The Magic Cafe Forum Index » » Puzzle me this... » » Two envelopes (0 Likes) Go to page [Previous]  1~2~3~4~5~6~7~8 [Next] Andrei Veteran user Romania 353 Posts  Posted: Mar 29, 2006 03:35 am 0 Thomas, nicely put. That's precisely it, much clearer than my own explanation. Andrei TomasB Inner circle Sweden 1143 Posts  Posted: Mar 29, 2006 03:41 am 0 A bieffect is that _if_ you can estimate Z (the highest possible amount he is willing to put in an envelope) and you discover something less than half of that estimation in your envelope your expected gain is positive if you switch to the other envelope. /Tomas Andrei Veteran user Romania 353 Posts  Posted: Mar 29, 2006 03:53 am 0 Yeah. The initial problem set no such limits, though. Andrei Daegs Inner circle USA 4283 Posts  Posted: Mar 29, 2006 04:06 am 0 Ah but can this tackle the problem of identically true statements and conflicting conclusions(thus there still is a paradox). refering to of course:Quote:On 2006-03-26 16:19, Daegs wrote: 1: Your envelope contains X. By switching you either lose X/2 or gain X, so the amount you gain is greater than the amount you lose. 2: The amounts in the Envelopes are Y and 2Y, by switching you either lose Y or gain Y, so the amount you gain is the same as the amount you lose. OTOH, it seems that there isn't a boundry in effect... if we can pick *any* number and know there is a possiblity of 2x in that envelope then that really means up to infinity, It seems that assuming there is a boundry can help things, but there is still the problem of the statements or if you try to tackle it unbounded... also, it seems that you are saying it IS 50/50 as long as X <= Z/2... so why not restrict the values of X to within Z/2, would that not force it to go back to 50/50 and thus the paradox?(or would Z then become Z/2 as we are restricting choices and thus screw us up?) anyways its way too late and I've been working on mind problems all day(unrelated to this) so I'll look at your solution in depth tommorw... but still I'd like to see this applied to the statement part because that has me really stumped.(or explain why all my averages on the sim of constantly changing the new envelope don't even out but instead see a great increase). magicjohn2278 Special user Isle of Man UK 538 Posts  Posted: Mar 29, 2006 04:16 am 0 So the logic and maths shows us that it is more favourable to switch. The question is why and what’s wrong? What it boils down to, is we are confusing numbers, “things” and functions. And using the wrong scale to evaluate the potential gain or loss. We usually see numbers and things on a linear scale. -5..-4..-3..-2..-1..0..1..2..3..4..5..6..7..8..9..10..11….. The numbers really have no significance! They are just there to identify the points on the scale, it is the spaces between the points that we are interested in. If we are at point 2 and we move (add) 4 spaces, we end up at point 6. The spaces between the numbers are equal and have a value of 1. To move right, we add 1 to move left we subtract 1. But to solve this puzzle we should be looking at rises and falls on an exponential scale 1/8..1/4..1/2..1..2..4..8..16..32..64….. The spaces between the numbers are equal and have a value of 2X to move right and 1/2X to move left. Looking at our exponential scale, getting double your original number is exactly the opposite of losing half. And the loss is equal to the gain. (Remember we are counting spaces to evaluate potential loss or gain.) Obviously, if you have X and double it you now have 2X, if you have 2X and half it then you have X, so your gain by doubling is identical to your loss by halving. – Your starting point on the scale must be the point that you are at at the time. (Conversely, if we have X and half it, we have X/2, if we double X/2 we have X again, loss equal to gain.) So in our problem if we switch and double, our potential loss is the same as switching and getting half. So there is no advantage in switching. “Ahh…”, I hear you say.. “that’s rubbish! Even on your exponential scale, if you have 2 and double you have 4, if you half you have 1, how can losing 1 be the same as gaining 2?” But that is looking at numbers linearly. On the exponential scale the difference between 2 and 1 is the same as the difference between 2 and 4. We should be looking at the spacing of the numbers, not the numbers themselves. (The numbers are there only to identify the points. (Perhaps I should have used letters?)) So why doesn’t it work with numbers on a linear scale? Because it’s not a linear question, it’s an exponential one. We know the value of X (our starting point) so why isn’t the difference between losing half and gaining 1 the same. (It is on our exponential scale spaces.) We (I) can’t convert between points on our exponential scale, to points on our usual linear scale. They are two different animals! Here is a choice that we think we can evaluate on our linear scale: You have X, you can switch for X+1 or X-1 is it favourable to switch? Obviously it doesn’t matter. Using our linear scale, potential loss and gain is equal. But having decided not to bother, I now tell you that the things are cubes of gold with a side length of X… So how is it that you can now “prove” mathematically, (erroneously) using your linear scale, that it is more favourable to switch. What changed? If the X ’s are cubes, you should be using a N^3 scale. So, the “solution” to the paradox, is that we can prove that it is more favourable to switch, only by using an incorrect basis for the maths. I anxiously await your scorn! TomasB Inner circle Sweden 1143 Posts  Posted: Mar 29, 2006 05:33 am 0 If it was unclear although I implied it twice: limes[Z->infinity](P(X <= Z/2) / P(X > Z/2)) = 2 since P(X <= Z/2) / P(X > Z/2) = 2 for ALL Z > 0. The error in "1: Your envelope contains X. By switching you either lose X/2 or gain X, so the amount you gain is greater than the amount you lose." is in the wording "so the". It just doesn't follow from "By switching you either lose X/2 or gain X". You have to weigh both those cases by their probabilities of happening. If finding X/2 has a probability of 0.9 and finding 2X has a probability of 0.1, the expected value of that envelope would be 0.9 * X/2 + 0.1 * 2X = 0.65X so that clearly shows that to answer if you gain on an average by switching you _need_ to state the actual probabilities for ALL outcomes. /Tomas magicjohn2278 Special user Isle of Man UK 538 Posts  Posted: Mar 29, 2006 05:45 am 0 .. but Tomas, the probability of switching for more is exactly equal to the probability that you hold the lesser envelope... which in turn is exactly equal to the probability that you hold the greater. To get anywhere on this tack you need to prove that your original selection was not a 50/50 deal.... and I don't think you can. TomasB Inner circle Sweden 1143 Posts  Posted: Mar 29, 2006 05:59 am 0 Not sure what you mean, John. The original selection is not a 50/50 deal. There are so many possible amounts that can be in that envelope before it is checked. But if you mean that there is a 50/50 deal of picking either envelope, that of course has to be assumed and that's not a bad assumption. The choice _is_ random and you do not favour picking any envelope over the other. /Tomas magicjohn2278 Special user Isle of Man UK 538 Posts  Posted: Mar 29, 2006 06:07 am 0 The probability of picking either envelope A or B is 0.5 each. You have a 50/50 chance of holding the greater or lesser amount. Therefore the probability that the other envelope contains more is 0.5 and the probability that it contains less is 0.5 .. and you can't alter that! (Unless you can prove a tendancy to pick A on more or less occasions than B.) TomasB Inner circle Sweden 1143 Posts  Posted: Mar 29, 2006 06:09 am 0 The probabilities you speak of, John, are _not_ equal. I showed in the earlier post that they are 2/3 for X/2 and 1/3 for 2X being in the other envelope and that was due to that the conditional probabilities were dependant on X. Of all possible games there are _not_ an equal probability of every possible value in the envelope. That's what does it. /Tomas magicjohn2278 Special user Isle of Man UK 538 Posts  Posted: Mar 29, 2006 06:15 am 0 Just been back to read that - won't pretend to understand it.. "It is reasonable to assume that the envelopes are filled as follows. In one envelope an amount Y is put which is squarely distributed from 0 to Z. We do not know Z, but enough is to know that such an upper boundary must exist. In the other envelope Y/2 is always put." However, why is it reasonable to assume that the envelopes are filled as follows...? It is just as reasonable to assume that Y is put in one envelope and 2Y is put in the other. (with Y being Z/3) Which should disprove your argument. (I can't disprove it because I don't understand it!) magicjohn2278 Special user Isle of Man UK 538 Posts  Posted: Mar 29, 2006 06:21 am 0 .... and what happens to your argument if we limit Z to say \$300.... ... or stipulate that one envelope always contains \$50....? magicjohn2278 Special user Isle of Man UK 538 Posts  Posted: Mar 29, 2006 07:29 am 0 Say envelope A contains the greater amount, B the lesser. If we accept that the total value of the envelopes is finite (and we do, neither can contain infinity) does that affect our chances of picking A over B? No. The contents of the envelopes are irrelevant when we come to pick them - we don't know what they are. Say the maximum total contents of the envelopes is \$300 and for simplicity, that the contents in \$ are always an integer. the chances of picking an envelope with \$100 in is double the chances of picking one with \$200 or \$1. But in one case, the \$100 is in envelope A and in the other in envelope B. And that is what we are trying to establish - the probability of picking A over B. For every possible A envelope, there is a corresponding B envelope. (containing half A) So for all the possible values of A, there is an identical number of possible values for B. Thus the chances of picking one or the other are 50/50 .......................................................... Which.. going back to my post about exponential scales raises an interesting question if A=2B and B=A/2 , both are integers and A+B is <= 300 how many possible values of A and B are there? B must be in the range 1 to 100 = 100 A must be in the range 2 to 200 = ? If you actually calculate the value of A given that it can be 2 to 200 you have to use an exponental scale of +2 - use a linear scale and you would deduce that there could be 200-2=180 possible values for A. There aren't, there are only 100. Values for A increase by 2 every time, not 1. Steve Martin Inner circle 1119 Posts  Posted: Mar 29, 2006 07:37 am 0 John, the explanation that Tomas (and Andrei) has given is a bit of a tricky one to get one's head round, but I am confident that they are right. I have not fully "clocked" it yet, but I need to sit down and think about it. Aside from the maths involved, I think that ultimately this is the sort of thing that some sort of diagram will clarify. If we can create a diagram and some really simple language to describe it, then we'll have a better bet of grasping it (and - which is the proof of the pudding - explaining it to someone else). I'll come back when I've looked at it some more! Any man who reads too much and uses his own brain too little falls into lazy habits of thinking. Albert Einstein TomasB Inner circle Sweden 1143 Posts  Posted: Mar 29, 2006 07:54 am 0 I'll rephrase what you wrote a bit then you can put in 300 instead of Z if you want. Say the maximum total contents of the envelopes is Z. If one envelope has X > 2/3*Z the probability of the other having X/2 is 1 and the probability of it having 2X is 0. Otherwise the total contents would be above Z. If on the other hand either envelope has X <= 2/3*Z there is a 50/50 chance of there being X/2 and 2X in the other. So what is the relative probabilities of X being above and below 2/3*Z? Well that clearly divides it in 2/3 and 1/3. You will by all right argue that the X I speak of is in one envelope and not neccesarily in the envelope I pick. So let's say it's picked only half of the time. There will still be double as many below 2/3*Z as there are above since you cut both in half with that so the relative probabilities are still 2/3 and 1/3. Expected value of the other envelope with your resoning of filling the envelopes: 2/3 * (0.5 * X/2 + 0.5 * 2X) + 1/3 * (1 * X/2 + 0 * 2X) = X /Tomas Steve Martin Inner circle 1119 Posts  Posted: Mar 29, 2006 08:24 am 0 Tomas - if Z is the maximum total of both envelopes, surely it is impossible for X to ever be *greater* than 2/3 of Z. Equal or less than, yes, but not greater. ?? Also, what happens if we say that the maximum total of both envelopes is infinty? i.e. each envelope contains a number written on a piece of paper, one of which is twice the other? Is that, then, a significantly different puzzle to the one we are considering (maybe it is)? ?? Any man who reads too much and uses his own brain too little falls into lazy habits of thinking. Albert Einstein TomasB Inner circle Sweden 1143 Posts  Posted: Mar 29, 2006 08:47 am 0 Ooops, I knew I should have stopped writing in this thread, Steve. Before I can answer I need John to clarify how he'd write the simulation that makes sure that the total contents is below 300. It makes all the difference. As for what happens when Z approaches infinity I showed that, with the reasoning in my earlier post, the relative probabilities was a constant regardless of Z hence you can let Z slide up to infinity and the quota between those probabilities will still be 2. That's the only way I can see how there is not a 50/50 chance of there being X/2 or 2X in the other envelope on an average. /Tomas Andrei Veteran user Romania 353 Posts  Posted: Mar 29, 2006 08:57 am 0 Also, we're using a simplistic arithmetic average, when the distribution is anything but simplistic. How about using a geometric average? That would certainly work, but is there any logic behind it? Andrei Steve Martin Inner circle 1119 Posts  Posted: Mar 29, 2006 09:05 am 0 The simulation will be obliged to constrain both numbers so that neither is ever more than 2/3 of Z. I see what you mean about extending Z up to infinity using the same reasoning. Any man who reads too much and uses his own brain too little falls into lazy habits of thinking. Albert Einstein magicjohn2278 Special user Isle of Man UK 538 Posts  Posted: Mar 29, 2006 09:06 am 0 I still think the error is using the wrong numeric "scale" to evaluate the potential gain or loss by switching. (Scale probably isn't the right word - there is probably a mathematical expression for it.) The linear scale you are using to argue that if you switch then you will either gain X or lose X/2 is the wrong scale. This scale only applies when values move up with addition and move down with subtraction i.e +Y or -Y Unfortunately, the contents of our envelopes don't change using this scale. Their value increases by multiplication and decreases by division by a factor of 2. If you use the wrong (linear) scale to assess whether switching will improve or reduce your current position you will get the wrong answer. You are using the "Plus and Minus" scale and wrongly concluding that gaining 1 is better than losing ½ because 1 is more than ½. Which is wrong. If you start with 1, you switch and are lucky, so you gain 1, you have 2, switch again and lose half, you have 1, so once you showed a gain of 1, and once you showed a loss of ½ , yet you have the same as you started with! So illogical as it seems, the value of gaining 1 is identical to the loss of ½. The scale should be thinking about is "Multiplication and division by 2" and the question you should be asking is "what is the probability of doubling X by switching compared to the chances of halving it?" The answer is 0. The Magic Cafe Forum Index » » Puzzle me this... » » Two envelopes (0 Likes) Go to page [Previous]  1~2~3~4~5~6~7~8 [Next]
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