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

@FormantMath

2 жыл бұрын

You are welcome, and thanks for the compliment!

@defnotnaruto222

2 жыл бұрын

At the beginning I thought "Grab the two envelops and run away"

@MrPerfs

2 жыл бұрын

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.

@simonskelly3060

2 жыл бұрын

It not a paradox it makes sense you just don’t understand it

@ProfessorDBehrman

2 жыл бұрын

Very good explanation! 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

@Ashenshugura

2 жыл бұрын

Yeah, just be happy you got some free cash.

@mhelvens

2 жыл бұрын

Or to switch, and be sure to never learn what was in your original envelope. 😉

@Oblivion1407

2 жыл бұрын

Just like the train and switch problem. If I never know, I won’t be bothered. xd

@Solrex_the_Sun_King

2 жыл бұрын

The best strategy is to check first, and then not switch, unless it’s 1$, in which case switch.

@boldCactuslad

2 жыл бұрын

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...

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.

@FormantMath

2 жыл бұрын

You are most welcome. Glad you enjoyed the video.

@mmercier0921

2 жыл бұрын

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.

@error.418

2 жыл бұрын

@@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.

@Pukukupu

2 жыл бұрын

@@error.418 I understand nothing but I'm glad there is a counter argument lol

@Cuorion

2 жыл бұрын

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.

@Zhengrui0

2 жыл бұрын

What is it, to know? What is knowledge?

@aceman0000099

Жыл бұрын

@@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

@Zhengrui0

Жыл бұрын

@@aceman0000099 please write some more paragraphs in the comments about how *I* need to get over myself

@jeffwells641

10 ай бұрын

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.

@Paul71H

10 ай бұрын

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.🙂

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.

@clintonweir7609

10 ай бұрын

Right. The identity of your counterparty, along with some intuition, tells you more about the possible values than does any kind of math. If you don't believe they would be so kind as to give you 10x the amount you find in the envelope, stay. For example, if its your middle-class parents and it's $10,000 then you should probably stay, because they probably aren't going to give you $100k. And if the first envelope IS $100k, you should definitely stay, because they definitely aren't giving you a cool million. If you don't believe your counterparty would be so cheap as to only give you what's in the envelope, you should switch. If it's Bill Gates and the envelope contains $1k, you can guess he's probably a little more generous than that. If it's $100, you can be really confident the other envelope is the right way to go.

@Xonatron

10 ай бұрын

That’s a good point.

@clintonweir7609

10 ай бұрын

@@The_Real_Grand_Nagus Yeah. If they're going to pretend 1 quadrillion dollars is a possible value, then why not 1 quadrillionth of a dollar? Both values are impossible, after all.

@madmathematician4458

10 ай бұрын

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.

@madmathematician4458

10 ай бұрын

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!!!!!

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

@jonatankelu

2 жыл бұрын

Haha! Yes.

@mikailvandartel

2 жыл бұрын

Good one

@epistax4

2 жыл бұрын

clap

@guythatdoesthings4935

2 жыл бұрын

👏

@monkeybusiness673

2 жыл бұрын

So you actually GAIN 1/12 of a dollar...Nice!

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.

@grantofat6438

9 ай бұрын

The right strategy is: Grab both envelopes and run.

@JebidiahKrackedyetagain-xv9hc

9 ай бұрын

@@grantofat6438 Best information by far!!🤣🤣

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.

@davidbroadfoot1864

9 ай бұрын

Correct. This utilizes the concept of "utility functions" .. where not every dollar is equally valuable.

@blueberrymuffin4921

9 ай бұрын

I agree. But with your example, I would argue that it is worth switching. Because with a value that high, 10% of it is still 4m. 1000% of it insane, 400m. But I think I'd probably chicken out, better safe than sorry. It's not the best to be greedy.

@RyanNash20

7 ай бұрын

​@blueberrymuffin4921 Yep, it's entirely dependent on the value of the amount to the person. 6,000 in the first envelope would be a lot tougher for the average person. 60k is a lot of money for most, while 6k is a solid bit of money that could be very helpful, while 600 is fairly insignificant in comparison. While a multi millionaire would find the 6k to be pocket change and switch everytime, someone with no savings and bills piling up would have a hard time switching. I feel like if you take the value out of the money, then the paradox begins to make sense, but the paradox is created by our imperfect understanding and calculation of probability, not some unexplainable phenomenon. Kinda why schrodingers cat is a paradox, we just can't calculate probability perfectly, so we chalk it up to being both outcomes in some kind of superposition. Like the paradox only exists in our failure to calculate probability, not reality. Common sense states there's a 50/50 chance you will benefit from switching. The failure seems to be in the way we arrive mathematically at a conclusion that it's not. Or maybe I'm totally wrong, and I'm entirely open to being told why. I'm not some probability wiz, and I'm sure I am ignorant to more than I even realize regarding methods of calculation.

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

@madmathematician4458

10 ай бұрын

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.

@mikeyforrester6887

10 ай бұрын

@@madmathematician4458 Well you'd be helping if you actually pointed out what was fale then, it seemed fine to me.

@madmathematician4458

10 ай бұрын

@@mikeyforrester6887 I’m really confused. Do you not understand my explanation? Please read my comment over again, It’s pretty clear that the issue with the other guys analysis is that he thinks the 2 envelopes are separate variables when in fact the are the same exact variable which when you realize it it’s a “no duh” moment that they share the same probability!

@daliasprints9798

10 ай бұрын

Came here to say exactly this.

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.

@Arigator2

2 жыл бұрын

Paradoxes can't exist. Every paradox is not really a paradox. No surprise.

@boldCactuslad

2 жыл бұрын

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

@pi6141

2 жыл бұрын

@@Arigator2 I feels like sometime people come up with mathematical paradoxes as an excuse to not properly solve the problem.

@JamesJoyce12

2 жыл бұрын

@@Arigator2 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

@SillyPutty125

2 жыл бұрын

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.

you know what they say. the grass is always greener in the unobserved envelope.

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.

@blueberrymuffin4921

9 ай бұрын

Yes. Or another way to put it is: switching gets you an average of 5.05x what you have. But you already are holding an average of 5.05 more than the other envelope. It cancels out.

@terryboland3816

9 ай бұрын

So why don't you swap to the envelope with a higher expected amount?

@Mythraen

2 ай бұрын

​@@terryboland3816 If we assign the lower value x, then the higher value is 10x, and the average value is 5.5x, which they _both_ have. There's no expected value in switching from 5.5x to 5.5x.

@terryboland3816

2 ай бұрын

@@Mythraen And when I open the envelope so I know the actual amount inside. Is it rational for me to take a 50:50 gamble of doubling my money or halving it?

@Mythraen

2 ай бұрын

@@terryboland3816 That's a new problem. You now have information you did not have, the amount in the envelope you're holding. From a value-based perspective, it comes down to what you're willing to sacrifice on the chance it will give better rewards. For example, at $10, you probably won't care if you lose $9, but you will care if you gain $90. At $0.10, you probably won't care either way. At $10,000,000,000, you probably also won't care either way (I mean, you'll care some, I suppose, but who isn't going to be happy walking away with one billion dollars?) From a rational perspective, I don't think anything has changed. Since your envelope contained 5.5x on average, that means whatever it contains can be substituted for 5.5x. The other envelope thus contains exactly the same amount of money as your envelope... on average. But, let's look at it from the other perspective. Let's say you found $200.There is a 50% chance that the other envelope contains $20 and a 50% chance the other envelope contains $2000. From this perspective, you're risking $180 for an equal chance at $1800. If you treat it as a logic problem where the goal is to maximize your chance to have the largest amount of money, it looks from this like you should always switch as soon as you open your envelope. This is really weird, but at least it isn't an infinite loop. If you switch and then have the option to switch back so long as you don't open your new envelope, you're holding an envelope with an average value of $1010, which is significantly higher than $200. By this logic, you should always switch after opening your envelope. That seems wrong, like I'm missing something, but at least we've reached an end-point. I'll just leave this comment as-is for now. If I have any more thoughts on it, I'll let you know. I'm also curious what you think. Note, currently we have two perspectives, one says it doesn't matter if you switch, and the other says it's beneficial to switch. As long as you ignore the value judgement portion, it currently makes the most sense to switch after opening your envelope.

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.)

@elmeradams8781

9 ай бұрын

I don't understand, but I think you're on to something.

@Mythraen

9 ай бұрын

@@elmeradams8781 To put it as simply as I can, anything you can say about the value of the envelope you left behind is equally true about the envelope in your hand.

@chrisg3030

2 ай бұрын

@@Mythraen Yes, that's all there really is to it. So why do people get confused? Why do they fail to appreciate that the amount you left behind could equally well be 1/10th or 10 times the amount in your hand, making the switch equally disadvantageous as it was advantageous when vice versa. In other words if you happen to start with $1 or $100 and switch that for $10, instead of the other way round, then you make a bad bargain instead of a good one. Maybe people don't think of that because while you can imagine opening an envelope, finding $10 in it, then switching it, what does it mean to find an amount called "$1 or $100"? What could you do with that? PS I just thought of the answer to my own question at the end of the above. People don't think of the opposite switch, from x/10 or 10x to x, because opening the envelope doesn't face you with the need to calculate any probabilities. Since only x can be in the other envelope, if I open mine and find x/10, of course I switch for x. But if I open and find 10x, of course I don't.

@Mythraen

2 ай бұрын

@@chrisg3030 I believe the video showed why people get confused. There is a way to present it that shows the switch as beneficial. It's basically just a logic/statistics trick. The main part of the trick is that it's shown in only one direction.

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

@lfb6087

2 жыл бұрын

1. Put bank cards instead of real money 2. Paradox is back

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

@Ivankarongrafema

2 жыл бұрын

I'm disappointed that it will never realistically happen..

@PWLfr

2 жыл бұрын

that particular situation of course not but the premise of having to choose something random or keep something known, it happens all the time

@trentvlak

10 ай бұрын

You could invite all your math oriented friends to your birthday party, and make a bunch of pairs of envelopes, and present it to them as a party favor. The amounts wouldn't have to be high. You could do $1/$10 and $1/$0.10.

@madmathematician4458

10 ай бұрын

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!!!!!

The first paradox is only confusing because it's presented by arbitrarily assigning "x" to the chosen envelope. If you assign "x" to the one you didn't pick then the one you chose has an EV of 5.05x so you should never swap. If you take every combination of picking envelopes and assigning "x" to envelopes, you will just get an EV of "x" for swapping, which means swapping changes nothing.

@ajreukgjdi94

10 ай бұрын

I think the real lesson here is 2-fold. Learning how to phrase your question so that the results are more helpful is one thing, but learning how to interpret the results you got is another. The first part says that the average value of the other envelope is higher than the first. It is assumed that means it is better to switch. What it actually is shown is that you stand to gain more than you lose. In other words, it doesn't matter if you switch or not, you can't walk away with nothing.

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.

@Mythraen

2 ай бұрын

I'm not sure what you mean. You can definitely create a real-world scenario where you can't tell the difference between two envelopes and one contains 10x the amount of the other.

@efence4713

Ай бұрын

⁠@@MythraenThe problem lies in your ability to pay out arbitrarily high numbers. If you can’t account for even the .037% chance of the participant winning a trillion dollars or more, the entire problem is broken and the solution becomes more obvious.

@Mythraen

Ай бұрын

@@efence4713 No, the problem lies in the equivalence of the two envelopes. They are equal. Although, I guess it depends on what you mean by "the problem."

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.

Elegantly done. I hope you do more in the future!

Excellent! Please keep explaining how we are so easily confused, or any other thing that comes to your mind. Thanks!.

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.

@soberhippie

2 жыл бұрын

> because there's nothing I need that costs that much Not exactly. There's nothing I need that costs that much YET

@GemstoneActual

2 жыл бұрын

It matters what your planned projects cost; if you've got no projects, then it never pays to switch, because free money is free money, but, if you've gotta have at least $10k, for a project, then it pays to switch, for any amount less than that. You can't lose what you never had, but you can gain what you've already conceptualized.

@arinroday302

2 жыл бұрын

you are absolutely correct, but that wouldn't make for a good math question if it were relative like that, depending on person to person/. A solvable question would have to be concrete ......

@brunolevilevi5054

2 жыл бұрын

I think the same way but my values are way lower, I think maybe 10k I wouldn't "mind" losing, put upwards of 100k I wouldn't want to risk because its life changing money (even if not instant retirement money)

@kazedcat

2 жыл бұрын

This does not apply on the first scenario where you are not allowed to look inside the envelop. Looking what is inside breaks the symmetry because now you have one opened envelop and one unopened envelop.

6:28 "Well, there is only one option..." *Hands you an envelope containing a dime*

Great video! I watched it around a year ago, but it was so good I watched it again now! Thanks for the excellent explanation!

@madmathematician4458

10 ай бұрын

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.

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).

@Reashu

2 жыл бұрын

That doesn't seem to follow from the rules, can you explain why it does?

@omamba5105

2 жыл бұрын

@@Reashu Look at the chart at 5:14 - If you pick an envelope and the number revealed is 100, there is a 1/8 chance it is the 10:100 set and a 1/16 chance it is the 100:1000 set. Hence, twice as likely to be the greater of the 2 envelopes.

@SarzaelX

2 жыл бұрын

@@omamba5105 However by not switching, you actually lost money compared to the average outcome of switching. If you have 100, losing means you lost 90 dollars - while winning means you won 900 dollars. Even with a lower chance to win, the potential profit is large enough that you'd always be better off switching.

@omamba5105

2 жыл бұрын

@@SarzaelX How do I lose money I never had in the first place? As I originally said, the potential profit isn't what I'm playing for. I would prefer to bet on the more likely outcome, rather than potential profit. Switching to a lower value envelop would make me regret the choice way more than not switching when I had the lower value to begin with.

@rickyrickster1303

2 жыл бұрын

27:44 switching or not doesn’t matter

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).

@albertgarde5008

2 жыл бұрын

You're assuming a lot about the distribution here. E.g. if there were two possibilities: 99% probability of small envelope with 1$ and large with 10$ or 1% probability of small envelope with 10$ and large with 100$, then if you open your envelope and see 10$ you should not switch. So your strategy of only not switching if you get the maximum value does not hold. However, if you don't get a chance to look in either envelope then I'm pretty sure that you could prove that switching makes no difference for expected value as long as it is defined.

@thischannelhasnocontent8629

2 жыл бұрын

What you said is true for any number that is greater than the maximum divided by k, where k is the ratio between the two envelopes' contents. Not just for the maximum itself.

@madmathematician4458

10 ай бұрын

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.

@madmathematician4458

10 ай бұрын

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!!!!!

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.)

@madmathematician4458

10 ай бұрын

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.

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)

@DawnMandel

Жыл бұрын

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.

For the first paradox, who cares if the average is higher; you still have a ½ chance of the smaller number and a ½ chance of the larger number.

@Winasaurus

2 жыл бұрын

Because optimal strategies are based on expected profit. So it's good to look at a game like this through the lens of expected profit and see what optimal strategy it creates, as has been done with money games for years.

@CollieDog70

10 ай бұрын

If I said to you: Give me $10, and I'll toss a coin. If it's heads you win $100. If it's tails you win $1. That's a good bet to take. Lose $9 or Gain £90.

@ramaraksha01

10 ай бұрын

Yeah i don't see how average works in this case

@MarkusBohunovsky

10 ай бұрын

@@CollieDog70 That's not the right example though. It would be: Give me $10. I throw a die and if it's 5 or 6 (⅓ chance) you win $100. In all other cases (⅔) I keep the $10 and give you $1. If you know we'll play this game many times the advice is definitely to take the bet, but if you play it only once, it's not so clear. Now $10 (actually $9 of loss risk) may still be worth the risk, but say the risk of loss is something significant, let's say $ 90,000. Given that you only have one play and your chance of loosing two years of salary is 66%, does it really matter that your potential win is almost a million? The best advice may well be to not take the bet. The example can be expanded: let's say only one chance in a thousand to win, but the winnings are so large that an average profit calculation makes it worthwhile: Again, there will be a point where you should not take the bet, simply because it is unlikely you will win, no matter the potential size of the winnings (unless you get to play the game many times).

@user-op3zd5ke5c

10 ай бұрын

But profit much bigger then loss

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.

@FormantMath

2 жыл бұрын

Thanks, I am glad you liked the video. I absolutely agree with what you wrote, but let me just add that you need X and Y to be *independent* Cauchy RVs, in order to have (X + Y)/2 ~ Cauchy. Cheers!

@stevenglowacki8576

2 жыл бұрын

Yeah, i thought of the same distribution where the tails were so fat that despite being symmetrical and thus an obvious expected value of zero, it turned out to be undefined. I just didn't remember the name of it, due to 20 years of neglect of studying this subject.

@mattethebest1

2 жыл бұрын

One small point I'd like to make is that the CLT still holds since it requires that the distribution has all moments (the cauchy has no moments)

@madmathematician4458

10 ай бұрын

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 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.

@damienasmodeus928

2 жыл бұрын

this is a thing that doesn't make sense to me. The paradox would be still there, even if the series is not infinite.

@viliml2763

2 жыл бұрын

@@damienasmodeus928 No, it won't. If the series is not infinite, there will be a maximum number that you can possibly get, and if you see it, you know you shouldn't switch. In the dual formulation, in which you open the *other* envelope and try to prove that it's always better *not* to switch, seeing that number would tell you that you *should* switch. That large number with the opposite conclusion balances everything out.

@damienasmodeus928

2 жыл бұрын

@@viliml2763 yes, it will not mean anymore that you should switch without looking, but the paradox is still there after you look.

@bluerizlagirl

2 жыл бұрын

Whenever there appears to be a paradox in mathematics, the most usual explanation is a casual attempt to cancel infinity with another infinity as though it were equal to itself. Here, whatever amount was in the envelope to begin with (unless it is £1) was always more likely to be the larger amount than the smaller troubles since you can always switch the one-pound envelope; so just be grateful you got anything at all, not disappointed because you could have had more.

@tolkienfan1972

2 жыл бұрын

That depends on a version of the problem never given

Best video seen all YEAR ! Subbed.

The condition ¨the other enveloppe contains a higher average¨ applies to both, so switching doesn´t provide better odds. And if you´re allowed to look inside one enveloppe, it comes down to whether you can afford to take a gamble.

@IronicHavoc

2 жыл бұрын

The fact that the choice should be symmetrical is established early on, the point of the video is addressing the apparent paradox in the formal mathematics.

@STARDRIVE

2 жыл бұрын

@@IronicHavoc If I have a 100,000 mortgage, I'd settle for that 100k. Although the average is about half a million, there's still the possibility of getting just 10k and leaving me in financial trouble. That's what I meant :)

@madmathematician4458

10 ай бұрын

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!!!!!

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!!!!!

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?

@Astrussy

10 ай бұрын

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

@imbaby5499

10 ай бұрын

I don't think this is a variation of Monty Hall problem, that's a completely different mathematical problem.

@Astrussy

10 ай бұрын

@@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.

@imbaby5499

10 ай бұрын

@@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.

@Astrussy

10 ай бұрын

@@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.

Amazing video. I still had trouble understanding clearly what was going on with this paradox before i saw it, thank you very much!

Amazing explanation! Everything well chained together. Thank you!

This beautiful explanation lighted up the candle of curiosity I had for statistics that was once put out by college, thank you good sir! Probability can be truly beautiful

@madmathematician4458

10 ай бұрын

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.

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.

@EebstertheGreat

2 жыл бұрын

Indeed. Of course, it can't possibly matter what we do here. Both envelopes have the same finite expected value in that case, however you do it, so there is never a preference for switching. That's unlike in the paradox, where both envelopes have infinite expected value, making the question ill-defined. Still, it might be interesting to see what happens to the sum. Let's say the problem is set up like in this video, except that the maximum amount in any envelope is $10^100. That is to say, we only have 100 columns instead of infinitely many (so the top left number is still $1, and the bottom right number is now $10^100). If we don't adjust the probabilities, they will only add up to 1 - 2^-100, so to normalize it, we multiply all probabilities by 1/(1-2^-100). There is a probability of 1/(4-2^-102) that your envelope contains $1, so the profit from switching is $9. And there is a probability of 1/(2^101 - 2) that your envelope contains $10^100, so the profit from switching is -$9*10^99. Otherwise, there is a 1/(2^(n+2)(1-2^-100)) probability your envelope is less than the other and contains $10^n with 0 So the overall expected profit is the sum of all of these 200 profits, weighted by their respective probabilities. Since it is a finite sum, it doesn't matter what order we add them in. First, we factor out the 1/(1-2^-100) and the dollar sign to make the arithmetic easier. This makes the overall profit $ 1/(1-2^-100) [ (1/4)*9 - (1/2^101)*9*10^99 + Σ 1/2^(n+2)*9*10^n - Σ 1/2^(n+1)*9*10^(n-1) ], where the sums run from n=1 to 99. Checking with a calculator (like Wolfram Alpha) shows this sum is exactly 0. If you're bored, you can also play around with these sums on paper. Wolfram Alpha is probably just summing all 200 terms here, but you can be more clever. Remember that they are finite sums, so rearranging terms in whatever way is fine. It's possible to manipulate these to make everything cancel out. Also note that the constant factor at the front (to normalize the probability) is irrelevant, since it will get multiplied by zero eventually anyway.

@ariaden

2 жыл бұрын

@@EebstertheGreat The interesting question for the version with maximum amount is: How much do you win on average when you look (and switch unless you know you have the maximum), compared to how much do you win when you do not look (and switch always or never or randomly).

@EebstertheGreat

2 жыл бұрын

@@ariaden The expected value of each envelope is $1/(1-2^-100) [1/4 + (1/2^101)*10^100 + Σ 1/2^(n+2)*10^n + Σ 1/2^(n+1)*10^n] = [(1/4)+(5^100)/2+(15/16)*(5^99 - 1)]/(1-2^-100) = $200437884620256797810769319085246392679302912314358440022410559003621646641106021293771194368/36957847963250199322625 ≈ $5.42 * 10^69. So if you don't look, that's what you should get on average. If you get to look in the envelope, switching whenever you don't have the maximum amount but keeping the maximum, then the calculation is pretty much the same. For all the numbers except the max, you just swap the top and bottom rows, which doesn't affect their probability. The only change is to the last column, where the top right square has probability 0 and the bottom right has twice its usual probability. So we have to subtract $10^99/(2^101 - 2) and add $10^100/(2^101 - 2), which you can combine to $9 * 10^99/(2^101 - 2) = $12000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000/3380401600608611737324541881 ≈ $3.54 * 10^69. That means by adopting the optimal strategy of "look then switch unless max," you can increase your expected earnings from $5.42 * 10^69 to $8.97 * 10^69 (the extra .01 comes from rounding). That's a relative increase of 65.5%.

@ariaden

2 жыл бұрын

@@EebstertheGreat Nice. So in the version with maximum amount, looking at the value gives a significant benefit, even if the probability of actually seeing the maximum amount is extremely low. The big amount compensates for the low probability. I believe that means the version without maximum amount can be taken as a limit. The probability of seeing "infinte" amount is zero, but it does not mean it is not worth looking. It is a paradox, but not a surprising one, as both looking and not looking gives infinite expected value of winnings.

@EebstertheGreat

2 жыл бұрын

@@ariaden Right, if we set up the finite version this way (and remember, there are alternative ways we could do it), then looking and having the option to switch is a big advantage. Because in the overwhelming majority of cases, switching will earn a profit on average, but in a tiny minority of cases (i.e. only if you have the maximum amount), switching will result in a guaranteed colossal loss. So you benefit a lot by checking which case it is before making your decision. In the infinite limit, the probability you have the maximum amount is zero (indeed, there is no maximum), so the probability that you should switch is 100% (in fact, it is guaranteed). That seems absurd, and it kind of is, but only because the idea of two envelopes with infinite expected value is absurd in the first place. As a simpler example of this, imagine there aren't two envelopes, just a lottery with infinite expected value. After winning a million dollars, you are given the option to keep the money or play again. You should decide to play again, since the expected value is infinite. But no matter _what_ you win, you should always choose to play again, which seems to suggest you shouldn't even bother checking your winnings in the first place. But that's absurd, because the replay will be identical to the original game, so there is no reason to think the second play will be any better than the first before you know what the first outcome was. There is a reason most books on probability and statistics focus on distributions with finite means.

Its never been clearer than your explanation to this problem.Thanks you!

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

@ve4edj

10 ай бұрын

I'd be happy to lose 100 for a chance at 1k, but probably not losing 1k for a chance at 10k

@madmathematician4458

10 ай бұрын

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!!!!!

@joey2320

10 ай бұрын

​@@ve4edjinteresting! For me, 100k is my no switch amount. I'd risk the 10k profit for a chance at 100k and happily take my 1k if I lost, but I wouldn't do it at 100k.

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

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!

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.

absolutely top notch high quality content, thank you

Beautifully explained. I've probably read the (incomplete) law of total expectation over a hundred times and have dealt with absolutely convegent series countless times, yet I never made the connection between the two. Moreover, I didn't even make the connection the original paradox had to series. It was a great video.

@madmathematician4458

10 ай бұрын

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.

This is awesome! About half way through I was getting frustrated (partly as I’d forgotten a lot of the maths I’d learn at school) and was about to write a grumpy comment! And then I saw the rest until the end.. and it was exactly what I thought it should be.. so 1.5 hours of videos, relearning forgotten maths, frustration ended in satisfied correctness.. thanks for that!! Loved it!

@Ten_Thousand_Locusts

2 жыл бұрын

You were going to leave a negative comment because you're too stupid to understand what's being said in the video? Really? I think forgetting maths is the least of your worries, Jesus.

@LOSFOUFOUP

2 жыл бұрын

@@Ten_Thousand_Locusts And yet they ended up not writing a negative comment, unlike you.

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.

@Winasaurus

2 жыл бұрын

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 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

@HummerEffect

10 ай бұрын

how did he come up with those specific values? I miss this part too.

@davidbroadfoot1864

9 ай бұрын

There is nothing wrong with what he did. He simply provided an explicit scenario that was similar,, but more specific.

@yurrskiiii

9 ай бұрын

​@@davidbroadfoot1864 Yeah but you cant say the x value is random and then provide all these probabilities for what x and 2x are. Like why are there certain probabilities for x to be 10, 100, 1000 and it cant be anything else other than multiples of 10 if its supposed to be random. Seems kinda pointless.

@davidbroadfoot1864

9 ай бұрын

@@yurrskiiii He never said that the x value is random.

@yurrskiiii

9 ай бұрын

@@davidbroadfoot1864 ????? watch again from 0:19 to 0:28

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.

@jonb4020

2 жыл бұрын

"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." No. That's the error at the base of this alleged paradox. It is not! It has exactly the same probability.

@monkeybusiness673

2 жыл бұрын

@@jonb4020 Gosh darnit, you're right....

@guy-sl3kr

2 жыл бұрын

Yeah the whole assigning probabilities thing seems so arbitrary because it caused the situation that you're describing. And even after watching the whole video, it still seems arbitrary to me. I guess it was only for the sake of making the scenario mathematically rigorous? It seems like a completely different scenario to the original paradox to me though.

@DerIntergalaktische

2 жыл бұрын

@@jonb4020 Can you explain that? I don't understand why the assigned probabilities were also wrong. It seem fine to define the probabilities like he did. It seems fine to compute conditional probabilities the way he did. So why when seeing a 10, do we not know that it is twice as likely to be the higher number as it is to be the lower number?

@jonb4020

2 жыл бұрын

@@DerIntergalaktische Because it isn't twice as likely! It is either a) the higher of 1 or 10 or b) the lower of 10 or 100. There are no other possibilities. The whole alleged paradox is based on flawed logic. :-) The common sense is correct, the maths is fake.

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

@madmathematician4458

10 ай бұрын

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.

@Mythraen

2 ай бұрын

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?

Simply outstanding video, Great Job!

If you're supposed to switch in all cases, then your second choice should always be your first choice, but that then means your first choice is your second choice, so it doesn't matter what you choose. But if you are never supposed to switch, then the other choice becomes irrelevant, so again, it doesn't matter what you choose. This is the best explanation of the two party system I've ever heard.

@conniepr

Жыл бұрын

Haha!

@AutPen38

10 ай бұрын

If the grass is always greener on the other side, you should sit on the fence.

@madmathematician4458

10 ай бұрын

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.

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.

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.

@MrPerfs

2 жыл бұрын

the situation is much worse than having an infinite number of possible values (which can be cleanly handled in certain cases), the envelopes have infinite average values. Think about what the expected combined value of both envelopes are: There's a .5chance it's $11, a .25 chance it's $110, a .125 chance it's $1110... etc. Summing them up gets you $11*.5+110*.25+$1100*.125+etc = $5.5+$27.5+$137.5+etc=infinity. When you play games where on average you get infinite money you're going to run into some issues..

@hansj4832

2 жыл бұрын

@@MrPerfs Yes, that's the main problem in the paradox, though a simulation would fix that too because both computers and real life trials can only put a finite amount of money in each envelope. Having a maximum value in the envelope resolves the paradox since there is a case, however tiny, that you should not switch after observing since you know that the observed amount must be in the larger envelope. (Technically, a computer could just say "this envelope has infinite money in it", but it can only do that after reaching the maximum finite value it can hold. And there would be no reason to switch away from an infinity dollar bill.)

@RipleySawzen

2 жыл бұрын

When simulated, the answer you get will be different every time. Wildly different. Let's say you pull 16 envelopes. There's a 1/16 chance of pulling the $1,000/$10,000 pair. That gives you a 50/50 chance of going +$9,000 or -$9,000 in the end. Some may go +-$18,000 as well. You may think "Well, it'll even out in the end." but if you do 16 more runs, now you're getting random +-$90,000. The more runs you do, the greater the chance of some big dollar amount completely throwing off everything. The more simulations you run, the more you realize the numbers are just all over the place.

@hansj4832

2 жыл бұрын

@@RipleySawzen The idea is that you'd simulate this billions of times. Because computers have finite precision, there's eventually a point where a big number completely throwing off everything stops being a risk and you can pick up a trend. If you use a 32-bit number to seed your envelopes, the maximum envelope amount you'd end up with is $10^32 in the larger envelope, which would happen in approximately 1 out of 4.3 billion runs.

@Comradez

2 жыл бұрын

@@hansj4832 But no, the more billions of times you run the simulation, the more chance you have of amounts in googleplex dollars showing up to completely swamp the effect of all of the previous runs combined. It's like being allowed to run a martingale strategy with infinite money.

Wow! Brilliant explanation. More videos, please! 😃

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.

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?

@Ockerlord

2 жыл бұрын

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.

@AdmiralJota

2 жыл бұрын

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?

@Celia_Dawn

2 жыл бұрын

@@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.

@Celia_Dawn

2 жыл бұрын

@@AdmiralJota See my other comment. "Probability shouldn't decrease" and "Highest possible value is unbounded" aren't necessarily synonymous.

@cr1216

2 жыл бұрын

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.

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.

@MrPerfs

2 жыл бұрын

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.

@randomnobody660

2 жыл бұрын

@@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.

@GameNationRDF

2 жыл бұрын

@@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".

@randomnobody660

2 жыл бұрын

@@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.

@GameNationRDF

2 жыл бұрын

@@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 really thought this was going off into a dead end at first - the reolution I've always heard to the original paradox is simply "if you don't know the distrubutions of the possible amounts then you can't say that the expected value of the other envelope is 5.05 times what you have in yours", which I think is a completely fair resolution for the original statement of the problem, so I was worried when this video kinda skipped that step, but I'm really glad I stayed with it becuase I had not realised there was this whole second layer of cheating going on.

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).

@MrNicePotato

2 жыл бұрын

Well if you say that, then a "paradox" is something that by definition doesn't exist, since our universe is supposed to be logical. The whole point of the paradox is that the difference in money is proportional to one another, not linear. Even if the probability of a "better" or a "worse" envelope are both 0.5, the benefit of getting the better one far outweighs the loss in getting the worse one. Imagine the difference is 1 million times. You got an envelope with 1 dollar. Yes you could have just as likely got the better one already, but if you switch, the most you could lose is just that 1 dollar, or you could gain 1 million. That's where the paradox is.

@violetfactorial6806

2 жыл бұрын

@@MrNicePotato The paradox is that you can argue that either decision is better than the other using the same logic both ways. It's not a paradox though, because the logic pretends that the expected value is defined when it not actually defined. A real paradox, in my opinion, requires that there is no "hidden mistake" like that. A good, classic example of one would be to ask if this statement is true or false: "This statement is false." There is no hidden trick there, it's a full paradoxical question that can't be resolved logically.

@onradioactivewaves

Жыл бұрын

@@violetfactorial6806 so by saying the statement is false, its just a statement. Theres no contradiction because the statement is false, which means the statement was true, the the statement said it was false when in fact it was true, and thus saying it was false is true that it was false. Or you could just take it as a statement and not a logical argument. Or perhaps it is not a binary logic.

@madmathematician4458

10 ай бұрын

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 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 ;)

@madmathematician4458

10 ай бұрын

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.

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!

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

@madmathematician4458

10 ай бұрын

Just because a variable can be continuous and expand to infinite does not mean the variable is undefined. 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.

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

kinda surprised that this channed only has 384 subs, it deserves more

Well explained. Will watch your other videos.

This was beautifully explained, and very helpful. Many thanks.

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.

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

This was an amazing video. Thanks for this

Thinking about this literally derailed my whole day until I saw this. Thank you.

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.

@bendowson3124

Жыл бұрын

The issue is not so much that that’s an infinite series but that there’s an infinite number of amounts that each envelope could have. If there were only a finite number of possible amounts, there would be no paradox.

@DawnMandel

Жыл бұрын

I agree, in my opinion, the video just shows you can't conclude that E(Y) > 0 but it doesn't solve the Paradox. the logical step of always switching is not based on E(X) but just on the fact that it's beneficial to switch for every x seen in the envelope.

@bendowson3124

Жыл бұрын

@@DawnMandel The real issue is under the scenario being presented, the expected amount in any given envolope is acutally infinity. This means that no matter what amount you end up with when you select an envelope, that amount will always be less than what we would "expect" to find and therefore it is always more likely that we will find a higher amount if we switch. However, not only is such a scenario unrealistic but also when a given "expected value" is infinite in a given scenario, any related expected value cannot be calculated using the usual means. In this case, the expected amount in a given envelope is infinite and therefore the expected profit in switching cannot be calculated using the usual means.

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!!

@ccreutzig

2 жыл бұрын

The ”expected value” (which often is a value that would be highly unexpected, such as rolling a 3.5 on a die), if it exists, is a completely deterministic value of a probability distribution. Whether or not it has any meaning outside the mathematical world is a different question - which friends on whether that distribution was a good choice to model some aspect of that nonmathematical world.

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.

An awesome explanation. Clear and to the point. Loved it!

@madmathematician4458

10 ай бұрын

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.

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.

@stevenglowacki8576

2 жыл бұрын

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.

@BR-lx7py

2 жыл бұрын

@@stevenglowacki8576 multiples of 2 is still infinite EV. For example 1/2*1 + 1/4*2 + 1/8*4 + ….

@BR-lx7py

2 жыл бұрын

@@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.

@madmathematician4458

10 ай бұрын

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 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.

@gregorymorse8423

10 ай бұрын

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...

I amazed by how you animated and articulating things so organize and well

@ww4011

29 күн бұрын

I study probability, I knew how hard to say those things clearly

another great video Yuval!

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

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.

@cr1216

2 жыл бұрын

Roughly speaking, I think a more concrete example would be that if you perform the experiment a million times and only record the times where you had 10 dollars in your envelope, you will find on average the other envelope will have 34 dollars. If you only record the times where you had 100 dollars you will find on average the other envelope will have 340 dollars and so on. However the critical part about the overall average is that, although you performed the experiment a million times, a million is much less than say 100 million and one single outcome of (10 million, 100 million) pair would overturn the overall average no matter what other smaller-envelope results are. There is nothing special about the number 1 million, you can perform the experiment a googleplex times and still facing the same problem. There simply isn't an overall average and after running the experiment a googleplex times the "overall average" can be any number and if you perform the same experiment another googleplex times the "overall average" will change dramatically. And that's why you cannot generalize "for any concrete envelope" to "overall expectation of switching"

@cr1216

2 жыл бұрын

Another way of understanding can be the following way of thinking: the (10, 100) pair will converge very quickly possibly in 1000 experiments. the (100,1000) pair will converge slower but still in like say 10 thousand experiments, the (1000,10000) pair will converge in like say a million experiments and so on. Any pair of concrete numbers will converge in a finite number of experiments, but there are an infinite amount of pairs and the number of experiments needed for each pair to converge grows with the dollars of each pair, therefore the number of experiments needed for "all pairs" to converge is also infinite, so there is a difference between "any concrete pair" and "all pairs" in the sense that "all pairs" will never converge in a finite number of experiments but "any concrete pair" will always converge in a finite number of experiments.

@bobbobby3515

2 жыл бұрын

Wow, perfect answer!

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.

Great video, really fascinating. The one thing I don‘t totally understand though, is what it means for a random variable to not have an expected value… what does happen, if you simulate both strategies? Will you get different results each time? And if so, isn‘t that alone proof that it does not matter if you switch?

@FormantMath

2 жыл бұрын

Thanks a lot. You asked a great question: "what it means for a random variable to not have an expected value?" I am currently working on a new video, that will explain (among other things) this point. Very briefly: if you simulate on a computer many independent realizations of a random variable that doesn't have an expected value, then the cumulative average of these realizations won't converge to anything, i.e., it will diverge. This is in contrast with the case where the random variable does have an expected value, in which the cumulative average will converge to that expected value.

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.

@timq6224

2 жыл бұрын

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.

@LLlap

10 ай бұрын

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.

Extremely clear. Thank you!

Brilliant video. This is best, most satisfying explanation of the paradox ever created.

@FormantMath

2 жыл бұрын

Thanks! I appreciate it.

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.

@mikkoekstrom8109

2 жыл бұрын

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.

If I was given a die and told I would get 1 million dollars if I rolled a 6, my average profit is $166,666.67 no matter what I roll. THAT DOESN'T MEAN I HAVE GOOD ODDS

@KpxUrz5745

2 жыл бұрын

Yes. They don't mention that if you do NOT roll a six, they throw you into a wood chipper.

I loved this, you deserve more subscribers

@FormantMath

2 жыл бұрын

@alecman95 Thank you! Bring your friends ;-)

Please do keep making these magnificent classes!

@madmathematician4458

10 ай бұрын

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.

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.

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

I can't believe I've watched the whole thing. Very well done.

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

@kazedcat

2 жыл бұрын

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.

@Beregorn88

2 жыл бұрын

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...

@autoboeia9594

2 жыл бұрын

@@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.

@DawnMandel

Жыл бұрын

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

I actually hate this problem a lot because normally it is presented without enough concrete details to make sense of it....you did a great job of making it concrete so it made perfect sense! I loved the video! One question: In the concrete example from the video, the expected value of X, the money you get from keeping the first envelope given to you, was +\infty (...since the values go up by powers of 10 but the probabilities only went down by powers of 2). Is it true that if instead X has finite expectation then the paradox is resolved? (i.e. E(Y)=0 no matter how you calculate it?)

@FormantMath

2 жыл бұрын

Thanks a lot. To answer your question: if we change in the story only the rate at which the probabilities (of getting the pair of envelopes) decrease, and make it faster, so that E(X) is finite, then indeed we'll get that E(Y) exists, and that it's equal to zero no matter how we calculate it. But then the "paradox" disappears: if you open your envelope and see $1, you are better off switching (because the other envelope has $10), whereas for any other observed amount, the expected profit from switching is *negative*, so you shouldn't switch. Hence, it's no longer true to say "you are better off switching regardless of what you observe."

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.

Love this explanation !!

This reminds me of a story series I’ve read that centers around paradoxes. I remember one story about a fraudster, which ended with a bet that went like this: You pay ¥1000000 (the story was in Japanese) to enter, and start flipping a coin. The jackpot doubles if you flip heads, and you win the jackpot if you flip tails. The jackpot starts... at one measly yen. Now normally you would rightfully judge this as a stupid bet, but since the villain of the story failed to remember the “if the expected value exists” part and calculated the expected winnings to be -1000000 + 1(1/2) + 2(1/4) + 4(1/8) + 8(1/16) + ... = ∞, he ends up taking the bet and losing all the money he swindled.

@tsawy6

2 жыл бұрын

Naw, this is just an unbounded sum, doesn't that just give us an unbounded expected value? Like, if you tweaked the game where you'd have to stop playing if you win 1000000 times in a row, then the expected outcome would be a perfectly well defined break even