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 Subject: Probability and proofs Category: Reference, Education and News > Homework Help Asked by: jellojoe-ga List Price: \$5.00 Posted: 13 Feb 2004 15:14 PST Expires: 14 Mar 2004 15:14 PST Question ID: 306575
 ```This was a question posed in my Discrete Mathematics class at MIT one day at the end of lecture... You have two envelopes marked A and B. You know each envelope contains a positive integer of dollars. One of the envelopes contains k dollars and the other envelope contains 2k dollars where k is a positive integer. You are allowed to open one of the envelopes so you choose envelope A at random and inside are 8 dollars. You have a choice of giving the money back and keeping what's in envelope B or just keeping the eight dollars that were in envelope A. Which is the best strategy for making the most money? Proove this method.```
 Subject: Re: Probability and proofs Answered By: hailstorm-ga on 20 Feb 2004 15:38 PST Rated:
 ```jellojoe, Assuming we cannot tell anything by the appearance of the envelopes (for example, the one have more money being thicker), there is an exactly 50% chance of an envelope having k dollars, and an envelope having 2k dollars. However, we've seen inside envelope A, and it has \$8. Now, if envelope A is 2k dollars, then envelope B is half that of A, or \$4. But if envelope A is k dollars, then envelope B is twice that of A, or \$16. As gabor mentions below, the probability of envelope B having more dollars than envelope A is exactly the same as envelope A having more dollars than envelope B. So neither strategy has a higher probability of being \$2k. However, suppose you are the betting type, and are willing to take a potential loss of money by choosing envelope B? Envelope A has only one potential state. We know it has \$8, so it's potential value is \$8. However, envelope B still has two potential states: \$4 and \$16. Each of these states has a 50% chance of being true. So we can find the potential value of Envelope B by multiplying the potential value of each state by the potential probability of each state, and adding them all together: (50%)\$4 + (50%)\$16 = \$2 + \$8 = \$10 The potential value of Envelope B, based on the possibilities and probabilities of what is in it, is \$10. In it's unopened state, the potential value of envelope B is higher than the potential value of envelope A. Personally, I would choose Envelope B, then, without opening it, sell it to somebody for \$9, thus assuring I made more money in the deal. ;-)``` Request for Answer Clarification by jellojoe-ga on 20 Feb 2004 20:32 PST ```Ok, that was exactly what I was thinking, but the above argument seems intuitively faulty, since it does not matter what amount of money is revealed to be in the envelope you picked, as the other envelope will always be expected to be worth more on average. Therefore, you might as well not look at how much money is in the first envelope and just pick the second. However, if you do not look in the first envelope, there is nothing to distinguish the first envelope from the second! We have the paradox that no matter which envelope is picked, the other one will be expected to contain more money. It does not hurt you to switch (and you could use your argument as the rationale), but it should really not make a difference.``` Clarification of Answer by hailstorm-ga on 21 Feb 2004 02:49 PST ```jellojoe, The extra information you have is what makes the difference. Knowing nothing about either envelope, envelope A and envelope B both have a potential value of 1.5k dollars: (50%)k + (50%)2k = .5k + k = 1.5k This is because both envelope A and envelope B have the same two potential states. In the problem above, this changes because our information changes. As a result, envelope A has but one potential state, whereas envelope B still has two. Because we can now differentiate between envelope A and B, we can make an informed decision based on what is known about them. Let's consider the case where one envelope has k dollars, and one envelope has ak dollars, where a is a real constant greater than 0. Envelope A has \$8, so it's potential is \$8. Envelope B's potential states are 8a dollars and 8(1/a) dollars. Envelope B's total potential value can be expressed as follows: (50%)(8a) + (50%)(8/a) = 4a + 4/a = 4(a + 1/a) 4(a + 1/a) is greater than 8 in the case where (a + 1/a) is greater than 2. The lowest possible value for a positive number a is the case where a=1, which gives us exactly 2. That is to say, the case where both envelope A and envelope B contain exactly the same amount. In any other case, where envelope A and envelope B contain differing positive amounts of money, and the value of envelope A is known, the potential value of envelope B is always higher. So it is the case that knowledge of either of the envelopes contents decreases its potential value as relates to the other envelope. Therefore, if you know the contents of one envelope, it is fundamentally a better strategy to choose the unknown envelope.```
 jellojoe-ga rated this answer: ```Very good answer, the question actually turned out to be a probability of expectations type of question. I hope this is what my professor is looking for. The answer given to me though is sound and proven, good work.```

 Subject: Re: Probability and proofs From: pinkfreud-ga on 13 Feb 2004 15:34 PST
 ```This might be useful to you: http://www.j-bradford-delong.net/movable_type/archives/001395.html```
 Subject: Re: Probability and proofs From: jellojoe-ga on 13 Feb 2004 15:59 PST
 ```thanks for the link, the solution offered though changes the problem around, so ultimately their answer doesn't suit my problem, but it raises some interesting thoughts.```
 Subject: Re: Probability and proofs From: gabor-ga on 20 Feb 2004 13:59 PST
 ```I wrote up my take on the answer here: http://n.ethz.ch/student/cselleg/ganswers/ganswer1.html Hope it helps.```
 Subject: Re: Probability and proofs From: gabor-ga on 21 Feb 2004 05:19 PST
 ```hailstorm: In my first answer, I disregarded that opening the first envelope does actually give you additional information - that makes my answer incorrect. :-( |Let X1 and X2 be random variables for the amount of money in envelopes 1 and 2. Now, before selecting and opening an envelope, the entropies are H(X1) = H(X2) = inf (as both can take any value). Let's say we pick envelope 1, so we know X1; obviously H(X1|X1) = 0. Now H(X2|X1) = 1, as it X2 could either be 2*X1 or X1/2. By having opened the first envelope, we have a different situation at the start.```
 Subject: Re: Probability and proofs From: gabor-ga on 21 Feb 2004 05:25 PST
 ```hailstorm: I saw your clarification just after I wrote the last comment. Like I said, I think you're right.```
 Subject: Re: Probability and proofs From: jellojoe-ga on 21 Feb 2004 06:22 PST
 ```Alright I definitely agree with your answer now hailstorm, the problem I now have is convincing my professors its the right answer. Thanks for your help. Those mit professors are picky as hell, so I better get my wording down exactly right.```
 Subject: Re: Probability and proofs From: fsiegel-ga on 21 Feb 2004 09:34 PST
 ```It is right that the probability of the envelope being worth twice or half as much is equal (as Gabor excellently pointed out), however.. the outcome is not symmetrical. That means: if you would take the other envelope, you would get (\$4 + \$16)/2 = \$10 on average, as hailstorm has already pointed out. If you do that 10 times, you will (statistically) get 5x \$4 and %x \$16, that's \$100 compared to \$80 if you hadn't changed. However: this is only correct under the false assumption that you will ALWAYS have \$8 in the envelope that you pick first. You might as well get \$4 or \$16 in the next envelopes.. let's look what differs now: let's assume, you get envelopes twice with \$4, \$8 and \$16: if you always keep it, you will get: 2x\$4 + 2x\$8 + 2x\$16 = \$48 if you always take the other envelope, you will get: (\$2 + \$8) + (\$4 + \$16) + (\$8 + \$32) = \$2 + \$4 + 2x\$8 + \$16 + \$32 Now what do we see here? The term 2x\$8 turns up again. This gives us a hint of what is happening here. If you do that for every even number between an upper and a lower limit, you will easily see that it doesn't matter whether or not you change the envelope. On the average it's the same. That is, of course, except for the limits. That means: if you're on the upper limit and change your envelope, you will loose massive amounts of money, while when you're at the lower limit, you will gain a small amount of money. So the intuitive solution turns out to be right: it doesn't matter! (except for the limits) Back to our little riddle: There is now upper limit given here. This is what startles us. If there were no limit given at all (i.e. including negative numbers) it would be clear that all gain is equal to all loss. But we've got a lower limit here at \$0. So the riddle works by suggesting that it won't average itself out (which it does, as we've proven). Of course, in the real world there will always be an upper limit. So, the logical strategy for gaining the biggest amount of money would be: 1. if you've got an odd amount of money, get the other envelope 2. if you've got an even amount of money, keep your envelope, it doesn't matter, you won't gain or loose money by choosing the other one. 3. (additional) if your envelope contains more that one third of the total amount of money on earth (or, alternatively: more than one third of the amount of money you think the test conductor can get). KEEP IT! KEEP IT! KEEP IT! Buy an island in the carribean and be sure to send me 10%! Fabian P.S.: I've written my answer before hailstorm posted his clarification. But I still think my answer is correct.```
 Subject: Re: Probability and proofs From: jellojoe-ga on 21 Feb 2004 10:05 PST
 ```very thoughtful reasoning fsiegel, but i do not understand how you conclude steps 1 and 2 of your strategy from the notion of the limits and its affect on "averaging" out your expected gain/loss.```
 Subject: Re: Probability and proofs From: fsiegel-ga on 21 Feb 2004 11:12 PST
 ```You're right, I made a bit of a jump there. I just wanted to sketch the concept of the proof. a) Let's assume I get two envelopes for each number n between 1 and L (the upper limit), once the other envelope contains n/2 Dollars, the other time, it contains 2n Dollars. (except for those cases when n is odd, because n/2 is not an integer) b) if I always keep my first envelope, I will get for any even value m: 2*m c) if I always trade my envelope, I will get m/2 + 2*m . BUT (as shown above) I will always get m (exactly once) when I traded my m/2 evelope with the second containing m, and I will get m (exactly once) when I traded my 2*m evelope with the second containing m. That sums up to EXACTLY 2*m, because m has exactly two neighbours that are either m/2 or 2*m. d) This is true except for two cases: i) m is bigger than L/2 (that's because 2*m would be larger than the limit L) ii) m is not even. e) as there is no upper limit L given, d.i) is not important to us, i.e. I will get all even values exactly twice. f) in all cases I get an odd value j, the result is easy, as there is no preceeding value j/2, the other envelope HAS to be 2*j, so I need to trade. Additionally, j also "contributes" so that 2*j again has exactly one predecessor (in that case j) and one successor 4*j. g) this also shows another facett for the proof of this argument: every odd number is the "foundingstone" for an infinite number of even values m, that all have one successor 2*m and one predecessor m/2. Also all even number have exaclty one such "foundingstone". h) we have now proven that for every even value, the difference between trading and not averages itself out with the help of i) the odd number at the beginning of the sequence ii) infinity and that every odd number doesn't have a predecessor and therefore is not a good value to keep. I hope this clarifies it a bit, it's not a perfect proof for writing down, but I think there's not much missing. I might have been a little bit repetitive, but I'm a bit in a hurry. If you won't believe it, writing a small computer program for simulating it, might clarify what I meant. Fabian Addendum: step 3 of my strategy: it's clear from the above that it really should be "more than half of the upper limit".. however.. in a real world, he needs to have all money present on the table, so this reduces to "more than one third of the total amount of money available"```
 Subject: Re: Probability and proofs From: hailstorm-ga on 22 Feb 2004 07:21 PST
 ```fsiegel-ga, I'm afraid that you are also guilty of a false assumption: that you will have to open envelope B. Nowhere in the question is this stated. And, as I have shown above, the potential value of envelope B in its unknown, unopened state is \$10, compared with the known \$8 in envelope A. So as I made mention to in my answer, if you can find someone to sell the envelope to, you have a 100% chance of making more money using envelope B than by keeping what you know is in envelope A.```
 Subject: Re: Probability and proofs From: fsiegel-ga on 22 Feb 2004 09:26 PST
 ```Hailstorm, you are right, but only under the assumption that you'll find soomeone who is willing to risk the 50% chance of losing money. (granted: there are many people out there who would risk this) If you indeed can find someone and you'll draw an envelope only once, then that's a correct solution. But you might need to convince someone to do so. Which would be very easy, as he would be guaranteed to make \$1/envelope over the time, if he did that for a statistically relevant number of turns. But.. if repeating this process really IS an option, then why not keep it and keep your potential \$10? The answer is: you should, because (it might not have been obvious from my proof) you *will get them*. My proof only showed that you will get them whether or not you trade the envelope. So, effectively, you will lose money by selling the envelope to another guy. Fabian```
 Subject: Re: Probability and proofs From: hailstorm-ga on 22 Feb 2004 14:53 PST
 ```Another false assumption you're making is that you'll have an endless number of chances at this. Although I would love to have mysterious benefactors give me infinite chances to choose between envelopes with money in them, I am assuming that this is a one shot deal. For this case, with my sell-the-envelope strategy, you have a 50% chance of making more money than exists in either envelope! The other 50% of the time, you still make more money than the lowest price envelope. Whereas if you choose to open the envelope, 50% of the time you will have the lowest money value you can possibly get. By the way, if you sell the envelope for under \$10, you are still selling that person something that is greater in value than the purchase price. When you consider the millions of people who purchase lottery tickets, bet on horses, or play the slot machines, all of which have potential values *less* than their purchase price, I think that finding a buyer shouldn't be too difficult.```
 Subject: Re: Probability and proofs From: fsiegel-ga on 23 Feb 2004 01:33 PST
 ```I totally agree with you, and that's what I've already said before. If you only get to pull an envelope once, you only need someone to buy the other potential \$10 envelope from you. Like I already said.. many people would do so. But to be honest, you would have to tell him: "Hello, I'm a maths student at MIT and I'm not willing to risk the 50% chance of losing \$4 versus winning \$8." (Because that's what it essentially is) "So I ask you, are YOU willing to risk the 50% chance of losing \$5 versus winning \$7?" Like I already said.. many people would do so.. no question. But on the other hand.. I'm no specialist in psychology or marketing, so I hope that said MIT-student (jellojoe) will tell us what his professor really wanted to hear. Fabian```
 Subject: Re: Probability and proofs From: hailstorm-ga on 23 Feb 2004 02:55 PST
 ```I certainly hope jellojoe comes back to tell us how it all turned out. I have enjoyed the mental excerises with him and everyone who has contributed to this question!```
 Subject: Re: Probability and proofs From: jellojoe-ga on 23 Feb 2004 05:22 PST
 ```This is great reading you guys going back and forth, I'm glad there is so much interest in this problem. I especially like it because my professor let me hang on to the money and the other envelope until he says he will reveal the answer. I am assuming within the next few weeks he will let us know what the answer is.....I think he is holding off on the answer until we learn some technique in class that will help us understand what the answer is. I figure if I drop the class I can make up to \$24, not too bad, haha, but I don't think I would do that. I was thinking of taking the money from the envelope too and leaving a note inside that reads "You forgot to take into account the probability of Joe taking the money"......then again, I don't see myself doing that either....so i guess I'll have to wait until he tells me to bring in the envelope one day.....I'll keep you posted. oh yeah, i also tried holding the envelope up to a light (its one of those yellow envelopes) and i guess he put an envelope inside an envelope so I couldn't see what is inside. I tried shaking it too, to get the bills to seperate, but no luck, I figure if i see three bills in there, then there is no way there are 4 dollars in the envelope, I thought i saw only three, but two might have been stuck together```
 Subject: Re: Probability and proofs From: jellojoe-ga on 23 Feb 2004 19:10 PST
 ```By dumb luck I was reading my probability textbook tonight and it contained the same problem posed in my discrete math proofs course....so just by chance I got an answer to this question from another class!!!! talk about dumb luck! I won't post the answer here because it is too long to type, but hailstorm was on the right track to the answer. The knowledge of a limit would also change the problem, so if you can guess an upper and lower limit the answer changes, but if there is no limit then it is always better to switch....if you want to look at the answer get your hands on "Introduction to Probability" by Dimitri Bertsekas and John N. Tsitsikilis on page 107. ISBN 1-886529-40-X```
 Subject: Re: Probability and proofs From: gabor-ga on 24 Feb 2004 12:10 PST
 ```jellojoe: Unfortunately, I don't have the book the book you mentioned, but I dug in my bookshelf for the example that had been on my mind all along: The St. Petersburg Paradox. The book I have is John A. Rice: "Mathematical Statistics", 2nd ed and you can find this example on page 113. "A gambler has the following strategy for playing a sequence of games: - he starts off betting \$1; - if he loses, he doubles his bet; and he continues to double his bet until he finally wins. To analyze this scheme, suppose that the game is fair and that he wins or loses the amount he bets. (For example, when he bets \$8, he wins another \$8 with p = 1/2 and loses the money with p = 1/2). At trial 0, he bets \$1; if he loses, he bets \$2 at trial 1, and if he has not won by the k-th trial, he bets \$2^k. When he finally wins, he will be \$1 ahead, which can be checked by going through the scheme for the first few values of k. This seems like a foolproof way to win \$1. What could be wrong with it?" The answer is fairly obvious when you think about it for a while.```
 Subject: Re: Probability and proofs From: racecar-ga on 27 Feb 2004 11:58 PST
 ```The fundamental issue behind this problem is the probability distribution underlying the amounts of money in the envelopes. In stating that, having picked one envelope and looked inside, there is a probability of exactly 0.5 that the other envelope contains half as much, and a probability of exactly 0.5 that the other envelope contains twice as much, there is a tacit assumption that the distribution is uniform. But if it is uniform over all integers from zero to infinity, the expectation value of the amount of money in the evelopes in infinite. That is, there is zero probability of finding a finite amount of money in an envelope. So, to be realistic, we have to pick a probability distibution function that eventually goes to zero as the amount of money increases. Once we do this, it is no longer true that the probabilities of halving or doubling your money are equal. So the answer to the question is that it is impossible to answer without more information on the probability distribution describing the possible amounts of money in the envelopes.```
 Subject: Re: Probability and proofs From: jellojoe-ga on 27 Feb 2004 14:04 PST
 ```There is no limit on the amount of money possible in the envelopes, but in this case we know what the possibilities are for dollar amounts, so i was asking what the best strategy is given that we know one envelope contains 8 dollars.```
 Subject: Re: Probability and proofs From: fsiegel-ga on 03 Mar 2004 02:25 PST
 ```jellojoe: nice to see that you seem to have solved it. But I'm curious.. obviously, my solution has to be false. At the moment I just can't figure out why. Does anyone care to help me with that? I supppose I handled infinity incorrectly. Unfortunately I can't get my hands on the textbook you mentioned. It's not available in the entire city. I also greatly enjoyed this mental exercise, it was fun discussing it. Fabian```
 Subject: Re: Probability and proofs From: racecar-ga on 03 Mar 2004 11:38 PST
 `http://www.math.hmc.edu/funfacts/ffiles/20001.6-8.shtml`
 Subject: Re: Probability and proofs From: hailstorm-ga on 03 Mar 2004 15:07 PST
 ```From the link given by racecar: "For instance, suppose that amount of money in the two envelopes is(\$2k,\$2k+1) with probabilty (1/2)k, for each integer k>0. It is a fun exercise to check that no matter what you have in your envelope, the other envelope has higher expected value, and you should switch!" I'll claim victory here. ;-)```
 Subject: Re: Probability and proofs From: jellojoe-ga on 03 Mar 2004 16:01 PST
 ```See, that answer is similar to the one given in the book i wrote down. The book goes into a few more cases and details than the website, but yet I am still confused.... first it says there is a paradox, but then it never gives a definitive answer, it seems like the one case the website talks about is you already know whats in both envelopes which makes no sense, how can you know what is in both envelopes?? the problem never stated that, and the second method with the 2^(something), whatever it was....well thats nice and all, but its not an overall answer.....grrr to mathematicians and their inability to explain things in english.....I swear, mathematicians will know that a solution exists and that is good enough for them, but not good enough for an engineer like me```