long run being up or down 10K in my total life time is that noticeable of a difference. Now if it was something like a million dollars then I'm not switching because even if there is a possibility of more it just isn't worth losing something that already can greatly improve your life
I don’t see how you have resolved the paradox? We have been left with:
If you don’t open any envelop the situation is symmetrically so there should be no difference between the two envelopes, but if you open your envelope and observe the amount then you are better off always switching.
They wouldn't ask me to switch unless I picked the good one. I'd keep it.
And people struggle with the Monty Hall problem...
Could you cover the Monty Hall problem?
Open one, don't switch, burn the other
I just want to know who's handing out envelopes full of cash. I want at least one of them, regardless of the amount.
Thanks for the captions
Isn't it simpler to just say the moment you put the two amounts in the envelope, it is a 50-50 scenario of an independent random variable... And end up at the same conclusion?
finally a statistics video that actually takes weight in to account
Here's a solution to the look first situation: first decide how much you want to win, then look.
If I want to win at least $500, then I should switch on seeing $100 and not on seeing $1,000, and no other value seen changes whether I win at least $1,000, so I don't switch on those.
Under these conditions, there is exactly one situation where my goal is reached by switching, and no situation where my goal is lost.
I think the solution to the paradox is to define a maximal value like 10^k and to evaluate the expected value. I did the math and the expected value of the gained money by switching is 0 independ from k. Now let k-->inf and there the paradox ist solved.
Was there any justification for only looking at the average amount? For example when you pick $10, you are more likely to have $1 in the other envelope than $100, so switching should return $1 more often than $100. Looking just at an average of +$34 profit only makes sense if we have unlimited pairs of envelopes that all give $10 as the one you pick. And how can this be possible if we only have a 50% chance of picking $10 from any pairing that contains $10. I think this is a fallacy disguised as a maths problem which is never a fun exercise.
More people MUST watch this.
another great video Yuval!