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The Two envelope Paradox
O so simple...

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The Two envelope Paradox

DoctorInWaiting
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Posted 05/02/08 - 12:02 PM:	#26
This is a variation of the Monty Hall Problem. Probability wise without any information (like the monty hall problem instigated) it is a 50% chance. There is no real way of instigating which is the right envelope I feel. DMeister, It depends on what cards you had and exactly how many cards could help you within the implied odds. You probably already knew that though! haha

PeteSF
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Posted 05/04/08 - 06:52 PM:	#27
Brian Bosse wrote: Model 1 Envelope A contains X dollars and Envelope B contains 2X dollars. You do not know what envelope is A or B. If you happen to pick the envelope that is A and switch, you will have 2X. If you happen to pick the envelope that is A and stay, you will have X. If you happen to pick the envelope that is B and switch, you will have X. If you happen to pick the envelope that is B and stay, you will have 2X. Hi Brian, I'm considering a slight variation on the original problem, in which after selecting an envelope, you look inside to find $100, and I'm getting stuck. Consider Model 1: Model 1 If you happen to pick the envelope that is A and switch, you will have 2X. X = $100 If you happen to pick the envelope that is A and stay, you will have X. X = $100 If you happen to pick the envelope that is B and switch, you will have X. X = $50 If you happen to pick the envelope that is B and stay, you will have 2X. X = $50 This exhausts all possibilities. Switching averages out to 1.5X dollars, and staying averages out to 1.5X dollars. As such, it makes no difference whether you stay or switch. Now we have a problem, because the value of X is not constant among the four possibilities. Switching averages out to $125, staying averages out to $100. What happened?

Lex
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Posted 05/04/08 - 10:33 PM:	#28
No problem. How in the world did you get something ridiculous like that? x = 50, always, and it can't be any other way. If you switch from A to B you now have 50 dollars, not 100, which is x (consistent). If you switch from B to A you now have 100 dollars, which is 2x (also consistent).

PeteSF
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Posted 05/05/08 - 03:23 AM:	#29
Lex wrote: No problem. How in the world did you get something ridiculous like that? x = 50, always, and it can't be any other way. If you switch from A to B you now have 50 dollars, not 100, which is x (consistent). If you switch from B to A you now have 100 dollars, which is 2x (also consistent). In the scenario I'm considering, you've looked in your envelope and seen $100. If you've picked envelope A (containing X dollars), then X is clearly $100. If you've picked envelope B (containing 2X dollars), then X is $50.

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Posted 05/05/08 - 04:42 AM:	#30
Try to avoid multiplying or considering the amount in hand. You should only be considering the value you gain or lose by and nothing else. Consider this variation: You have 2 envelopes, one is empty, the other has $100. What is the expected value of switching verses standing? If you start saying X -> 2X or X -> X/2, then you see the obvious problem that you are treating both envelopes as empty, (2X = 0 and X/2 = 0), when they are not. Instead, it is +$100 if you switch to* the envelope with the money, and -$100, if you switch from it. The expected value is .5(+$100) + .5(-#100) = $50 - $50, meaning the expected value of switching/swapping is both $0. Maybe this is the same kind of fallacy that advertisers use when they claim something like "double your savings", (makes it sound like a lot, even if it is saving 2 cents instead of just 1 cent)? There may be another bias in play, as well -- the "grass is greener", which makes it seem reasonable to overestimate the value of switching???
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Posted 05/05/08 - 06:21 PM:	#31
PeteSF wrote: In the scenario I'm considering, you've looked in your envelope and seen $100. If you've picked envelope A (containing X dollars), then X is clearly $100. If you've picked envelope B (containing 2X dollars), then X is $50. No matter how I think about it sounds completely ridiculous. If you have picked envelope A (2x) and that is $100 then envelope B (x) will contain $50. If you have picked B (x) and that is $100 then A will naturally have $200. Yes in those two cases x is not the same but why in the world should it be? You are talking about two different sets of envelopes, one where 2x = 100 and x = 50, and another where x = 100 and 2x = 50. Seriously I have no idea what you are talking about, you need to straighten out your reasoning there is a huge hole somewhere.

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Posted 05/06/08 - 12:45 AM:	#32
PeteSF, Hi Brian, I'm considering a slight variation on the original problem, in which after selecting an envelope, you look inside to find $100, and I'm getting stuck. Consider Model 1: Model 1 If you happen to pick the envelope that is A and switch, you will have 2X. X = $100 If you happen to pick the envelope that is A and stay, you will have X. X = $100 If you happen to pick the envelope that is B and switch, you will have X. X = $50 If you happen to pick the envelope that is B and stay, you will have 2X. X = $50 This exhausts all possibilities. Switching averages out to 1.5X dollars, and staying averages out to 1.5X dollars. As such, it makes no difference whether you stay or switch. Now we have a problem, because the value of X is not constant among the four possibilities. Switching averages out to $125, staying averages out to $100. What happened? You are making the unjustified assumption that the conditional probability of having picked the envelope with more money in it, based on the fact that you observe it has $100 in it, is exactly 0.5. If you do not know the distribution of possible values of X before opening the envelope, then observing what is in the envolope adds no new information. If you do know this, then of course observing that the envelope contains $100 changes the odds, but exactly how it changes the odds depends entirely on that distribution. The example you cite would only obtain if the probabity of X = 50 and X = 100 were exactly equal. Now, you may intuitively think that equal probabilities for all possible values of X should be assumed, in the absense of any other a priori information. But this intuition is wrong. After all, no bounds have been placed on the possible range. A uniform distribution from 0 to infinity makes no sense. And if you do place a limit, then that changes things too. For example, if I know that the value of X ranges from $1 to $500, and assume that it is uniform, then I would definitely be better off switching if I see the envelope contains anything less than or equal to $500. But I definitely won't want to switch if it contains more than $500. These two effects balance out, so that if I don't see what is in the envelope, then on average it will make no difference whether I switch or not (as symmetry requires). DM
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Posted 07/13/08 - 05:58 PM:	#33
Death Monkey wrote: PeteSF, You are making the unjustified assumption that the conditional probability of having picked the envelope with more money in it, based on the fact that you observe it has $100 in it, is exactly 0.5. DM I find $100 inside the envelope I have picked. Then I know that the possible distributions of money in the envelopes are only two: A=$100/B=$200 and A=$100/B=$50. Since I have no further information, I should apply the principle of indifference and assign equal probability to each possible distribution. So I have: p(A=$100)=1 p(B=$200)=1/2 p(B=$50)=1/2 The conditional probabilities are: p(B=$200\|A=$100) = p(B=$200 & A=$100)/p(A=$100) = p(B=$200) = 1/2 p(B=$50\|A=$100) = p(B=$50 & A=$100)/p(A=$100) = p(B=$50) = 1/2 Or should the principle of indifference not apply here?

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Posted 07/14/08 - 12:32 AM:	#34
LuaLuna, I find $100 inside the envelope I have picked. Then I know that the possible distributions of money in the envelopes are only two: A=$100/B=$200 and A=$100/B=$50. Since I have no further information, I should apply the principle of indifference and assign equal probability to each possible distribution. So I have: p(A=$100)=1 p(B=$200)=1/2 p(B=$50)=1/2 The conditional probabilities are: p(B=$200\|A=$100) = p(B=$200 & A=$100)/p(A=$100) = p(B=$200) = 1/2 p(B=$50\|A=$100) = p(B=$50 & A=$100)/p(A=$100) = p(B=$50) = 1/2 Or should the principle of indifference not apply here? I think that what you are calling the "principle of indifference" here is not any logical rule, but in fact a common logical fallacy. It is not valid to say that if you have N possibilities, and do not know anything about their relative probabilities, that you can reasonably assign to them each a probability of 1/N. On the contrary. This simply is not the case. Always when working with probabilities, you must carefully consider the real question being asked. I emphasize "real" here, because we have a tendency to think of probabilities in an abstract way, where we imagine that there are some fixed "rules" of probability that we can just blindly apply to problems. That isn't how it works. In the case of the envelope problem, when you open the envelope and see $100 in it, strictly speaking, the probability of the other envelope having $200 in it is either 0 or 1. We just do not know which. It is already a fixed event. Either it does or it doesn't. There is no probability. The real question here is, "based on what I know, would I be better off switching or not?" It is only when you try to incorrectly cast this in terms of a probability question, that you end up with the apparent paradox. So how should we solve the problem? Well, as we have already seen, the easiest way is to just use symmetry. If you truly do not know anything about the possible values you could find in the envelopes, then seeing what is in one of them adds no relevant information. Therefore, by symmetry, it does not matter if you switch. If we want to present it as a probability problem, then we have to carefully formulate the corresponding problem. In this case the question becomes something like: "If we were to repeat the following experiment many times, and I always switch, will I end up with more money than if I never switch?" Now we have to specify the experiment. Here is where we run into trouble. I can't specify the experiment without making some assumption about the distribution of values in the envelopes. Note that the symmetry solution I described requires no such assumption. Now let's consider the assumption you have suggested. Forget about specific numbers here, because we are looking for a general rule. Call whatever value you find in the first envelope X. Now you are assuming that the combination (X, 2X) has the same probability as the combination (0.5X, X). This cannot possibly be true for all X. The result is simply not a valid distribution. It could work for specific values of X. For example, it is possible that the original distribution is such that the combination (50, 100) has the same probability as the combination (100, 200). But to assume (just because you have seen the number 100), that this specific distribution is the correct one, is a very strong and unjustified assumption. DM
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Posted 07/15/08 - 05:26 AM:	#35
Hello DM and all others. Death Monkey wrote: I think that what you are calling the "principle of indifference" here is not any logical rule, but in fact a common logical fallacy. It is not valid to say that if you have N possibilities, and do not know anything about their relative probabilities, that you can reasonably assign to them each a probability of 1/N. On the contrary. This simply is not the case. DM But this is the standard formulation, isn't it? Whether it really applies or when it does, is another question. I suspect it doesn't apply here. Death Monkey wrote: In the case of the envelope problem, when you open the envelope and see $100 in it, strictly speaking, the probability of the other envelope having $200 in it is either 0 or 1. We just do not know which. It is already a fixed event. Either it does or it doesn't. There is no probability. DM Imagine I flip a coin. The coin happens to roll past the door into the next room, so that no one can still see how it has come up. Then I propose a bet on heads and tails. Someone objects: 'no rational bet is possible for there is no possible probability calculus once the coin has fallen and stopped; I would certainly have played before the coin was thrown or before it was lying on the ground but I can no longer do so now'. I think most people would reply that the odds are still 1/2 and the game makes just as sense as before. Probability calculus is not rendered rational by some indetermination in the facts themselves but by our lack of information about the facts; otherwise probability calculus would be nonsensical in a deterministic universe. What is the difference between the envelopes and the coin just described?

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Posted 07/15/08 - 06:18 AM:	#36
LauLuna, It is not valid to say that if you have N possibilities, and do not know anything about their relative probabilities, that you can reasonably assign to them each a probability of 1/N. On the contrary. This simply is not the case. But this is the standard formulation, isn't it? Whether it really applies or when it does, is another question. I suspect it doesn't apply here. I'm not sure what you mean by it being "the standard formulation". It is an assumption. And while it is an assumption that is commonly made when applying statistics to real-world problems, that is only because the context of those problems makes it an appropriate assumption to make. There are many cases where it is not an appropriate assumption, and then, of course, it is not used. Imagine I flip a coin. The coin happens to roll past the door into the next room, so that no one can still see how it has come up. Then I propose a bet on heads and tails. Someone objects: 'no rational bet is possible for there is no possible probability calculus once the coin has fallen and stopped; I would certainly have played before the coin was thrown or before it was lying on the ground but I can no longer do so now'. I think most people would reply that the odds are still 1/2 and the game makes just as sense as before. Probability calculus is not rendered rational by some indetermination in the facts themselves but by our lack of information about the facts; otherwise probability calculus would be nonsensical in a deterministic universe. I think you missed my point. My point was not that probability theory is somehow rendered irrational or inapplicable to problems where the outcome is already fixed. That is not true. My point was that in order to properly apply probability theory to such problems, you must clearly specify the corresponding thought experiment, which can then be considered probabilistically. For example, in the case of a coin being flipped, but hidden, and then asking how likely it is to be heads or tales, the correspoding thought experiment is quite simple: Imagine I were to flip the coin many times. Each time I hide the result, choose either heads or tails, and then look at the result. What percentage of the time would I be correct if I always chose heads? What about if I always chose tails? What about if I randomly choose either heads or tails (50/50) each time? If it is a fair coin, then we know that the answer in all three cases is 50/50. Based on this, we can establish that we are no better off choosing heads for our real case (where the flip is already determined, and we just don't know the result), than we would be choosing tails. It is an even bet. What is the difference between the envelopes and the coin just described? I assume that by this you mean "why can we assign the same odds to the coin being heads as it being tails, but not assign the same odds to the other envelope having $50 as it having $200"? If so, then the answer is that in the case of the coin, we have good physical reasons to assign the two outcomes equal probability. In fact, we usually even specify our justification for this assumption, by saying something along the lines of "we will assume it is a fair coin". We are making an assumption about the distribution of outcomes in our thought experiment. Namely that each time we flip, we are equally likely to get heads as we are to get tails. This assumption directly follows from the fact that we are assuming it is a fair coin, which is an assumption about the physical system being discussed, not some sort of a priori logical principle that can always be applied. There is no such justification for assuming that $50 and $200 are equally probable values for the second envelope, because nothing about the problem in away indicates that they should be. On the contrary, we can see quite easily that if we do make such an assumption, we run into contradictions. DM
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Posted 07/16/08 - 03:44 AM:	#37
Death Monkey wrote: I think you missed my point. My point was not that probability theory is somehow rendered irrational or inapplicable to problems where the outcome is already fixed. That is not true. My point was that in order to properly apply probability theory to such problems, you must clearly specify the corresponding thought experiment, which can then be considered probabilistically. For example, in the case of a coin being flipped, but hidden, and then asking how likely it is to be heads or tales, the correspoding thought experiment is quite simple: Imagine I were to flip the coin many times. Each time I hide the result, choose either heads or tails, and then look at the result. What percentage of the time would I be correct if I always chose heads? What about if I always chose tails? What about if I randomly choose either heads or tails (50/50) each time? If it is a fair coin, then we know that the answer in all three cases is 50/50. Based on this, we can establish that we are no better off choosing heads for our real case (where the flip is already determined, and we just don't know the result), than we would be choosing tails. It is an even bet. It is easy to devise a similar thought experiment for the envelopes. Assume we have a billion pairs of envelopes, 50% of them contain $100/$200, 50% contain $100/$50. Someone, who only knows that one of the envelopes contains twice as much as the other, makes a billion choices and gets 25% of his envelopes containing $200, 25% containing $50, and 50% containing $100. If he never switches, he obtains $112.5 * 1 billion. Now he computes the odds of switching applying the indifference principle: I got $200 25% of times, then: $4000.5 + $1000.5 = $250. I would get $2500.25 billion I got $50 25% of times, then: $1000.5 + $250.5 =$62.5. I would get $62.50.25 billion I got $100 50% of times, then: $2000.5 + $500.5 = $125. I would get $1250.5 billion This adds up to $140.6251 billion. Gain of switching in all cases: 25%, the same as in the case when only one choice was made. So, this seems not to be the point. I agree that the point is that the principle of indifference applies for the coin but not for the envelopes. This seems clear but I am not clear about why. To this you say: Death Monkey wrote: I assume that by this you mean "why can we assign the same odds to the coin being heads as it being tails, but not assign the same odds to the other envelope having $50 as it having $200"? If so, then the answer is that in the case of the coin, we have good physical reasons to assign the two outcomes equal probability. In fact, we usually even specify our justification for this assumption, by saying something along the lines of "we will assume it is a fair coin". We are making an assumption about the distribution of outcomes in our thought experiment. Namely that each time we flip, we are equally likely to get heads as we are to get tails. This assumption directly follows from the fact that we are assuming it is a fair coin, which is an assumption about the physical system being discussed, not some sort of a priori logical principle that can always be applied. This is the crucial point. Do we really assign heads 50% odds, relying on how the relevant physical processes must go? How the coin ends up does not uniquely depend on the coin being fair but on a big amount of physical circumstamces: the weight of the coin, the given impulse, the wind, the floor ... Circumstances that we cannot compute but that are as physically determined (once the coin has landed) as the distribution of the money in the envelopes; we have no apriori reason to believe these circumstances will scheme to make the coin come up heads exactly half the time. If we know the coin is fair, we reason this way: 'the only known causal factor suggests equal chances; I can't compute most of the factors but, in absence of knowledge, I will apply the principle of indifference and assign also 50% to the global causes leading to heads'. This is exactly the same type of application of the principle of indifference as in the case of the envelopes, an application based on the lack of knowledge. But perhaps things are not really so. Could it be that we have implicit reasons to expect the causal factors in the coin toss to favor heads and tails equal number of times throughout a sufficiently large number of experiments? Just having no reason to believe otherwise wouldn't make a difference between the coin and the envelopes! This sounds odd but... what else? Regards

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Posted 07/16/08 - 06:55 AM:	#38
LuaLuna, If he never switches, he obtains $112.5 * 1 billion. That's also what he gets if he always switches. The only way he would get something different is if he decided whether or not to switch based on how much money he saw in the first envelope. And to do that he would have to already know something about the distribution of possible values, which he doesn't. Now he computes the odds of switching applying the indifference principle: I got $200 25% of times, then: $4000.5 + $1000.5 = $250. I would get $2500.25 billion I got $50 25% of times, then: $1000.5 + $250.5 =$62.5. I would get $62.50.25 billion I got $100 50% of times, then: $2000.5 + $500.5 = $125. I would get $1250.5 billion I don't get your math here. On one hand, you seem to be using your a priori knowledge of the probabilities of each value when you specify that the first envelope with contain $100 50% of the time and so on, but then you promptly ignore this information when you include in your computations the possibility of the other envelope containing $25 or $200. Put simply, the computation you are doing does not correspond to any valid probabilistic model of the actual problem. This is the crucial point. Do we really assign heads 50% odds, relying on how the relevant physical processes must go? Absolutely. For example, if the coin was designed differently, for example being rounded with one side much heavier than the other, it would never even occur to us to assign them equal probabilities just because there are only two of them. Even if the geometry and mass distribution were so complex that we had no idea how to compute the relative probabilities, it would be naive for us to then simply assume that there is not going to be any bias one way or the other. How the coin ends up does not uniquely depend on the coin being fair but on a big amount of physical circumstamces: the weight of the coin, the given impulse, the wind, the floor ... Circumstances that we cannot compute but that are as physically determined (once the coin has landed) as the distribution of the money in the envelopes; we have no apriori reason to believe these circumstances will scheme to make the coin come up heads exactly half the time. But if the coin is fair, then none of these other factors will result in any bais towards heads or tails. That is what matters. If we know the coin is fair, we reason this way: 'the only known causal factor suggests equal chances; I can't compute most of the factors but, in absence of knowledge, I will apply the principle of indifference and assign also 50% to the global causes leading to heads'. Incorrect. Once we establish that the coin is fair, we do know that none of the other factors will bias it one way or the other. That is, in fact, precisely what we mean* by saying the coin is fair. If we did not actually know that none of the various factors in determining how the coin will land will tend to favor heads or tails, then we could not claim that the coin is fair. This is exactly the same type of application of the principle of indifference as in the case of the envelopes, an application based on the lack of knowledge. An invalid application of it, in both cases. But perhaps things are not really so. Could it be that we have implicit reasons to expect the causal factors in the coin toss to favor heads and tails equal number of times throughout a sufficiently large number of experiments? I'm not sure what you mean by implicit here. Like I said, if we did not have good reason to expect that, we would not call it a "fair coin". Let's put it another way. If I simply present a coin, and tell you that I am going to flip it a bunch of times, and that each time it comes up heads you give me a dollar, and each time it comes up tales I'll give you two dollars, would you play my game? Or would you demand to inspect the coin first? DM
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Posted 07/17/08 - 02:59 PM:	#39
Death Monkey wrote: LuaLuna, Once we establish that the coin is fair, we do know that none of the other factors will bias it one way or the other. DM What about the wind? What about the force the hand exerts on the coin in tossing it? What about the shape of the floor and its irregularities? How can we know all these factors will add up precisely in the way required to compensate over time any partial slant coming from any of them? Is no physical theory required to sustain such a belief? Of course the fairness of the coin decisively matters but how can it be enough to exclude any other possible bias? Regards

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Posted 07/17/08 - 03:04 PM:	#40
Death Monkey wrote: LuaLuna, Once we establish that the coin is fair, we do know that none of the other factors will bias it one way or the other. DM What about the wind? What about the force the hand exerts on the coin in tossing it? What about the shape of the floor and its irregularities? How can we know all these factors will add up precisely in the way required to compensate over time any partial slant coming from any of them? Is no physical theory required to sustain such a belief? Of course the fairness of the coin decisively matters but how can it be enough to exclude any other possible bias? Regards

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Posted 07/18/08 - 12:54 AM:	#41
LauLuna, Once we establish that the coin is fair, we do know that none of the other factors will bias it one way or the other. What about the wind? What about the force the hand exerts on the coin in tossing it? What about the shape of the floor and its irregularities? How can we know all these factors will add up precisely in the way required to compensate over time any partial slant coming from any of them? Pretty simple, actually. By testing the coin. Of course, from a physics point of view, we can check its geometry and mass distribution. For any of the factors you mention to produce a bias, there would need to be a significant asymetry. Is no physical theory required to sustain such a belief? Of course a physical theory is required. After all, it is a belief about the physical behavior of a physical object. That is exactly my point. If we have good physical reasons to to believe the coin is not going to be biased, we call it a "fair coin", and are justified in assuming equal probabilities. If we do not, then we don't call it a fair coin, and do not assume equal probabilities. Of course the fairness of the coin decisively matters but how can it be enough to exclude any other possible bias? I don't understand your question. The "fairness" of the coin is not some independant property it has. The coin is defined to be "fair" if and only if there are no biases. That is what "fair coin" means. DM
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Posted 07/18/08 - 01:09 AM:	#42
DoctorInWaiting wrote: This is a variation of the Monty Hall Problem. Probability wise without any information (like the monty hall problem instigated) it is a 50% chance. There is no real way of instigating which is the right envelope I feel. I'm not sure what this is supposed to mean, but in the standard Monty Hall offer, he provides new information that does change to balance of subjective probabilities.
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Posted 07/18/08 - 08:06 AM:	#43
I think Lauluna is onto something. In the case where ( I know) there are 2 pairs of evelopes, $50 & $100, and $100 & $200, then the swap would be justified in the case where I open my envelope and find $100. There is all sorts of 'extra information' that I have compared to the original problem. I know that I haven't got either the lowest of the low pair, or the highest of the high pair, and if I did have either of those - $50 or $200, then there would be no question what to do. I think this is the (mis)calculation that suggests in the original problem that it is worth swapping.
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Posted 07/18/08 - 10:03 AM:	#44
I have heard a similar situation to this, but it is not the same. I do think, however it will lend something to why this phenomenon occurs, and I apologize if it is too off topic. It is called the Monty Hall Paradox Say there are 3 envelopes, two contain X dollars and one contains 2X dollars. You pick one envelope. You then get to choose whether you keep that envelope or choose another envelope. Your original chance of picking the evelope that contains 2X dollars is 33%, but after you make your initial choice, one of the envelopes is opened to display its amount is not 2X dollars. Then given the choice to switch, you increase your chances to selecting the envelope with 2X dollars to 2/3. It might first appear that the probability is 1/2, but doing that ignores that your first 1/3 choice influenced which door would in fact be opened. The scenario you have proposed is similar. If you look at the overall amount of of money you can earn as a rate money/probability then the value proposed is different than if you were to use the situational rate of money/probability.
"Before enlightenment - chop wood, carry water. After enlightenment - chop wood, carry water." ~Zen Buddhist Proverb I have this sneaking suspicion that I am really God and have invented everyone and everything as a means of fooling myself into not being lonely. I'm not sure if it is working yet.

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Posted 07/19/08 - 05:07 AM:	#45
Death Monkey wrote: I don't understand your question. The "fairness" of the coin is not some independant property it has. The coin is defined to be "fair" if and only if there are no biases. That is what "fair coin" means. DM I really think you don't get my point. My fault surely. Let me try again. The fairness of the coin refers only to the coin's physical construction. But there can be other biases. Suppose the one who flips the coin has a special way to do it that favors heads, which may happen even if all involved agents are completely unaware of the fact. Imagine that the interaction of the specific way in which the coin is tossed once and once again with the wind direction is what favor heads, all agents being unaware thereof. Imagine... Why are we entitled to believe that a fairly constructed coin will not encounter systematic bias arising from the environment? On what grounds are we to expect that heads will tend to come up 50% of times? Is this not plainly a case of application of the principle of indifference justified by the absence of further information? Or have we any physical theory that entitles us to believe that the sum of all environmental factors will tend to favor equal distribution of chances? So, again: what is the ultimate difference between the case of the coin and the case of the envelopes? Regards

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Posted 07/19/08 - 05:56 AM:	#46
cckcckcc wrote: Say there are 3 envelopes, two contain X dollars and one contains 2X dollars. You pick one envelope. You then get to choose whether you keep that envelope or choose another envelope. Your original chance of picking the evelope that contains 2X dollars is 33%, but after you make your initial choice, one of the envelopes is opened to display its amount is not 2X dollars. Then given the choice to switch, you increase your chances to selecting the envelope with 2X dollars to 2/3. It might first appear that the probability is 1/2, but doing that ignores that your first 1/3 choice influenced which door would in fact be opened. This is not exactly the Monty Hall paradox. For your chances to improve when switching it is required that: 1. the envelope you opened does not contain 2X (this was implicit, I presume) 2. more importantly: the one who opened the second envelope KNEW where the 2X were and CHOSE the other envelope. It is the information used by the one who opens the second envelope that changes the probability from 1/3 to 2/3. And this is what makes this case different from the two envelope paradox. Regards

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Posted 07/20/08 - 12:48 AM:	#47
LauLuna The fairness of the coin refers only to the coin's physical construction. But there can be other biases. Suppose the one who flips the coin has a special way to do it that favors heads, which may happen even if all involved agents are completely unaware of the fact. Imagine that the interaction of the specific way in which the coin is tossed once and once again with the wind direction is what favor heads, all agents being unaware thereof. Imagine... If any of these "what-ifs" were considered to be significantly likely to happen, we would not assume equal probabilities for heads and tails anymore. Again, we do not just simply assume that no bias is possible. That assessment is based on experience and evidence. Coin-flipping has become a common way of making random choices because it has been found to be the case that when the coin itself is not biased, the resulting flips are 50/50. Also, if the coin is geometrically symmetric, there is no possible way that bias can occur without one of the participants being involved, either knowingly, or by not flipping properly. For example, everybody knows that you cannot simply drop the coin. Why are we entitled to believe that a fairly constructed coin will not encounter systematic bias arising from the environment? On what grounds are we to expect that heads will tend to come up 50% of times? Again, both experience and physics. Is this not plainly a case of application of the principle of indifference justified by the absence of further information? Or have we any physical theory that entitles us to believe that the sum of all environmental factors will tend to favor equal distribution of chances? The latter. Again, it is easy to see when you consider the issue of symmetry. Physical factors cannot conspire to bias the results if there is no significant physical difference between the two sides. Also, like I said before, one need not (and usually will not), rely just on theory. If you really are worried that the coin might be biased, you test it. And if you really believe that a person could deliberately force a bias by the way he flips it, then obviously you would not assume that the flips are going to be equally distributed then. So even then, it's not based on lack of information, but on the positive belief that a person cannot influence the result of the flip without obviously flipping it improperly. If one did not already hold this belief, they would no more assume that the flip is going to be random than they would if they were just letting the person choose the result. DM
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Posted 07/20/08 - 05:55 AM:	#48
Death Monkey wrote: Also, if the coin is geometrically symmetric, there is no possible way that bias can occur without one of the participants being involved, either knowingly, or by not flipping properly. For example, everybody knows that you cannot simply drop the coin. Not so. The bias can come from natural events in the environement. Death Monkey wrote: Again, it is easy to see when you consider the issue of symmetry. Physical factors cannot conspire to bias the results if there is no significant physical difference between the two sides. But they could yield 100% heads by virtue of the chain of physical causes that leads to the result. I know we usually overlook this but the question is on what grounds are we justified to do so. Experience is no good candidate for mathematical justification in computing probabilities. Let me show you how the two envelope paradox is no different from the coin toss case by presenting a strengthened version of the paradox. We could call it 'the strengthened Kraitchik paradox' (from Maurice Kraitchik, the French mathematician who proposed the original paradox for the first time). The following is a physically impossible idealized case but such cases are admissible in mathematical reasoning. A person P makes, say, one billion choices between envelopes A and B in each pair, and gets $10 in, say, 10 occasions. P opens all chosen envelopes and sees what lies in them. Now P is informed that the choice between all possible distributions has been made by tossing a fair coin (an infinite number of times; this is an idealized thought experiment). Then P is allowed to switch the envelopes in those cases in which the chosen envelope contains $10. P reasons this way: if I never switch, I’ll get $10·10 = $100 from the envelopes with $10. But if I always switch, the probabilities are that I will get $20 half the times (which already amounts to $100) and $5 the other half (which sums up to %25); in total %125 instead of the $100 that I would get if I always kept the chosen envelope. The problem is that P could have reasoned similarly if he had chosen the other envelopes in those ten cases. So the paradox arises again. But now the distribution of money into the envelopes depends on the toss of a fair coin. Edited by LauLuna on 07/20/08 - 06:38 AM. Reason: need to render it more understandable

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Posted 07/22/08 - 04:39 AM:	#49
LauLuna, Also, if the coin is geometrically symmetric, there is no possible way that bias can occur without one of the participants being involved, either knowingly, or by not flipping properly. For example, everybody knows that you cannot simply drop the coin. Not so. The bias can come from natural events in the environement. How do you figure? Any natural event in the environment would have to couple to the coin differently depending on its position/orientation, in order to introduce a bias towards heads or tails. That requires some asymmetry in the coin. Again, it is easy to see when you consider the issue of symmetry. Physical factors cannot conspire to bias the results if there is no significant physical difference between the two sides. But they could yield 100% heads by virtue of the chain of physical causes that leads to the result. I know we usually overlook this but the question is on what grounds are we justified to do so. Experience is no good candidate for mathematical justification in computing probabilities. On the contrary, it is pretty much the only justification we can have. After all, the mathematical equations we use to compute probabilities are just models. The applicability of a particular model is based on observation. If you don't think that testing the coin tells you anything about the odds of it coming up heads or tails in future flips, then I really don't know what to say, besides "you're wrong". Let me show you how the two envelope paradox is no different from the coin toss case by presenting a strengthened version of the paradox. We could call it 'the strengthened Kraitchik paradox' (from Maurice Kraitchik, the French mathematician who proposed the original paradox for the first time). The following is a physically impossible idealized case but such cases are admissible in mathematical reasoning. A person P makes, say, one billion choices between envelopes A and B in each pair, and gets $10 in, say, 10 occasions. P opens all chosen envelopes and sees what lies in them. Now P is informed that the choice between all possible distributions has been made by tossing a fair coin (an infinite number of times; this is an idealized thought experiment). This is, of course, nonsense, since there is no distribution of all possible distributions. To constrain the thought experiment so that there is a distribution of possible distributions of monetary values, provides you with exactly the distribution of possible monetary values that you need to be able to decide, based on what you saw in the first envelope, whether or not to switch. Also note that in such a scenario, whatever that presupposed distribution of possible monetary distributions is, it will definitely not be one such that, for any value seen in the first envelope, there will be a 50/50 conditional probability of it being the smaller of the two envelopes. Then P is allowed to switch the envelopes in those cases in which the chosen envelope contains $10. P reasons this way: if I never switch, I’ll get $10·10 = $100 from the envelopes with $10. But if I always switch, the probabilities are that I will get $20 half the times (which already amounts to $100) and $5 the other half (which sums up to %25); in total %125 instead of the $100 that I would get if I always kept the chosen envelope. The problem is that P could have reasoned similarly if he had chosen the other envelopes in those ten cases. So the paradox arises again. But now the distribution of money into the envelopes depends on the toss of a fair coin. As I pointed out, the flaw here is that if you do construct the experiment as you suggest, whereby some set of possible distributions is constructed, and then for each trial you use coin-flipping to choose one distribution for that trial, then it will absolutely not be the case that conditional probability of having opened the smaller of the two envelopes does not depend on the value of that envelope. No matter what the overal distribution of possible distributions you allow for the experiment is, there will be a resulting probability distribution for combinations of values in the envelopes, and that distribution will necessarily be one for which it is not always 50/50 for all observed values of the first envelope to be the smaller of the two values. Let me illustrate. Let's constrain the set of distributions as follows. We will consider all cases where the value of the smaller envelope ranges from $1 to $X, in $1 increments, with equal probabilities for all values. Now for each value of X, we have a possible distribution. We then use coin-flipping, as you suggested, to choose the value of X for a particular trial. We now need a distribution of X. Let's say that we set up our coin-flipping decision such that X is uniformly distribution from $10 to $10000, in $1 increments. This gives us a conditional probability distribution for whether the first envelope is the larger or smaller, that clearly does depend on the value seen. For example, in the case cited above, since $10 is pretty low in the overal range, it is more likely to be the smaller value. But if you saw something like $9000, it would be less likely. And for any value over $10000, you could be 100% sure that it is the larger envelope. And as I said above, if you do not constrain the set of possible distributions, then the whole idea is nonsense. There is no distribution of possible distributions for you to then choose from, using coin-flipping or otherwise. It would be like suggesting a thought experiment where you uniformly select a real number from positive to negative infinity by performing coin-flips. You just can't do it, theoretically or otherwise. DM
Pseudoscience makes Baby Jesus cry.

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