There are 3 equally likely possibilities for the first envelope that you choose: it contains 𝓍/2, 2𝓍, or 𝓍. So 1 chance in 3 for each. If it contains 𝓍/2 and you swap, you'll either get an envelope with 𝓍 or one with 2𝓍, with equal probability. If it contains 2𝓍 and you swap, you'll either get an envelope with 𝓍/2 or one with 𝓍, with equal probability. If it contains 𝓍 and you swap, you'll either get an envelope with 𝓍/2 or one with 2𝓍, with equal probability. If you average out the 3 equally probable first envelope contents, it comes to 7𝓍/6. If you average out the 6 equally probable returns on swapping, it's also 7𝓍/6. So no advantage or disadvantage to swapping overall.
@Milan Velky Thanks for your reply. Yes, there are two possible equally probable (50/50) different outcomes for every swap. But there are three possible equally probable different amounts in the first envelope before you swap. Thinking there's only one is the big mistake that most people make. Maybe because when you say "let the amount in the first envelope be x" it sounds like you aren't allowed to consider any other possibility for that same envelope. In fact you have to.
just keep switching between envelopes, the expected value will increase by a factor of 1¼ every time, soon enough you’ll have as much money as you could ever want
Indeed. That’s 1 way to definitively debunk this method of ”expected value”, in this case. Other, than just using common sense. The problem is assigning probabilistic value, to the magnitude of the prize. There could be 3 times, or 10 more, or Googolplex times more money, in 1 envelope, than in the other; and it doesn’t affect the probability of choosing the better envelope. It’s 50/50, anyways. The switching strategy works for the Monty Hall -Problem, where there’s 3 options to choose from, and 1 wrong option is revealed to you, upon your initial choice. Here, it’s, like the amount of money, in each envelope, somehow affects the probability of you having chosen the better envelope; which is obvious nonsense. 👍🏻
Some commenters are saying this is like the Monty Hall problem-but it's not. It only appears that way. In the Monty Hall problem, after you make your first choice, the host gives you additional information by opening one of the other two doors and revealing a goat. In light of this additional information, it makes sense for you to switch. In the 2-envelope problem, after you choose your envelope the first time, you don't get any additional information. So it makes no difference whether you switch or not.
The Monty Hall problem works as it does because Monty Hall knows something you don't: he knows which doors contain a goat, and he knowingly takes a goat. At that point, you know there are fewer goats to choose from, and the math proceeds from there.
@@Qermaq Exactly. The only similarity to the Monty Hall Problem is that the question is, "is it worth switching", but the underlying math of the problem is nothing alike. In MHP, there is additional information given after your first choice, AND that additional information isn't independent from your first selection.
I see an easier solution. If both envelopes contain at least A money, we could pretend that we start with A money to begin with, and the envelopes in front of us either contain A or 0 money. If we choose an envelope with value x its expected value is obviously A/2, while the other envelope also has the expected value of A/2. No paradox
As far as I understand the real problem is that the envelopes have an “unknown” amount of money. The problem is in the statement. Without a probability distribution on the possible amounts we intuitively assume that any amount is “equally likely”, but this is not a distribution. There *is no* uniform distribution on ℝ or even on ℕ. Without a known distribution our best guess is something that is not an actual distribution. This leads to the expected value E[x] being undefined. The idea that the second envelope has “y” amount of money *does* lead to E[y] = 1.25 E[x], but only because we assumed that E[x] exists. If you don’t like undefined expected values you can also think of E[x] being infinite, this will also lead to E[y] being infinite and E[y]=1.25E[x] but this time it’s a calculation with infinity, which doesn’t give any useful results either.
Respectfully, this points out the problem perfectly, but is still a misdirection: the amounts in the envelopes are 100% irrelevant. This confuses [inherent value] with [probability]. If one envelope were empty and the other contained a $100 bill or a billion dollar lottery ticket would make no difference in the posited inquiry: do you better your odds for "success" if you switch.
@@xyz.ijk. No, the amounts and the distribution are completely relevant. For example, say the envelopes have numbers which are all positive integer values. Then if you see $1 in your envelope there is a 100% chance the other envelope has $2 in it, it's not a 50% chance. The expected value of switching entirely depends on the distribution used and the amount you see.
@@Bodyknock IFF you know them. I wrote a longer response in a level-1 answer. I believe you are answering a question to a problem not posed; in other words, I believe that the way Brian posed the question does not lead to the Monty Hall problem or even a derivative, but I will go watch the video again to be sure. I appreciate your response, but I think we are addressing two different things. Again, I will go back and review the video.
@@xyz.ijk. On a tangent, an interesting way to look at this distribution issue is consider the scenario where the envelopes can have amounts in any integer power of 2 (so 1/4, 1/2, 1, 2, 4, etc) Notice they all have exactly one non-zero digit in their expansion (e.g. 0.01, 0.1, 1, 10, 100, etc). Altogether that means in this setup an envelope having "twice the amount" means the 1's digit is shifted to the left one space, and "half the amount" means the 1's digit is shifted to the right one space. Now the question is, can you construct a probability distribution for where to place the 1s digit in such a binary expansion so that all possible places have equal value? In other words, you need 10 to be as likely as 1 or 100 for example, and 0.001 to be as likely as 0.01 or 0.0001. The answer is no, you can't, because doing so is equivalent to assigning a uniform random distribution on the integers which isn't possible. Some of these numbers will be more likely than others. So even looking at this case where you have only countably infinite many possible amounts, and every possible amount does have a corresponding pair of values which are double it and half it, you still can't even in that case construct a probability distribution where, no matter what value you see in your envelope, the other envelope must have a 50/50 chance of being double or half the value you see.
The trick here is the "stealing from infinity" part. The mathematical view does not consider that there can be certain absolute greatest amount of money that can be in the envelope and if it is, then the other envelope can only contain half of that amount, not a double. As soon as you add this fact to your calculation - regardless where you set the top amount - the expected value of switching drops to zero too, because there is non-zero probability you selected envelope with this absolutely greatest amount of money and you need to account for that.
This is exactly why the “increased expected value” is 25% 100%-50% divided by 2 is a sloppy analysis for calculating an expected value. That is why when the total value 3x is considered, no stealing from infinity occurs. The two unknown amounts exist within a bounded region instead of mystically existing along the number line somewhere on which x is half of 2x
STOP.DOING.MATH NUMBERS WERE NOT SUPPOSED TO BE GIVEN NAMES YEARS OF COUNTING yet NO REAL-WORLD USE FOUND for going higher than your FINGERS. Wanted to go higher anyway for a laugh? We had a tool for that: It was called "GUESSING" "Yes please give me ZERO of something. Please give me INFINITY of it." LOOK at what Mathematicians have been demanding your Respect for all this time,with all the calculators & abacus we built for them. (This is REAL Math,done by REAL Mathematicians): ????? ??????? ???????????????? "Hello I would like =@-$'&#%' apples please" They have played us for absolute fools
The philosophical resolution is fundamentally flawed. As stated in the video, it essentially claims that it is disingenuous to assign probabilities to events that already happened. Such an objection betrays a lack of understanding of how probability theory works and what purpose it serves. See, randomness has nothing to do with whether a universe is deterministic or not. That is a misconception. To prove that point irrefutably, we have coin-flipping machines. In practice, we study coin flips by using probability theory, and doing this gives us statistically accurate results that are empirically verifiable. However, we all know that it is possible to uniquely determine the outcome of a coin flip with sufficient information of the initial conditions, and coin-flipping machines demonstrate this quite effectively. Coin flips are a deterministic physical process, yet due to the chaotic nature of the process, randomness models the process very accurately. Why is that? Because randomness has nothing to do with whether a process is deterministic or not. Randomness has to do with information. We talk about randomness in situations where we have imperfect information, and we abandon randomness when we have perfect information. If I buy a brand new deck of cards to play poker, I can determine, with exact 100% probability, where a given card will be in the deck. Why? Because if I buy it from the appropriate brand, it will be the case that all decks sold by the brand always come with the cards in the exact same order. This gives me perfect information, so there is nothing actually random about the system. Randomness becomes relevant when you shuffle cards without counting them: the shuffling makes you lose information. It is this loss of information that results in imperfect information about the system, and thus, randomness. If you could keep track of each individual shuffle, though, you would never lose information, and after any given shuffle, you would still be able to uniquely determine the position of any given card with 100% accuracy. Realistically, a human could not do this, but a computer could. We apply probability theory to "events that already happened" all the time in chemistry and physics, and we always end up with empirically verifiable results, and with theories that match the data to a high degree of confidence and precision. So there actually is nothing disingenuous about it. We use probability in those cases due to imperfect information, not because randomness is somehow "not applicable to deterministic systems," which is false.
I think that switching your choice before actually making the choice means there is still only a single choice ever made. So there is one outcome and one probability, which is not changed at all during the decision process, no matter how many non-choices are made.
@@oddlyspecificmath That's a seemingly rational suggestion! I really like it. But it is interesting to look at how it might be controversial too. As a programmer it is easy to think about actions, or function evaluations, as equivalent to their definitions. The function has a data type defining the range and the domain parametrically, and so it is easy to think of f(x) as equal to the number y. So our discussion is about the difference between the number y at x, and the definition of y=f(x). I believe that real number theory is based upon the presumed equivalence of these two ideas; but a finitist would say, the definition is metaphysical if the function does not terminate with a static result unaffected by further iteration of the functional process. This is just the meaning of certain real numbers being defined by the apparent convergence of functional numerical expressions. Linear algebra and combinatorial number theory are very interesting though in that matrix multiplications can be understood as the activity of one matrix upon another compatible matrix. Even though on its surface a matrix is just a grid of numbers, their ordering and the actions performed on them provide an outline of generalized combinatorial analysis and productions in the form of lists of numbers. But of course the action of matrix operations is not dictated by the list of numbers, only the memory heuristic of the system performing the operations. I have no conclusions about these ideas, my previous ideas having been overturned too often to encourage my attachment to the newer ones. But it's a really interesting paradox, action versus choice.
@@haniamritdas4725 Nice response. I'm not traditionally disciplined in field rigor so I apologize ahead of time for where that shows. ¶2: To see if I have it by analogy, I'm imaging a group of mutually-attractive particles floating in space. Their final, object form (when they've all converged into a central mass) is the output, while the paths taken could be the definition of "the assembly function". Implicit and explicit expressions come to mind here too, perhaps as a _process statement_ vs a wrapped instance of _output-specified_ . ¶3: Are you alluding to irrationals and transcendentals? Recursion and infinite sequences? ¶4: I'm currently reviewing Mathemaniac's recent video _"Matrix transpose isn't just swapping rows and columns"_ which seems relevant here. Taken with ¶2, moving the basis vectors seems to be related to changing how one arrives at certain outputs, though my intuition hints this doesn't change the "truth" of some "master" function resulting in that output. [reviewing what I just wrote, I am relying on some personal research that would help make my case...but it probably doesn't belong in a comment. I hope what I've written isn't confusing / is worth stating regardless] ¶5: "memory heuristic" : A...heuristically-developed personal interpretation _(I, ah, Googled to make a better guess)_ leads me to think you mean...:? a procedural/stored/imperative process, vs a declarative one. To use coding: C++ is imperative -- like a matrix having fixed parameters, static algorithm and deterministic output -- while SQL/Gradle are declarative...perhaps even stochastic both in parameters and algorithm...where as long as the result is a permutation of the goal, it's acceptable. In case I went down a rabbit hole, I'll stop there and inquire if I'm close. ¶6: I like what you wrote here. At first I read: "discouraged by the flux" but now I think it's more like "it is what it is". ~ Thanks for giving me something to think about.
3:09 It's funny thinking Bri is using a Canadian accent for this example. "If one envelope contains an amount, eh? and the other contains twice as much, two, eh?"
The problem with this is that the value of x is being treated as though it is predetermined before you make your choice, which it is not. x depends on your choice. If the amounts of money in the envelopes are A and 2A, you have a 0.5 chance of picking the one with A in it, and assigning the value of A to x. Likewise for the one with 2A, you have a 0.5 chance of assigning x:=2A. When you do your probability calculation to calculate the expected value of the money you get from switching, the x in the first term is not the same number as the x in the second term, yet you treated it is if it was. I was rather surprised to hear there is no widely accepted resolution to this supposed paradox. It's just a case of rigorously checking all the implicit assumptions.
I like these probability situations, but they are not good for my mental health. I have spent years pondering situations considerably simpler than this one. William James might say, "You're going to get money either way, so don't ruin it by twisting your brain in a knot."
In turn, the expected value based on geometric average is x It isn't obvious which average is more reliable in this case: some averages produce more optimistic predictions, some - less optimistic
It all depend what are the rules. If two values are generate at the same moment before you choose,switching may give or take the same value. But if the first value is generated before you choose and then second value is generated, you should always switch to the second option.
I'm intrigued by what Wiktor says, but admit I don't understand. If "two values are generated at the same moment before you choose" does that mean for example that someone puts $10 in one envelope and $20 in the other so you can't see which is which, and then asks you to choose. Yes, in this case choosing then switching "may give or take the same value", by which I think you mean that you're just as likely to lose $10 as gain $10. I agree with that. But what does it mean if the second value is generated 𝘢𝘧𝘵𝘦𝘳 you choose? For example, does the person who puts the money in the envelopes know what you chose and uses that information to decide what to put in the second envelope before you switch?
@@chrisg3030 yeah, basically it means "after your first choice" Say you choose A and so $10 are put into it. If only now a coin is flipped to decide whether $5 or $20 are put into B, you should switch.
@@petersievert6830 So after you've chosen an envelope A, another person puts $10 into it. Do you know this?. If you know, then it's the same as starting with $10 that's not in any envelope at all. Yes, I agree you should switch it for envelope B, if that has either $5 or $20 in it with equal probability. Have I understood you? But with these rules it's really a one envelope game. What if you never know directly what's in envelope A? But if you know that B has $5 or $20 in it, you can infer that A has $10. So you're in the same position. But if you don't know what's in A and don't know what's in B, only that B has half or double A, then you're in a different position. If B has double and you switch, you gain, and if it has half, you lose, but by the same amount. So the advantage and disadvantage to switching cancel each other.
Imagine the information not shared to us was that the envelopes have ₹100 and ₹50 (for example) If we choose ₹100 envelope, then saying that by switching we have 50-50 chance of having either 2x or x/2 money is just wrong The envelopes don't contain ₹100 and ₹200, so there is no 2x In other words, assigning the other envelope x/2 and 2x money actually uses two different x's Its not the same variable Another way to think about this is by definition, allotting the heavier envelope as x, then saying we have 50-50 chance of either choosing x or x/2; Which would be correct since there is no 2x here, same as our example
The variable, x, is itself a random variable, which isn't accounted for. We have v1 = x, v2 = {2x, 1/2x}, but we don't admit that x = {A, 2A} So v2 = 2x only when x=A, and v2 = 1/2x only when x=2A, thus we have conditional probabilities. EV(v2) = P(v2=2x)*2x + P(v2=1/2x)*1/2x = P(v2=2x|x=A)*P(x=A)*2A + P(v2=2x|x=2A)*P(x=2A)*2(2A) + P(v2=1/2x|x=A)*P(x=A)*1/2A + P(v2=1/2x|x=2A)*P(x=2A)*1/2(2A) = 1*1/2*2A + 0 + 0 + 1*1/2*1A = 3/2 A Comparing that to the EV(v1), we see there's no benefit of switching.
5:09 What's the problem with assigning probabilities to events which have already been determined? Isn't that a major part of Bayesian probability? If I flip a fair coin and don't look at it, can I now no longer assign a probability to the result because it's already determined? Bayesian probability is based on the knowledge I have, and since I haven't gained any knowledge by flipping the coin and not looking at it, the probability of heads/tails is still 50/50.
the initial calculation of the expected value is more like the situation of "you have a bucket of money and you get to flip a coin heads u double the money, tails u half the money" which of course given the choice you just keep flipping that coin and enjoy the sweet rewards of slightly better odds of gaining money
Here's how see the problem: One of the envelopes has the money x, and the other has 2x. We select one envelope and we have a 50% chance of choosing x and a 50% of choosing 2x. Let's consider both cases individually: CASE I (if our randomly selected envelope has amount 'x') We can either opt to stay with our envelope or switch, thus we have 2 possible outcomes (which I'll denote by R) R1 (we switch) -> we get 2x R2 (we don't switch) -> we get x CASE II (if our randomly selected envelope contains '2x') Again we can switch or keep R3 (we switch) -> we get x R4 (we don't switch) -> we get 2x So basically there are 4 outcomes, and we still have an equal chance of winning x or 2x. Then how is switching better? Am I missing something?
The mistake is separating by "x is the smaller value". Because both times you call it x, but one time it's less than the other. This leads to losing half of x being the same amount as gaining a full x. Because one time x is 2x, basically. You should have separated by "I'm on the smaller value". This would allow to factor in the relative sizes. This would basically lead to resolution 1, but I wanted to point out the mistake in the initial way of viewing this.
This isn't a paradox. It's a misunderstanding that comes from reusing 'x' to mean different things in two different situations, and then trying to combine expressions as though they represented the same thing. The other envelope only contains x/2 if x is the larger value. The other envelope only contains 2x if x is the smaller value. Those aren't the same x. Label the envelopes in a consistent manner and say they always have x and 2x. Then you have four situations: 1) you picked x and did not choose to switch: you end with x 2) you picked 2x and did not choose to switch: you end with 2x 3) you picked x and chose to switch: you end with 2x 4) you picked 2x and chose to switch: you end with x The expected value of what you picked is the mean: (x+2x+2x+x)/4 = 1.5x. Asking for E(x) when x itself is dependent on what you picked is what leads to the confusion. If you call the money value x, then what you want is E(envelope) in terms of x.
I heartily agree when you say the envelopes always have x and 2x. Those are the two possible contents of either envelope you pick. Or you can say the envelopes always have x and x/2 if you like. But to talk about either of the two envelopes containing x, 2x, or x/2 is to pose a different problem. That problem isn't necessarily one with two values for x. It could keep the same value for x, but be better called a three rather than two envelope problem, which I'll be happy to discuss with you later if you like. Where I diverge now is when you (and many others) only average out 𝘳𝘦𝘵𝘶𝘳𝘯𝘴 from switching and not switching. If you also average out 𝘨𝘢𝘪𝘯𝘴 from switching and not switching then then the four situations are 1) you picked x and did not choose to switch: you end with x. Gain: 0 2) you picked 2x and did not choose to switch: you end with 2x. Gain 0 3) you picked x and chose to switch: you end with 2x. Gain x 4) you picked 2x and chose to switch: you end with x. Gain -x (that is, you lose x) Mean gain: (0 + 0 + x + -x)/4 = 0. And in particular the mean gain from switching is 0. Nothing to be expected from switching.
Assigning 2x and x/2 is logically correct without any problem. The real explanation is E[XY] = E[X]*E[Y] only works when X and Y are independent. X = Your original choice = 1 or 2 with equal chance. Y = The other envelope / the envelope you originally choose = 2 or 1/2 with equal chance. It is true that E[Y] = 1.25, but what you get is E[XY] not E[X]*E[Y]. When Y=2, X always=1. When y=1/2, X always=2. So E[XY]=1/2*2*1+1/2*1/2*2=1.5 same as E[X] = 1.5 Summary: Original choice E[X]=1.5 Switching multiplier E[Y]=1.25 After switching E[XY]=1.5 E[XY] =/= E[X]*E[Y] because X, Y are dependent.
I like to think this going to the extreme scenario. Lets say that in the first envelope there are 10$. In the second envelope could be twice or half, 20$ or 5$. If you are lucky with the change, you will win 10$ extra, lose 5$ otherwise (win -5$). Now lets take a bigger step. In the first envelope still 10$, but in the second one will be 100$ or 1$. Changing envelopes will gives you 90$ or -9$. Now the risk seems lower, right? Finally go to the extremest case. As always are 10$ in the first envelope, but in the second one can have an infinity amout of money with the risk of losing all the 10$. Who in his mind would'nt take that risk? As you can see, in this three cases the odds are equal, but the bigger numbers shows how worth it can be.
I like this because you clearly are forcing us to recognize the real didactic. I don’t believe you would unintentionally confuse probability and value. It poses an independent event as a weighted event. It doesn’t matter whether the second envelope has 2x or [random]x. What if there was 23 times as much in one envelope as the other, or one envelope contained a $1 bill and the other contained a $100 bill, or one had a piece of paper that read 0.577 (or goat) and the other read 3.141 (or car)? It would make no difference. The amount inside the envelope is irrelevant. I agree this is not the Monty Hall problem for many reasons. And it is also not phrased as a psychology problem. But it’s also not a paradox. The problem intentionally confuses [inherent value] with [probability]. Here is an answer by analogy. You have two fair die, one is marked with only one dot on each side and the other is marked with only two dots on each side. You can’t see either die before or while you throw it and you can’t see the result of the first throw. (It’s irrelevant whether you see the result of the second throw.) Do you improve your odds of “success” by throwing the second die? Does it matter whether one of die has sides of two dots of 200 or 2,000? It doesn’t matter whether you “pick a hand “, choose an envelope, throw a die of this type, or any other choice between fair “a or b” items. The amount contained in or on each is irrelevant and that’s where the problem falls off the rails.
At 1:56 Bri says "In other words our expected value of switching is 25% more than keeping the envelope we originally chose" Yes but only if we happened to originally choose the envelope with x in it. If instead we happen with equal probability to pick the one with x/2 then the expected value of switching is 100% more than the original envelope, doing the same math. Wow! But here comes the nemesis. Suppose the original choice of envelope was the third possibility, 2x. Then calculated the same way, the EV of switching is 125% less, yes, less than keeping that envelope. I'm happy to go into how I got those numbers and what they mean with anybody, but for the time being just note that 25% + 100% - 125% = 0% Once again we come down to a point I've been driving at in other comments, you've got to consistently assess all the possibilities you enumerated.
Alice: Hey Bob. Here's an envelope with some money in it. It's yours. Bob: Thanks. How much? Alice: I'm not going to tell you. But before you open it you can switch it for this second envelope which has either half or double yours. Bob: But if the second envelope has half or double mine, doesn't that mean mine has half or double the second? So where's the advantage? Alice: Good thinking Bob. I'm impressed. Look at it this way. Let's call the amount in your envelope "x". Bob: I'm with you. Alice: So the second has either half x or double x. Bob: So if I switch I could get only half x in return, which means I lost half x. Alice: Yes, but you're equally likely to get double x in return, which means you gained x. You stand to gain double what you stand to lose with equal probability. Bob: Cool. Alice: What's more, I'll let you switch and switch back as often as you like. So you stand to gain double what you stand to lose each time! So if half the switches cause you to lose half x each time, you nevertheless gain twice as much each time with the other half of the switches. Bob: So say I do 100 switches and I lose $1 with each of 50 switches, I gain $2 with each of the other 50. Mega cool. Let's start now. Alice: Not so fast Bob. Seeing as I'm giving you such a great deal I think I deserve a small something in return. Bob: Like what? Alice: Just a trivial sum per switch. Let's negotiate . . .
I came in here champing at the bit, but it seems you might agree with my initial assessment: no matter how you dance with the numbers, you can pick "less" or "more", and if you don't know which you have, switching isn't doing a damn to your chances.
Another issue here is the statement "given you see X in your envelope there is a 50% chance the other envelope has 2X and a 50% chance the other envelope has X/2" isn't actually possible unless the probability distribution of the random Xs is such that it could happen. For example, there is no uniform random distribution of the Natural numbers such that every number has equal probability of being chosen. So if you were to assume that the envelopes have positive integer amounts of money then if you see $1 in your envelope then there is a 100% chance the other envelope has $2, not a 50% chance. And no matter what distribution you use for the Natural numbers it will always be the case some numbers are more likely than others so you can never have a situation where literally every Natural number in your envelope results in the same 50/50 chance of the other number being twice its value or half its value. For a similar reason there's no uniform probability density of the positive real numbers either, so even if you allow the envelopes to have non-integer positive or even irrational amounts you still run into an issue where not every number can have a 50% chance of being twice or half the other number in the other envelope.
I literally did a million test runs on this and it turns out that there is no difference between switching and not switching, which is exactly what you'd expect.
Kind of reminds me of Schrödinger’s cat in a way Both have the greater and both have the lesser amount of money at the same time until it is known which is which
I think the problem is when you assign x to the money in one envelope, in the case where you stand to gain, that x is smaller than in the case where you stand to lose, so while you stand to gain 100% vs lose 50%, you're potentially gaining 100% of a smaller amount of money. The problem is that you assume x to be the same in both cases, when it isn't.
Before getting philosophic, I propose stating the problem more clearly: There is a total amount of money, x, unevenly distributed into two envelops, one envelop containing twice as much as the other. Let y be the amount in the envelop with less money, then the other envelop contains 2y, such that x = 2y +y = 3y. Now, let's consider two probabilistic experiments: A) A person can pick any of the two envelops with equal probability and keep the amount of money in such envelop. B) A person can pick any of the two envelops with equal probability and then interchange the envelops, keeping the amount of money in the second envelop. The question is: In each of the experiments A) and B), what's the expected value in terms of the money kept by the person picking the envelop? For case A) we have the following independent events with equal probabilities: - A1) Choosing the envelop containing amount y with probability 1/2, the event having associated value y - A2) Choosing the envelop containing amount 2y with probability 1/2, the event having associated value 2y Therefore the expected value of experiment A) is E_A = 1/2 (y) + 1/2 (2y) = 1/2 (3y) = 3/2 y. Since y = x/3, E_A = (3/2) (x/3) = x/2. In other words, the expected value in case A is getting half of the total amount. If you think about it, it makes total sense because there are exactly two envelops, and even when the money is unevenly distributed, since picking any of the two envelops has the same probability, the higher amount obtained when choosing the envelop with more money is compensated by the lower amount obtained when choosing the other. This is true for any other proportion used to divide the money among the envelops, as long as their sum continues to be x. Guess what happens if instead of two envelops you divided the x dollars into three different envelops, again irrespective of the proportion? Back to the topic of the video, for case B) we have, "on the other hand", the following independent events with equal probabilities: - B1) Choosing the envelop containing amount y with probability 1/2 and then switching to the other envelop, therefore the event having associated value 2y. - B2) Choosing the envelop containing amount 2y with probability 1/2 and then switching to the other envelop, therefore having associated value y. The expected value for experiment B) is thus E_B = 1/2 (2y) + 1/2 (y), but addition is commutative, right? Therefore E_B = E_A; Q.E.D.
How does this problem change if there are more than 2 envelopes? And an uneven number of outcomes overall? eg. 70% of the envelopes contain twice what you originally selected, 30% contain the same amount as you chose initially? What if you choose one envelope initially and then a choice of 0,1 or 2 further envelopes?
I think the important point is that the probability choice happened when you picked the 1st envelope. After that, there is no probability to the second choice. You are definitely picking between what you chose and the opposite. Whichever envelope you choose at 1st, the second choice is between that or its opposite. So your 1st choice has a 50/50 for the better or worse option. If you switch, you are just switching to the other 50%. So the probabilities calculations are like this: The probability that you chose A is 50%, while the probability you chose 2A is 50%. If you choose to switch: The probability you chose A is 50%, times the probability the envelope you switch to is 2A is 100%, yields 50%. The probability you chose 2A is 50%, times the probability the envelope you switch to is A is 100%, yields 50%. Either way, it's 50/50.
One small error - there is no expected loss... If you start with 0 dollars and pick an envelope, regardless of which one you chose, you come out on top!!
In the first analysis, it's not 1/2 of more and 1/2 of less; it has to be 1/3 of more and 2/3 of less. Whatever the distribution of values is, the probability has to decrease as the value approaches infinity. P.S. I'm not sure how to fully make sense of this...
Or you can realize that probability and statistics deal with a set of lots of events, not a single one. The 1.25x as a expected value means that If you repeat this event a lot of time, like a really lot, and you sum the amount of everyone that switched to be close to 1.25 times the amount in regards to people who didn't, because If you switch and gain, you gain more than you lose, If you switch and lose. This adds up on the long run...
hi i think this problem is from vsause2 and you add a little bit of your ideas in it, I liked this problem because it tells us how to use probability correctly
Before I watch the video, I will attempt to find the fault in what he said. The problem is that x isn't constant. Choosing the smaller one means x is a smaller value and choosing the larger one means x is a larger value. Let's say one has 50$ and the other has 100$. The average between choosing either of these is 75, and we can see that 50 +100/4 is indeed 75. However, we use two different values for x. So a better way to show it is, let's say one has x and the other has 2x. You will always have the probability x + x/2 no matter what. That's what I think isk
When u pick between 2 envelopes there are 2 porbabilities so 50% probability and when u add the switch or not the probabilities change to 4 probabilities so 25% so would you switch or not
now if you pick the other envelope, a will give you a new envelope. This new envelope will have either 2* the amount in that envelope or 1/2 the amount in that envelope. If you change your mind, i will give you another envelope. What is the most optimal money gaining strategy?
the problem with the x-solution is that your x changes its value and you showed it with the A-solution. You cannot set x to $1000 and $2000 in the same equation. x has to be either 1000 or 2000 but not both at the same time :) when 1000 = 2000 = x everything is possible and provable, even that 0 = 1.
what if the envelopes had money of 0 value (monopoly money for example) and 2 × 0 = 0 which would make it a 0% gain either way and we were trolled all along
The problem with the formula is that when switching to the lesser envelope you assume that the amounts where x and x/2 while when switching to the higher envelope you assumed that the values were x and 2x. And then all is put in one formula. So the formula implies that when you picked the lesser amount, both values were half as high as otherwise. Short: You only can lose x/2 but you can win 2x. Of course then it looks like it is better to switch! Say, your mate put in 4€ and 8€. You choose one. Now the math says when you switch there is a 50% chance to get 2x or 0.5x. If you picked the 4€ then x=4€ otherwise x=8€. So you can't have one formula like E=0.5*(x/2) + 0.5*(2x) where you mix both different values for x. The x in (x/2) actually is 8€ and the x in 2x is 4€. So the expected value is 0.5*(8€/2) + 0.5*(2*4€) = 6€ which is correct and shows that it does not make a difference if you switch or not. Another way would be to say that the amounts are x and 2x. so switching gives you 0.5*(x) + 0.5*(2x) = 1.5x. For x=4€ that is again 6€. You could say now that the rewards are not fixed at 4€ and 8€ so no matter if you chose the smaller or the bigger, x could be the same. This is not true. Because you pick after the money was put. So x can't be the same no matter which envelope you picked. But we can change the experiment a little bit! You get the envelope with 4€ and your mate secretly puts 8€ or 2€ in the other. So the expected reward for switching is 0.5*8€+0.5*2€ = 5€. So indeed it is better for you to switch. And it really is.
Another game to make it even more clear: There are 3 marbles. You can pick 2 or 1. Then you get the rest. What is better to pick in the beginning? So let's say you picked x marbles. Then in the second step you get 2x if you chose 1 marble and only x/2 if you chose 2 marbles. So if you pick 1 marble you end up with 3x otherwise only 1.5x. So you should pick one marble! That is of course wrong. x is different depending on what you chose, so you can't compare 3x and 1.5x. In the video both x were then even put in one formula.
@@Clyntax I see merit in thinking of 3 marbles because that's the number of equally probable possibilities for the first envelope in the original problem. That's right, not 1, not 2, but 3. They are x, x/2, 2x. So if it contains x, then the second contains x/2 or 2x. If you switch then you stand to lose x/2 or gain x. if it contains x/2, then the second contains x or 2x. If you switch then you stand to gain x/2 or gain 3x/2. if it contains 2x, then the second contains x or x/2. If you switch then you stand to lose x or lose 3x/2. The sum of all possible losses is x/2+x+3x/2=3x. The sum of all possible gains is x+x/2+3x/2=3x. So no advantage in switching overall. The mistake is to consider only the first of the 3 possibilities, namely that the first envelope contains x. It's the only one where you can have either a gain or a loss on switching, and since the gain is bigger you naturally go with the switch. But consider all 3 together, then possible gains and losses are balanced. Why do we make that mistake? Maybe because once we stipulate the first envelope contains x, then we think we aren't allowed to consider any other possibilities for that same envelope. In fact we have to.
People do Bayesian calculations all the time. Let's say you opened your first envelop and you found a $10 bill. Then the state of the world can either be ($5, $10) or ($10, $20). Now what's the posterior probability of the state of the world being ($10, $20) and what's the posterior of the state of the world being ($5, $10)?
This is siiikar to schrodinger a cat, until you open the other envelope you both have and don’t have the largest amount of money. You need to account for type 1 and type 2 error, that you switch and end up with less or that you don’t switch and end up with less. The cost to switching and being wrong is 50% whatever was in your envelope, the cost of not switching and having the lesser amount is costing you hundred percent. Not really supposed to use percentages with negative numbers but hey. …
How about the ratio of the money in two envelopes unknown so that A is partitioned two parts of a with an unknown fraction should it effect the choice strategy or make a difference?
No wait, here's the problem. When we're doing the probabilities and we hit the operation "2x + x/2", the first x is not the same as the second x, because x is however much money is in the envelope you have chosen. Let's say the envelopes contain 10 dollars and 20 dollars: then "2x + x/2" is 2*10 + 20/2.
The correct way to do it is to also factor in the amount in the envelope we're holding. So it's not "2x + x/2", it's more like "2a + b/2", where b = 2a.
@@Cobalt_Spirit Yes. But what you don't know is whether you have the envelope with "a" or the one with "b" (which equals "2a"). 1.5a is 1.5 times the lower value, regardless of which envelope you have. We could just as correctly say that the average value is 0.75b.
I agree that's one way to look at it. Another way is to acknowledge each envelope has three, not two possible amounts, they are x, x/2, 2x. Here x keeps the same value whatever that is. If the first envelope contains x, then the other contains x/2 or 2x. If you switch then you lose either x/2 or gain x. But you have to consider the other two possibilities for the first envelope: If it contains x/2, then the other contains x or 2x. If you switch then you gain either x/2 or gain 3x/2. If it contains 2x, the the other contains x/2 or x. If you switch then you lose x/2 or you lose 3x/2. Total gains and losses with the 6 possible outcomes are equal, they cancel. No advantage to be expected from switching. You could call this a 3-posssibility 2-envelope problem. But I think it's simpler if you just put each of the three possible amounts into its own envelope and call it a 3 envelope problem. If you switch one envelope you get one of the other two at random. What's the two envelope problem then? I have my idea, but I'd like to hear anyone else's.
5:13 and yet, all it serves to do is summarize what we don't know about our current situation, why would it be disingenuous? Is it just to criticize the very attempt at summarizing it into probability?
I think the problem here is confusing x as being both a random variable and a definite but arbitrary number: Given the description, it's quite obvious that the sample space in this scenario has just two events: you picked the envelope with less money, or the one with more (call these events a and b respectively). Let X be a R.V. corresponding to the amount of money in the envelope you picked (recall a random variable is a function that maps the elements in the event space to real numbers). We can define the contents of the other envelope in terms of X: Let Y be an R.V. corresponding to the amount of money in the other envelope, then Y(a)=2X(a), Y(b)=X(b)/2. The expected value of switching E(Y) = Pr(a)Y(a) + Pr(b)Y(b) = Pr(a)2X(a) + Pr(b)X(b)/2 =X(a)+X(b)/4 You may notice that we cannot combine the fractions in this state, since X(a) =/= X(b). There is a way to solve this properly from this point though, we can say that X(b)=2X(a) from the problem definition (the amount of money in the envelope you picked if you picked the one with more money is twice as much as if you picked the one with less money). Then, E(Y) = X(a)+2X(a)/4 =1.5X(a) , and E(X) = X(a)/2 + X(b)/2 = X(a)/2 + 2X(a)/4 =1.5X(a) Therefore E(X) = E(Y), both expected values are the same. You may still be left wondering "well what step precisely is false if we let x be a number rather than a random variable?". Let's walk through the steps and lay out the statements formally: 0:49 Let X = x, where X is a R.V. corresponding to the amount of money in the envelope we picked, and x is an arbitrary element in the range of X 0:53 Let Y be an R.V. corresponding to the amount of money in the other envelope, defined as follows, Y(a)=2x if event a occured, Y(b)=x/2 if event b occured 1:05 Pr(a) = 1/2, Pr(b) = 1/2 1:24 E(X) = x 1:30 Pr(Y=2x) = 1/2 and Pr(Y=x/2) = 1/2 1:37 E(Y) = Pr(a)Y(a) + Pr(b)Y(b) 1:44 = 1/2 * 2x + 1/2 * x/2 = x + x/4 1:48 = 5x/4 = 1.25x Did you catch that? Recall that x is an arbitrary element in the range of X, so it must be either X(a) or X(b) At 0:53 we had different conditions for Y(a) and Y(b). If event a occured, then Y(a) = 2x, but if event a occured, then x must be X(a), and Y(a) = 2X(a). Similarly, If event b occured, then Y(b) = x/2, but if event b occured, then x must be X(b), and Y(b)=X(b)/2. But at 0:13 we were told that X(b) = 2X(a), and at 0:07 that X(a) =/= 0 and X(b) =/= 0, so it must be that X(a) =/= X(b) Thus, the two x variables we were working with are not actually the same number, and their sum is not 1.25x but rather 1.5*X(a), or alternatively 0.75*X(b). The last step is wrong. Essentially the mistake comes from seemingly defining Y in terms of x (a fixed arbitrary variable) when in fact the definition given is in terms of X (a random variable). Then somewhere along the way we forgot that we had left conditions on x, and assumed that wherever x appeared it would be the same. Given the problem description, you cannot simply drop the conditions. If you drop the conditions, you are working with a different problem entirely. The analysis would be accurate if the problem were instead something more like: "Here is an envelope, it contains some cash. Now, I have two other envelopes: one with twice as much cash as the one I gave you, and one with half as much. Would you like to trade me your envelope for one of the others?" (in which case take the deal)
Solution: Punch the guy offering the envelopes. While he's trying to figure out what's going on , take the envelopes and run off. This guarantees more than X dollars. Math, for some reason, doesn't like these lateral thinking solutions. But it's just as valid as any other solution.
I think I know what's going on. Those 50-50 probabilities are actually conditional probabilities. If we call the respective values in envelopes 1 and 2 E1 and E2, using notation, the affirmative P(E2=2x) = P(E2=x/2) = 0.5 is actually wrong, or at least doesn't make any sense. What's the probability that envelope 1 has 5 dollars for example? It isn't defined. What should be written is P(E2=2x | E1=x) = P(E2=x/2 | E1=x) = 0.5. Other than that, all you have are probabilities for relations, namely P(E2=2*E1) = P(E1=2*E2)=0.5. So the correct reasoning for the expected value of money you would get by switching to envelope 2 would be: E2*P(E2=2*E1)+E2*P(E1=2*E2)=E2*0.5+E2*0.5 = E2 (which leads you nowhere). Now if you choose envelope 1, and THEN you were told: "THIS ENVELOPE CONTAINS x DOLLARS" Then the situation changes. now you have additional information and you can calculate the expected value you'd get by switching, and indeed it would be 1.25x, i.e., you have an advantage by switching. To understand that switching given that you know the amount contained in the envelope you selected is advantageous, consider the analogous situation: You have two options: 1. select an envelope with a known amount x, or 2. select a "lottery" envelope that has either 2x or x/2 with equal probability, you'd expect to gain more by selecting the lottery envelope.
"you'd expect to gain more by selecting the lottery envelope." Yes, but what's almost always omitted is that if you don't know what's in either envelope, then there's an equal chance that it's E1 that has x/2 or 2x, and E2 that contains x, and so the advantage of switching - however you calculated it - must surely be exactly cancelled or reversed. My balance sheet uses possible net gains and losses on the switch rather than expected value: A) If E1 has x then switching to E2 will get you x/2 or 2x. So you stand to gain -x/2 (that is, incur a net loss of x/2) or x. B) E1 has x/2or 2x then switching to E2 will get you x. So you stand to gain x/2 or -x. Possible gains obviously sum to 0. Gains and losses cancel. No advantage to switching. Why do we tend to make the mistake of only considering case A? Maybe because once we've said "let the contents of E1 be x", then we think we can't logically or consistently consider any alternative. In fact we have to.
@@chrisg3030 yes. I think we fool ourselves by thinking just because we named something as "x" it remains an indeterminate quantity. Actually, when we say E1=x it's just as determined as saying E1=50 cents.
@@Felipe-sw8wp Yep, they're both as equally determinate or indeterminate as each other. It's as if the x is also some unit of currency in some fictional country just like the dollar or euro. Its algebraic use is irrelevant here. You could write x on a piece of paper and put it in an envelope, and then put another piece of paper with x/2 or 2x written on it in another, and it would work just as if you did the same with $10, $5, and $20 bills instead. This brings me to another possible reason for not considering case B. People can visualize setting the game up by putting, say, $10 in an envelope. But how do you put $20 or $5 in one envelope? There's no sum of money called "20 or 5 dollars". A pizza parlor doesn't have that as a price on the menu board unless they mean big and regular or something. I guess there are work-arounds for that, like putting each sum into its own envelope, shuffling, and picking at random.
But then you have a 1.25 probability of being incorrect also, by the same reasoning applied to choosing the lowest envelope instead. Instead of saying, "What is the probability this is the higher envelope," we ask, "What is the probability this is the lower envelope," the probability would be the same, as the question is the same. So, you have a 125% chance of it being the higher and 125% chance of it being the lower, whatever your guess. So, you're still stuck with the base of 1:1 whether you switch or not.
Why doesn’t anyone just run a computer sim where person A switches 100% of the time, and person B never switches? Run the sim 10,000 times and see whichever person gets more money at the end.
@@roeekawaz1372 to my shame, I made this comment before he got to the explanation, and then forgot about this comment Still, it’s weird that I’ve watched probably 5 videos on this ‘paradox’ and not one even mentions using computer simulations to model it
Head to squarespace.com/brithemathguy to save 10% off your first purchase of a website or domain using code BRITHEMATHGUY
This video uploaded today but pinned comment 6 days ago? Wth
@@marcusscience23 maybe the video was private for us, but visible for members of the channel.
There are 3 equally likely possibilities for the first envelope that you choose: it contains 𝓍/2, 2𝓍, or 𝓍. So 1 chance in 3 for each.
If it contains 𝓍/2 and you swap, you'll either get an envelope with 𝓍 or one with 2𝓍, with equal probability.
If it contains 2𝓍 and you swap, you'll either get an envelope with 𝓍/2 or one with 𝓍, with equal probability.
If it contains 𝓍 and you swap, you'll either get an envelope with 𝓍/2 or one with 2𝓍, with equal probability.
If you average out the 3 equally probable first envelope contents, it comes to 7𝓍/6. If you average out the 6 equally probable returns on swapping, it's also 7𝓍/6. So no advantage or disadvantage to swapping overall.
@Milan Velky Thanks for your reply. Yes, there are two possible equally probable (50/50) different outcomes for every swap. But there are three possible equally probable different amounts in the first envelope before you swap. Thinking there's only one is the big mistake that most people make. Maybe because when you say "let the amount in the first envelope be x" it sounds like you aren't allowed to consider any other possibility for that same envelope. In fact you have to.
@Milan Velky No, it makes it simple.
just keep switching between envelopes, the expected value will increase by a factor of 1¼ every time, soon enough you’ll have as much money as you could ever want
🤯
Actually that converges
Indeed. That’s 1 way to definitively debunk this method of ”expected value”, in this case. Other, than just using common sense. The problem is assigning probabilistic value, to the magnitude of the prize. There could be 3 times, or 10 more, or Googolplex times more money, in 1 envelope, than in the other; and it doesn’t affect the probability of choosing the better envelope. It’s 50/50, anyways. The switching strategy works for the Monty Hall -Problem, where there’s 3 options to choose from, and 1 wrong option is revealed to you, upon your initial choice. Here, it’s, like the amount of money, in each envelope, somehow affects the probability of you having chosen the better envelope; which is obvious nonsense. 👍🏻
wow
Some commenters are saying this is like the Monty Hall problem-but it's not. It only appears that way.
In the Monty Hall problem, after you make your first choice, the host gives you additional information by opening one of the other two doors and revealing a goat. In light of this additional information, it makes sense for you to switch.
In the 2-envelope problem, after you choose your envelope the first time, you don't get any additional information. So it makes no difference whether you switch or not.
The Monty Hall problem works as it does because Monty Hall knows something you don't: he knows which doors contain a goat, and he knowingly takes a goat. At that point, you know there are fewer goats to choose from, and the math proceeds from there.
I think they meant the Monty Hall problem backwards
They’re similar, but different in calculation
I'd say the similarity to the MHP is a part of its construction - a red herring.
@@Qermaq Exactly. The only similarity to the Monty Hall Problem is that the question is, "is it worth switching", but the underlying math of the problem is nothing alike. In MHP, there is additional information given after your first choice, AND that additional information isn't independent from your first selection.
This switching reminds me of exam days when I write the right option but at the last moment I switch into the wrong option.
This video will make me more expressible as p/q where p and q are integers and q is nonzero.
I personally see this as either a 100% gain or a 200% gain.
I see an easier solution.
If both envelopes contain at least A money, we could pretend that we start with A money to begin with, and the envelopes in front of us either contain A or 0 money.
If we choose an envelope with value x its expected value is obviously A/2, while the other envelope also has the expected value of A/2. No paradox
As far as I understand the real problem is that the envelopes have an “unknown” amount of money. The problem is in the statement.
Without a probability distribution on the possible amounts we intuitively assume that any amount is “equally likely”, but this is not a distribution. There *is no* uniform distribution on ℝ or even on ℕ.
Without a known distribution our best guess is something that is not an actual distribution. This leads to the expected value E[x] being undefined. The idea that the second envelope has “y” amount of money *does* lead to E[y] = 1.25 E[x], but only because we assumed that E[x] exists.
If you don’t like undefined expected values you can also think of E[x] being infinite, this will also lead to E[y] being infinite and E[y]=1.25E[x] but this time it’s a calculation with infinity, which doesn’t give any useful results either.
Respectfully, this points out the problem perfectly, but is still a misdirection: the amounts in the envelopes are 100% irrelevant. This confuses [inherent value] with [probability]. If one envelope were empty and the other contained a $100 bill or a billion dollar lottery ticket would make no difference in the posited inquiry: do you better your odds for "success" if you switch.
@@xyz.ijk. No, the amounts and the distribution are completely relevant. For example, say the envelopes have numbers which are all positive integer values. Then if you see $1 in your envelope there is a 100% chance the other envelope has $2 in it, it's not a 50% chance. The expected value of switching entirely depends on the distribution used and the amount you see.
@@Bodyknock IFF you know them. I wrote a longer response in a level-1 answer. I believe you are answering a question to a problem not posed; in other words, I believe that the way Brian posed the question does not lead to the Monty Hall problem or even a derivative, but I will go watch the video again to be sure. I appreciate your response, but I think we are addressing two different things. Again, I will go back and review the video.
@@xyz.ijk. I didn't say anything about the Monty Hall problem because that's not relevant here.
@@xyz.ijk. On a tangent, an interesting way to look at this distribution issue is consider the scenario where the envelopes can have amounts in any integer power of 2 (so 1/4, 1/2, 1, 2, 4, etc) Notice they all have exactly one non-zero digit in their expansion (e.g. 0.01, 0.1, 1, 10, 100, etc). Altogether that means in this setup an envelope having "twice the amount" means the 1's digit is shifted to the left one space, and "half the amount" means the 1's digit is shifted to the right one space.
Now the question is, can you construct a probability distribution for where to place the 1s digit in such a binary expansion so that all possible places have equal value? In other words, you need 10 to be as likely as 1 or 100 for example, and 0.001 to be as likely as 0.01 or 0.0001. The answer is no, you can't, because doing so is equivalent to assigning a uniform random distribution on the integers which isn't possible. Some of these numbers will be more likely than others.
So even looking at this case where you have only countably infinite many possible amounts, and every possible amount does have a corresponding pair of values which are double it and half it, you still can't even in that case construct a probability distribution where, no matter what value you see in your envelope, the other envelope must have a 50/50 chance of being double or half the value you see.
The trick here is the "stealing from infinity" part. The mathematical view does not consider that there can be certain absolute greatest amount of money that can be in the envelope and if it is, then the other envelope can only contain half of that amount, not a double. As soon as you add this fact to your calculation - regardless where you set the top amount - the expected value of switching drops to zero too, because there is non-zero probability you selected envelope with this absolutely greatest amount of money and you need to account for that.
This is exactly why the “increased expected value” is 25%
100%-50% divided by 2 is a sloppy analysis for calculating an expected value. That is why when the total value 3x is considered, no stealing from infinity occurs. The two unknown amounts exist within a bounded region instead of mystically existing along the number line somewhere on which x is half of 2x
Easy Solution : Dont do maths and pick any, simplest thing ever!
STOP.DOING.MATH
NUMBERS WERE NOT SUPPOSED TO BE GIVEN NAMES YEARS OF COUNTING yet NO REAL-WORLD USE FOUND for going higher than your FINGERS.
Wanted to go higher anyway for a laugh? We had a tool for that: It was called "GUESSING"
"Yes please give me ZERO of something. Please give me INFINITY of it."
LOOK at what Mathematicians have been demanding your Respect for all this time,with all the calculators & abacus we built for them.
(This is REAL Math,done by REAL Mathematicians):
????? ??????? ????????????????
"Hello I would like =@-$'&#%' apples please"
They have played us for absolute fools
🙄😉
Occam's razor
@Maksymilian Czerniak bro, u are getting da money, and that's what matters, haha
steal both and run away
The philosophical resolution is fundamentally flawed. As stated in the video, it essentially claims that it is disingenuous to assign probabilities to events that already happened. Such an objection betrays a lack of understanding of how probability theory works and what purpose it serves. See, randomness has nothing to do with whether a universe is deterministic or not. That is a misconception. To prove that point irrefutably, we have coin-flipping machines. In practice, we study coin flips by using probability theory, and doing this gives us statistically accurate results that are empirically verifiable. However, we all know that it is possible to uniquely determine the outcome of a coin flip with sufficient information of the initial conditions, and coin-flipping machines demonstrate this quite effectively. Coin flips are a deterministic physical process, yet due to the chaotic nature of the process, randomness models the process very accurately. Why is that? Because randomness has nothing to do with whether a process is deterministic or not. Randomness has to do with information. We talk about randomness in situations where we have imperfect information, and we abandon randomness when we have perfect information. If I buy a brand new deck of cards to play poker, I can determine, with exact 100% probability, where a given card will be in the deck. Why? Because if I buy it from the appropriate brand, it will be the case that all decks sold by the brand always come with the cards in the exact same order. This gives me perfect information, so there is nothing actually random about the system. Randomness becomes relevant when you shuffle cards without counting them: the shuffling makes you lose information. It is this loss of information that results in imperfect information about the system, and thus, randomness. If you could keep track of each individual shuffle, though, you would never lose information, and after any given shuffle, you would still be able to uniquely determine the position of any given card with 100% accuracy. Realistically, a human could not do this, but a computer could.
We apply probability theory to "events that already happened" all the time in chemistry and physics, and we always end up with empirically verifiable results, and with theories that match the data to a high degree of confidence and precision. So there actually is nothing disingenuous about it. We use probability in those cases due to imperfect information, not because randomness is somehow "not applicable to deterministic systems," which is false.
I think that switching your choice before actually making the choice means there is still only a single choice ever made. So there is one outcome and one probability, which is not changed at all during the decision process, no matter how many non-choices are made.
@@xyz.ijk. Okay. You can ride the lawnmower to the corner store if you want to, it makes no difference... to me. ;)
@@haniamritdas4725 Sounds like an adventure!
Perhaps it's more clear to call the selected choice _the action_ to delineate an occurrence vs an option?
@@oddlyspecificmath That's a seemingly rational suggestion! I really like it. But it is interesting to look at how it might be controversial too.
As a programmer it is easy to think about actions, or function evaluations, as equivalent to their definitions. The function has a data type defining the range and the domain parametrically, and so it is easy to think of f(x) as equal to the number y. So our discussion is about the difference between the number y at x, and the definition of y=f(x).
I believe that real number theory is based upon the presumed equivalence of these two ideas; but a finitist would say, the definition is metaphysical if the function does not terminate with a static result unaffected by further iteration of the functional process. This is just the meaning of certain real numbers being defined by the apparent convergence of functional numerical expressions.
Linear algebra and combinatorial number theory are very interesting though in that matrix multiplications can be understood as the activity of one matrix upon another compatible matrix. Even though on its surface a matrix is just a grid of numbers, their ordering and the actions performed on them provide an outline of generalized combinatorial analysis and productions in the form of lists of numbers.
But of course the action of matrix operations is not dictated by the list of numbers, only the memory heuristic of the system performing the operations.
I have no conclusions about these ideas, my previous ideas having been overturned too often to encourage my attachment to the newer ones. But it's a really interesting paradox, action versus choice.
@@haniamritdas4725 Nice response. I'm not traditionally disciplined in field rigor so I apologize ahead of time for where that shows.
¶2: To see if I have it by analogy, I'm imaging a group of mutually-attractive particles floating in space. Their final, object form (when they've all converged into a central mass) is the output, while the paths taken could be the definition of "the assembly function". Implicit and explicit expressions come to mind here too, perhaps as a _process statement_ vs a wrapped instance of _output-specified_ .
¶3: Are you alluding to irrationals and transcendentals? Recursion and infinite sequences?
¶4: I'm currently reviewing Mathemaniac's recent video _"Matrix transpose isn't just swapping rows and columns"_ which seems relevant here. Taken with ¶2, moving the basis vectors seems to be related to changing how one arrives at certain outputs, though my intuition hints this doesn't change the "truth" of some "master" function resulting in that output. [reviewing what I just wrote, I am relying on some personal research that would help make my case...but it probably doesn't belong in a comment. I hope what I've written isn't confusing / is worth stating regardless]
¶5: "memory heuristic" : A...heuristically-developed personal interpretation _(I, ah, Googled to make a better guess)_ leads me to think you mean...:? a procedural/stored/imperative process, vs a declarative one. To use coding: C++ is imperative -- like a matrix having fixed parameters, static algorithm and deterministic output -- while SQL/Gradle are declarative...perhaps even stochastic both in parameters and algorithm...where as long as the result is a permutation of the goal, it's acceptable. In case I went down a rabbit hole, I'll stop there and inquire if I'm close.
¶6: I like what you wrote here. At first I read: "discouraged by the flux" but now I think it's more like "it is what it is".
~ Thanks for giving me something to think about.
3:09
It's funny thinking Bri is using a Canadian accent for this example.
"If one envelope contains an amount, eh? and the other contains twice as much, two, eh?"
@@vteegalapalli r/woooosh
The problem with this is that the value of x is being treated as though it is predetermined before you make your choice, which it is not. x depends on your choice. If the amounts of money in the envelopes are A and 2A, you have a 0.5 chance of picking the one with A in it, and assigning the value of A to x. Likewise for the one with 2A, you have a 0.5 chance of assigning x:=2A. When you do your probability calculation to calculate the expected value of the money you get from switching, the x in the first term is not the same number as the x in the second term, yet you treated it is if it was.
I was rather surprised to hear there is no widely accepted resolution to this supposed paradox. It's just a case of rigorously checking all the implicit assumptions.
I was seeing if someone already said this. This is the one which I agree with
This! I thought the same. I'm surprised this is an issue at all...
I like these probability situations, but they are not good for my mental health. I have spent years pondering situations considerably simpler than this one. William James might say, "You're going to get money either way, so don't ruin it by twisting your brain in a knot."
if one of the envelopes says "geico" you should probably switch ...
Very fun problem.
Keep up your great work.
Thanks, will do!
3rd scenario: Each envelope contains $0 therefore the choice is meaningless
My friend did this with me but they were 2 empty envelopes
I feel like assigning the value of the other envelope relative to the envelope you have is a mathematical curiosity but isn't really germane.
In turn, the expected value based on geometric average is x
It isn't obvious which average is more reliable in this case: some averages produce more optimistic predictions, some - less optimistic
RMS >= AM >= GM >= HM
AM Inequality = Aunty Man Inequality
GM Inequality = Girl Man Inequality
HM Inequality = Housewife Man Inequality
My first vid of yours is dividing by 0 it was a year ago it was very awesome cheers brian!
trick is simple, one choice is unbalanced. Try to solve for x and y not not 2x or x/2 giving either two options non 50 % value.
The Monty Hall problem is most mind boggling problem. Dependent Events and Independent Events in statistics.
I once had this as homework in statistical mechanics😂
It all depend what are the rules. If two values are generate at the same moment before you choose,switching may give or take the same value. But if the first value is generated before you choose and then second value is generated, you should always switch to the second option.
That's a nice one, because it is an actual resolution for the "wrong" probability instead of only presenting an approach that suits intuition.
I'm intrigued by what Wiktor says, but admit I don't understand.
If "two values are generated at the same moment before you choose" does that mean for example that someone puts $10 in one envelope and $20 in the other so you can't see which is which, and then asks you to choose. Yes, in this case choosing then switching "may give or take the same value", by which I think you mean that you're just as likely to lose $10 as gain $10. I agree with that.
But what does it mean if the second value is generated 𝘢𝘧𝘵𝘦𝘳 you choose? For example, does the person who puts the money in the envelopes know what you chose and uses that information to decide what to put in the second envelope before you switch?
@@chrisg3030 yeah, basically it means "after your first choice"
Say you choose A and so $10 are put into it. If only now a coin is flipped to decide whether $5 or $20 are put into B, you should switch.
@@chrisg3030 peter sievert is right
@@petersievert6830 So after you've chosen an envelope A, another person puts $10 into it. Do you know this?. If you know, then it's the same as starting with $10 that's not in any envelope at all. Yes, I agree you should switch it for envelope B, if that has either $5 or $20 in it with equal probability. Have I understood you? But with these rules it's really a one envelope game.
What if you never know directly what's in envelope A? But if you know that B has $5 or $20 in it, you can infer that A has $10. So you're in the same position.
But if you don't know what's in A and don't know what's in B, only that B has half or double A, then you're in a different position. If B has double and you switch, you gain, and if it has half, you lose, but by the same amount. So the advantage and disadvantage to switching cancel each other.
Imagine the information not shared to us was that the envelopes have ₹100 and ₹50 (for example)
If we choose ₹100 envelope, then saying that by switching we have 50-50 chance of having either 2x or x/2 money is just wrong
The envelopes don't contain ₹100 and ₹200, so there is no 2x
In other words, assigning the other envelope x/2 and 2x money actually uses two different x's
Its not the same variable
Another way to think about this is by definition, allotting the heavier envelope as x, then saying we have 50-50 chance of either choosing x or x/2;
Which would be correct since there is no 2x here, same as our example
The variable, x, is itself a random variable, which isn't accounted for.
We have v1 = x, v2 = {2x, 1/2x}, but we don't admit that x = {A, 2A}
So v2 = 2x only when x=A, and v2 = 1/2x only when x=2A, thus we have conditional probabilities.
EV(v2) = P(v2=2x)*2x + P(v2=1/2x)*1/2x
= P(v2=2x|x=A)*P(x=A)*2A + P(v2=2x|x=2A)*P(x=2A)*2(2A) + P(v2=1/2x|x=A)*P(x=A)*1/2A + P(v2=1/2x|x=2A)*P(x=2A)*1/2(2A)
= 1*1/2*2A + 0 + 0 + 1*1/2*1A
= 3/2 A
Comparing that to the EV(v1), we see there's no benefit of switching.
5:09 What's the problem with assigning probabilities to events which have already been determined? Isn't that a major part of Bayesian probability? If I flip a fair coin and don't look at it, can I now no longer assign a probability to the result because it's already determined? Bayesian probability is based on the knowledge I have, and since I haven't gained any knowledge by flipping the coin and not looking at it, the probability of heads/tails is still 50/50.
Finally 45670,50$ that teacher owes us
the initial calculation of the expected value is more like the situation of
"you have a bucket of money and you get to flip a coin heads u double the money, tails u half the money"
which of course given the choice you just keep flipping that coin and enjoy the sweet rewards of slightly better odds of gaining money
Love this! Any more problems/paradoxes like this?
Here's how see the problem:
One of the envelopes has the money x, and the other has 2x. We select one envelope and we have a 50% chance of choosing x and a 50% of choosing 2x. Let's consider both cases individually:
CASE I (if our randomly selected envelope has amount 'x')
We can either opt to stay with our envelope or switch, thus we have 2 possible outcomes (which I'll denote by R)
R1 (we switch) -> we get 2x
R2 (we don't switch) -> we get x
CASE II (if our randomly selected envelope contains '2x')
Again we can switch or keep
R3 (we switch) -> we get x
R4 (we don't switch) -> we get 2x
So basically there are 4 outcomes, and we still have an equal chance of winning x or 2x. Then how is switching better? Am I missing something?
The mistake is separating by "x is the smaller value". Because both times you call it x, but one time it's less than the other. This leads to losing half of x being the same amount as gaining a full x. Because one time x is 2x, basically. You should have separated by "I'm on the smaller value". This would allow to factor in the relative sizes.
This would basically lead to resolution 1, but I wanted to point out the mistake in the initial way of viewing this.
This isn't a paradox. It's a misunderstanding that comes from reusing 'x' to mean different things in two different situations, and then trying to combine expressions as though they represented the same thing.
The other envelope only contains x/2 if x is the larger value. The other envelope only contains 2x if x is the smaller value. Those aren't the same x.
Label the envelopes in a consistent manner and say they always have x and 2x. Then you have four situations:
1) you picked x and did not choose to switch: you end with x
2) you picked 2x and did not choose to switch: you end with 2x
3) you picked x and chose to switch: you end with 2x
4) you picked 2x and chose to switch: you end with x
The expected value of what you picked is the mean: (x+2x+2x+x)/4 = 1.5x.
Asking for E(x) when x itself is dependent on what you picked is what leads to the confusion. If you call the money value x, then what you want is E(envelope) in terms of x.
I heartily agree when you say the envelopes always have x and 2x. Those are the two possible contents of either envelope you pick. Or you can say the envelopes always have x and x/2 if you like. But to talk about either of the two envelopes containing x, 2x, or x/2 is to pose a different problem. That problem isn't necessarily one with two values for x. It could keep the same value for x, but be better called a three rather than two envelope problem, which I'll be happy to discuss with you later if you like.
Where I diverge now is when you (and many others) only average out 𝘳𝘦𝘵𝘶𝘳𝘯𝘴 from switching and not switching. If you also average out 𝘨𝘢𝘪𝘯𝘴 from switching and not switching then then the four situations are
1) you picked x and did not choose to switch: you end with x. Gain: 0
2) you picked 2x and did not choose to switch: you end with 2x. Gain 0
3) you picked x and chose to switch: you end with 2x. Gain x
4) you picked 2x and chose to switch: you end with x. Gain -x (that is, you lose x)
Mean gain: (0 + 0 + x + -x)/4 = 0. And in particular the mean gain from switching is 0. Nothing to be expected from switching.
Absolutely, I wrote a similar comment before I read yours
Assigning 2x and x/2 is logically correct without any problem. The real explanation is E[XY] = E[X]*E[Y] only works when X and Y are independent.
X = Your original choice = 1 or 2 with equal chance.
Y = The other envelope / the envelope you originally choose = 2 or 1/2 with equal chance.
It is true that E[Y] = 1.25, but what you get is E[XY] not E[X]*E[Y]. When Y=2, X always=1. When y=1/2, X always=2. So E[XY]=1/2*2*1+1/2*1/2*2=1.5 same as E[X] = 1.5
Summary:
Original choice E[X]=1.5
Switching multiplier E[Y]=1.25
After switching E[XY]=1.5
E[XY] =/= E[X]*E[Y] because X, Y are dependent.
I like to think this going to the extreme scenario. Lets say that in the first envelope there are 10$. In the second envelope could be twice or half, 20$ or 5$. If you are lucky with the change, you will win 10$ extra, lose 5$ otherwise (win -5$). Now lets take a bigger step. In the first envelope still 10$, but in the second one will be 100$ or 1$. Changing envelopes will gives you 90$ or -9$. Now the risk seems lower, right? Finally go to the extremest case. As always are 10$ in the first envelope, but in the second one can have an infinity amout of money with the risk of losing all the 10$. Who in his mind would'nt take that risk?
As you can see, in this three cases the odds are equal, but the bigger numbers shows how worth it can be.
I like this because you clearly are forcing us to recognize the real didactic.
I don’t believe you would unintentionally confuse probability and value.
It poses an independent event as a weighted event. It doesn’t matter whether the second envelope has 2x or [random]x.
What if there was 23 times as much in one envelope as the other, or one envelope contained a $1 bill and the other contained a $100 bill, or one had a piece of paper that read 0.577 (or goat) and the other read 3.141 (or car)?
It would make no difference. The amount inside the envelope is irrelevant. I agree this is not the Monty Hall problem for many reasons.
And it is also not phrased as a psychology problem.
But it’s also not a paradox. The problem intentionally confuses [inherent value] with [probability].
Here is an answer by analogy. You have two fair die, one is marked with only one dot on each side and the other is marked with only two dots on each side.
You can’t see either die before or while you throw it and you can’t see the result of the first throw. (It’s irrelevant whether you see the result of the second throw.)
Do you improve your odds of “success” by throwing the second die? Does it matter whether one of die has sides of two dots of 200 or 2,000?
It doesn’t matter whether you “pick a hand “, choose an envelope, throw a die of this type, or any other choice between fair “a or b” items. The amount contained in or on each is irrelevant and that’s where the problem falls off the rails.
2:04 What the heck is going on? How does that make any sense?!
At 1:56 Bri says "In other words our expected value of switching is 25% more than keeping the envelope we originally chose"
Yes but only if we happened to originally choose the envelope with x in it. If instead we happen with equal probability to pick the one with x/2 then the expected value of switching is 100% more than the original envelope, doing the same math. Wow!
But here comes the nemesis. Suppose the original choice of envelope was the third possibility, 2x. Then calculated the same way, the EV of switching is 125% less, yes, less than keeping that envelope. I'm happy to go into how I got those numbers and what they mean with anybody, but for the time being just note that 25% + 100% - 125% = 0%
Once again we come down to a point I've been driving at in other comments, you've got to consistently assess all the possibilities you enumerated.
Plot twist: the envelopes both contain 0 money and you wasted 2 weeks thinking which you are gonna pick
I think this is a weirder variation of Monty Hall puzzle. Can you guide us about it (plus your view?)? (Short please)
Alice: Hey Bob. Here's an envelope with some money in it. It's yours.
Bob: Thanks. How much?
Alice: I'm not going to tell you. But before you open it you can switch it for this second envelope which has either half or double yours.
Bob: But if the second envelope has half or double mine, doesn't that mean mine has half or double the second? So where's the advantage?
Alice: Good thinking Bob. I'm impressed. Look at it this way. Let's call the amount in your envelope "x".
Bob: I'm with you.
Alice: So the second has either half x or double x.
Bob: So if I switch I could get only half x in return, which means I lost half x.
Alice: Yes, but you're equally likely to get double x in return, which means you gained x. You stand to gain double what you stand to lose with equal probability.
Bob: Cool.
Alice: What's more, I'll let you switch and switch back as often as you like. So you stand to gain double what you stand to lose each time! So if half the switches cause you to lose half x each time, you nevertheless gain twice as much each time with the other half of the switches.
Bob: So say I do 100 switches and I lose $1 with each of 50 switches, I gain $2 with each of the other 50. Mega cool. Let's start now.
Alice: Not so fast Bob. Seeing as I'm giving you such a great deal I think I deserve a small something in return.
Bob: Like what?
Alice: Just a trivial sum per switch. Let's negotiate . . .
Schrödinger's envelope.
I came in here champing at the bit, but it seems you might agree with my initial assessment: no matter how you dance with the numbers, you can pick "less" or "more", and if you don't know which you have, switching isn't doing a damn to your chances.
Another issue here is the statement "given you see X in your envelope there is a 50% chance the other envelope has 2X and a 50% chance the other envelope has X/2" isn't actually possible unless the probability distribution of the random Xs is such that it could happen. For example, there is no uniform random distribution of the Natural numbers such that every number has equal probability of being chosen. So if you were to assume that the envelopes have positive integer amounts of money then if you see $1 in your envelope then there is a 100% chance the other envelope has $2, not a 50% chance. And no matter what distribution you use for the Natural numbers it will always be the case some numbers are more likely than others so you can never have a situation where literally every Natural number in your envelope results in the same 50/50 chance of the other number being twice its value or half its value.
For a similar reason there's no uniform probability density of the positive real numbers either, so even if you allow the envelopes to have non-integer positive or even irrational amounts you still run into an issue where not every number can have a 50% chance of being twice or half the other number in the other envelope.
I literally did a million test runs on this and it turns out that there is no difference between switching and not switching, which is exactly what you'd expect.
I remember watching this in a movie about casinos I forget what it called lol
Mathematically the probability adds but in reality its always 50/50 unless you get some sort of extra info before you switch
Schrödinger's envelope
Kind of reminds me of Schrödinger’s cat in a way
Both have the greater and both have the lesser amount of money at the same time until it is known which is which
I think the problem is when you assign x to the money in one envelope, in the case where you stand to gain, that x is smaller than in the case where you stand to lose, so while you stand to gain 100% vs lose 50%, you're potentially gaining 100% of a smaller amount of money. The problem is that you assume x to be the same in both cases, when it isn't.
Nobody:
Me: *weighs envelopes*
Everybody: 🤯
Before getting philosophic, I propose stating the problem more clearly:
There is a total amount of money, x, unevenly distributed into two envelops, one envelop containing twice as much as the other. Let y be the amount in the envelop with less money, then the other envelop contains 2y, such that x = 2y +y = 3y.
Now, let's consider two probabilistic experiments:
A) A person can pick any of the two envelops with equal probability and keep the amount of money in such envelop.
B) A person can pick any of the two envelops with equal probability and then interchange the envelops, keeping the amount of money in the second envelop.
The question is: In each of the experiments A) and B), what's the expected value in terms of the money kept by the person picking the envelop?
For case A) we have the following independent events with equal probabilities:
- A1) Choosing the envelop containing amount y with probability 1/2, the event having associated value y
- A2) Choosing the envelop containing amount 2y with probability 1/2, the event having associated value 2y
Therefore the expected value of experiment A) is E_A = 1/2 (y) + 1/2 (2y) = 1/2 (3y) = 3/2 y. Since y = x/3, E_A = (3/2) (x/3) = x/2.
In other words, the expected value in case A is getting half of the total amount. If you think about it, it makes total sense because there are exactly two envelops, and even when the money is unevenly distributed, since picking any of the two envelops has the same probability, the higher amount obtained when choosing the envelop with more money is compensated by the lower amount obtained when choosing the other. This is true for any other proportion used to divide the money among the envelops, as long as their sum continues to be x. Guess what happens if instead of two envelops you divided the x dollars into three different envelops, again irrespective of the proportion?
Back to the topic of the video, for case B) we have, "on the other hand", the following independent events with equal probabilities:
- B1) Choosing the envelop containing amount y with probability 1/2 and then switching to the other envelop, therefore the event having associated value 2y.
- B2) Choosing the envelop containing amount 2y with probability 1/2 and then switching to the other envelop, therefore having associated value y.
The expected value for experiment B) is thus E_B = 1/2 (2y) + 1/2 (y), but addition is commutative, right? Therefore E_B = E_A; Q.E.D.
just take both and run away
Plot twist: there was no money in the envelope
*Ok, Now Give Me Money*
I have some very definite views on this problem. But I keep switching them.
What's more I expect their value to increase each time.
How does this problem change if there are more than 2 envelopes? And an uneven number of outcomes overall? eg. 70% of the envelopes contain twice what you originally selected, 30% contain the same amount as you chose initially? What if you choose one envelope initially and then a choice of 0,1 or 2 further envelopes?
Damn, you are obligating me to subscribe.
Welcome aboard :)
I think the important point is that the probability choice happened when you picked the 1st envelope. After that, there is no probability to the second choice. You are definitely picking between what you chose and the opposite. Whichever envelope you choose at 1st, the second choice is between that or its opposite. So your 1st choice has a 50/50 for the better or worse option. If you switch, you are just switching to the other 50%.
So the probabilities calculations are like this:
The probability that you chose A is 50%, while the probability you chose 2A is 50%.
If you choose to switch:
The probability you chose A is 50%, times the probability the envelope you switch to is 2A is 100%, yields 50%.
The probability you chose 2A is 50%, times the probability the envelope you switch to is A is 100%, yields 50%.
Either way, it's 50/50.
One small error - there is no expected loss... If you start with 0 dollars and pick an envelope, regardless of which one you chose, you come out on top!!
In the first analysis, it's not 1/2 of more and 1/2 of less; it has to be 1/3 of more and 2/3 of less. Whatever the distribution of values is, the probability has to decrease as the value approaches infinity.
P.S. I'm not sure how to fully make sense of this...
couldnt this be empirically tested with simulations??
Or you can realize that probability and statistics deal with a set of lots of events, not a single one. The 1.25x as a expected value means that If you repeat this event a lot of time, like a really lot, and you sum the amount of everyone that switched to be close to 1.25 times the amount in regards to people who didn't, because If you switch and gain, you gain more than you lose, If you switch and lose. This adds up on the long run...
you don't know that this isn't quantum entangled money which materialises at the instant you open your envelope
hi i think this problem is from vsause2 and you add a little bit of your ideas in it, I liked this problem because it tells us how to use probability correctly
vsause did not invent the envelope paradox...
@@MeButOnTheInternet ik but he covered it before bri did
I think VSauce talked about Monte, no? Do we have a link?
What if they coin flip after you switch, heads double, tails half?
Before I watch the video, I will attempt to find the fault in what he said.
The problem is that x isn't constant. Choosing the smaller one means x is a smaller value and choosing the larger one means x is a larger value.
Let's say one has 50$ and the other has 100$. The average between choosing either of these is 75, and we can see that 50 +100/4 is indeed 75. However, we use two different values for x.
So a better way to show it is, let's say one has x and the other has 2x. You will always have the probability x + x/2 no matter what.
That's what I think isk
So around 31.25% to 50%. 40.625% you pick the right envelope.
OK now how do i remember?
The number pi:_am I a joke to you?
It seems like your logic is...A-ok!
Coincidence is the law of the universe
When u pick between 2 envelopes there are 2 porbabilities so 50% probability and when u add the switch or not the probabilities change to 4 probabilities so 25% so would you switch or not
now if you pick the other envelope, a will give you a new envelope. This new envelope will have either 2* the amount in that envelope or 1/2 the amount in that envelope. If you change your mind, i will give you another envelope. What is the most optimal money gaining strategy?
the problem with the x-solution is that your x changes its value and you showed it with the A-solution. You cannot set x to $1000 and $2000 in the same equation. x has to be either 1000 or 2000 but not both at the same time :) when 1000 = 2000 = x everything is possible and provable, even that 0 = 1.
what if the envelopes had money of 0 value (monopoly money for example) and 2 × 0 = 0
which would make it a 0% gain either way and we were trolled all along
Plot twist: both envelopes have $0
The problem with the formula is that when switching to the lesser envelope you assume that the amounts where x and x/2 while when switching to the higher envelope you assumed that the values were x and 2x. And then all is put in one formula. So the formula implies that when you picked the lesser amount, both values were half as high as otherwise. Short: You only can lose x/2 but you can win 2x. Of course then it looks like it is better to switch! Say, your mate put in 4€ and 8€. You choose one. Now the math says when you switch there is a 50% chance to get 2x or 0.5x. If you picked the 4€ then x=4€ otherwise x=8€. So you can't have one formula like E=0.5*(x/2) + 0.5*(2x) where you mix both different values for x. The x in (x/2) actually is 8€ and the x in 2x is 4€. So the expected value is 0.5*(8€/2) + 0.5*(2*4€) = 6€ which is correct and shows that it does not make a difference if you switch or not.
Another way would be to say that the amounts are x and 2x. so switching gives you 0.5*(x) + 0.5*(2x) = 1.5x. For x=4€ that is again 6€.
You could say now that the rewards are not fixed at 4€ and 8€ so no matter if you chose the smaller or the bigger, x could be the same. This is not true. Because you pick after the money was put. So x can't be the same no matter which envelope you picked. But we can change the experiment a little bit! You get the envelope with 4€ and your mate secretly puts 8€ or 2€ in the other. So the expected reward for switching is 0.5*8€+0.5*2€ = 5€. So indeed it is better for you to switch. And it really is.
Another game to make it even more clear: There are 3 marbles. You can pick 2 or 1. Then you get the rest. What is better to pick in the beginning? So let's say you picked x marbles. Then in the second step you get 2x if you chose 1 marble and only x/2 if you chose 2 marbles. So if you pick 1 marble you end up with 3x otherwise only 1.5x. So you should pick one marble!
That is of course wrong. x is different depending on what you chose, so you can't compare 3x and 1.5x. In the video both x were then even put in one formula.
@@Clyntax I see merit in thinking of 3 marbles because that's the number of equally probable possibilities for the first envelope in the original problem. That's right, not 1, not 2, but 3. They are x, x/2, 2x. So
if it contains x, then the second contains x/2 or 2x. If you switch then you stand to lose x/2 or gain x.
if it contains x/2, then the second contains x or 2x. If you switch then you stand to gain x/2 or gain 3x/2.
if it contains 2x, then the second contains x or x/2. If you switch then you stand to lose x or lose 3x/2.
The sum of all possible losses is x/2+x+3x/2=3x. The sum of all possible gains is x+x/2+3x/2=3x. So no advantage in switching overall.
The mistake is to consider only the first of the 3 possibilities, namely that the first envelope contains x. It's the only one where you can have either a gain or a loss on switching, and since the gain is bigger you naturally go with the switch. But consider all 3 together, then possible gains and losses are balanced. Why do we make that mistake? Maybe because once we stipulate the first envelope contains x, then we think we aren't allowed to consider any other possibilities for that same envelope. In fact we have to.
Which envelope is heavier?
People do Bayesian calculations all the time. Let's say you opened your first envelop and you found a $10 bill. Then the state of the world can either be ($5, $10) or ($10, $20). Now what's the posterior probability of the state of the world being ($10, $20) and what's the posterior of the state of the world being ($5, $10)?
This is siiikar to schrodinger a cat, until you open the other envelope you both have and don’t have the largest amount of money.
You need to account for type 1 and type 2 error, that you switch and end up with less or that you don’t switch and end up with less. The cost to switching and being wrong is 50% whatever was in your envelope, the cost of not switching and having the lesser amount is costing you hundred percent. Not really supposed to use percentages with negative numbers but hey. …
How about the ratio of the money in two envelopes unknown so that A is partitioned two parts of a with an unknown fraction should it effect the choice strategy or make a difference?
more Monty Hall than Monty Hall
It has a similar feel!
The thumbnail and Video title keeps updating.
No wait, here's the problem. When we're doing the probabilities and we hit the operation "2x + x/2", the first x is not the same as the second x, because x is however much money is in the envelope you have chosen. Let's say the envelopes contain 10 dollars and 20 dollars: then "2x + x/2" is 2*10 + 20/2.
Which means the correct way to do it is…
The correct way to do it is to also factor in the amount in the envelope we're holding. So it's not "2x + x/2", it's more like "2a + b/2", where b = 2a.
@@kingbeauregard So 2a+2a/2=2a+a=3a
And then 3a/2 because 50% chance.
So 1.5a
@@Cobalt_Spirit Yes. But what you don't know is whether you have the envelope with "a" or the one with "b" (which equals "2a"). 1.5a is 1.5 times the lower value, regardless of which envelope you have. We could just as correctly say that the average value is 0.75b.
I agree that's one way to look at it. Another way is to acknowledge each envelope has three, not two possible amounts, they are x, x/2, 2x. Here x keeps the same value whatever that is.
If the first envelope contains x, then the other contains x/2 or 2x. If you switch then you lose either x/2 or gain x.
But you have to consider the other two possibilities for the first envelope:
If it contains x/2, then the other contains x or 2x. If you switch then you gain either x/2 or gain 3x/2.
If it contains 2x, the the other contains x/2 or x. If you switch then you lose x/2 or you lose 3x/2.
Total gains and losses with the 6 possible outcomes are equal, they cancel. No advantage to be expected from switching. You could call this a 3-posssibility 2-envelope problem. But I think it's simpler if you just put each of the three possible amounts into its own envelope and call it a 3 envelope problem. If you switch one envelope you get one of the other two at random.
What's the two envelope problem then? I have my idea, but I'd like to hear anyone else's.
5:13 and yet, all it serves to do is summarize what we don't know about our current situation, why would it be disingenuous? Is it just to criticize the very attempt at summarizing it into probability?
I think the problem here is confusing x as being both a random variable and a definite but arbitrary number:
Given the description, it's quite obvious that the sample space in this scenario has just two events: you picked the envelope with less money, or the one with more (call these events a and b respectively).
Let X be a R.V. corresponding to the amount of money in the envelope you picked (recall a random variable is a function that maps the elements in the event space to real numbers).
We can define the contents of the other envelope in terms of X: Let Y be an R.V. corresponding to the amount of money in the other envelope, then Y(a)=2X(a), Y(b)=X(b)/2.
The expected value of switching
E(Y) = Pr(a)Y(a) + Pr(b)Y(b)
= Pr(a)2X(a) + Pr(b)X(b)/2
=X(a)+X(b)/4
You may notice that we cannot combine the fractions in this state, since X(a) =/= X(b).
There is a way to solve this properly from this point though, we can say that X(b)=2X(a) from the problem definition (the amount of money in the envelope you picked if you picked the one with more money is twice as much as if you picked the one with less money). Then,
E(Y) = X(a)+2X(a)/4
=1.5X(a)
, and
E(X) = X(a)/2 + X(b)/2
= X(a)/2 + 2X(a)/4
=1.5X(a)
Therefore E(X) = E(Y), both expected values are the same.
You may still be left wondering "well what step precisely is false if we let x be a number rather than a random variable?".
Let's walk through the steps and lay out the statements formally:
0:49 Let X = x, where X is a R.V. corresponding to the amount of money in the envelope we picked, and x is an arbitrary element in the range of X
0:53 Let Y be an R.V. corresponding to the amount of money in the other envelope, defined as follows, Y(a)=2x if event a occured, Y(b)=x/2 if event b occured
1:05 Pr(a) = 1/2, Pr(b) = 1/2
1:24 E(X) = x
1:30 Pr(Y=2x) = 1/2 and Pr(Y=x/2) = 1/2
1:37 E(Y) = Pr(a)Y(a) + Pr(b)Y(b)
1:44 = 1/2 * 2x + 1/2 * x/2 = x + x/4
1:48 = 5x/4 = 1.25x
Did you catch that?
Recall that x is an arbitrary element in the range of X, so it must be either X(a) or X(b)
At 0:53 we had different conditions for Y(a) and Y(b). If event a occured, then Y(a) = 2x, but if event a occured, then x must be X(a), and Y(a) = 2X(a). Similarly, If event b occured, then Y(b) = x/2, but if event b occured, then x must be X(b), and Y(b)=X(b)/2.
But at 0:13 we were told that X(b) = 2X(a), and at 0:07 that X(a) =/= 0 and X(b) =/= 0, so it must be that X(a) =/= X(b)
Thus, the two x variables we were working with are not actually the same number, and their sum is not 1.25x but rather 1.5*X(a), or alternatively 0.75*X(b). The last step is wrong.
Essentially the mistake comes from seemingly defining Y in terms of x (a fixed arbitrary variable) when in fact the definition given is in terms of X (a random variable). Then somewhere along the way we forgot that we had left conditions on x, and assumed that wherever x appeared it would be the same.
Given the problem description, you cannot simply drop the conditions. If you drop the conditions, you are working with a different problem entirely.
The analysis would be accurate if the problem were instead something more like:
"Here is an envelope, it contains some cash. Now, I have two other envelopes: one with twice as much cash as the one I gave you, and one with half as much. Would you like to trade me your envelope for one of the others?" (in which case take the deal)
its all fun and joy until A = 0
Solution: Punch the guy offering the envelopes. While he's trying to figure out what's going on , take the envelopes and run off. This guarantees more than X dollars.
Math, for some reason, doesn't like these lateral thinking solutions. But it's just as valid as any other solution.
monty hall vibes
Yes, especially from that one prince claimed to be from Nigeria
I think I know what's going on. Those 50-50 probabilities are actually conditional probabilities. If we call the respective values in envelopes 1 and 2 E1 and E2, using notation, the affirmative P(E2=2x) = P(E2=x/2) = 0.5 is actually wrong, or at least doesn't make any sense. What's the probability that envelope 1 has 5 dollars for example? It isn't defined.
What should be written is P(E2=2x | E1=x) = P(E2=x/2 | E1=x) = 0.5. Other than that, all you have are probabilities for relations, namely P(E2=2*E1) = P(E1=2*E2)=0.5.
So the correct reasoning for the expected value of money you would get by switching to envelope 2 would be:
E2*P(E2=2*E1)+E2*P(E1=2*E2)=E2*0.5+E2*0.5 = E2 (which leads you nowhere).
Now if you choose envelope 1, and THEN you were told: "THIS ENVELOPE CONTAINS x DOLLARS"
Then the situation changes. now you have additional information and you can calculate the expected value you'd get by switching, and indeed it would be 1.25x, i.e., you have an advantage by switching.
To understand that switching given that you know the amount contained in the envelope you selected is advantageous, consider the analogous situation:
You have two options:
1. select an envelope with a known amount x, or
2. select a "lottery" envelope that has either 2x or x/2 with equal probability,
you'd expect to gain more by selecting the lottery envelope.
"you'd expect to gain more by selecting the lottery envelope." Yes, but what's almost always omitted is that if you don't know what's in either envelope, then there's an equal chance that it's E1 that has x/2 or 2x, and E2 that contains x, and so the advantage of switching - however you calculated it - must surely be exactly cancelled or reversed.
My balance sheet uses possible net gains and losses on the switch rather than expected value:
A) If E1 has x then switching to E2 will get you x/2 or 2x. So you stand to gain -x/2 (that is, incur a net loss of x/2) or x.
B) E1 has x/2or 2x then switching to E2 will get you x. So you stand to gain x/2 or -x.
Possible gains obviously sum to 0. Gains and losses cancel. No advantage to switching.
Why do we tend to make the mistake of only considering case A? Maybe because once we've said "let the contents of E1 be x", then we think we can't logically or consistently consider any alternative. In fact we have to.
@@chrisg3030 yes. I think we fool ourselves by thinking just because we named something as "x" it remains an indeterminate quantity. Actually, when we say E1=x it's just as determined as saying E1=50 cents.
@@Felipe-sw8wp Yep, they're both as equally determinate or indeterminate as each other. It's as if the x is also some unit of currency in some fictional country just like the dollar or euro. Its algebraic use is irrelevant here.
You could write x on a piece of paper and put it in an envelope, and then put another piece of paper with x/2 or 2x written on it in another, and it would work just as if you did the same with $10, $5, and $20 bills instead.
This brings me to another possible reason for not considering case B. People can visualize setting the game up by putting, say, $10 in an envelope. But how do you put $20 or $5 in one envelope? There's no sum of money called "20 or 5 dollars". A pizza parlor doesn't have that as a price on the menu board unless they mean big and regular or something. I guess there are work-arounds for that, like putting each sum into its own envelope, shuffling, and picking at random.
Cool Problem😁
4:20, chosing? i think you meant choosing
But then you have a 1.25 probability of being incorrect also, by the same reasoning applied to choosing the lowest envelope instead.
Instead of saying, "What is the probability this is the higher envelope," we ask, "What is the probability this is the lower envelope," the probability would be the same, as the question is the same. So, you have a 125% chance of it being the higher and 125% chance of it being the lower, whatever your guess. So, you're still stuck with the base of 1:1 whether you switch or not.
2.5=2.5
Why doesn’t anyone just run a computer sim where person A switches 100% of the time, and person B never switches?
Run the sim 10,000 times and see whichever person gets more money at the end.
It's pretty obvious it doesn't contribute.
The money in the envelopes is predetermined therefore changing your choice wont help u
@@roeekawaz1372 to my shame, I made this comment before he got to the explanation, and then forgot about this comment
Still, it’s weird that I’ve watched probably 5 videos on this ‘paradox’ and not one even mentions using computer simulations to model it
Does it differ if the ratio of money is unknown in the envelopes like an unknown fraction I mean?
you select either one you dont get any money so whyy bother, but what if A + 0