### RE: Observation selection effects

Brent Meeker and Jesse Mazer and others wrote: Well, lots and lots of complex mathematical argument on the two envelope problem... But no-one has yet pointed out a flaw in my rather simplistic analysis: (1) One envelope contains x currency units, so the other contains 2x currency units; (2) If you stop at the first envelope you choose, expected gain is: 0.5*x + 0.5*2x = 1.5x; (3) If you open the first envelope then switch to the second, your expected gain is: 0.5*2x + 0.5*x = 1.5x - as above, just in a different order, obviously; (4) If, in a variation, the millionaire flips a coin to give you double or half the amount in the first envelope if you switch envelopes, expected gain is: 0.25*2x + 0.25*0.5x + 0.25*x + 0.25*4x = 1.875x. In the latter situation you are obviously better off switching, but it is a mistake to assume that (4) applies in the original problem, (3) - hence, no paradox. Is the above wrong, or is it just so obvious that it isn't worth discussing? (I'm willing to accept either answer). Stathis Papaioannou _ Searching for that dream home? Try http://ninemsn.realestate.com.au for all your property needs.

### RE: Observation selection effects

-Original Message- From: Stathis Papaioannou [mailto:[EMAIL PROTECTED] Sent: Thursday, October 14, 2004 7:36 AM To: [EMAIL PROTECTED]; [EMAIL PROTECTED]; [EMAIL PROTECTED] Subject: RE: Observation selection effects Brent Meeker and Jesse Mazer and others wrote: Well, lots and lots of complex mathematical argument on the two envelope problem... But no-one has yet pointed out a flaw in my rather simplistic analysis: (1) One envelope contains x currency units, so the other contains 2x currency units; (2) If you stop at the first envelope you choose, expected gain is: 0.5*x + 0.5*2x = 1.5x; (3) If you open the first envelope then switch to the second, your expected gain is: 0.5*2x + 0.5*x = 1.5x - as above, just in a different order, obviously; (4) If, in a variation, the millionaire flips a coin to give you double or half the amount in the first envelope if you switch envelopes, expected gain is: 0.25*2x + 0.25*0.5x + 0.25*x + 0.25*4x = 1.875x. In the latter situation you are obviously better off switching, but it is a mistake to assume that (4) applies in the original problem, (3) - hence, no paradox. Is the above wrong, or is it just so obvious that it isn't worth discussing? (I'm willing to accept either answer). Stathis Papaioannou It's not wrong - I just don't think it addresses the paradox. To resolve the paradox you must explain why it is wrong to reason: I've opened one envelope and I see amount m. If I keep it my gain is m. If I switch my expected gain is 0.5*m/2 + 0.5*2m = 1.25m, therefore I should switch. To say that in another, similiar game (4) this reasoning is correct, doesn't explain why it is wrong in the given case. Your (2) and (3) aren't to the point because they don't recognize that after opening one envelope you have some information that seems to change the expected value. In my analysis, it is apparent that the trick of showing the expected value doesn't change depends on the feature of the problem statement that the distribution of the amount of money is scale free - i.e. all amounts are equally likely. If you accept this, then a Bayesian analysis of your rational belief shows that the expected value doesn't change when you open the envelope and see amount m. Intuitively, observing a value from a distribution that is flat from zero to infinity *doesn't* give you any information. Solving the paradox is to show explicitly why this is so. As Jesse and others have pointed out this scale-free (all amounts are equally likely) aspect of the problem as stated is unrealistic and in any real situation your prior estimate of the scale of the amounts will cause you to modify your expected value after you see the amount in first envelope. This modification may prompt you to switch or not - but it's a different problem. Brent Meeker

### RE: Observation selection effects

-Original Message- From: Jesse Mazer [mailto:[EMAIL PROTECTED] Sent: Tuesday, October 05, 2004 11:01 PM To: [EMAIL PROTECTED]; [EMAIL PROTECTED] Subject: RE: Observation selection effects -Original Message- From: Jesse Mazer [mailto:[EMAIL PROTECTED] Sent: Tuesday, October 05, 2004 8:45 PM To: [EMAIL PROTECTED]; [EMAIL PROTECTED] Subject: RE: Observation selection effects If the range of the smaller amount is infinite, as in my P(x)=1/e^x example, then it would no longer make sense to say that the range of the larger amount is r times larger. Sure it does; r*inf=inf. P(s)=exp(-x) - P(l)=exp(-x/r) But it would make just as much sense to say that the second range is 3r times wider, since by the same logic 3r*inf=inf. In other words, this step in your proof doesn't make sense: In other words, the range of possible amounts is such that the larger and smaller amount do not overlap. Then, for any interval of the range (x,x+dx) for the smaller amount with probability p, there is a corresponding interval (r*x, r*x+r*dx) with probability p for the larger amount. Since the latter interval is longer by a factor of r P(l|m)/P(s|m) = r , In other words, no matter what m is, it is r-times more likely to fall in a large-amount interval than in a small-amount interval. As for your statement that P(s)=exp(-x) - P(l)=exp(-x/r), that can't be true. It doesn't make sense that the value of the second probability distribution at x would be exp(-x/r), since the range of possible values for the amount in that envelope is 0 to infinity, but the integral of exp(-x/r) from 0 to infinity is not equal to 1, so that's not a valid probability distribution. Also, now that I think more about it I'm not even sure the step in your proof I quoted above actually makes sense even in the case of a probability distribution with finite range. What exactly does the equation P(l|m)/P(s|m) = r mean, anyway? For any give amount of money, m, found in the first envelope, it is more probable by a factor of r that it came from the Larger envelope - where probable means degree of rational belief, not fraction in a statistical ensemble. It can't mean that if I choose an envelope at random, before I even open it I can say that the amount m inside is r times more likely to have been picked from the larger distribution, since I know there is a 50% chance I will pick the envelope whose amount was picked from the larger distribution. Is it supposed to mean that if we let the number of trials go to infinity and then look at the subset of trials where the envelope I opened contained m dollars, it is r times more likely that the envelope was picked from the larger distribution on any given trial? This can't be true for every specific m--for example, if the smaller distribution had a range of 0 to 100 and the larger had a range of 0 to 200, But whole point is that there is no specific m from which you can reason. if I set m=150, then in every single trial where I found 150 dollars in the envelope it must have been selected from the larger distribution. You could do a weighted average over all possible values of m, like integral over all possible values of m of P('I found m dollars in the envelope I selected')*P('the envelope I selected had an amount taken from the smaller distribution' | 'I found m dollars in the envelope I selected'), which you could write as integral over m of P(m)*P(s|m), but I don't think it would be true that the ratio integral over m of P(m)*P(l|m)/integral over m of P(m)*P(s|m) would be equal to r, in fact I think both integrals would always come out to 1/2 so the ratio would always be 1...and even if I'm wrong, replacing P(l|m)/P(s|m) with this ratio of integrals would mess up the rest of your proof. Jesse No, it doesn't depend on assuming a flat distribution for the money, only for our knowledge (or on our acceptance of problem as stated). Here's the more explicit (but less intuitive) proof - I hope the formatting doesn't get chopped up too much by your mail reader. Without loss of generality, we can describe our prior density functions for the amounts in the two envelopes in terms of a density function, fo(x), the ratio r of the larger amount to the smaller, and a scale factor, k. Let L be the event that the evelope with the larger amount is picked and S the event that the envelope with the smaller amount if picked. Then our prior density functions for the amount m in the envelope is: For the smaller amount our prior is:f(m|S k) = k fo(km) and for the larger amount: f(m|L k) = (k/r) fo(km/r) Our uncertainity about the scale factor, k, is described by a density g(k). So f(m|S) = INT k fo(km) g(k) dk ,where INT is integral zero-to-infinity f(m|L) = INT (k/r) fo(km/r) g(k) dk Now in the first equation make a change of variable in the integral by y=km f(m|S) = INT (y/m) fo(y) g(y/m) dy/m = (1/m^2) INT y fo(y) g(y/m) dy

### Re: observation selection effects

Thanks, Kory, that takes care of my confusion. The same to Jesse's post. John Mikes - Original Message - From: Kory Heath [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Sunday, October 10, 2004 7:17 PM Subject: Re: observation selection effects At 02:57 PM 10/10/2004, John M wrote: Then it occurred to me that you made the same assumption as in my post shortly prior to yours: a priviledge of ME to switch, barring the others. I think this pinpoints one of the confusions that's muddying up this discussion. Under the Flip-Flop rules as they were presented, the Winning Flip is determined before people switch, and the Winning Flip doesn't change based on how people switch. In that scenario, my table is correct, and there is no paradox. We can also consider the variant in which the Winning Flip is determined after people decide whether or not to switch. But that game is functionally identical to the game where there is no coin-toss at all - everyone just freely chooses Heads or Tails, then the Winning Flip is determined and the winners are paid. Flipping a coin, looking at it, and then deciding whether or not to switch it is identical to simply picking heads or tails! The coin-flips only matter in the first variant, where they determine the Winning Flip *before* people make their choices. In this variant, it doesn't matter whether you switch or not (i.e. whether you choose heads or tails) - you are more likely to lose than win. We can use the same 3-player table we've been discussing to see that there are eight possible outcomes, and you only win in two of them. Once again, there's no paradox, although you might *feel* like there is one. You might reason that the Winning Flip is equally likely to be heads or tails, so no matter which one you pick, your odds of winning will be 50/50. What's missing from this logic is the recognition that no matter what you pick, your choice will automatically decrease the chances of that side being in the minority. -- Kory

### Re: observation selection effects

At 04:47 PM 10/10/2004, Jesse Mazer wrote: If I get heads, I know the only possible way for the winning flip to be heads would be if both the other players got tails, whereas the winning flip will be tails if the other two got heads *or* if one got heads and the other got tails. I agree with this, but I want to add a subtle point: it's correct to switch *even if I haven't looked at my own coin*. That's because, despite the fact that I don't know whether my own coin is heads or tails, I know that, whichever it is, it's more likely to be in the majority than the minority. That's not to say that *nobody* needs to look at my coin. In order to determine whether or not my choice to switch puts me in the heads or the tails group, *someone's* going to have to look at my coin. But the rules of the game allow me to pass off this act of looking to someone else - they essentially allow me to tell the casino worker hey, take a look at my coin, will ya, and assign me to the opposite. The important point is that I can safely pass off this instruction to switch without even knowing the result of my coin-flip, because I know that, whatever my coin is, it's more likely to be in the majority group. If we change the rules of the game slightly, and say that, instead of choosing whether or not to switch, you have to actually choose heads or tails, then, of course, you yourself do need to see the result of your own coin-flip. -- Kory

### re: observation selection effects

You're right, as was discussed last week. It seems I clicked on the wrong thing in my email program and have re-sent an old post. My apologies for taking up the bandwidth! --Stathis From: Kory Heath [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: re: observation selection effects Date: Sat, 09 Oct 2004 18:17:50 -0400 At 10:35 AM 10/9/2004, Stathis Papaioannou wrote: From the point of view of typical player, it would seem that there is not: the Winning Flip is as likely to be heads as tails, and if he played the game repeatedly over time, he should expect to break even, whether he switches in the final step or not. That's not correct. While it's true that the Winning Flip is as likely to be heads as tails, it's not true that I'm as likely to be in the winning group as the loosing group. Look at the case when there are only three players. There are eight possible outcomes: Me: H Player 1: H Player 2: H - WF: T Me: H Player 1: H Player 2: T - WF: T Me: H Player 1: T Player 2: H - WF: T Me: H Player 1: T Player 2: T - WF: H Me: T Player 1: H Player 2: H - WF: T Me: T Player 1: H Player 2: T - WF: H Me: T Player 1: T Player 2: H - WF: H Me: T Player 1: T Player 2: T - WF: H I am in the winning group in only two out of these eight cases. So my chances of winning if I don't switch are 1/4, and my chances of winning if I do switch are 3/4. There's no paradox here. -- Kory _ Enter our Mobile Babe Search and win big! http://ninemsn.com.au/babesearch

### Re: observation selection effects

At 07:17 PM 10/10/2004, Kory Heath wrote: We can also consider the variant in which the Winning Flip is determined after people decide whether or not to switch. In a follow-up to my own post, I should point out that your winning chances in this game depend on how your opponents are playing. If all of your opponents are playing randomly, then you have a negative expectation no matter what you do. If your opponents are not playing randomly, then you may be able to exploit patterns in their play to generate a positive expectation. -- Kory

### re: observation selection effects

Here is a similar paradox to the traffic lane example: In the new casino game called Flip-Flop, an odd number of players pay $1 each to gather in individual cubicles and flip a coin (so no player can see what another player is doing). The game organisers tally up the results, and the result which is in the minority is said to be the Winning Flip, while the minority result is said to be the Losing Flip. For example, if there are 101 players and of these 53 flip heads while 48 flip tails, tails is the Winning Flip and heads is the Losing Flip. Before the result of the tally is announced, each player must commit to either keep the result of their original coin flip, whether heads or tails, or switch to the opposite result. The casino then announces what the Winning Flip was, and players whose final result (however it was obtained) corresponds with this are paid $2, while the rest get nothing. The question now: is there anything to be gained by switching at the last step of this game? From the point of view of typical player, it would seem that there is not: the Winning Flip is as likely to be heads as tails, and if he played the game repeatedly over time, he should expect to break even, whether he switches in the final step or not. On the other hand, it seems clear that if nobody switches, the casino is ahead, while if everbody switches, the players are ahead; so switching would seem to be a winning strategy for the players. This latter result is not due to any cooperation effect, as only those players who switch get the improved (on average) outcome. Stathis Papaioannou

### Re: observation selection effects

Stathis, in this new FLip-Flop I see some slight merit beyond the symmetry of switching from one unknown to another unknown: If I got heads,I can THINK of the majority getting heads. (No justifiacation, however, but a slight idea that I am the 'average guy', not the exceptional minority). In that case it is allowable to switch, to get into the less likely group, the paying minority. I would not include any calculations: the conditions are not quantizable in my opinion. In the moment you start quantizing, the situation reverts into a perfect symmetry with the opposite.UNLESS there are hidden parameters pointing to a NON-50-50 situation. The 'insufficient number of sampling' is unsatisfactory it works both ways. John Mikes - Original Message - From: Stathis Papaioannou To: [EMAIL PROTECTED] Sent: Saturday, October 09, 2004 10:35 AM Subject: re: observation selection effects Here is a similar paradox to the traffic lane example: In the new casino game called Flip-Flop, an odd number of players pay $1 each to gather in individual cubicles and flip a coin (so no player can see what another player is doing). The game organisers tally up the results, and the result which is in the minority is said to be the Winning Flip, while the minority result is said to be the Losing Flip. For example, if there are 101 players and of these 53 flip heads while 48 flip tails, tails is the Winning Flip and heads is the Losing Flip. Before the result of the tally is announced, each player must commit to either keep the result of their original coin flip, whether heads or tails, or switch to the opposite result. The casino then announces what the Winning Flip was, and players whose final result (however it was obtained) corresponds with this are paid $2, while the rest get nothing. The question now: is there anything to be gained by switching at the last step of this game? From the point of view of typical player, it would seem that there is not: the Winning Flip is as likely to be heads as tails, and if he played the game repeatedly over time, he should expect to break even, whether he switches in the final step or not. On the other hand, it seems clear that if nobody switches, the casino is ahead, while if everbody switches, the players are ahead; so switching would seem to be a winning strategy for the players. This latter result is not due to any cooperation effect, as only those players who switch get the improved (on average) outcome. Stathis Papaioannou

### re: observation selection effects

At 10:35 AM 10/9/2004, Stathis Papaioannou wrote: From the point of view of typical player, it would seem that there is not: the Winning Flip is as likely to be heads as tails, and if he played the game repeatedly over time, he should expect to break even, whether he switches in the final step or not. That's not correct. While it's true that the Winning Flip is as likely to be heads as tails, it's not true that I'm as likely to be in the winning group as the loosing group. Look at the case when there are only three players. There are eight possible outcomes: Me: H Player 1: H Player 2: H - WF: T Me: H Player 1: H Player 2: T - WF: T Me: H Player 1: T Player 2: H - WF: T Me: H Player 1: T Player 2: T - WF: H Me: T Player 1: H Player 2: H - WF: T Me: T Player 1: H Player 2: T - WF: H Me: T Player 1: T Player 2: H - WF: H Me: T Player 1: T Player 2: T - WF: H I am in the winning group in only two out of these eight cases. So my chances of winning if I don't switch are 1/4, and my chances of winning if I do switch are 3/4. There's no paradox here. -- Kory

### Re: Observation selection effects

Hal Finney writes: Not to detract from your main point, but I want to point out that sometimes there is ambiguity about how to count worlds, for example in the many worlds interpretation of QM. There are many examples of QM based world-counting which seem to show that in most worlds, probability theory should fail. I'm not sure what examples you have in mind here, but this is actually the general point I was trying to make: probability theory doesn't seem to work the same way in a many worlds cosmology, due to complications such as observers multiplying and then not being able to access the entire probability space after the event of interest. Consider these three examples: (A) In a single world cosmology, I claim that using my magic powers, I have bestowed on you, and you alone, the ability to pick the winning numbers in this week's lottery. If you then buy a lottery ticket and win the first prize, I think it would be reasonable to concede that there was probably some substance to my claim (if not magic powers, then at least an effective way of cheating). (B) In a single world cosmology, I announce that using my magic powers, I have bestowed on some lucky gambler the ability to pick the winning numbers in this week's lottery. Now, someone does in fact win the first prize this week, but that is not surprising, because there is almost always at least one winner each week. I cannot reasonably claim to have helped the winner unless I had somehow tagged him or otherwise uniquely identified him before the lottery was drawn, as in (A). (C) In a many worlds cosmology, I seek you out as in (A) and make the same claim about bestowing on you the ability to pick the winning numbers in this week's lottery. You buy a ticket, and win first prize. Should you thank me for helping you win, as in (A)? In general, no; this situation is actually more closely analogous to (B) than to (A). For it is certain that at least one future version of you will win, just as it is very likely that at least one person will win in the single world example. I can only claim that I helped you win if I somehow identified which version in which world is going to win before the lottery is drawn, and that is impossible. Stathis Papaioannou _ Check out Election 2004 for up-to-date election news, plus voter tools and more! http://special.msn.com/msn/election2004.armx

### Re: Observation selection effects

Stathis Papaioannou writes: Hal Finney writes: Not to detract from your main point, but I want to point out that sometimes there is ambiguity about how to count worlds, for example in the many worlds interpretation of QM. There are many examples of QM based world-counting which seem to show that in most worlds, probability theory should fail. I'm not sure what examples you have in mind here, The specific kind of example goes like this. Suppose you take a vertically polarized photon and pass it through a polarizer that is tilted slightly from the vertical. Quantum mechanics predicts that there is a high chance, say 99%, that the photon will pass through, and a low chance, 1%, that it will not make it and be absorbed. Now, the many worlds interpretation can be read to say that the universe splits into two when this experiment occurs. There are two possible outcomes: either it passes through or it is absorbed. So there are two universes corresponding to the two results. However, the universes are not of equal probability, according to QM. One should be observed 99% of the time and the other only 1% of the time. The discrepancy gets worse if we imagine repeating the experiment multiple times. Each time the multiverse splits again in two. If we did it, say, 20 times, there would be 2^20 or about 1 million universes. In only one of those universes did the photon take the 99% chance each time, yet that is the expected result. By a counting argument, the chance of getting that result is only one in a million since only one world out of a million sees it. This is the apparent contradiction between the probability predictions of orthodox quantum mechanics and the MWI, assuming that we count worlds in this way. but this is actually the general point I was trying to make: probability theory doesn't seem to work the same way in a many worlds cosmology, due to complications such as observers multiplying and then not being able to access the entire probability space after the event of interest. Consider these three examples: (A) In a single world cosmology, I claim that using my magic powers, I have bestowed on you, and you alone, the ability to pick the winning numbers in this week's lottery. If you then buy a lottery ticket and win the first prize, I think it would be reasonable to concede that there was probably some substance to my claim (if not magic powers, then at least an effective way of cheating). (B) In a single world cosmology, I announce that using my magic powers, I have bestowed on some lucky gambler the ability to pick the winning numbers in this week's lottery. Now, someone does in fact win the first prize this week, but that is not surprising, because there is almost always at least one winner each week. I cannot reasonably claim to have helped the winner unless I had somehow tagged him or otherwise uniquely identified him before the lottery was drawn, as in (A). (C) In a many worlds cosmology, I seek you out as in (A) and make the same claim about bestowing on you the ability to pick the winning numbers in this week's lottery. You buy a ticket, and win first prize. Should you thank me for helping you win, as in (A)? In general, no; this situation is actually more closely analogous to (B) than to (A). For it is certain that at least one future version of you will win, just as it is very likely that at least one person will win in the single world example. I can only claim that I helped you win if I somehow identified which version in which world is going to win before the lottery is drawn, and that is impossible. I'm afraid I don't agree with the conclusion in (C). I definitely should thank you. To see this, let's make my thanks a little more sincere, in the form of a payment. Suppose I agree in advance to pay you $1000 if you succeed in helping me win the lottery. I say that is a wise decision on my part. It doesn't cost me anything if you don't help, and if you do have some way of rigging the lottery then I can easily afford to pay you this modest sum out of my winnings. But I think your reasoning suggests that it is unwise, since I will win anyway, so why should I pay anything to you? I don't need to thank you in this way. Do you agree that this follows from your reasoning? Hal Finney

### Re: Observation selection effects

Jesse Mazer wrote: I don't think that's a good counterargument, because the whole concept of probability is based on ignorance... No, I don't agree! Probability is based in a sense on ignorance, but you must make full use of such information as you do have. If you toss a fair coin, is Pr(heads)=0.5? According to your argument, it could actually be anything between zero and one, because it is possible I am lying about it being a fair coin! Here is another two envelope example: Two envelopes, A and B, contain two doses of the drug Lifesavium, the Correct Dose and the Half Dose. If you give the patient more than 1.5 times the Correct Dose you will certainly kill him, while if you give him the Half Dose you will save his life, although he won't make an immediate recovery as he would if you gave him the Correct Dose. If you don't give him any medication at all, again, he will surely die. Once you open an envelope, the medication in in such a form that you must give the full dose, or nothing. You are faced with the two envelopes, the above information and the sick patient, with no other help, on a desert island. There is one further complication: if you open the first envelope, and then decide to open the second envelope, you must destroy the contents of the first envelope in order to get to the second envelope. OK: so you open envelope A and find that it contains 10mg of Lifesavium. You don't know whether this is The Correct Dose or the Half Dose; so envelope B may have either 5mg or 20mg, right? And if 10mg is the Correct Dose, then if you discard envelope A in favour of envelope B, there is a 50% chance that envelope B will have double the Correct Dose and you will kill the patient - so you had better stick with envelope A, right? I think you can see the error in the above argument. You already know that the amount in each envelope is fixed, so even though you have no idea of the actual dosages involved, or which envelope contains which dose, even after opening the first envelope, there is NO WAY you can give the patient an overdose. There is no way envelope B can contain 20mg of Lifesavium, but even though you cannot know this, you can use the above reasoning to deduce that there is no expected benefit from choosing a strategy of switching or not switching - as you can also see intuitively from the symmetry of the situation, whether you choose envelope A or B first. In the game with the envelopes and the money, the analogous error is to think that there is a possibility of doubling your money when you have actually picked the envelope containing the larger sum first. As I said in my previous post, if this assumption is valid, then you are playing a different game in which our eccentric millionaire flips a coin to decide (without telling you which) if he will put double or half the sum you find on opening envelope A into envelope B. You would then certainly be better off, on average, if you switched envelopes. Stathis Papaioannou _ Check out Election 2004 for up-to-date election news, plus voter tools and more! http://special.msn.com/msn/election2004.armx

### Re: Observation selection effects

Stathis Papaioannou wrote: Jesse Mazer wrote: I don't think that's a good counterargument, because the whole concept of probability is based on ignorance... No, I don't agree! Probability is based in a sense on ignorance, but you must make full use of such information as you do have. Of course--I didn't mean it was based *only* on ignorance, you must incorporate whatever information you have into your estimate of the probability, but no more. Your argument violates the but no more rule, since it incorporates the knowledge of an observer who has seen how much money both envelopes contain, while I only know how much money one envelope contains. If you toss a fair coin, is Pr(heads)=0.5? According to your argument, it could actually be anything between zero and one, because it is possible I am lying about it being a fair coin! My argument implies nothing of the sort. But your argument would seem to imply that if I am watching a videotape of a fair coin toss, then if someone else has already watched the tape, it would be permissible to incorporate their knowledge of the outcome of the toss into a probability calculation, even if I myself don't have this knowledge. Here is another two envelope example: Two envelopes, A and B, contain two doses of the drug Lifesavium, the Correct Dose and the Half Dose. If you give the patient more than 1.5 times the Correct Dose you will certainly kill him, while if you give him the Half Dose you will save his life, although he won't make an immediate recovery as he would if you gave him the Correct Dose. If you don't give him any medication at all, again, he will surely die. Once you open an envelope, the medication in in such a form that you must give the full dose, or nothing. You are faced with the two envelopes, the above information and the sick patient, with no other help, on a desert island. There is one further complication: if you open the first envelope, and then decide to open the second envelope, you must destroy the contents of the first envelope in order to get to the second envelope. OK: so you open envelope A and find that it contains 10mg of Lifesavium. You don't know whether this is The Correct Dose or the Half Dose; so envelope B may have either 5mg or 20mg, right? And if 10mg is the Correct Dose, then if you discard envelope A in favour of envelope B, there is a 50% chance that envelope B will have double the Correct Dose and you will kill the patient - so you had better stick with envelope A, right? I think you can see the error in the above argument. You already know that the amount in each envelope is fixed, so even though you have no idea of the actual dosages involved, or which envelope contains which dose, even after opening the first envelope, there is NO WAY you can give the patient an overdose. There is no way envelope B can contain 20mg of Lifesavium, but even though you cannot know this, you can use the above reasoning to deduce that there is no expected benefit from choosing a strategy of switching or not switching - as you can also see intuitively from the symmetry of the situation, whether you choose envelope A or B first. This case is not analogous to the two-envelope problem, because in this case it is part of *my* knowledge that one envelope contains the Correct dose and the other contains the Half dose, and neither contains a Double dose. In contrast, your analysis of the two-envelope problem relied on information I don't have, namely whether the two envelopes contained $50 and $100 or $25 and $50. Suppose I know that the envelope-stuffer flipped a fair coin to decide whether to put $25 or $50 in one envelope, then put double that amount in the other. I randomly choose one envelope and open it, and find $50. Do you disagree that my average expected return from switching would now be (0.5)(25) + (0.5)(100) = 62.5? If this experiment was repeated many times and we looked only at the subset of cases where the first envelope I opened contained $50 and I chose to switch, wouldn't my average winnings in this subset of cases be $62.50? In the game with the envelopes and the money, the analogous error is to think that there is a possibility of doubling your money when you have actually picked the envelope containing the larger sum first. But *you* don't know that the envelope you picked was the one with the larger sum. This is akin to arguing that it is an error to think there is a possibility of winning if you bet $100 on heads in a videotaped coin toss, since someone who's already watched the tape knows it comes up tails, even though you don't know that. Would you indeed say it's an error to believe my average expected return is $50 in this case? Jesse

### Re: Observation selection effects

Jesse Mazer wrote: I don't think that's a good counterargument, because the whole concept of probability is based on ignorance... No, I don't agree! Probability is based in a sense on ignorance, but you must make full use of such information as you do have. Of course--I didn't mean it was based *only* on ignorance, you must incorporate whatever information you have into your estimate of the probability, but no more. Your argument violates the but no more rule, since it incorporates the knowledge of an observer who has seen how much money both envelopes contain, while I only know how much money one envelope contains. Sorry Jesse, I can see in retrospect that I was insulting your intelligence as a rhetorical ploy, and we shouldn't stoop to that level of debate on this list. You say that you must incorporate whatever information you have, but no more in the envelopes/money example. The point I was trying to make with my envelope/drug example is that you need to take into account the fact that the amount in each envelope is fixed, but again you are right, it was not exactly analogous. But you have passed over the final point in my last post, which I now restate: (1) The original game: envelope A and B, you know one has double the amount of the other, but you don't know which. You open A and find $100. Should you switch to B, which may have either $50 or $200? (2) A variation: everything is the same, up to the point where you are pondering whether to switch to envelope B, when the millionaire walks in, and hidden from view, flips a coin to decide whether to replace whatever was originally in envelope B with either double or half the sum in envelope A, i.e. either $50 or $200. Now, which game would you prefer to play, (1) or (2)? They are not the same. In game (1), if the $100 in A is actually the higher amount, if you switch you will get $50 for sure; but in game (2) if the $100 is actually the higher amount you have a 50% chance of getting $200 if you switch. It works in the reverse way if the $100 is the lower amount - you could lose $50 rather than gain $100 - but the possible gain outweighs the possible loss. Look at it another way: game (2) is actually asymmetrical. The amount you win if you play it many times will be different if you switch, because you really do have more to gain than to lose by switching (and the millionaire will have to pay out more on average). On the other hand, intuitively, you can see that your expected gains in game (1) should be the same whether you switch or not. The paradox comes from reasoning as if you are playing game (2) when you are really playing game (1). Stathis Papaioannou _ Discover how everyone everything in our world's connected: http://www.onebigvillage.com.au?obv1=hotmail

### Re: Observation selection effects

Addition to my last post: (1) The original game: envelope A and B, you know one has double the amount of the other, but you don't know which. You open A and find $100. Should you switch to B, which may have either $50 or $200? (2) A variation: everything is the same, up to the point where you are pondering whether to switch to envelope B, when the millionaire walks in, and hidden from view, flips a coin to decide whether to replace whatever was originally in envelope B with either double or half the sum in envelope A, i.e. either $50 or $200. Say one envelope contains $x and the other $2x. If you keep the first envelope in game (2), and if you keep the first one OR switch in game (1), you should expect to win $1.5x. If you switch in game (2), you should expect to win 0.25*($0.5x + $2x + $x +$4x) = $1.875x. Stathis Papaioannou _ Enter our Mobile Babe Search and win big! http://ninemsn.com.au/babesearch

### Re: Observation selection effects

This has been an interesting thread so far, but let me bring it back to topic for the Everything List. It has been assumed in most posts to this list over the years that our current state must be a typical state in some sense. For example, our world has followed consistent laws of physics for as long as anyone has been able to determine - the old no white rabbit worlds observation. In the face of ensemble type theories such as the MWI of QM, this is seen as presenting a problem: if anything that can happen, does happen, why does our experience of the world include only a very limited, orderly subset of this anything? There have been many attempts to answer the above question, eg. see Russell Standish' paper Why Ockham's Razor? But does our current orderly world imply that most possible worlds are orderly? It seems to me that there is an asymmetry between (a) what we can expect for the future, and (b) what we can deduce about the probability implicit in (a) from what actually does happen. Suppose that according to X-Theory, in the next minute the world will split into one million different versions, of which one version will be the same sort of orderly world we are used to, while the rest will be worlds in which it will be immediately obvious to us that very strange things are happening, eg. dragons materialising out of thin air, furniture levitating, the planet Jupiter hurling itself into the sun, etc. I think it is reasonable to expect that if X-Theory is correct, we will very likely see these bizarre and frightening things happen in the next minute. Now, here we are, a minute later, and nothing bizarre has happened after all. Does this mean that X-Theory is probably wrong? Perhaps not. After all, the theory did predict with 100% certainty that one version of the world will continue as before. The objection to this will no doubt be, yes, but how likely is it that WE end up in that particular version? And the answer to this objection is, it is 100% certain that WE end up in that particular version; just as it is 100% certain that 999,999 copies of us end up in the bizarre versions. Those 999,999 copies are not continuing to type away as I am, because they are running around in a panic. The above is simply a version of the Anthropic Principle as applied to intelligent life in the universe. A particular ensemble theory may predict that it is overwhelmingly unlikely that a particular universe will allow the development of intelligent life. Does the fact that we are here, appparently intelligent and alive, count as evidence against that theory? No, because the theory predicts that although unlikely, it is certain to happen in at least ONE universe, and obviously that universe will be the one we find ourselves in. Stathis Papaioannou _ In the market for a car? Buy, sell or browse at CarPoint: http://server-au.imrworldwide.com/cgi-bin/b?cg=linkci=ninemsntu=http://carpoint.ninemsn.com.au?refid=hotmail_tagline

### Re: Observation selection effects

Stathis Papaioannou wrote: Sorry Jesse, I can see in retrospect that I was insulting your intelligence as a rhetorical ploy, and we shouldn't stoop to that level of debate on this list. No problem, I wasn't insulted... You say that you must incorporate whatever information you have, but no more in the envelopes/money example. The point I was trying to make with my envelope/drug example is that you need to take into account the fact that the amount in each envelope is fixed Well, I think that's like saying that in the videotaped coinflip example, you need to take into account the fact that the outcome of the flip is already fixed. I don't think it matters whether it's really fixed or not, since probabilities are about your knowledge rather than objective reality (they're epistemological, not ontological), and since you are equally ignorant of the outcome regardless of whether the flip happens in realtime or on video, your probabilistic reasoning should be the same. But you have passed over the final point in my last post, which I now restate: (1) The original game: envelope A and B, you know one has double the amount of the other, but you don't know which. You open A and find $100. Should you switch to B, which may have either $50 or $200? (2) A variation: everything is the same, up to the point where you are pondering whether to switch to envelope B, when the millionaire walks in, and hidden from view, flips a coin to decide whether to replace whatever was originally in envelope B with either double or half the sum in envelope A, i.e. either $50 or $200. Well, my argument about the two-envelope paradox all along has been that you need to think about the probability distribution the millionaire uses to stuff the two envelopes, and that once you do that the apparent paradox disappears. So we need to consider what probability distributions the millionaire used in these examples. Let's say, for example, that the millionaire flips a coin to decide whether to put $50 or $100 in one envelope, and then puts double that amount in the other. In that case, if I pick an envelope randomly, there is a 1/4 chance I'll find $50 inside, a 1/2 chance I'll find $100 inside, and a 1/4 chance I'll find $200 inside. If I find either $50 or $200, I know with complete certainty how much the other envelope contains; but if I find $100, then from my point of view there's a 1/2 chance the other envelope contains $50 and a 1/2 chance the other envelope contains $200. Now, if you assume that in game (2) the millionaire *only* replaces the amount in the second envelope with a new amount based on a coinflip *if* I found $100 in the first envelope, but doesn't mess with the second envelope if I found $50 or $200 in the first one, then both games are exactly equal, from my point of view. In both cases, whenever I find $100 in the first envelope, my average expected return from switching would be (0.5)($50) + (0.5)($200) = $125, so it's better to switch. On the other hand, if in game (2) the millionaire replaced the amount in the second envelope with a new amount based on a coinflip *regardless* of how much I found in the first envelope, this would change my strategy if I found either $50 or $200 in the first envelope. In this case, it will be to my advantage to switch no matter how much I find in the first envelope, since my average expected return from switching will always be 1.25 times however much I found in that envelope. In contrast, in game (1) my average expected return from switching would be $100 if I found $50 in the first envelope and $100 if I found $200 in the first envelope, while my average expected return from switching if I found $100 would still be $125, so my total average expected return from switching regardless of what I find in the first envelope is (1/4)($100) + (1/2)($125) + (1/4)($100) = $112.50, while my average expected return from sticking with my first choice regardless of how much I find is (1/4)($50) + (1/2)($100) + (1/4)($200) = $112.50 as well. Again, in game (1) the resolution of the apparent paradox must be that for any possible probability distribution the millionaire uses to pick the amounts in the envelopes, your average expected return from sticking with your first choice if you don't open it to see how much is inside must always be equal to your average expected winnings if you decide to switch without first checking how much was inside your first choice. Now, which game would you prefer to play, (1) or (2)? They are not the same. With the conditions I mentioned--the millionaire flips a coin to decide whether to put $50 or $100 in one envelope, then puts double in the other, and in game (2) he only replaces the amount in the second envelope if you find $100 in the first envelope you choose--then the two games actually are exactly the same, in terms of probabilities and average expected returns. If you want to suggest some

### Re: Observation selection effects

Norman Samish writes: QUOTE- Assume an eccentric millionaire offers you your choice of either of two sealed envelopes, A or B, both containing money. One envelope contains twice as much as the other. After you choose an envelope you will have the option of trading it for the other envelope. Suppose you pick envelope A. You open it and see that it contains $100. Now you have to decide if you will keep the $100, or will you trade it for whatever is in envelope B? You might reason as follows: since one envelope has twice what the other one has, envelope B either has 200 dollars or 50 dollars, with equal probability. If you switch, you stand to either win $100 or to lose $50. Since you stand to win more than you stand to lose, you should switch. -ENDQUOTE The problem is that you are reasoning as if the amount in each envelope can vary during the game, whereas in fact it is fixed. Suppose envelope A contains $100 and envelope B contains $50. You open A, see the $100, and then reason that B may contain either $50 or $200, each being equally likely. In fact, B cannot contain $200, even though you don't know this yet. It is easy enough for an external observer (who does know the contents of each envelope) to calculate the probabilities: if you keep the first envelope, your expected gain is 0.5*$100 + 0.5*$50 = $75. If you switch, your expected gain is 0.5*$100 (if you open B first) + 0.5*$50 (if you open A first) = $75, as before. Ignorance of the actual amounts may lead you to speculate that one of the envelopes may contain $200, but it won't make the money magically materialise! And even if you don't know the actual amounts, the above analysis should convince you that nothing is to be gained by switching envelopes. If the game changes so that, once you have opened the first envelope, the millionaire decides by flipping a coin whether he will put half or double that amount in the second envelope, then you are actually better off switching. Stathis Papaioannou _ Discover how everyone everything in our world's connected: http://www.onebigvillage.com.au?obv1=hotmail

### RE: Observation selection effects

-Original Message- Norman Samish: The Flip-Flop game described by Stathis Papaioannou strikes me as a version of the old Two-Envelope Paradox. Assume an eccentric millionaire offers you your choice of either of two sealed envelopes, A or B, both containing money. One envelope contains twice as much as the other. After you choose an envelope you will have the option of trading it for the other envelope. Suppose you pick envelope A. You open it and see that it contains $100. Now you have to decide if you will keep the $100, or will you trade it for whatever is in envelope B? You might reason as follows: since one envelope has twice what the other one has, envelope B either has 200 dollars or 50 dollars, with equal probability. If you switch, you stand to either win $100 or to lose $50. Since you stand to win more than you stand to lose, you should switch. But just before you tell the eccentric millionaire that you would like to switch, another thought might occur to you. If you had picked envelope B, you would have come to exactly the same conclusion. So if the above argument is valid, you should switch no matter which envelope you choose. Therefore the argument for always switching is NOT valid - but I am unable, at the moment, to tell you why! Of course in the real world you have some idea about how much money is in play so if you see a very large amount you infer it's probably the larger amount. But even without this assumption of realism it's an interesting problem and taken as stated there's still no paradox. I saw this problem several years ago and here's my solution. It takes the problem as stated, but I do make one small additional restrictive assumption: Let: s = envelope with smaller amount is selected. l = envelope with larger amount is selected. m = the amount in the selected envelope. Since any valid resolution of the paradox would have to work for ratios of money other than two, also define: r = the ratio of the larger amount to the smaller. Now here comes the restrictive assumption, which can be thought of as a restrictive rule about how the amounts are chosen which I hope to generalize away later. Expressed as a rule, it is this: The person putting in the money selects, at random (not necessarily uniformly), the smaller amount from a range (x1, x2) such that x2 r*x1. In other words, the range of possible amounts is such that the larger and smaller amount do not overlap. Then, for any interval of the range (x,x+dx) for the smaller amount with probability p, there is a corresponding interval (r*x, r*x+r*dx) with probability p for the larger amount. Since the latter interval is longer by a factor of r P(l|m)/P(s|m) = r , In other words, no matter what m is, it is r-times more likely to fall in a large-amount interval than in a small-amount interval. But since l and s are the only possibilities (and here's where I need the non-overlap), P(1|m) + P(s|m) = 1 which implies, P(s|m) = 1/(1+r) and P(1|m) = r/(1+r) . Then the rest is straightforward algebra. The expected values are: E(don't switch) = m E(switch) = P(s|m)rm + P(l|m)m/r = [1/(1+r)]rm + [r/(1+r)]m/r = m and no paradox. Brent Meeker

### [Fwd: RE: Observation selection effects]

I always forget to reply-to-all in this list. So below goes my reply which went only to Hal Finney. -Forwarded Message- From: Eric Cavalcanti [EMAIL PROTECTED] To: Hal Finney [EMAIL PROTECTED] Subject: RE: Observation selection effects Date: Tue, 05 Oct 2004 12:57:14 +1000 On Tue, 2004-10-05 at 10:20, Hal Finney wrote: Stathis Papaioannou writes: In the new casino game Flip-Flop, an odd number of players pays $1 each to individually flip a coin, so that no player can see what another player is doing. The game organisers then tally up the results, and the result in the minority is called the Winning Flip, while the majority result is called the Losing Flip. Before the Winning Flip is announced, each player has the opportunity to either keep their initial result, or to Switch; this is then called the player's Final Flip. When the Winning Flip is announced, players whose Final Flip corresponds with this are paid $2 by the casino, while the rest are paid nothing. Think about if the odd number of players was exactly one. You're guaranteed to have the Winning Flip before you switch. Then think about what would happen if the odd number of players was three. Then you have a 3/4 chance of having the Winning Flip before you switch. Only if the other two players' flips both disagree with yours will you not have the Winnning Flip, and there is only a 1/4 chance of that happening. Exactly. It is interesting to note that, even though you are more likely to be in the Winning Flip, there is no disadvantage in Switching. To understand that, we can look at the N=3 case, and see that if I am in the Winning Flip with someone else, then if I change I will still be in the Winnig Flip with the other person. As opposed to Stathis initial thought, even though the Winning Flip is indeed as likely to be Heads as Tails, each individual is more likely to be in the Winning Flip as in the Losing Flip in any given run. So that this would never make it into a Casino game, because the house would lose money in the long run. Eric.

### RE: Observation selection effects

Brent Meeker wrote: Of course in the real world you have some idea about how much money is in play so if you see a very large amount you infer it's probably the larger amount. But even without this assumption of realism it's an interesting problem and taken as stated there's still no paradox. I saw this problem several years ago and here's my solution. It takes the problem as stated, but I do make one small additional restrictive assumption: Let: s = envelope with smaller amount is selected. l = envelope with larger amount is selected. m = the amount in the selected envelope. Since any valid resolution of the paradox would have to work for ratios of money other than two, also define: r = the ratio of the larger amount to the smaller. Now here comes the restrictive assumption, which can be thought of as a restrictive rule about how the amounts are chosen which I hope to generalize away later. Expressed as a rule, it is this: The person putting in the money selects, at random (not necessarily uniformly), the smaller amount from a range (x1, x2) such that x2 lt; r*x1. In other words, the range of possible amounts is such that the larger and smaller amount do not overlap. Then, for any interval of the range (x,x+dx) for the smaller amount with probability p, there is a corresponding interval (r*x, r*x+r*dx) with probability p for the larger amount. Since the latter interval is longer by a factor of r P(l|m)/P(s|m) = r , In other words, no matter what m is, it is r-times more likely to fall in a large-amount interval than in a small-amount interval. But since l and s are the only possibilities (and here's where I need the non-overlap), P(1|m) + P(s|m) = 1 which implies, P(s|m) = 1/(1+r) and P(1|m) = r/(1+r) . Then the rest is straightforward algebra. The expected values are: E(don't switch) = m E(switch) = P(s|m)rm + P(l|m)m/r = [1/(1+r)]rm + [r/(1+r)]m/r = m and no paradox. This is right, but it's a pretty special case--there are an infinite number of possible probability distributions the millionaire could use when deciding how much money to put in one envelope, even if we assume he always puts double in the other. For example, he could use a distribution that gives him a 1/2 probability of putting between 0 and 1 dollars in one envelope (assume the dollar amounts can take any positive real value, and he uses a flat probability distribution to pick a number between 0 and 1), a 1/4 probability of putting in between 1 and 2 dollars, a 1/8 probability of putting in between 2 and 3 dollars, and in general a 1/2^n probability of putting in between n-1 and n dollars. This would insure there was some finite probability that *any* positive real number could be found in either envelope. The basic paradox is that the argument tries to show that the average expected payoff from picking the second envelope is higher than the average expected payoff from sticking with the first one, *regardless of what amount you found in the first envelope*--in other words, even without opening the first envelope you'd be better off switching to the second, which doesn't make sense since the envelopes are identical and your first pick was random. But it's not actually possible that, regardless of what you found in the first envelope, there would always be a 50% chance the other envelope contained half that and a 50% chance it contained double that...for that to be true, the amount in the first envelope would have to be picked using a flat probability distribution which is equally likely to give any number from 0 to infinity, and as I said that's impossible. But my argument was not really sufficiently general either, because it doesn't rule out other possibilities like a 55% chance the other envelope contained half what was found in the first envelope and a 45% chance it contained double, in which case your average expected payoff would still be higher if you switched. A truly general argument would have to show that, for any logically possible probability distribution the millionaire uses to pick the amounts in the envelopes, the average expected payoff from switching will always be exactly equal to the average expected winnings from sticking with your first choice. There are two different ways this can be true: Possibility #1: it may be that you know enough about the probability distribution that opening the envelope and seeing how much is inside allows you to refine your evaluation of the average expected payoff from switching. I gave an example of this in my post, where the millionaire picks an amount from 1 to a million to put in one envelope and puts double that in the other; in that case, if you open your first pick and find an amount greater than a million, the average expected payoff from switching is 0. But even if the average expected payoff may vary depending on what you find in the

### Re: Observation selection effects

Dear Brent, I enjoyed your description which was quite a show for my totally qualitative (and IN-formally intuitive) logic - how you, quanti people substitute common sense for those letters and numbers. Norman's paradox is an orthodox paradox and you made it into a metadox (no paradox). Have a good day John Mikes PS: to excuse my lingo: my 1st Ph.D. was Chemistry-Physics-Math. J - Original Message - From: Brent Meeker [EMAIL PROTECTED] To: Everything-List [EMAIL PROTECTED] Sent: Monday, October 04, 2004 6:19 PM Subject: RE: Observation selection effects -Original Message- Norman Samish: The Flip-Flop game described by Stathis Papaioannou strikes me as a version of the old Two-Envelope Paradox. Assume an eccentric millionaire offers you your choice of either of two sealed envelopes, A or B, both containing money. One envelope contains twice as much as the other. After you choose an envelope you will have the option of trading it for the other envelope. Suppose you pick envelope A. You open it and see that it contains $100. Now you have to decide if you will keep the $100, or will you trade it for whatever is in envelope B? You might reason as follows: since one envelope has twice what the other one has, envelope B either has 200 dollars or 50 dollars, with equal probability. If you switch, you stand to either win $100 or to lose $50. Since you stand to win more than you stand to lose, you should switch. But just before you tell the eccentric millionaire that you would like to switch, another thought might occur to you. If you had picked envelope B, you would have come to exactly the same conclusion. So if the above argument is valid, you should switch no matter which envelope you choose. Therefore the argument for always switching is NOT valid - but I am unable, at the moment, to tell you why! Of course in the real world you have some idea about how much money is in play so if you see a very large amount you infer it's probably the larger amount. But even without this assumption of realism it's an interesting problem and taken as stated there's still no paradox. I saw this problem several years ago and here's my solution. It takes the problem as stated, but I do make one small additional restrictive assumption: Let: s = envelope with smaller amount is selected. l = envelope with larger amount is selected. m = the amount in the selected envelope. Since any valid resolution of the paradox would have to work for ratios of money other than two, also define: r = the ratio of the larger amount to the smaller. Now here comes the restrictive assumption, which can be thought of as a restrictive rule about how the amounts are chosen which I hope to generalize away later. Expressed as a rule, it is this: The person putting in the money selects, at random (not necessarily uniformly), the smaller amount from a range (x1, x2) such that x2 r*x1. In other words, the range of possible amounts is such that the larger and smaller amount do not overlap. Then, for any interval of the range (x,x+dx) for the smaller amount with probability p, there is a corresponding interval (r*x, r*x+r*dx) with probability p for the larger amount. Since the latter interval is longer by a factor of r P(l|m)/P(s|m) = r , In other words, no matter what m is, it is r-times more likely to fall in a large-amount interval than in a small-amount interval. But since l and s are the only possibilities (and here's where I need the non-overlap), P(1|m) + P(s|m) = 1 which implies, P(s|m) = 1/(1+r) and P(1|m) = r/(1+r) . Then the rest is straightforward algebra. The expected values are: E(don't switch) = m E(switch) = P(s|m)rm + P(l|m)m/r = [1/(1+r)]rm + [r/(1+r)]m/r = m and no paradox. Brent Meeker

### [Fwd: Re: Observation selection effects]

Original Message Subject: Re: Observation selection effects Date: Sat, 04 Sep 2004 02:29:54 -0400 From: Danny Mayes [EMAIL PROTECTED] To: [EMAIL PROTECTED] References: [EMAIL PROTECTED] These problems remind me of the infamous Monty Hall problem that got Marilyn vos Savant in some controversy. Someone wrote and asked the following question: You are on "lets make a deal", and are chosen to select a door among 3 doors, one of which has a car behind it. You randomly select door 1. Monty, knowing where the car is, opens door 2 revealing an empty room, and asks if you want to stay with door one. The question was: Is there any benefit in switching from door 1 to door 3. Common sense would suggest Monty simply eliminated one choice, and you have a 50-50 chance either way. Marylin argued that by switching, the contestant actually increases his odds from 1/3 to 2/3. The difference coming about through the added information of the car not behind door 2. This example is discussed in the book "Information: The New Language of Science" by Hans Christian von Baeyer, which I am trying to read, but only getting through bits and pieces as usual due to my work schedule. According to the book, vos Savant still gets mail arguing her position on this matter. It seems to me it would be very easy to resolve with a friend, letting one person play Monty and then keeping a tally of your success in switching vs. not switching (though I haven't tried this- my wife didn't find it intriguing enough, unfortunately). I think these games provide good examples of how our common sense often works against a deep understanding of what is really going on around here. I also think they point to a very fundamental level of importance of the role of information in understanding the way our world (or multiverse) works Jesse Mazer wrote: Norman Samish: The "Flip-Flop" game described by Stathis Papaioannou strikes me as a version of the old Two-Envelope Paradox. Assume an eccentric millionaire offers you your choice of either of two sealed envelopes, A or B, both containing money. One envelope contains twice as much as the other. After you choose an envelope you will have the option of trading it for the other envelope. Suppose you pick envelope A. You open it and see that it contains $100. Now you have to decide if you will keep the $100, or will you trade it for whatever is in envelope B? You might reason as follows: since one envelope has twice what the other one has, envelope B either has 200 dollars or 50 dollars, with equal probability. If you switch, you stand to either win $100 or to lose $50. Since you stand to win more than you stand to lose, you should switch. But just before you tell the eccentric millionaire that you would like to switch, another thought might occur to you. If you had picked envelope B, you would have come to exactly the same conclusion. So if the above argument is valid, you should switch no matter which envelope you choose. Therefore the argument for always switching is NOT valid - but I am unable, at the moment, to tell you why! Basically, I think the resolution of this paradox is that it's impossible to pick a number randomly from 0 to infinity in such a way that every number is equally likely to come up. Such an infinite flat probability distribution would lead to paradoxical conclusions--for example, if you picked two positive integers randomly from a flat probability distribution, and then looked at the first integer, then there would be a 100% chance the second integer would be larger, since there are only a finite number of integers smaller than or equal to the first one and an infinite number that are larger. For any logically possible probability distribution the millionaire uses, it will be true that depending on what amount of money you find in the first envelope, there won't always be an equal chance of finding double the amount or half the amount in the other envelope. For example, if the millionaire simply picks a random amount from 0 to one million to put in the first envelope, and then flips a coin to decide whether to put half or double that in the other envelope, then if the first envelope contains more than one million there is a 100% chance the other envelope contains less than that. For a more detailed discussion of the two-envelope paradox, see this page: http://jamaica.u.arizona.edu/~chalmers/papers/envelope.html I don't think the solution to this paradox has any relation to the solution to the flip-flop game, though. In the case of the flip-flop game, it may help to assume that the players are all robots, and that each player can assume that whatever decision it m

### RE: Observation selection effects

Brent Meeker wrote: -Original Message- From: Jesse Mazer [mailto:[EMAIL PROTECTED] Sent: Tuesday, October 05, 2004 6:33 PM To: [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Subject: RE: Observation selection effects Brent Meeker wrote: On reviewing my analysis (I hadn't looked at for about four years), I think it works without the restrictive assumption that the range of distirbutions not overlap. It's still necessary that P(l|m)+P(s|m)=1 and P(l|m)/P(s|m)=r, which is all that is required for the proof to go thru. Hmm, I think I misread your analysis, I had somehow gotten the idea that you were assuming a uniform probability of the envelope-stuffer picking any number between x1 and x2 for the first envelope, but I see this isn't necessary, so your proof is a lot more general than I thought. Still not completely general though, because the envelope-stuffer can also use a distribution which has no upper bound x2 on possible amounts to put in the first envelope, like the one I mentioned in my last post: For example, he could use a distribution that gives him a 1/2 probability of putting between 0 and 1 dollars in one envelope (assume the dollar amounts can take any positive real value, and he uses a flat probability distribution to pick a number between 0 and 1), a 1/4 probability of putting in between 1 and 2 dollars, a 1/8 probability of putting in between 2 and 3 dollars, and in general a 1/2^n probability of putting in between n-1 and n dollars. This would insure there was some finite probability that *any* positive real number could be found in either envelope. Likewise, he could also use the continuous probability distribution P(x) = 1/e^x (whose integral from 0 to infinity is 1). And if you want to restrict the amounts in the envelope to positive integers, he could use a distribution which gives a 1/2^n probability of putting exactly n dollars in the first envelope. Jesse That doesn't matter if I can do without the no-overlap assumption - which I think I can. Do you see a flaw? When I first did it I was drawing pictures of distributions and I thought I needed non-overlap to assert that P(l|m)/P(s|m)=r and P(l|m)+P(s|m)=1. But now it seems that the last follows just from the fact that the amount was either from the larger or the smaller. The ratio doesn't depend on the ranges not overlapping, it just depends on the fact that the larger amount's distribution must be a copy of the smaller amount's distribution stretched by a factor of r. But in order for the range of the larger amount to be double that of the smaller amount, you need to assume the range of the smaller amount is finite. If the range of the smaller amount is infinite, as in my P(x)=1/e^x example, then it would no longer make sense to say that the range of the larger amount is r times larger. Also, what if the envelope-stuffer is only picking from a finite set of numbers rather than a continuous range? For example, he might have a 1/3 chance of putting 100, 125 or 150 dollars in the first envelope, and then he would double that amount for the second envelope. In this case, your assumption no matter what m is, it is r-times more likely to fall in a large-amount interval than in a small-amount interval wouldn't seem to be valid, since there are only six possible values of m here (100, 125, 150, 200, 250, or 300) and three of them are in the smaller range. Jesse

### RE: Observation selection effects

-Original Message- From: Jesse Mazer [mailto:[EMAIL PROTECTED] Sent: Tuesday, October 05, 2004 8:45 PM To: [EMAIL PROTECTED]; [EMAIL PROTECTED] Subject: RE: Observation selection effects Brent Meeker wrote: -Original Message- From: Jesse Mazer [mailto:[EMAIL PROTECTED] Sent: Tuesday, October 05, 2004 6:33 PM To: [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Subject: RE: Observation selection effects Brent Meeker wrote: On reviewing my analysis (I hadn't looked at for about four years), I think it works without the restrictive assumption that the range of distirbutions not overlap. It's still necessary that P(l|m)+P(s|m)=1 and P(l|m)/P(s|m)=r, which is all that is required for the proof to go thru. Hmm, I think I misread your analysis, I had somehow gotten the idea that you were assuming a uniform probability of the envelope-stuffer picking any number between x1 and x2 for the first envelope, but I see this isn't necessary, so your proof is a lot more general than I thought. Still not completely general though, because the envelope-stuffer can also use a distribution which has no upper bound x2 on possible amounts to put in the first envelope, like the one I mentioned in my last post: For example, he could use a distribution that gives him a 1/2 probability of putting between 0 and 1 dollars in one envelope (assume the dollar amounts can take any positive real value, and he uses a flat probability distribution to pick a number between 0 and 1), a 1/4 probability of putting in between 1 and 2 dollars, a 1/8 probability of putting in between 2 and 3 dollars, and in general a 1/2^n probability of putting in between n-1 and n dollars. This would insure there was some finite probability that *any* positive real number could be found in either envelope. Likewise, he could also use the continuous probability distribution P(x) = 1/e^x (whose integral from 0 to infinity is 1). And if you want to restrict the amounts in the envelope to positive integers, he could use a distribution which gives a 1/2^n probability of putting exactly n dollars in the first envelope. Jesse That doesn't matter if I can do without the no-overlap assumption - which I think I can. Do you see a flaw? When I first did it I was drawing pictures of distributions and I thought I needed non-overlap to assert that P(l|m)/P(s|m)=r and P(l|m)+P(s|m)=1. But now it seems that the last follows just from the fact that the amount was either from the larger or the smaller. The ratio doesn't depend on the ranges not overlapping, it just depends on the fact that the larger amount's distribution must be a copy of the smaller amount's distribution stretched by a factor of r. But in order for the range of the larger amount to be double that of the smaller amount, you need to assume the range of the smaller amount is finite. I only had to assume it was finite to avoid overlap in the ranges. Dropping that assumption the range of the lower amount can be infinite. If the range of the smaller amount is infinite, as in my P(x)=1/e^x example, then it would no longer make sense to say that the range of the larger amount is r times larger. Sure it does; r*inf=inf. P(s)=exp(-x) - P(l)=exp(-x/r) Also, what if the envelope-stuffer is only picking from a finite set of numbers rather than a continuous range? For example, he might have a 1/3 chance of putting 100, 125 or 150 dollars in the first envelope, and then he would double that amount for the second envelope. In this case, your assumption no matter what m is, it is r-times more likely to fall in a large-amount interval than in a small-amount interval wouldn't seem to be valid, since there are only six possible values of m here (100, 125, 150, 200, 250, or 300) and three of them are in the smaller range. Yes, that's true. The proof depends on smooth, integrable distributions. Brent

### RE: Observation selection effects

-Original Message- From: Jesse Mazer [mailto:[EMAIL PROTECTED] Sent: Tuesday, October 05, 2004 8:45 PM To: [EMAIL PROTECTED]; [EMAIL PROTECTED] Subject: RE: Observation selection effects If the range of the smaller amount is infinite, as in my P(x)=1/e^x example, then it would no longer make sense to say that the range of the larger amount is r times larger. Sure it does; r*inf=inf. P(s)=exp(-x) - P(l)=exp(-x/r) But it would make just as much sense to say that the second range is 3r times wider, since by the same logic 3r*inf=inf. In other words, this step in your proof doesn't make sense: In other words, the range of possible amounts is such that the larger and smaller amount do not overlap. Then, for any interval of the range (x,x+dx) for the smaller amount with probability p, there is a corresponding interval (r*x, r*x+r*dx) with probability p for the larger amount. Since the latter interval is longer by a factor of r P(l|m)/P(s|m) = r , In other words, no matter what m is, it is r-times more likely to fall in a large-amount interval than in a small-amount interval. As for your statement that P(s)=exp(-x) - P(l)=exp(-x/r), that can't be true. It doesn't make sense that the value of the second probability distribution at x would be exp(-x/r), since the range of possible values for the amount in that envelope is 0 to infinity, but the integral of exp(-x/r) from 0 to infinity is not equal to 1, so that's not a valid probability distribution. Also, now that I think more about it I'm not even sure the step in your proof I quoted above actually makes sense even in the case of a probability distribution with finite range. What exactly does the equation P(l|m)/P(s|m) = r mean, anyway? It can't mean that if I choose an envelope at random, before I even open it I can say that the amount m inside is r times more likely to have been picked from the larger distribution, since I know there is a 50% chance I will pick the envelope whose amount was picked from the larger distribution. Is it supposed to mean that if we let the number of trials go to infinity and then look at the subset of trials where the envelope I opened contained m dollars, it is r times more likely that the envelope was picked from the larger distribution on any given trial? This can't be true for every specific m--for example, if the smaller distribution had a range of 0 to 100 and the larger had a range of 0 to 200, if I set m=150, then in every single trial where I found 150 dollars in the envelope it must have been selected from the larger distribution. You could do a weighted average over all possible values of m, like integral over all possible values of m of P('I found m dollars in the envelope I selected')*P('the envelope I selected had an amount taken from the smaller distribution' | 'I found m dollars in the envelope I selected'), which you could write as integral over m of P(m)*P(s|m), but I don't think it would be true that the ratio integral over m of P(m)*P(l|m)/integral over m of P(m)*P(s|m) would be equal to r, in fact I think both integrals would always come out to 1/2 so the ratio would always be 1...and even if I'm wrong, replacing P(l|m)/P(s|m) with this ratio of integrals would mess up the rest of your proof. Jesse

### RE: [Fwd: RE: Observation selection effects]

On Tue, 2004-10-05 at 19:31, Brent Meeker wrote: I always forget to reply-to-all in this list. So below goes my reply which went only to Hal Finney. -Forwarded Message- From: Eric Cavalcanti [EMAIL PROTECTED] Think about if the odd number of players was exactly one. You're guaranteed to have the Winning Flip before you switch. No, you're guranteed NOT to be in the winning flip. Then think about what would happen if the odd number of players was three. Then you have a 3/4 chance of having the Winning Flip before you switch. Only if the other two players' flips both disagree with yours will you not have the Winnning Flip, and there is only a 1/4 chance of that happening. Exactly. It is interesting to note that, even though you are more likely to be in the Winning Flip, there is no disadvantage in Switching. To understand that, we can look at the N=3 case, and see that if I am in the Winning Flip with someone else, then if I change I will still be in the Winnig Flip with the other person. As opposed to Stathis initial thought, even though the Winning Flip is indeed as likely to be Heads as Tails, each individual is more likely to be in the Winning Flip as in the Losing Flip in any given run. So that this would never make it into a Casino game, because the house would lose money in the long run. I think you've confused the definitions of winning flip and losing flip. The winning flip is the *minority at the time of the flip* For N=3 you can't be in the winning flip with someone else at the time of the flip - but you can switch to it. Yes, you're right. Hal and I have confused the definitions. It is still not a paradox, though. You are more likely to be in the Losing Flip. So that this could indeed be a Casino game. Eric.

### Re: Observation selection effects

Norman Samish: The Flip-Flop game described by Stathis Papaioannou strikes me as a version of the old Two-Envelope Paradox. Assume an eccentric millionaire offers you your choice of either of two sealed envelopes, A or B, both containing money. One envelope contains twice as much as the other. After you choose an envelope you will have the option of trading it for the other envelope. Suppose you pick envelope A. You open it and see that it contains $100. Now you have to decide if you will keep the $100, or will you trade it for whatever is in envelope B? You might reason as follows: since one envelope has twice what the other one has, envelope B either has 200 dollars or 50 dollars, with equal probability. If you switch, you stand to either win $100 or to lose $50. Since you stand to win more than you stand to lose, you should switch. But just before you tell the eccentric millionaire that you would like to switch, another thought might occur to you. If you had picked envelope B, you would have come to exactly the same conclusion. So if the above argument is valid, you should switch no matter which envelope you choose. Therefore the argument for always switching is NOT valid - but I am unable, at the moment, to tell you why! Basically, I think the resolution of this paradox is that it's impossible to pick a number randomly from 0 to infinity in such a way that every number is equally likely to come up. Such an infinite flat probability distribution would lead to paradoxical conclusions--for example, if you picked two positive integers randomly from a flat probability distribution, and then looked at the first integer, then there would be a 100% chance the second integer would be larger, since there are only a finite number of integers smaller than or equal to the first one and an infinite number that are larger. For any logically possible probability distribution the millionaire uses, it will be true that depending on what amount of money you find in the first envelope, there won't always be an equal chance of finding double the amount or half the amount in the other envelope. For example, if the millionaire simply picks a random amount from 0 to one million to put in the first envelope, and then flips a coin to decide whether to put half or double that in the other envelope, then if the first envelope contains more than one million there is a 100% chance the other envelope contains less than that. For a more detailed discussion of the two-envelope paradox, see this page: http://jamaica.u.arizona.edu/~chalmers/papers/envelope.html I don't think the solution to this paradox has any relation to the solution to the flip-flop game, though. In the case of the flip-flop game, it may help to assume that the players are all robots, and that each player can assume that whatever decision it makes about whether to switch or not, there is a 100% chance that all the other players will follow the same line of reasoning and come to an identical decision. In this case, since the money is awarded to the minority flip, it's clear that it's better to switch, since if everyone switches more of them will win. This problem actually reminds me more of Newcomb's paradox, described at http://slate.msn.com/?id=2061419 , because it depends on whether you assume your choice is absolutely independent of choices made by other minds or if you should act as though the choice you make can cause another mind to make a certain choice even if there is no actual interaction between you. Jesse

### RE: Observation selection effects

Hal Finney writes: Stathis Papaioannou writes: Here is another example which makes this point. You arrive before two adjacent closed doors, A and B. You know that behind one door is a room containing 1000 people, while behind the other door is a room containing only 10 people, but you don't know which door is which. You toss a coin to decide which door you will open (heads=A, tails=B), and then enter into the corresponding room. The room is dark, so you don't know which room you are now in until you turn on the light. At the point just before the light goes on, do you have any reason to think you are more likely to be in one room rather than the other? By analogy with the Bostrom traffic lane example you could argue that, in the absence of any empirical data, you are much more likely to now be a member of the large population than the small population. However, this cannot be right, because you tossed a coin, and you are thus equally likely to find yourself in either room when the light goes on. Again the problem is that you are not a typical member of the room unless the mechanism you used to choose a room was the same as what everyone else did. And your description is not consistent with that. This illustrates another problem with the lane-changing example, which is that the described mechanism for choosing lanes (choose at random) is not typical. Most people don't flip a coin to choose the lane they will drive in. Yes, this is correct. The typical observer must be typical in the way he makes the choice of room or lane. With the traffic example, given that there are slower and faster lanes on most roads, even in the absence of road works or accidents, this may mean that for whatever reason the typical driver on that day is more likely to choose the slower lane on entering the road. If this is so, then a winning strategy for getting to your destination faster could be to pick the lane with the most immediate appeal, then reflect on this (having participated in the present discussion) and choose a _different_ lane. This is analogous to counter-cyclical investing in the stock market, where you deliberately try to do the opposite of what the typical investor does. But there may be a problem with the above argument. Suppose everyone really did flip a perfectly fair coin to decide which lane of traffic to enter. It is then still very most that one lane would be more crowded than the other at any given time, purely through chance. Now, every driver might reason, everyone including me has flipped a coin to decide which lane to enter, so there is nothing to be gained by changing lanes. However, most of the drivers reasoning thus would, by chance, be in the more crowded lane, and therefore most would in fact be better off changing lanes. --Stathis Papaioannou _ Protect yourself from junk e-mail: http://microsoft.ninemsn.com.au/protectfromspam.aspx

### RE: Observation selection effects

Eric Cavalcanti writes: QUOTE- And this is the case where this problem is most paradoxical. We are very likely to have one of the lanes more crowded than the other; most of the drivers reasoning would thus, by chance, be in the more crowded lane, such that they would benefit from changing lanes; even though, NO PARTICULAR DRIVER would benefit from changing lanes, on average. No particular driver has basis for infering in which lane he is. In this case you cannot reason as a random sample from the population. -ENDQUOTE I find this paradox a little disturbing, on further reflection. You enter the traffic by tossing a coin, so you are no more likely to end up in one lane than the other, and you would not, on average, benefit from changing lanes. Given that you are in every respect a typical driver, what applies to you should apply to everyone else as well. This SHOULD be equivalent to saying that if every driver decided to change lanes, on average no particular driver would benefit - as Eric states. However, this is not so: the majority of drivers WOULD benefit from changing. (The fact that nobody would benefit if everyone changed does not resolve the paradox. We can restrict the problem to the case where each driver individually changes, and the paradox remains.) It seems that this problem is an assault on the foundations of probability and statistics, and I would really like to see it resolved. Stathis Papaioannou _ FREE* Month of Movies with FOXTEL Digital: http://adsfac.net/link.asp?cc=FXT002.7542.0

### RE: Observation selection effects

On Mon, 2004-10-04 at 10:42, Stathis Papaioannou wrote: Eric Cavalcanti writes: QUOTE- And this is the case where this problem is most paradoxical. We are very likely to have one of the lanes more crowded than the other; most of the drivers reasoning would thus, by chance, be in the more crowded lane, such that they would benefit from changing lanes; even though, NO PARTICULAR DRIVER would benefit from changing lanes, on average. No particular driver has basis for infering in which lane he is. In this case you cannot reason as a random sample from the population. -ENDQUOTE I find this paradox a little disturbing, on further reflection. You enter the traffic by tossing a coin, so you are no more likely to end up in one lane than the other, and you would not, on average, benefit from changing lanes. Given that you are in every respect a typical driver, what applies to you should apply to everyone else as well. This SHOULD be equivalent to saying that if every driver decided to change lanes, on average no particular driver would benefit - as Eric states. However, this is not so: the majority of drivers WOULD benefit from changing. (The fact that nobody would benefit if everyone changed does not resolve the paradox. We can restrict the problem to the case where each driver individually changes, and the paradox remains.) It seems that this problem is an assault on the foundations of probability and statistics, and I would really like to see it resolved. I found the answer of why you should be more likely to enter in the crowded lane in this case. The answer came after I tried to think about an example for few people (which turned out not to work as I thought it would) Suppose a coin is toss for N people, which enter one of two rooms according to the result. Suppose first N=3. Then it is more likely that I will be in the crowded room, even though there was no particular bias in each coin toss. But still, if I am given the option to change, and if I am in the crowded room, I'll probably still be in the crowded room after I change! Now as N grows large, it is still more likely that I will be in the crowded room, only it is less so. I was neglecting the effect that you make yourself when you enter the room/lane. When N is large and even, it is equally likely that the lane I enter is slower/faster. But it may be that both lanes have same numbers, so my entering will make that lane be the slower, and that's where the effect comes from. If it is odd, and I enter the fast lane, it is possible that they become equal. If I enter the slower lane, it will become even slower. A minute of thought shows that my changing lanes does not affect the result, though, as much as changing rooms does not make me more likely to be in the less crowded when N=3. Therefore it is not a good advice for people to change lanes in this case, even though it is more likely that they are in the slower lane! Eric.

### RE: Observation selection effects

Eric Cavalcanti writes: From another perspective, I have just arrived at the road and there was no particular reason for me to initially choose lane A or lane B, so that I could just as well have started on the faster lane, and changing would be undesirable. From this perspective, there is no gain in changing lanes, on average. Here is another example which makes this point. You arrive before two adjacent closed doors, A and B. You know that behind one door is a room containing 1000 people, while behind the other door is a room containing only 10 people, but you don't know which door is which. You toss a coin to decide which door you will open (heads=A, tails=B), and then enter into the corresponding room. The room is dark, so you don't know which room you are now in until you turn on the light. At the point just before the light goes on, do you have any reason to think you are more likely to be in one room rather than the other? By analogy with the Bostrom traffic lane example you could argue that, in the absence of any empirical data, you are much more likely to now be a member of the large population than the small population. However, this cannot be right, because you tossed a coin, and you are thus equally likely to find yourself in either room when the light goes on. --Stathis Papaioannou _ FREE pop-up blocking with the new MSN Toolbar get it now! http://toolbar.msn.click-url.com/go/onm00200415ave/direct/01/

### RE: Observation selection effects

Stathis Papaioannou writes: Here is another example which makes this point. You arrive before two adjacent closed doors, A and B. You know that behind one door is a room containing 1000 people, while behind the other door is a room containing only 10 people, but you don't know which door is which. You toss a coin to decide which door you will open (heads=A, tails=B), and then enter into the corresponding room. The room is dark, so you don't know which room you are now in until you turn on the light. At the point just before the light goes on, do you have any reason to think you are more likely to be in one room rather than the other? By analogy with the Bostrom traffic lane example you could argue that, in the absence of any empirical data, you are much more likely to now be a member of the large population than the small population. However, this cannot be right, because you tossed a coin, and you are thus equally likely to find yourself in either room when the light goes on. Again the problem is that you are not a typical member of the room unless the mechanism you used to choose a room was the same as what everyone else did. And your description is not consistent with that. Suppose we modify it so that you are handed a biased coin, a coin which will come up heads or tails with 99% vs 1% probability. You know about the bias but you don't know which way the bias is. You flip the coin and walk into the room. Now, I think you will agree that you have a good reason to expect that when you turn on the light, you will be in the more crowded room. You are now a typical member of the room so the same considerations that make one room more crowded make it more likely that you are in that room. This illustrates another problem with the lane-changing example, which is that the described mechanism for choosing lanes (choose at random) is not typical. Most people don't flip a coin to choose the lane they will drive in. Instead, they have an expectation of which lane they will start in based on their long experience of driving in various conditions. It's pretty hard to think of yourself as a typical driver given the wide range of personality, age and experience among drivers on the road. Hal Finney

### Re: Observation selection effects

Eric Cavalcanti writes regarding http://plus.maths.org/issue17/features/traffic/index.html: I agree with the general conclusion: when we randomly select a driver and ask her whether she thinks the next lane is faster, more often than not we will have selected a driver from the lane which is in fact slower and more densely packed. ... From another perspective, I have just arrived at the road and there was no particular reason for me to initially choose lane A or lane B, so that I could just as well have started on the faster lane, and changing would be undesirable. From this perspective, there is no gain in changing lanes, on average. That's a good question. One thing I would note is that if everyone entering the road chose between the two lanes with equal probability, and stayed in their lane, then neither lane would be more crowded than the other. So to some extent your premises are contradictory. If everyone behaved like this, one lane wouldn't be faster than the other. Extending the argument, suppose I drive for a couple of miles, and get to another point where I want to decide if I should change lanes. Since I had no reason to change lanes a couple of miles ago, I still have no reason to do so now. Unless, of course, I can clearly see that the next lane is faster, but adding that assumption changes the problem completely. I think this is true as well, assuming you have not changed lanes yet. Let's go on and suppose that you drive for a while and change lanes occasionally based on how the traffic seems to be moving at that moment, and that you are a typical driver in this regard. Then your alarm clock rings and you ask yourself, am I more likely to be in the more crowded lane. I think you will agree that in that case, the answer is yes. Does this resolve the paradox? Hal Finney