# Double Or Half?

Today’s post is going to be a riddle; I figured it was about time for another one.

You and I both have envelopes filled with money. My envelope contains either double or half the amount of money that’s in yours. If you want, I’m going to let you switch envelopes. Should you stay, switch, or does it not matter?

The riddle itself isn’t actually the riddle (you’ll see what I mean in a minute). I’m going to solve this riddle for you in two different ways.

Solution 1

Suppose your envelope contains $100. That means my envelope contains either$200 or $50. If you switch, half the time you will gain$100, and half the time you will lose $50. The expected value that you will end up with is$125. It is therefore in your favor to switch envelopes because, on average, you will come out $25 richer. Assuming that little x is the amount of money in your envelope, and big X is the amount of money you get when you switch, below is the simple calculation: $E[X] = \frac{1}{2}(2x) + \frac{1}{2}(\frac{1}{2}x) = 1.25x$ Solution 2 Suppose the amounts of money in the two envelopes are$100 and $200. That means that half the time when you switch you will gain$100 (going from $100 to$200), and the other half of the time you will lose $100 (going the other way, from$200 to $100). Therefore your gains and losses cancel out on average and it doesn’t matter if you stay or switch because the expected value is$150 either way. Assuming the values in the envelopes are x and 2x, and that big X is the expected value of your envelope whether you switch or not, the simple calculation is below:

$E[X] = \frac{1}{2}(2x) + \frac{1}{2}(1x) = 1.5x$

Obviously, the two solutions are contradictory and can’t both be right. The first solution proves that it’s better to switch envelopes, and the second solution proves that it’s not.

Here is the real riddle: which is the correct solution? Or, alternatively, is there a third solution that I neglected to mention? Also, just determining the correct solution isn’t enough, you must also explain why the other solution(s) is/are wrong.

Have fun!

By the way, there is a good discussion about this on Reddit. For those of you who are stumped, the answer can be found below in my first comment, encoded so as not to spoil it.

## 23 Comments on Double Or Half?

1. WrongMath says:

Come on, you can’t be serious about “Solution” 2 !

2. Philip says:

Below is my “official” answer encoded in rot13 so as not to spoil it.

Vg qbrf abg znggre vs lbh fgnl be fjvgpu. Fbyhgvba 2 vf gur pbeerpg fbyhgvba.

Jr pna ehyr bhg fbyhgvba 1 jvgu n cebbs ol pbagenqvpgvba. Cergraq gurer jnf fbzrguvat gb or tnvarq ol fjvgpuvat rairybcrf. Ng gur fgneg, lbhe rairybcr pbagnvaf rvgure qbhoyr be unys jung’f va zl rairybcr. Nsgre fjvgpuvat, lbhe rairybcr FGVYY pbagnvaf rvgure qbhoyr be unys jung’f va zl rairybcr. Hfvat gur fnzr ernfbavat, vg vf abj va lbhe orfg vagrerfg gb fjvgpu onpx. Guvf pbagenqvpgf gur bevtvany pbapyhfvba gung vg jnf orggre gb fjvgpu va gur svefg cynpr.

Gur dhrfgvba vf abj gb haqrefgnaq jul fbyhgvba 1 vf vapbeerpg. Fbyhgvba 1 pnfhnyyl fyvcf va n guveq rairybcr haqre gur enqne. Vg gevpxf lbh vagb guvaxvat gung gurer ner guerr cbffvoyr inyhrf: k, 2k, naq 0.5k, jura va ernyvgl gurer ner bayl gjb inyhrf. Nygubhtu lbh qba’g xabj jung gur inyhrf ner, bar bs gjb fpranevbf zhfg or gur pnfr: (N) gur gjb rairybcrf ner k naq 2k, be (O) gur gjb rairybcrf ner k naq 0.5k. Gurer vf ab guveq pnfr jurer nyy guerr inyhrf ner va cynl ng bapr. Vg vf gura rnfl gb frr gung obgu pnfrf N naq O qrtrarengr vagb gur frpbaq fbyhgvba.

Fb, jul vf gur frpbaq fbyhgvba pbeerpg? Vg’f pbeerpg fvzcyl orpnhfr lbh unir ab vasbezngvba. Vg’f yvxr zr gryyvat lbh gung V unir n pbva va zl unaq naq lbh gelvat gb thrff vs vg’f urnqf-hc be gnvyf-hc, ohg V’z tvivat lbh gur bcgvba gb ghea zl pybfrq svfg rvgure hc be qbja. Vg qbrfa’g znggre vs lbh ghea zl svfg bire be abg orpnhfr lbh unir nofbyhgryl ab vasbezngvba hcba juvpu gb onfr n fgengrtl.

Jvxvcrqvn tbrf vagb guvf ceboyrz va terngre qrcgu: uggc://ra.jvxvcrqvn.bet/jvxv/Gjb_rairybcrf_ceboyrz

V jbhyq or erzvff vs V qvqa’g nqqerff gur pbagebirefl fheebhaqvat guvf ceboyrz, juvpu onfvpnyyl obvyf qbja gb crbcyr pbzcynvavat gung gur ceboyrz vfa’g jryy fgngrq. Ner gurer guerr rairybcrf be abg? Ner lbh nyybjrq gb bcra lbhe rairybcr? Ubj rknpgyl qvq V pubbfr gur nzbhagf va gur rairybcrf? Ner gur rairybcrf unaqrq bhg ng enaqbz, be qvq V checbfrshyyl tvir lbh gur k rairybcr naq gura syvc n pbva gb frr vs V tbg 2k be 0.5k? Gurfr ner nyy tbbq dhrfgvbaf, ohg V jbhyq nethr gung va gur nofrapr bs vasbezngvba gur ernqre fubhyq vasre nf yvggyr nf cbffvoyr.

Abguvat jnf fnvq nobhg n guveq rairybcr, be nobhg fcrpvsvp cebprqherf hfrq gb svyy gurz, be nobhg ubj gurl jrer unaqrq bhg, be nobhg bcravat gurz, fb lbh fubhyq ernfbanoyl nffhzr gung V fvzcyl jrag vagb gur arkg ebbz, gbbx bhg gjb rairybcrf, svyyrq gurz jvgu jungrire zbarl V unq ba zr, fuhssyrq gurz hc, naq gura tnir lbh bar naq cbfrq gur dhrfgvba. V srry vg jbhyq or n qvffreivpr gb gur evqqyr gb nffhzr zber guna gung, orpnhfr fheryl fhpu vzcbegnag guvatf jbhyq unir orra zragvbarq.

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3. Sigh says:

Problem 1: You have $100 and the chance to trade for a 50/50 chance of getting$50 dollars or $200. Solution 1 is correct. Problem 2: Your envelope contains either$100 or $200 with a chance to trade for which ever value it doesn’t have. Solution 2 is correct. These are different problems, there is nothing to see here. 4. Venkat says: I could see solution 1 is right. As soon as I started reading solution 2, I was like “whoa! why do you suppose all that?!”. Solution 2 is for the problem where the other guy seems to have exactly the same amount of money (on average) as you. Hence switching is a waste of time. 5. andy says: The problem description actually allows for an infinite number of solutions. So let’s break down the problem. There are two outcomes, one is that you have 2x the money as the other guy, the other is that you have 1/2 the money as the other guy. You didn’t explicitly mention that these two outcomes have equal probability but we’ll assume that they do. So here we go: Solution 3: There are four sets of envelopes with secret contents. #1 has$1, #2 has $2, #3 has$1000 and #4 has $2000. You flip a coin, if it comes up heads then you get envelope #1 and the other guy gets envelope #2. Otherwise, you get envelope #4 and the other guy gets #3. This fits the description of the problem: there’s a 50% chance that you have double his money, and a 50% chance that you have half. In this world, it’s much better to hold on to your envelope, because you risk losing$1000 for a chance to gain $1. Solution 4: Similar to the above setup, except if the coin is heads, you get #2 and he gets #1, and otherwise you get #3 and he gets #4. This also fits the problem description. In this world it’s much better to switch. And so on. Lots of possible universes fit the problem description. 6. Philip says: I think I’ve found a good explanation for the controversy: people keep trying to change the problem to suit their own assumptions. It’s funny to read over the arguments and watch people change the number of envelopes, make decisions about the probability distributions, and so on and so forth. You can’t do that! You can only assume what is specifically stated by the problem. I submit to you that there is no difference between the following two answers: (A) There is not enough information to solve the problem. (B) It doesn’t matter if you switch envelopes. It would be like me asking you if there is a winning strategy for picking lottery numbers, and then you complaining that, “There’s not enough information because you haven’t told us how the winning numbers are picked.” Right, so there ISN’T a winning strategy. If I had said something like, “The winning numbers are going to be increasing primes,” then maybe you could come up with a strategy, but there is nothing in the riddle to suggest that, so you must not assume anything and just go with randomness. Solution 2 must be the correct solution based on the information given. It simply isn’t possible to come up with a winning strategy involving any amount of switching without either making unfounded assumptions or somehow gathering more information. I think justrasputin hit the nail on the head: http://www.reddit.com/r/math/c.....ey/c0e0kx8 “The problem with solution 1 is that it treats the second envelope’s worth as a result of what envelope 1 is, that is, as a dependent variable. Meanwhile, they are related but independent variables.” Based on what the problem says, you cannot make the assumption that solution 1 does, namely about the other envelope being dependent upon what’s in yours, because you don’t know which envelope was filled with money first. Also, the riddle is clearly a lesson in symmetry. Both envelopes contain either double or half the value of the other, so it’s impossible for either player to have the advantage. 7. mike says: The answer to the riddle is, It Does Not Matter. My proof is as follows: (x > p) = (X > p) = : ) Where x is the money in your envelope, X is the money in the other envelope and p is how much money you had in your (p)ocket before adding x or X. In other words, you’re better off than you were by keeping either envelope. Don’t let math cloud the issue, this is a greed-based riddle. Take the money and hope life continues to serve up problems like this. 8. Yat says: Hello One needs to be very careful with the original statements for this kind of problems. Lots of times I saw the one with the two childred stated the wrong way, such as the actual solution is that the other one has a 1/2 probability to be a girl. There is a problem here too : The statements are not the same as the one in the original problem, the one you link from wikipedia. Here, you give no information about how the enveloppes are picked, you only state that yours contains either double or half the money in mine. Nothing implies that this binary choice affects the amount in my enveloppe. If I translate your statement into a protocole, I have been given an enveloppe with a random amount (whatever the distribution is), and then a coin was flipped, to determine wether your enveloppe should contain double or half. The expected value in my enveloppe (even if I don’t know what it is) is exactly the same in both cases. In this problem, solution 1 is correct : I switch. Just once, because when I consider the odds for a second switch, I am not in this situation anymore : If my new enveloppe is the big one, the distribution of the amount has now doubled, and if I have the small one it has been halved. Even if I don’t know this distribution, I know that I will loose more if I switch back from the big one to the small one than I would gain if I switch back from the small one to the big one. Just run a simulation, following litterally your problem statements : Chose randomly an amount for my enveloppe, and then chose randomly to either put twice this amount or half of it in your enveloppe. Of course, the first amount distribution can be whatever you want, we can’t make assumptions on an amount distribution if it is not stated in the problem. But the double/half distribution, as it is a binary choice, has to be even. The only way that this does not work is if you assume that your enveloppe being half or double of mine affects the amount in my enveloppe. Nothing in the statements allows to assume this. The original problem is stated slightly differently but this changes everything : I pick randomly from two enveloppes, one containing twice as the other. Whatever the distribution of the small enveloppe, the one of the big enveloppe has every amount doubled. In other words, the expected value of my enveloppe is higher if I have the big one that if I have the small one. This solves the problem easily, no matter what the distributions are I lose exactly the same thing if I switch in the wrong way than what I gain if I swith in the good way. Switching or not does not matter. Simulation would work here too : randomly chose the contents of the enveloppes in any way that pleases you, but my picking has to be even. 9. John Stolzmann says: Solution 2 is the correct way to look at the problem. People who favor switching will argue that once you open an envelope, and know its amount (x), you have a 50/50 chance of 2x and a 50/50 chance of .5x by switching. Mathematically, you would be better off switching if this were the case. However, this is not actually what is happening. Some simple logic will show that switching doesn’t increase expected value. Suppose that I give you two envelopes and you choose one but don’t open it. I tell you that one envelope has twice as much as the other one. You know that your envelope has some amount in it, but don’t know how much…We’ll call that value “x.” Now that you know that you have “x,” you know that the other envelope has either 2x or .5x. So by switching, you have an E.V. of 1.25x. So logically you switch and now know that your envelope has a value of 1.25x. But the other envelope is either double or half that value…So it either has 2.5x or .625x, so the value of the other envelope is worth 1.5625x…and you keep switching back and forth ad infinitum. I ask, how does knowing what the value of x is change the problem? 10. Philip says: Yat, I’m assuming you’re referring to the following very famous problem: “A couple has two children, one of whom is a boy. What is the probability that the other child is also a boy?” The answer is a counterintuitive 1/3 because the problem does not state WHICH child is a boy. If the problem had instead stated, “the first child is a boy,” then the answer would be the expected 1/2. It’s hard to follow your terminology between “my statement of the problem” and “the statement on Wikipedia”. In fact, you seem to contradict yourself. First you say that I “give no information about how the envelopes are picked,” but then a few sentences later you say, “I have been given an envelope with a random amount, and then a coin was flipped to determine whether your envelope should contain double or half.” This is, of course, an unfounded assumption, and is exactly why the first answer is wrong. This is the exact phrase used in my statement of the problem: “My envelope contains either double or half the amount of money that’s in yours.” It’s a statement about VALUE, not about how the money got there, nor about how the envelopes were handed out. Solution 1 is wrong through a simple proof by contradiction (as John Stolzmann says above): after switching envelopes, my envelope still contains either double or half the amount of money that’s in yours. Any strategy that applies before switching also applies after switching. If you don’t believe me, fill the envelopes as you mentioned above: put some amount of money in envelope A and then flip a coin to determine whether or not envelope B gets double or half. Regardless of which envelope I then give to you — the person playing the game, A or B, I can always make the statement that my envelope contains either double or half the amount of money in yours. Because of this, it is absolutely, 100% impossible to determine which envelope you received or how they were filled. The statement that my envelope contains either double or half is perfectly symmetrical. Solution 1 would be correct only if the problem had stated something along the lines of, “Your envelope is filled first, and then my envelope is filled according to the following rules.” PS I’m just being a nitpicky dick and arguing with you for fun, as it’s clear that you know what you’re talking about, Yat 11. [...] problem reminded me of how unintuitive probabilities can be. There is a fundamental flaw with how the original post worded the problem: You and I both have envelopes filled with money. My envelope contains either [...] 12. Yat says: Well, in fact my way of formulating the protocole was precisely to avoid doing any assumptions, I just follow the statements litterally : as you talk about the amount in your enveloppe relatively to the one in mine, first I have to consider the amount in mine, and then the fact that the amount in yours is either double or half. Considering that I picked randomly from two enveloppes would be making an assumption. That’s why when I read your statements and try not to make any assumptions, it is equivalent for me to :“Your envelope is filled first, and then my envelope is filled according to the following rules.” I guess the interpretation on when one makes an assumption is pretty subjective, I just like being an “original statement nazi”. If you really want to remove every possible assumptions, then you can just consider that both amounts are random, whatever the distribution is, and then you look inside both enveloppes and one of them _happens_ to be the double of the other. In that case we are in situation 2 too. But it would again be assuming that you _had_ to make this statement. If I follow the same kind of reasonning you explain on the Monty Hall problem, you are the one giving me the opportunity to switch, so you are probably trying to screw with me. Your enveloppe is probably half of mine, I should definitely keep mine. Man, you messed with my head big time ! Concerning the two children riddle, yes that is the one I am talking about. I’ve seen many people who need to formulate this problem by introcucing a meeting with one of the childs. This can be done, but only with a good reason to assume that given the circumstances of the meeting, the child you met could only be a boy. Indeed, this is more impressive because it really seems to be equivalent to saying that the first child is a boy and asking the sex of the second one, but it is slightly different and it still works. You can even say which child is a boy, because it’s the one you met. 13. Veggie says: I did the experiment in 10000 trials (read: rows) in Excel. There did not appear to be any advantage between the strategy of switching or staying. 14. Philip says: Cool! It’s always nice to have some empirical evidence. You should upload your Excel file to one of the many free online file hosting services and then post it here for people to see. http://www.filesavr.com/ 15. Clay says: “My envelope contains either double or half the amount of money that’s in yours.” If the two alternatives presented in the above sentence are equally likely, then Solution 1 applies. On the other hand, if you _assume_ the envelopes were first filled with x and 2x dollars and then randomly assigned to two people, then yes, that’s very symmetrical, and Solution 2 applies. But such a process is nowhere mentioned in the problem statement. (Well, I am overstating my position a bit. Actually I think it’s an ambiguous problem statement. But if I had to choose, I would say Interpretation/Solution 1 is better supported by the literal meaning of the words in the problem statement.) 16. Cristiano says: Well… what I think about this? Both are ways of viewing the case. In solution 1 it is obviously a bennefit to swich and in solution 2 it doesn’t matter if you do or not so…. ultimately, the best solution would be switching because in any of the cases you don’t lose and you might just get an advantage 17. Abhi says: i feel switching would be the right option…cause once u switch ,you actually don’t know whether the money in your envelope is doubled or halved …..:)so be optimistic,that the profit is yours .(p.s: this is only if the other person does not allow you to check his envelope after switching) 18. Kevin says: It does not matter Pick ½ 1 2 Switch #1 1 2 ½ Switch #2 2 ½ 1 as there appears to be equal chance, it does not matter. 19. Zarel says: Hello! The Wikipedia article you link to on the two envelopes paradox goes into more detail, including the correct answer, which has not been discussed yet. First, I’d like to respond to a few statements here: Sigh says: These are different problems, there is nothing to see here. The two envelopes contain random values. Assigning them values of$50, $100, and$200 as in the examples is done without loss of generality, and is a standard method of making a proof or explanation more readable.

Philip says: I submit to you that there is no difference between the following two answers:

(A) There is not enough information to solve the problem.

(B) It doesn’t matter if you switch envelopes.

I submit to you that this is wrong. These two are mutually exclusive. The question of which is correct isn’t clear, though, because your problem is phrased in English and English can be ambiguously interpreted. If I had to make a call, the way I would interpret the question would make ‘A’ the correct answer.

Veggie says: I did the experiment in 10000 trials (read: rows) in Excel. There did not appear to be any advantage between the strategy of switching or staying.

No, you didn’t. The problem as stated is “two envelopes filled with money”. What you probably simulated was “two envelopes filled with a random integer amount of money between 1 and 4,294,967,296″ or something like that.

Your problem statement is: “two envelopes filled with money”, but you do not specify how much money.

There are several different interpretations of this statement, but given the context it is unlikely that this is a question about our socioeconomic condition and how much a game show is likely to afford to put in an envelope, the assumption is that the envelopes are filled with a random amount of money.

Now, at this point, astute readers should be asking themselves “Is that a rigorous statement?”

Well, usually, it is. If you say “random”, that means a uniformly random distribution. “Randomly pick an integer from 1 to 10″ means that there’s a 10% probability each for 1, 2, etc.

Uniformly random distributions even exist for uncountably infinite sets. You can choose a random real number between 0 and 1. It’ll have a 90% chance of being greater than 0.1, etc.

However: You cannot choose a random number greater than 0, since continuous uniform distributions exist only for bounded sets.

Thus, “two envelopes filled with a random amount of money” is meaningless unless you either set the bounds or specify the distribution you’re talking about.

Philip says: The question is now to understand why solution 1 is incorrect. Solution 1 casually slips in a third envelope under the radar. It tricks you into thinking that there are three possible values: x, 2x, and 0.5x, when in reality there are only two values. Although you don’t know what the values are, one of two scenarios must be the case: (A) the two envelopes are x and 2x, or (B) the two envelopes are x and 0.5x. There is no third case where all three values are in play at once. It is then easy to see that both cases A and B degenerate into the second solution.

This is… wrong.

A mentor once told me: “There’s no such thing as a counterproof.” By this, he meant that if someone can prove that a “proof” is wrong, that “proof” was never a real proof in the first place – it was wrong.

What you wrote isn’t the flaw in Solution 1. It’s… speculation, and it’s incorrect at that. It still doesn’t answer the question “What’s wrong with Solution 1?”

So what’s the flaw in Solution 1? As it turns out, absolutely nothing.

Solution 1 is completely correct. There does indeed exist a rigorous logical proof of “If the amount of money in both envelopes is uniformly distributed and one contains twice the amount the other contains, then it is advantageous to switch.”

So how can that be possible? The answer is by ex falso quodlibet, also known as the principle of explosion.

That’s why you have to be careful with proofs of contradiction. All a proof by contradiction will tell you is that one of your assumptions is incorrect. It won’t tell you which.

Here’s a classic example: The proof that 1 is the largest positive integer:

Let n be the largest positive integer. By contradiction, assume n != 1. n^2 is larger than n. But n is the largest positive integer. Contradiction! Therefore, n = 1.

There’s nothing wrong with the proof itself. But it turns out, the contradicted assumption isn’t that n != 1 (which is correct), it’s the assumption that the largest positive integer even exists.

The same thing is going on here: The only thing wrong about Solution #1 is the assumption that the uniform distribution exists for unbounded sets.

20. Philip says:

In response to Zarel:

The two envelopes contain random values. Assigning them values of $50,$100, and $200 as in the examples is done without loss of generality, and is a standard method of making a proof or explanation more readable. Absolutely, except in this case assigning the envelopes values comes WITH loss of generality. So, you can’t do it. It would be like if I said, “Prove that all even numbers are divisible by two,” and you started your proof with, “Okay, assume the even number is 12…” Philip says: I submit to you that there is no difference between the following two answers: (A) There is not enough information to solve the problem. (B) It doesn’t matter if you switch envelopes. Zorel says: I submit to you that this is wrong. These two are mutually exclusive. If you were God, you would know which envelope contained more money and know whether to switch. By the problem’s very statement there is “not enough information” to make the right choice, so the question is how to make the best GUESS. If there is “not enough information” to make a good guess, then the answer to the question obviously must be that it doesn’t matter what your guess is. Thus, “two envelopes filled with a random amount of money” is meaningless unless you either set the bounds or specify the distribution you’re talking about. It’s not meaningless, the envelopes were obviously filled somehow, but since you DON’T KNOW how, you can’t use it to try to solve the problem. There’s no such thing as a counterproof. Of course there is. You can nitpick the finer points of grammar if you want to waste your time, but if somebody shows me a proof that is wrong, what am I supposed to do, NOT point out the inconsistency? What you wrote isn’t the flaw in Solution 1. It’s… speculation, and it’s incorrect at that. It still doesn’t answer the question “What’s wrong with Solution 1?” So what’s the flaw in Solution 1? As it turns out, absolutely nothing. Solution 1 is completely correct. There does indeed exist a rigorous logical proof of “If the amount of money in both envelopes is uniformly distributed and one contains twice the amount the other contains, then it is advantageous to switch.” It seems you’ve committed an egregious logical error: claiming that something is wrong WITHOUT offering a counter-proof, or indeed any reason at all. The problem with solution 1 is simple: once you switch envelopes, your envelope STILL contains either double or half the other envelope, which means that if it was advantageous to switch before, it is now advantageous to switch back, which is a contradiction. If you want to claim that solution 1 is correct, you MUST address that contradiction. Otherwise, your argument carries no weight. Also, you mentioned the “uniform distribution” of the money in the envelopes, which is an assumption that was NEVER stated, and therefore you are using information that you don’t have, which is another logical error. That’s why you have to be careful with proofs of contradiction. All a proof by contradiction will tell you is that one of your assumptions is incorrect. It won’t tell you which. Proofs by contradiction are often used to prove things wrong (for obvious reasons, ha ha). If you want to prove that 1 is the largest positive integer, I can simply say the word “2″ and I have successfully contradicted your assertion. Likewise, as mentioned above, there is a very simple contradiction in solution 1. If you want to claim that solution 1 is the correct solution, you’re going to HAVE to address that contradiction and get around it somehow. 21. Neoncito says: Defining the problem is where all is about! Problem 2 is well defined. Having a$100 envelope and a $200 second envelope (only two), limits the set of possibilities and conditions the behavior; if you start with any envelope, the sequence is defined, there will be no randomness, that’s why you win/loose half of the time, if we represent that in percentage of winning instead of amount, we will have a 50% winning/losing chance (which that 1.5 stands for). Then solution 2 is actually more a description than a problem. Gain and loss is being analyzed. In the other hand, problem 1 is poorly defined and contains bad assumptions. Considering the condition “gain A or lose A/2″ that will not only leave us with$100 (initial), $200 or$50 envelopes (look,2 possibilities already), this will imply going down to $25,$12.5, $6.25… and upb to$200, $400,$800… The possibilities wining/losing-50% are then split by the fact you can chose between 2 different envelopes apart from the one you have in your hand (your baseline on each time you select to stay or switch) that’s what 1.25 stands for.

22. Mark says:

The chance to win or lose is always 1:1 per exchange. The interessting thing starts, when you do this exchange multiple times. If your envelope _always_ contains 100$, you’ll win in the end, cause in average you’ld win 25$ per exchange. So the contradiction given is just a special case where the amounts in the envelopes and the win/lose-rates alterate in a very specific way. You can’t say anything about the probabilties, if your envelope contains different amounts of money each time, because e.g. in the next step you might lose everything you won before. This is where the problem starts and where solution 2 is invalid.

23. jonahtas says:

guys, just feel the weight and choose the heavier one!