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
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
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! Norman Samish - Original Message - From: "Stathis Papaioannou" <[EMAIL PROTECTED]> To: <[EMAIL PROTECTED]> Cc: <[EMAIL PROTECTED]> Sent: Monday, October 04, 2004 5:43 PM Subject: RE: Observation selection effects Here is another version of the paradox, where the way an individual chooses does not change the initial probabilities: 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. The question: if you participate in this game, is there any advantage in Switching? On the one hand, it seems clear that the Winning Flip is as likely to be heads as tails, so if you played this game repeatedly, in the long run you should break even, whether you Switch or not. On the other hand, it seems equally clear that if all the players Switch, the casino will end up every time paying out more than it collects, so Switching should be a winning strategy, on average, for each individual player. I'm sure there is something wrong with the above conclusion. What is it? And I haven't really thought this through yet, but does this have any bearing on the self sampling assumption as applied in the Doomsday Argument etc.? Stathis Papaioannou
RE: Observation selection effects
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. Hal Finney
RE: Observation selection effects
Here is another version of the paradox, where the way an individual chooses does not change the initial probabilities: 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. The question: if you participate in this game, is there any advantage in Switching? On the one hand, it seems clear that the Winning Flip is as likely to be heads as tails, so if you played this game repeatedly, in the long run you should break even, whether you Switch or not. On the other hand, it seems equally clear that if all the players Switch, the casino will end up every time paying out more than it collects, so Switching should be a winning strategy, on average, for each individual player. I'm sure there is something wrong with the above conclusion. What is it? And I haven't really thought this through yet, but does this have any bearing on the self sampling assumption as applied in the Doomsday Argument etc.? Stathis Papaioannou _ Searching for that dream home? Try http://ninemsn.realestate.com.au for all your property needs.
Re: Use of Three-State Electronic Level to Express Belief
At 11:59 29/09/04 -0700, George Levy wrote: Bruno Marchal wrote: Hi George, [out-of-line message] perhaps you could try to motivate your "qBp == If q then p". I don't see the relation with "if q is 1 then p is known, and and if q is 0 then p is unknown". How do you manage the "known" notion.Imagine a three port device such as an electrically controlled switch. Let's say that this device has three lines connected to it: an input connected to p, a control connected to q and an output that we'll call qBp. If the control sets the switch to OFF (ie. q=0) , the output is not connected to the input. Therefore for anyone observing the output, the value of p is unknown, i.e., qBp = x. The electronic value of x can be any arbitrary value except 0 and 1 which are reserved for the possible known binary values. If the control sets the switch to ON (ie. q=1), the output is connected to the input. Therefore for anyone observing the output, the value of p is known. It is either 0 or 1 depending on what the input p is. Giving any logic L1, it is always interesting to look if there is no other (perhaps better known) logic L2 such that you can interpret L1 in L2. Now it can be shown that most modal logic cannot be easily or directly represented by a multi-valued logic, so I doubt your proposal could work. You can always try, but let us be sure we agree on the "intuitive" meaning of Bp, in the case of the Smullyan's "self- rererential" interpretation of the "B". So we have a machine M. The machine M print propositions from time to time. Bp means that the machine M print p. The machine could print Bp. In that case the machine prints the proposition that she prints p. A machine is self-referentially correct (SRC) if her use of B is correct. Examples: The following machine (programs) is SRC: Begin print "hello" print "B hello" End The following machine (programs) is not SRC: Begin print "B hello" End because the machine pretend that she print hello, but will never do it. OK? Of course, we will add conditions; mainly that the machine' set of proposition will be closed for modus ponens, i.e. that if she print (one day, soon or later) X, and if she prints X->Y, then she will print Y. Etc. To sum up: Bp means "M asserts p". Bp is true (resp. false) if and only if M asserts p (resp. does not assert p). Bruno http://iridia.ulb.ac.be/~marchal/
