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Title: Two letters containing money, but... Post by zennehoy on Aug 18th, 2006, 8:26am Considering the amount of discussion we have had over this puzzle at work it should probably go in hard, but then again maybe we're just no good at puzzles... Your uncle presents you with two letters for your birthday, and asks you to choose and open one of them. You decide to humor him, and having opened your chosen letter discover $100. With a roguish grin, your uncle now tells you that one of the letters contains twice the amount in the other (i.e. the other letter contains either $50 or $200), and asks if you would like to switch. Would you do so? Why? My apologies if this puzzle or a similar one has been discussed before; I couldn't find it. Even if it has, perhaps there's a few of you who haven't encountered it. Enjoy! Zen |
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Title: Re: Two letters containing money, but... Post by honkyboy on Aug 18th, 2006, 9:07am The answer depends on what value you place on the money. If you're in great need of that $100, perhaps you've had your eye on a $95 dollar video game and can now afford it, you may not want to risk switching. On the other hand if it were a $195 game, switching might be the only option to have a change at affording it. [hideb]If you are strictly a gambling man, you should always switch as you will on average end up with %25 percent more money.[/hideb] |
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Title: Re: Two letters containing money, but... Post by zennehoy on Aug 18th, 2006, 9:59am At this point your uncle would jump up and shout with glee: "Hah, gotcha!" (at least to your hidden answer, the remainder is perfectly accurate) I'll let you think on it some more before demonstrating why your answer can't be right, Zen |
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Title: Re: Two letters containing money, but... Post by zennehoy on Aug 18th, 2006, 10:15am Actually, I'll demonstrate it now, and make it hidden: [hide] Let's say your uncle had told you that one letter contains twice as much as the other before you opened the first. We'll assume the letter you chose first has x money in it, meaning that the letter you discarded contains either x/2 or x*2. By your reasoning, switching would increase your payoff to 1.25*x. If you switch, nothing in your situation has really changed however. The letter you now hold has y money in it (by definition of y), the original has either y/2 or y*2 in it. Switching again, our payoff increases to 1.25*y. Rinse, repeat, we have a millionaire![/hide] |
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Title: Re: Two letters containing money, but... Post by SMQ on Aug 18th, 2006, 10:36am I disagree (and so agree with honkyboy) -- the two situations are not the same. Consider: 1) You are given two envelopes and told: one of these envelopes contains $100; the other contains either $50 or $200 with equal probability. What are the expected values of the envelopes? 2) You are given two envelopes and told: this envelope contains $100; the other contains either $50 or $200 with equal probability. What are the expected values of the envelopes? In the first case, both envelopes have the same expected value, namely ($50 + $100 + $100 + $200) / 4 = $112.50 In the second case, this envelope has an expected value of $100, and the other envelope has an expected value of ($50 + $200) / 2 = $125. So in the original scenario you gave the odds favor switching. --SMQ |
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Title: Re: Two letters containing money, but... Post by Sameer on Aug 18th, 2006, 11:09am this looks very similar to the Monte Hall three door problem!!! |
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Title: Re: Two letters containing money, but... Post by zennehoy on Aug 18th, 2006, 11:16am First of all, you are not told what either envelope contains. You are only given a relationship of the amount contained within them. Thus opening the first envelope will give you no usefull information on the state of the system (other than that the other envelope contains exactly twice or half the amount you now hold, but the number itself is irrellevant since you can always define the envelope you are holding to contain 1, whether you've opened it or not). I don't think it makes a difference whether you gain the relationship knowledge before or after opening (or even selecting) the first envelope either. I'll admit that the statement of the other containing either $50 or $200 was misleading [hide](however, note that no probabilities are given)[/hide], Zen on 08/18/06 at 11:09:45, Sameer wrote:
That's what I thought too, at first. |
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Title: Re: Two letters containing money, but... Post by honkyboy on Aug 18th, 2006, 12:36pm Suppose that instead your uncle had said "one of the letters contains ONE MILLION times the ammount in the other". You would switch in a heartbeat, but the x and y example you gave still says it doesn't matter. That logic only works if the ammount you stand to gain is equal to the amount you stand to lose. Since |x-x/2)| < |x-2x| you still always want to switch. |
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Title: Re: Two letters containing money, but... Post by zennehoy on Aug 18th, 2006, 1:32pm Ah, but you could have picked the envelope with a million already. That way the other envelope would contain some shavings from a cent... |
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Title: Re: Two letters containing money, but... Post by honkyboy on Aug 18th, 2006, 1:40pm on 08/18/06 at 13:32:31, zennehoy wrote:
certainly - I would take that risk. |
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Title: Re: Two letters containing money, but... Post by SMQ on Aug 18th, 2006, 1:53pm I agree there's something interesting going on here, so let's take the uncle out of the equation for the moment and just look at the odds. (In real life, it all depends on what kind of guy your uncle is...) How about this scenario: You have agreed to participate in an experiment at a federally-funded research institution (i.e. they have plenty of somebody-else's-money to throw around.) You are given two identical envelopes. You are told that: - Each envelope contains an amount of money - One of the envelopes contains twice (or ten-thousand times) as much money as the other, and - After chosing and opening one envelope you can either keep that amount or switch and keep the amount in the other envelope. (Note the slight difference from the original riddle in that you are told all of this up front.) You open one of the envelopes and find $100. Do you stay or switch? Why? I'm beginning to doubt my earlier position on this riddle, as the probabilistic argument honkyboy and I have been giving earlier would seem to say that it is always better to switch... It's definitely not a Monte Hall problem, as the new information (that one of the envelopes contains $100) is of no use to you; in fact, it feels more like "Deal or No Deal (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_general;action=display;num=1141752315)" than "Let's Make a Deal". --SMQ |
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Title: Re: Two letters containing money, but... Post by honkyboy on Aug 18th, 2006, 2:18pm I am also doubting myself. My first impulse always seems to be wrong on these types of problems. ??? A strategy of always switching can't be advantageous. :-/ It is reminding me of the missing dollar riddle, as I am trying to find out what happened to the missing 25%. |
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Title: Re: Two letters containing money, but... Post by Astrix on Aug 18th, 2006, 2:44pm Simple. I know my uncle would never give me $200 for my birthday, so I know the other envelope must contain $50. Besides, he decided what to say after I opened it. Maybe if I had picked the lesser one he'd have just said, Happy Birthday. But because I picked the bigger one he's roguishly trying to trick me into changing my choice. Probabilistically, the problem is that an envelope can not contain an infinite amount of money. Therefore there is some limit, which you don't know, and if the amount in one envelope is more than half of that limit, then you are guaranteed to lose. And the amount you'll lose in those cases is enough to compensate for that odd .25 gain that is so tempting in the lesser cases. Now if one envelope always contains $100, and the other always $50 or $200, and you always pick the $100 first, well then the question is whether the amount in the other envelope is actually random, or if the roguish uncle is choosing an amount based on his devious understanding of how you'll act. |
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Title: Re: Two letters containing money, but... Post by zennehoy on Aug 18th, 2006, 3:02pm on 08/18/06 at 14:44:56, Astrix wrote:
Oh dear, I have once again worded a riddle too eloquently ;) Let's take the wording of SMQ's impersonal federally-funded research institution instead then. That also works well because they have much more money than I could ever place a limit on (or they'd just figure out ways of making more). As SMQ already mentioned, the riddle reads differently in that you have all the information before you start, but I maintain that the point in time at which you acquire this information is irrelevant, Zen |
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Title: Re: Two letters containing money, but... Post by honkyboy on Aug 18th, 2006, 3:42pm Thanks Astrix, I see my error. There must be an upper limit for the amount of money possible to be in an envelope. Obviously zero must be the lower limit. If the larger dollar amount is chosen randomly, then 25% of the time you will be picking an envelope that has a dollar ammount less than 1/2 the upper limit. Without knowing the upper bound, as Astrix pointed out, this is where you lose the extra 25 by always switching. However, if the upper bound is known then you are able to make an informed decision by opening one envelope. If by doubling your dollar amount wont go over the bound, then switch envelopes, otherwise do not switch. So . . . back to the original problem, if you know for a fact that your uncle easily could have forked over $200, you should switch. -yes Regis, this is my final answer. |
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Title: Re: Two letters containing money, but... Post by SWF on Aug 18th, 2006, 6:16pm on 08/18/06 at 13:53:47, SMQ wrote:
This is even more like Envelope Gample (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_hard;action=display;num=1027910317). If you know the probability distribution used to the select the values in each envelope, then the expectation on switching can be computed. Not knowing this makes it a guessing game on what the probabiltity distribution for choosing X was. |
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Title: Re: Two letters containing money, but... Post by Icarus on Aug 18th, 2006, 8:20pm on 08/18/06 at 11:16:01, zennehoy wrote:
on 08/18/06 at 08:26:21, zennehoy wrote:
The fact that you DO know the contents of one of the envelopes when you make your decision makes a difference. But it really matters how the values were chosen in the first place. Your symmetry argument assumes that, before the contents of either envelope are revealed, all possible values are equally likely. But this is impossible - you cannot have a uniform distribution over an unbounded range. Therefore as the values get larger, they must also eventually become less likely to be picked (even if your uncle is Midas). Exactly how this distribution tapers off will determine whether or not switching is a good idea once a value has been determined. In my opinion, the two most realistic scenarios for how the envelope values are picked (assuming Uncle wishes to be fair) are: (1) Uncle decided that $100 was an appropriate size gift, so he chose that for one of his values. Then he flipped a coin to see if the other should hold $50 or $200. (2) Uncle decided that either $50 or $100 would be appropriate for the lower value, so he flipped a coin to see which it would be. He then placed double the amount in the other envelope. In either of these scenarios, honkeyboy and SMQ's analysis holds. By expectation, you would be better off to switch. |
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Title: Re: Two letters containing money, but... Post by zennehoy on Aug 19th, 2006, 12:08am on 08/18/06 at 18:16:44, SWF wrote:
Yes, that's basically it. Not sure why I didn't find it, I'm pretty sure I searched for envelope on the hard page. :-/ on 08/18/06 at 20:20:40, Icarus wrote:
I still don't see how this can be the case. You know the value in the envelope you picked first whether you've opened it or not, namely one unit. Knowing the conversion rate of unit to $ shouldn't make any difference to the problem. on 08/18/06 at 20:20:40, Icarus wrote:
Considerations of how your uncle (or SMQ's FFRI) picked the values in the envelopes are futile, or rather not part of the problem. Taking your first example, why couldn't he just as well have picked $200 first, and then randomly chosen $100 or $400? Without any absolute knowledge of what he can afford, both initial choices for the initial value are equally likely from your perspective. What I'm getting at here: there's a simple solution not making assumptions about the bounds of the values in either envelope. Zen |
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Title: Re: Two letters containing money, but... Post by Icarus on Aug 19th, 2006, 12:23pm Let me ask you this, to see if I can shed some light on it: Suppose you know that your uncle is a rich and eccentric man who has given out gifts from $2 to $2,000,000 before apparently on whims. Upon finding that $100, I would switch. I am near the low end of his giving range, and have good reason to believe it is nearly as likely to get a higher value as a lower, so as has been calculated before, the expectation is in my favor. Now suppose instead I found a check for $1,000,000,000. I know that my uncle's assets are not infinite, no matter how rich he is. And at $1 billion ($1 milliard, for those folks east of the pond), it is likely I am at the top of them. Hence higher values are unlikely, and the other envelope is almost certainly lower. So I am sticking. Somewhere in the middle, I may not be sure. There can be other factors to consider. But the simple fact that these two values give different responses tells me that the value I have already seen is important. And if you next argument is that I don't know anything about the assets of my uncle, then you are mistaken. Even if I had never heard of the guy before, I know upper and lower limits for how much money he has, because I have general knowledge about wealth and how it is distributed. honkeyboy & SMQ were generally correct in their analysis of the $100 envelope. The mistake was assuming that just because this applies to the $100 envelope, that it applies to all envelopes. |
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Title: Re: Two letters containing money, but... Post by honkyboy on Aug 19th, 2006, 12:30pm on 08/19/06 at 00:08:44, zennehoy wrote:
You are correct that without knowledge about the bounds of the values there is just one simple solution. However I don't feel it is correct to say that "Considerations of how your uncle (or SMQ's FFRI) picked the values in the envelopes are futile, or rather not part of the problem" The question is "Would you stay or switch? Why?" Well the answer to that question depends on what assumtions you are able to make about the range and distribution of the money your uncle put in the envelopes. If you can make no assumptions then sure stick with your first envelope. There's no point in switching. However, with some knowledge of the distribution you can gain some information from the first envelope picked. Just for example: If your uncle has been known to gift ammounts anywhere from $10 to $10,000, then a good assumption in this case would be that switching envelopes is beneficial. If you know for a your stingy uncle has never given over $150 to anyone, then don't switch. My first reply was certainly wrong, there is no way to know that switching gives you an advantage. It only may or may not be the case. Though I don't think you can correctly say that switching doesn't help you any more than you can say that it does. The true answer depends on methods behind the choices your uncle made. The problem statement doesn't say anything to allow you to conclude anythingabout them, but in a real life situation I would sure try to. |
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Title: Re: Two letters containing money, but... Post by jollytall on Aug 19th, 2006, 1:20pm Let me give you my take on this. If we take the detailed wording where you know in advance the 1:2 ratio is different from the original puzzle. In the original version you here the extra information after the first is open. But even in this case you still have to know the intention behind. He might be greedy and save on you and give this extra information if you took the nhigher (so stick). Or he is a nice chap and wants to give you the higher amount anyway, so offers you the swap option if you took the smaller (so stick). If we have critical winning limits (to by a new game), then I agree with the conclusions made re that. If we start establishing limits and we assume the uncle want to have a fair game, then we end up in the surprise date of the exam problem. If the maximum is $1000000, he will never give 900000 and 450000, because choosing the higher you will never stop. So nothing between 500001 and 1000000 is ever given. But if we know this then nothing between 250001 and 500000 will be given and so on. The same goes for the lower end, and as a consequence we will figure out that this game cannot be played :-) But, if we know that the amounts are "realistic", we do not have critical "I need that amount" limits and no extra information is received after the first envelope is opened, then I do not see any preference between stick and swap. Let's assume that there is a better strategy (either stick or swap is giving a higher EV). If this would be the "stick" strategy, then we take the first envelope and go. So our expected value is the average of the two envelopes. If it is the swap strategy, we can immediately throw away the first envelope and leave with the second. The expected value is the same. So, in my view there is no preference between the two... ... unlike in the Monte Hall, where the conditions change between the first window opened and the option to swap. |
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Title: Re: Two letters containing money, but... Post by honkyboy on Aug 19th, 2006, 1:33pm jollytail, Any range of number must have an upper and lower bound. This is what makes this problem 'realistic'. It is also what make it 'matter' whether or not a switch is made. What the range of numbers is and how evenly they are distributed within the range are unknown for this problem. For this reason the only definate answer you can give for the problem is that, with the information given, the correct choice can't be known. One is always free to make an educated guess. |
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Title: Re: Two letters containing money, but... Post by jollytall on Aug 19th, 2006, 9:28pm It's nice. I have never had a tail before (except...). I agree with you, that under the precise wording of the riddle there is no right choice. Even with the original wording, you need to know your uncle's motives. But regarding the fixed range, my comment was not that we do or do not know the distribution. The emphasis was on the unlce wanting a fair game. Fair game in this context would mean that you cannot know what you MUST do (to maximise your EV) after you open the first one. If he can put in money between A and B (where B/A is a large number), he theoretically cannot put in any amount less than 2A (since then finding that, you are sure that the other cannot have the half of it, i.e. you must swap) and more than B/2 (since then you know that the other cannot be double of it, i.e. you must stick). So the real range is 2A..B/2. And now go back to step 1 and soon you will have no range at all. |
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Title: Re: Two letters containing money, but... Post by Grimbal on Aug 20th, 2006, 9:23am The mistake is to believe that because an envelope was choosen randomly, the probability that the other envelope has a larger amount is at 50%. This is true before you open the envelope. But after you opened one envelope and discovered X, the question becomes: what is the relative probability that the envelopes contain X and X/2 as compared to X and 2X. If the probabilities were always equal, that would mean that the distribution is uniform from 0 to infinity, which of course cannot be true. So, there must be cases where the probability that the other envelope is X/2 is higher than for 2X. If you have some idea of how much your oncle would give, you can make a better-than-average decision to switch or not. If you have no clue at all, you are not better off than if you didn't open the envelope. |
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Title: Re: Two letters containing money, but... Post by Icarus on Aug 20th, 2006, 2:37pm But you are not worse off, either. And it is unlikely you would have no clue. The situation entirely changes when you open the first envelope, even if you knew about the doubling before hand. |
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Title: Re: Two letters containing money, but... Post by Eigenray on Aug 20th, 2006, 6:03pm This has all been said before, so I'll be brief: For this game to be well-defined, there must be some probability distribution on the values in the envelopes: say the pair {x, 2x} occurs with probability p(x). The expected value of switching, given your envelope contains x, is x*[2p(x) + 1/2 p(x/2)]/[p(x) + p(x/2)], while the expected value of staying is of course x. In order for the "always switch" strategy to maximize the expected value, we need p(x) > 1/2 p(x/2) for all x. This is certainly possible, but note that if this happens, then the expected value of the unopened envelope in your hand is already infinite (as is the other envelope, of course). What we have here are two random variables X,Y, with E(Y|X=x) > x, and E(X|Y=y) > y. But this just means [sum]y yp(x,y) > [sum]y xp(x,y) and [sum]x xp(x,y) > [sum]x yp(x,y), and summing the first, say, gives only E(Y) = [sum]x([sum]y yp(x,y)) > [sum]x([sum]y xp(x,y)) = E(X), not E(Y) > E(X), which would indeed be a contradiction (but the only way to for equality to hold is if both sides are infinite). Whatever X is, Y is better, and whatever Y is, X is better, but there's no contradiction*. On the other extreme, for "always stay" to maximize your expected value, we need p(x) < 1/2 p(x/2) for all x, but this is impossible: p(x/2n) > 2np(x), and so [sum] p(x) will either be infinite or 0, never 1. So for any instance of the game, there is always at least one x for which you should switch given x, and if the expected value of the game is finite, there's at least one x for which you should stay given x. But of course you can't know the optimal strategy without some information about p. For example, suppose that for some fixed a>0, 0<r<1, p(2na) = rn(1-r), n>0. If r<1/2, you should switch iff x=a (and the expected value is a(1-r)(2-r)/(1-2r)). Otherwise, always switch (and the expected value is infinite). Of course, this is just to maximize the expected value. Your utility function is quite nonlinear. Consider the following "game": You flip a coin until either (1) you get tails or (2) you decide to stop. If you got a tail, you get $0, otherwise if you got n heads, you get $3n. Now, the expected value of continuing after n heads is at least 3n+1/2 > 3n, so you should never stop . . . which is obviously a pretty bad strategy. -- *But it is a bit odd. The following is an argument against the continuum hypothesis: Let f : [0,1] -> omega1 be a bijection (omega1 being the smallest uncountable ordinal), and consider the following game. Players X and Y pick real numbers x,y in [0,1] with uniform probability distribution. X wins iff f(x) < f(y), otherwise Y wins. For each x, the set of y for which Y wins has cardinality |f(x)| < omega1, which is countable, hence has Lebesgue measure 0. So no matter what X picks, he wins with probability 1. But no matter what Y picks, Y wins with probability 1 too. (By Fubini's theorem, this must be because {(x,y) : f(x)<f(y)} is not measurable, even though each cross-section is.) |
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Title: Re: Two letters containing money, but... Post by Grimbal on Aug 21st, 2006, 4:08am It changes only insofar that you can compare that amount to what your uncle is likely to give. It doesn't change just because you know the amount. |
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