A Weblog of Centrist Voices in American Politics
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A Weblog of Centrist Voices in American Politics | |
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March 28, 2006Comments
It's been TOO QUIET around here lately. Somebody start talking. I did my part by starting a topic on TEEN SEX, so come on, people, step up to the plate! Posted by: PatHMV at March 28, 2006 12:19 PMYou stole my best one for the day already, Pat, by posting while I was starting this thread! Posted by: Tully at March 28, 2006 12:27 PMSince the sentiment seems to be that things are very slow, I'll offer an apolitical math question. You're playing the classic "shell game." Three shells, one of which has a ball under it, are shuffled around. After the shuffle, you have no idea which shell the ball is under. You randomly pick one of the shells. After you pick, the person running the game lifts one of the two shells you did not pick, revealing that there was no ball under it. He then offers you the option of switching your pick to the other shell, which you did not pick and was not lifted. Is it better to stay put, to switch, or does it even matter either way probabilistically? Posted by: WHQ at March 28, 2006 01:19 PMActually turned on the cable news (an infrequent occurence) to watch the water-hosing in France, and heard an ambassador on Fox mention that the protestors were "chanting the French word for malaise." Um, isn't malaise the French word for malaise? :-) Posted by: Tully at March 28, 2006 01:40 PMIs it better to stay put, to switch, or does it even matter either way probabilistically? Makes no difference. If the game is straight, then the odds of the ball being under either remaining shell are still 50/50, and making a 50/50 decision on re-choosing will not change that. And of course, if the game is rigged, your odds are zero either way. Posted by: Tully at March 28, 2006 01:55 PMWhat Tully said, on both counts. With the shells, you're 50-50 at that point. According to the Oxford English Dictionary: [French malaise discomfort (c1135 in Old French), uncomfortable sense of imperfect health (1587 in Middle French), also fig. (of a society, social grouping, etc., from early 19th cent.) The present word is a reborrowing of the French etymon: cf. earlier MALEASE n.]It's been in English usage meaning vague, non-specific physical discomfort since about 1768, and meaning general uneasiness of mind or spirit or the unhealthy state of an institution, organization activity, or situtation since about 1877. WHQ, did any of us ever manage to answer the last mathematics quiz you posted, way back when? Posted by: PatHMV at March 28, 2006 02:21 PMMakes no difference. If the game is straight, then the odds of the ball being under either remaining shell are still 50/50, and making a 50/50 decision on re-choosing will not change that. Not exactly. Consider this - the shell shown to have nothing under it was chosen under restriction. It could not be the shell you originally chose, and it could not be the shell with the ball under it. What was the probability that your original choice was correct? Pat, if it was the one I'm thinking of (i.e. the bird flying back and forth between the head-on approaching trains), pretty much yes. I don't think anyone was assertive in their explanation, but I think bk and maybe Tully had considered the correct answer in their responses. Posted by: WHQ at March 28, 2006 02:39 PMLong time readers know my aversion to the use of profanity in political discourse. I am happy to report that I am not alone in my feelings, by a long shot. A recent AP-Ipsos poll found that 87% of Americans were bothered at least "a little" by the use of profanity: 3. How much does it bother you when people use profanity or swear words?Also, 67% of people believe that the use of swear words is more common compared to 20 years ago. So, posters, ask yourself, as you decide whether to use profanity in your post, "do I really want to bother 87% of my audience by using this swear word? Do I really want to bother 36% of my audience 'a lot'? Is using the word THAT important to getting my point across?" (Hint: the answer is almost always no.) Posted by: PatHMV at March 28, 2006 02:41 PMBut WHQ, that fact gives no information about which of the other 2 shells the ball is under. At that moment in time, there are 2 shells, one with a ball and one without, with no information to increase the odds of the ball being under one or the other of them. The odds of having picked the right shell, as between those 2 which are left, is 50-50. Posted by: PatHMV at March 28, 2006 02:48 PMI included this little puzzle in my masters project, in the cointext of "Let's Make a Deal," a game show from the 70s. Tully and Pat are wrong. If you do trials on it and test it, you'll see why. Now, here's where I quote my master's project: When this question is posed, most people answer to the effect that, “It doesn't matter if you switch. There are two doors, and the prize is either behind one, or behind the other, so the probability is 1/2 whether you switch or not. It's that simple.” If you don't believe this, try it. An easy way to do it is by using 3 sugar packets, one with an X on the bottom. Posted by: bk at March 28, 2006 03:06 PM"Let's Make a Deal" is arguably the best game show of all time. Posted by: WHQ at March 28, 2006 03:20 PMThat's utter crap, Brian. Let's say I'm the contestant on Let's Make a Deal. I pick door #3. Monty opens door #1, and it's empty. One of the other two doors (2 and 3) has the duck, and the other has the new car. Ok. Now, I suddenly collapse of a heart attack and am rushed to the hospital. They call my brother on the phone to finish the game. He hasn't seen my pick and knows nothing about the game to that point. All he is told and asked is to pick either door #2 or door #3. What are his odds of picking the right door? 50-50. His odds and my odds cannot be different simply because I know what my first choice was. What changes the odds from 1/3 to 50-50 is that we suddenly know more information about the 3 shells, which of 2 of them are empty. Now, if you do a long random sample, then over time you will see the proper statistical result SO LONG as you include in your sample the times you randomly removed a shell and revealed the pea instead of empty. But if you know ahead of time that Monty will always remove an empty shell, then the odds were never 1/3 to begin with, because the setup would always reduce to a 50-50 option anyway. Posted by: PatHMV at March 28, 2006 03:41 PMFurther: There is a significant difference in the shell hypothetical (which I read to say the dealer would always remove an empty shell) and the Let's Make a Deal scenario, where each door has something, good, bad, or indifferent. You say: Now what if you don't switch? If you don't switch, then everything that happens after the first choice is moot. You picked one of the three doors, and your chances of that choice being correct are only one in three.The odds are not something inherent in you, personally. The odds are based on the information available about the options. Before the first door was revealed, the odds of your choice were indeed 1 in 3. Your odds go up once you find out that door #1 is not it BECAUSE you are no longer dealing with the same set of choices. Door #1 COULD have been the car or the donkey when you made your choice. That's why the odds were 1 in 3 at the time. Now that there are only 2 options left, the odds are 50-50. You haven't "increased the odds" of your first pick." You are making a SECOND pick, which is necessarily independent of the first. What exactly was your master's in? Posted by: PatHMV at March 28, 2006 03:48 PMHere's a little history of this problem. I remember when this little conundrum was first addressed by the fraud, "Marilyn Vos Savant", supposedly the "smartest woman in the world." There are a number of postings and articles about it, which include a discussion of the actual "Let's Make a Deal" experiment which Monty conducted. The linked article describes the history of the problem, and traces it back to Martin Gardner's Three Prisoner Problem article in Scientific American in 1959. Posted by: PatHMV at March 28, 2006 04:16 PMbk's right. You had a 1/3 chance of being right and a 2/3 chance of being wrong in the first choosing. If you were right (1/3), the unrevealed, unchosen shell will be empty. If you were wrong (2/3), the unrevealed unchosen shell will have a ball under it. It's that simple. If you were to play a shell game that only had two shells to begin with and were shown nothing, switching your pick would be neutral. That you started with three shells, and restricted which two of the three the revealer had to choose from by yourself choosing, it's very different. Posted by: WHQ at March 28, 2006 04:23 PMPat, If it's any consolation, I agreed precisely with your view until I did a simulation. After you do an honest simulation, and find out that you are wrong, I'll expect an apology for impugning my master's degree and field expertise. As with all things probabilistic, the proof is in the experimentation. My account is objectively verifiable. I've done it, and others have done it. I'm right, and you are wrong. See, my master's is in critical thinking, so I tested first, and then I explained the verifiable results. Now you're a lawyer, so you are used to arguing your way out of things and not having to admit you are wrong. But in this case, you most assuredly are wrong, except for this part, which is true but not relevant to the question which is being posed: I suddenly collapse of a heart attack and am rushed to the hospital. They call my brother on the phone to finish the game. He hasn't seen my pick and knows nothing about the game to that point. All he is told and asked is to pick either door #2 or door #3. What are his odds of picking the right door? 50-50. His odds and my odds cannot be different simply because I know what my first choice was. His odds are different because he is not making the same "choice" that you are. He is being offered to choose one of two. You have been asked to choose 1 from 3, and then whether you'd like to switch after an unchosen option has been revealed as incorrect. REMEMBER, the question is "does it matter whether you stick with your original choice, or switch?" And the answer is yes. If it's a fair game (meaning in this case that you are always offered the option to switch after being shown that one of the unchosen options is incorrect) then you will experience 2/3 success by switching, and 1/3 success is you choose not to switch. Your brother has a 50% chance of choosing the right door if he has no data to base his choice on. But if he knows which door you have chosen, and knows that the game is a fair game, then he has additional data to help him make the decision. This is an infamous math controversy, and I'm a math textbook editor. I know what I'm talking about here. Once again, I challenge you to do a simulation. What you will find is that when you do many trials, the 2nd choice (whether to switch) is a choice to switch the right shell (or door) 2/3 of the time, and a choice to switch to the wrong shell only 1/3 of the time. The odds are not something inherent in you, personally. The odds are based on the information available about the options. Before the first door was revealed, the odds of your choice were indeed 1 in 3. Your odds go up once you find out that door #1 is not it BECAUSE you are no longer dealing with the same set of choices. Door #1 COULD have been the car or the donkey when you made your choice. That's why the odds were 1 in 3 at the time. Now that there are only 2 options left, the odds are 50-50. You haven't "increased the odds" of your first pick." You are making a SECOND pick, which is necessarily independent of the first. Sorry, but you are misconceptualizing. You had a 1 in 3 chance of being right when you made your choice. If you don't switch, then you still have a 1 in 3 chance of having been right. Lifting up an empty shell doesn't affect the odds of your original choice having been correct. See, you'll have chosen the right shell only 1/3 of the time. Showing you an empty shell doesn't affect the original probability. Posted by: bk at March 28, 2006 04:30 PMBefore I finish even reading your post, Brian, I really didn't mean for that master's degree comment to come out quite as obnoxious as it sounded. My apologies. Had I really intended to insult you, I would have included a reference to degrees coming out of CrackerJack boxes or something. Now back to your post. Posted by: PatHMV at March 28, 2006 04:43 PMI love the passion. It's beautiful. Posted by: WHQ at March 28, 2006 04:45 PMI will concede that, if the person running the game can choose to or not to offer the revealing and switch, the probability applying to switching versus not is indeterminate. He may only offer the switch when he knows you originally chose correctly. Posted by: WHQ at March 28, 2006 04:55 PMOn second thought, I'm not conceding. I'm acknowledging. I would never have disagreed with that. Implicit in my proposition was that the person running the game would have to reveal an empty shell and offer a switch under any circumstances. Posted by: WHQ at March 28, 2006 05:00 PMThat one-third two-thirds analysis assumes several things not stated in the original problem, though. Namely, that the person turning the shells already knows exactly what's under them. If they do not know, if the two remaining shells are truly "random choices," then odds truly are 50/50 after the turn. You are being offered a new choice between the two remaining shells, only one of which has the pea. Of the three possibilities, one has been randomly eliminated, and there is NO reason for the odds of either of the remaining shells to have (or not have) the pea to have changed. It's still 1/3 against 1/3, "even money." Now, if the turner knows what's under the shells and will never turn over the shell with the pea under it, there is ZERO probability that the pea will be under the shell first turned, thus ending the simulation and preventing a second choice. That in turn affects the probability of the outcome. The set of two unchosen shells then remains at 2/3 probability for having the pea, as the truner themselves cannot turn over the shell with the pea. Human agency has altered the odds. That's where the apparent paradox arises. But that assumption wasn't stated, so I discounted it. I've seen the problem before, used to use it to beat up on students about implicit assumptions and human agency. But we did it playing three card monte. :-D Posted by: Tully at March 28, 2006 05:06 PMIf some are looking for a topic, I have an observation. Most here probably support evolutionary theory in the Great Debate over Intelligent Design. Many Americans probably understand the growing need for increased co operation and enforcement of international law given the emergent dangers of proliferation and radicalism abroad. One area of evolutionary research has been understanding the enigma of "co operation" in light of the popularized Darwinian prime directive of "survivial of the fittest". Most evolutionary "experts" agree that the basis of such co operation goes beyond kin selection and enters the world of "altruistic" behavior. The political debate over Iran and indeed whether a "hegemony" of sorts can increase global co operation is an interesting proposition in light of evolutionary paradigms of co operation and punishment. The following links certainly posit the existence of the genetic basis for altruistic "punishment", as well as altruistic co operation. I link a few articles (some abstracts) below for any to consider when speculating on the future of "hegemony" in relation to what science ponders about human co operation and its limits. Perhaps the "Ecological Dialectic" (Evolution in motion) might one day integrate both the "Orwellian Dialectic" and the "Clash of Cvilization Dialectic" that represent the two political extremes of late. Such an evolutionary approach seems befitting a non-ideological center...... first off, something about monkeys The origins of altruistic punishment and co operation the suprising asymetry of punishment both an example of "punishment" and a lack of an "international co operation" to punish enjoy the links if you're really bored......... Posted by: maxtrue at March 28, 2006 05:09 PMLet the 3 shells, regardless of their position on the game board, be X, Y, and Z. Let Z always represent the removed shell, which is known to be empty. In that case, the player is never actually choosing 1 of 3 shells. He is always, from the beginning, choosing 1 of 2 shells, either X or Y. Therefore, his odds are always 50-50 to begin with. Posted by: PatHMV at March 28, 2006 05:13 PMAnd in the time it took to tap that out, more comments show up! Oh well. Posted by: Tully at March 28, 2006 05:15 PMPat, IF the turner of the first shell knows where the pea is, he will NEVER reveal the pea on the first turn. So that set of two unchosen shells remains at 2/3 probability, which is why you should switch. The first shell has NOT been eliminated randomly, but has been selectively eliminated with full knowledge of location. So the 2/3's set remains at 2/3's, even though it's down to one shell. Posted by: Tully at March 28, 2006 05:27 PMTully, that's what my assumption was in my last post, that the turner knows what's under the shells and will only reveal an empty one. It only appears that the player is being given a choice between 3 shells (a, b, and c). But because the turner knows and will reveal an empty shell, the player is actually only being offered a choice between 2 shells (x and y, where z is the empty shell which will be revealed by the turner). Posted by: PatHMV at March 28, 2006 05:39 PMTully, you said: The set of two unchosen shells then remains at 2/3 probability for having the pea, as the truner themselves cannot turn over the shell with the pea. At that point, the set of two unchosen shells has a 100% probability for having the pea, since by definition the chosen, turned-over shell does not have the pea. Posted by: PatHMV at March 28, 2006 05:43 PMExpanding on my x, y analysis: Z equals the revealed shell. X equals the leftmost unrevealed shell. Y equals the next leftmost unrevealed shell. A, B, and C represent the positional spaces of the 3 shells. Thus, if we have BAC for the position of the shells, and the turner reveals A as empty, then X equals the shell at spot B and Y equals the shell at spot C. Or if we have CBA for the position of shells, and the turner reveals C as empty, then X equals the shell at spot B and Y equals the shell at spot A. Thus, every pick made by the player will correspond to either X or Y. Assuming the player makes completly random selections, then there will be an equal distribution of choices over time between X and Y, just as there is an equal chance that the pea is under either X or Y. The trick is that it looks like the player is choosing among 3 shells, when in fact they are only choosing among 2 to begin with. It is a fallacy to say that there is a 2/3 chance that the pea was in either the shell the player didn't choose and the shell revealed by the turner BECAUSE THERE WAS NEVER ANY CHANCE AT ALL that the pea would be under the shell revealed by the turner. Originally, there was a 1/3 chance it was under A, a 1/3 chance it was under B, and a 1/3 chance it was under C. Reveal C. We now know that there was NEVER a chance that it was under C. There is now a 100% chance that it is under either A or B. Posted by: PatHMV at March 28, 2006 06:12 PMMan, I haven't had this much fun since I got a 35 on the ACT math section... Posted by: PatHMV at March 28, 2006 06:17 PMMy bad at terminology and clarity. Please note the following applies ONLY when the turner knows where the pea is and has two shells while you have one, and MUST turn an empty shell and MUST offer you a choice betweeen the two remaining, yours and his. The turner begins with a set of two, with a 2/3 probability of his "set" having the pea. Since the turner KNOWS where the pea is, and thus can ALWAYS turn an empty shell, that set of two REMAINS at 2/3 odds, even when they turn one over and show it to be empty. Your shell still has only 1/3 chance of having the pea. That one of their shells is empty and that they have shown you that empty shell does not change their odds of having had the pea in the first place. Now, the turner has a 100% probability of having an empty shell to turn, because they have two shells and there's only one pea. Since they KNOW where the pea is, they can ALWAYS turn over an empty shell, whether they have one empty shell or two in their 2/3's set. But their "set odds" remain the same, as do yours. What has been eliminated (versus a random draw & turn) are the 1/3 odds that the first turn will reveal a pea, AND the odds that your shell will be turned first. They had a 2/3 chance to get the pea, and since they know where the pea actually is, that does not change when they turn the first shell to show it's empty, as there is NO chance that the first turn will reveal the pea. They still get the pea on the initial draw, two out of three--and they still have an empty shell to show on that first turn, every single time. If nobody knows where the pea is, and one of the 2/3's set is randomly turned and shown to be empty, then the odds the pea is under either of the two remaining shells is indeed 50/50, because the turn was actually random and might actually have revealed the pea, but did not. That the turned shell was part of a subset is then irrelevant. The two remaining shells are no different, each having had initial odds of 1/3, and now having odds of 1/2 after the random elimination. Human agency and knowledge does not guarantee you the win, but it boosts the heck out of your odds. They are no longer random odds, but selected odds. But if you wanna stick with the other line, let's get together and play poker sometime..... Posted by: Tully at March 28, 2006 06:49 PMI think framing the question differently would help. Assume three marbles are in a bag, two red, one blue. I get $5 and a 1.75 of Macallan's 18 year if I pick the blue one. I draw one out but I'm not allowed to look at it. The guy at the distillery then turns around, looks into the bag, removes a red one and shows it to me. In this example, he has no idea which one it was until after I made my decision. Should I switch to the one remaining in the bag? I think it's a little clearer that I should now switch. I know that there is a true 2/3rds chance that I have a red one in my hand. By removing one of the red marbles, my chances are now 50-50 but these chances are directly tied to the previous 1 out of 3 choice. God, I'm ready to win that 1.75 now. Posted by: Scotch Drinker at March 28, 2006 06:57 PM Damn it, Tully, every time I think I've got you beat, suddenly you appear right beforehand. I've got to improve my reaction time. Posted by: Scotch Drinker at March 28, 2006 06:58 PMOK, I'm confused now. Are we assuming that the probability is still 1/3 after the first choice, or that the probability for the first is always 1/3 right, and 2/3 wrong? If the player picks a shell, and then is asked to switch or stay AFTER the empty other is revealed, isn't that choice eliminated, thus making the odds 50-50? The first choice's odds are left unchanged, but how can the odds be still 1/3, with only two choices left? Forgive me if I've missed the obvious. This is a great discussion, BTW. Posted by: Rafique Tucker at March 28, 2006 07:02 PMOK, Rafique, here it goes. If the choices of all shells are random, including the choice of the shell first turned, and that first shell is empty, THEN the odds that either remaining shell has the pea are 50/50. Pick either, no diff. But if the other person (with two shells) KNOWS which shell has the pea and will NOT expose the pea on the first turn, their turning over an empty shell out of their two does not alter thier original odds of 2/3. All they've done is show you an empty shell that they would inevitably have, and they already know it's empty. So if you get the choice to take their remaining shell at that point, knowing they purposely chose to expose an empty shell, you should switch. Their "set" odds are still 2/3 to have the pea under their remaining shell. There was never any chance that they would expose the pea on the first turn, so the narrowing of odds is completely illusory. Posted by: Tully at March 28, 2006 07:10 PMIOW, all the two-shell person did was consolidate their odds to one shell by eliminating a known dud. You've still got a 2/3 chance of your shell being a dud. They've still got just a 1/3 chance of having a dud left. Posted by: Tully at March 28, 2006 07:13 PMThey sell Macallan's 18 yr old in a 1.75L? My budget aches just thinking about it, and I'm a sipper not a drinker. Posted by: Tully at March 28, 2006 07:43 PMYes. I get it now. Thanks Tully. That was a good one. Posted by: Rafique Tucker at March 28, 2006 07:48 PMThank WHQ. I was just being a smartass in ignoring the implied assumptions. Posted by: Tully at March 28, 2006 08:29 PMRegardless of the odds, if somebody offers you a chance to guess at three card monte, (no relation to Monty) don't play. ;-) Posted by: Blue Jean at March 28, 2006 08:57 PMLOL. Very very true. That's a case where the pea isn't even on the table any more. The problem only works based on the presumption of a fiar game where the person turning shells will always show you an empty opn eand offer you a choose. Pat, you're wrong about the "illusion" that there was a 1 in 3 chance initially. The 1 in 3 chance is there...3 shells, one pea. Let's use that scenario...3 shells, one pea, the turner always reveals an empty shell and offers you a choice. Now Pat, lets go back to the brother scenario. You choose shell A from shells A, B, and C. At this point there's a 1/3 chance that your choice, A, is correct. The turner reveals that shell C is empty, leaving you to choice to stay with A or switch to B. Now, you brother is invited on stage. He is asked to choose between A and B, without any knowledge of which one was your initial choice. His chances of being right are indeed only 50-50. But you have additional data, so your choice can be more informed. You know that there's a 2/3 chance that you first choice was wrong. And trust me on this, the revealing of an empty shell doesn't change the probability of your first choice having been right. It remains at 2/3. As Tully says, the turner can always show you an empty shell, because there is only 1 pea. Think about it this way. Suppose there were 1 million briefcases, and one of them has a million bucks inside. You pick briefcase number 86. The chance that you chose the right case in one in 1 million. Same deal, the choice is always offered after revealing empty briefcases Now, the briefcase opener starts opening empty briefcases. Are you going to maintain that seeing empty briefcases makes it more likely that your initial choice was the right one? Suppose the briefcase opener opens 999,998 cases and only two are left. Would you switch? You better. When you chose briefcase 86, you chose from one million briefcases. Now there are only two left... • the one you chose when there were 1 million to choose from •and the one that's left after 999,998 of the other 999,999 briefcases were shown to be empty Only a fool would stick with their first choice in a fairly run game. By the way, we're watching "deal or no deal," which involves money in briefcases. this reasoning doesn't really directly relate to this show, because each of the briefcases has a different amount of money, so each briefcase revealed does add to the amount of data you have at hand, and briefcases are opened randomly. In the scenario we've been discussing, the shell reveal is not random, the turner chooses an empty shell with purpose. DoND is an amazing study in the misunderstanding of probability and the willingness to press your luck. I saw someone turn down 189k with only 6 or 7 cases left . and end uop with only 5 grand. Dumb. Posted by: bk at March 28, 2006 09:00 PMI've been a little scarce lately--absolutely slammed at work, so down time to browse has been quite limited. We've talked about the French marches (riots) in the past, but what do we think about all of the marches within our own borders this past weekend? I guess I'm a little torn as to what the solution is right now, but it's obvious we have a problem. While I recognize the fact that, for the most part, illegals often seem to be a hard-working bunch who do contribute to the American economy, it rubs me a little wrong to see "illegals" march for their "rights." I'm not one to espouse throwing them out, but I guess I'm not quite sure where non-citizens who entered the country illegally seem to feel they are entitled to the same rights as legal citizens. I'm certainly not advocating violence, but it just doesn't rub me the right way. Posted by: AR at March 28, 2006 09:17 PMYou want experimental, I'll give you experimental. I put together an Excel table to try this out. I populated 10,000 rows with random numbers from 1 to 3, designated the "pea" column. That represents the shell under which the pea is (duh). A second column of 10,000 was populated with another random number from 1 to 3. That's the "player's choice" column. So far so good, right? A count reveals that the player won 3,379 times and lost 6,621 times, just as expected. Now we must choose the reveal shell. The reveal shell must meet 2 criteria: it cannot have the pea under it, and it cannot be the shell picked by the player. In lieu of writing a complicated formula to pick a random number which was NOT either the picked shell or the pea shell, I decided to just pick a random shell to reveal. Then I eliminated all rows where the revealed shell number matched either the picked shell or the pea shell. That shouldn't affect the randomness of the remaining rows, right? This left 4,486 rows where an empty shell was removed. This is a crucial step, so if my methodology here is incorrect, please let me know. Then I counted those rows where the player won with his original pick, and the number of rows where the player lost (i.e., should have switched). The result? Player won by sticking with the original pick 2,259 times. Player should have switched 2,227 times. Switch, don't switch? Your odds are 50-50, just like I said. Sometimes common sense is right. The Excel file is available on my website here. Posted by: PatHMV at March 28, 2006 09:44 PMBrian, the logical fallacy occurs here, where you said: But you have additional data, so your choice can be more informed. You know that there's a 2/3 chance that you first choice was wrong. THAT IS NOT NEW INFORMATION. You knew before you picked that the reveal would be of an empty shell. Revealing it provides no new information, at all. Posted by: PatHMV at March 28, 2006 09:47 PMYour methodology is wrong, Pat, in the initial assumption of ANY randomness in the reveals. You shouldn't be selectively eliminating reveals in your model at all. You should be ignoring them entirely. By making that assumption in your model, your results include "reveals" where the shell was picked randomly, but was empty, and "reveals" where the one-set shell was picked, and the cross-set of the two. But the shell "revealed" is NOT random. Not even remotely. ALL of the potential positive reveals have been eliminated by the base assumptions, while not altering the distributions. Since they're not informative, they don't count. Re-play: You get a choice of one out of three. Now, having chosen and without a reveal, what are your odds? Would you swap your one shell right now for the other two? You know at least one of them is empty, and nothing more. Then one of those is turned and is empty, but the turner knows where the pea is, and will NEVER turn a positive. What useful info have you gained that changes the odds? None. You're still choosing between the one-shell and two-shell sets. Posted by: Tully at March 28, 2006 11:29 PMNo, you're not. Because you know that, no matter which one you pick, the turner is going to reveal an empty shell. My methodology accounted for the selectivity of the reveal by eliminating all random reveals which revealed the pea, along with all random reveals which resulted in picking the shell already chosen by the user. In other words, in the end, the randomness of the reveals in my methodology was to randomly pick between one of the two empty shells. I did eliminate all of the potentially positive reveals. My spreadsheet looks at the actual odds of picking the right shell WHEN a shell is revealed meeting 2 criteria: (1) the reveal is not the shell with the pea, and (2) the reveal is not the shell chosen by the player. If you can show me a spreadsheet with a different methodology and a different result, please do. The players choice, from the beginning, is only between 2 shells: the shell with the pea in it, and the empty shell which will not be revealed by the turner. Go back and look at my X, Y, Z argument above. Posted by: PatHMV at March 29, 2006 12:09 AMPat, As Tully point out, you simulation doesn't give the right results because it doesn't properly simulate the scenario. Your problem is not really a misunderstanding of simple probability, it's a problem of conceptualization. I hereby challenege you to find a willing friend to do this following simulation. I will even drive down to Lousisiana and do it with you, if you'll buy me dinner. Bear in mind that this is what I had to do myself to convince me that the reasoning your are sticking with is indeed mistaken. Here's how to do the simulation. Find a friend. Get 3 unmarked playing cards, 2 kings and 1 queen. Play "find the queen." Do a minimum of 100 trials, each of them as follows: •Your friend mixes up the 3 cards without looking at them and places them face down. You pick a card, setting it to one side. •Then your friend picks up the other 2 cards, and shows you a king from this group. Then he asks you if you'd like to switch, or stick •after you choose to stick or switch, your chosen card is revealed That's one trial. Record the results in a four column table that tracks If you do this, you will discover that when you switch, you find the queen about 2/3 of the time, and then when you stick with your original choice you only pick the queen one third of the time. Ity does matter if you switch. You will NOT find the queen half the time regardless of whether you switch or stick, you'll do twice as well when you switch. Would you care to placer a wager on the outcome? Posted by: bk at March 29, 2006 09:33 AMSorry, but you're still wrong, Pat. The player's initial choice is one shell in three (Set A), and the other two shells form Set B. They cannot be considered as individual shells for purposes of probability, because the reveal changes nothing about the two-shell set. The second choice, after the reveal, is still between the initial results of a one-shell set and a two-shell set distribution, not between two one-shell sets. In lieu of writing a complicated formula to pick a random number which was NOT either the picked shell or the pea shell, I decided to just pick a random shell to reveal. Then I eliminated all rows where the revealed shell number matched either the picked shell or the pea shell. That shouldn't affect the randomness of the remaining rows, right? There's your problem. It doesn't affect the randomness of the remaining rows, but it DOES affect the count by eliminating all random positives. But there are no random positives--there are no positives at all. Go back to your spreadsheet, and check your initial distribution. Your odds of having the pea in Set A are one in three, in Set B two in three. Set B must always show an empty shell in the reveal. The reveal is not random, but manipulated. Since the results at this point are manipulated, the entire line count for Set B is the appropriate set to use for your "choice" set denominator in the second decision. Instead, you've eliminated random reveals from Set B in your methodology for your "second pick" count. This affects both your numerator and your denominator. But Set A is never tested, and Set B has no random reveals--all reveals are manipulated to be negatives regardless of Set B contents. Yet you subtracted those lines from your "second choice" dataset, eliminating them from both numerator and denominator, skewing your count. Since positives are not even possible (the reveal is forced, not random) those lines have to be counted in the dataset for the second choice. Eliminating random positives when no positives are even possible alters your count. The correct count to use for the second choice is that of the initial dataset, as it's the only count that was actually random. See the problem? The reveal does not actually eliminate ANY lines from your database, because the reveal is forced and is enforced only against Set B. The correct data set for when Set B holds the pea is your initial distribution. Set B's odds never change, regardless of the reveal. The reveal does not change the numbers. Human agency--the game is rigged. In this case it's actually rigged in your favor, IF you realize the set probabilities have not been changed by the reveal, and that you are being offered the choice of Set A versus Set B, and shell A versus B or C. Posted by: Tully at March 29, 2006 09:53 AMBrian, how can I adjust the Excel model to properly model the scenario? The picks are all a matter of random chance. I don't see the difference between my Excel model and the scenario you describe. It seems more efficient to model it in the computer rather than actually play it out. But if I have modeled the question wrong in the computer, I'd really like to change it so I can see the light as you have. You say that my brother does indeed have a 50-50 chance of being right (after my mid-show heart attack). If my odds are different from his, than the distribution of my picks between the two non-revealed cards must be non-random. If my picks are evenly distributed between the two shells just as his are, then we cannot have different odds. That said, I'll be happy to buy you dinner any time you'd like down here. We have good steaks, good seafood, and even some decent barbecue here. Posted by: PatHMV at March 29, 2006 10:00 AMTully, as I see it, one can say that "Set A" is the set of the two peas which will remain unrevealed. You don't know when you pick which 2 they are, but you know, by definition, that Set A must contain both the pea and the unrevealed empty shell, while Set B, the one shell revealed, will always be empty. Therefore, my choice is between the two shells in set A, 50-50. The reveal must meet 2 conditions, not just one. It cannot be the shell with the pea, and it cannot be the shell you picked. And that's the only selections I tossed out, just as the turner would know to toss out those picks before even making them. I really don't see how I've eliminated random reveals in your Set B. I have removed only those which do not meet the necessary 2 conditions for the reveal. Posted by: PatHMV at March 29, 2006 10:10 AMYou say that my brother does indeed have a 50-50 chance of being right (after my mid-show heart attack). If my odds are different from his, than the distribution of my picks between the two non-revealed cards must be non-random. If my picks are evenly distributed between the two shells just as his are, then we cannot have different odds. Wrong! And right! IF you pick randomly, your long-term odds are breakeven. But you have inside information about the distribution, so you can indeed have different odds than an uninformed stranger. Your odds cannot be worse than 50/50 picking randomly, but correct information can make them better. In this case some one walking up to the table after the reveal would have a 1/1 shot, if they had NO other information at all. But YOUR odds, if you paid attention, are to choose between a 1/2 bet and a 2/1 bet. Picking randomly gives you 1/2 * 2/1 = 2/2 = 1/1 in long run. 50/50. Part of the illusion is that a 50/50 choice would be an improvement over your odds at table from your first pick. It is--50/50 would exactly match your long-term results from RANDOM picks in the second choice. BUT, since we're talking odds, you have to total the long-term results. If you stubbornly and consistently hold your original shell, you'll lose two bets out of three and get hosed. If you pick randomly between Set A and Set B, you'll break even in the long run. If you stubbornly and consistently pick Set B, you'll consistently win twice as often as you lose. Posted by: Tully at March 29, 2006 10:17 AMIOW, you're confusing payoffs with probabilities. Pat, you can't eliminate the gross numbers from the dataset, as there is NO condition which will create a positive reveal. Eliminating lines to adjust for positive reveals is stepping past the rules. There are no eliminations, as there are no positive reveals. All potential reveals are nulled by manipulation, but that does NOT reduce them out of your gross trial count. And I'm using Set A as your one-shell pick, Set B for the two remainders. That's your initial distribution. Tully, I'm just not seeing it. We're talking past each other. If I wrote a complicated set of if-then statements to pick the reveal (if picked=A and pea=A, then reveal either B or C, if picked =A and pea=B, then reveal C, etc.), would that make for an accurate model? Posted by: PatHMV at March 29, 2006 10:44 AMI could show you the Perl Script we used to use as a long-run trial demo, but I quit playing with prorgramming long ago so I'd have trouble expalining it nowadays. The actual math isn't quite as simple as 1-2-3. Correct odds from switching are not precisely 2/3 for various reasons having to do with small percentages from a possible one-shell set reveal (false or otherwise) being excluded. More like 13/20 versus 7/20, more or less. 65% versus 35% with some random variation, instead of 66.6% versus 33.3%. But that's a quibble. It so happens there's an ongoing experimental dataset for this very thing. Play away. Let me know how you do. Posted by: Tully at March 29, 2006 10:45 AMAnd I still don't see why your grouping of Sets A and B is any more valid than my approach of saying that Set A is the 2 shells which will remain unrevealed. Posted by: PatHMV at March 29, 2006 10:46 AMAnswer: No, it wouldn't. The only relevant model is the initial distribution. The reveal is meaningless, because it's manipulated. It does not change the odds for either Set A or Set B having the pea. It only demonstrates that the two-shell set must have at least one empty shell, which you already knew. It provides no new information. It's an information thing, but the reveal is not the information that's relevant. Only the initial distribution is relevant. You get a choice between the 1/3 that you hold, and the 2/3 that they hold. Someone walking up and choosing at random after the reveal is choosing at 50/50, because to them the sets of A & B are neither distinct nor ordered. They start out not holding either set, and so have no preference, implied or othwerwise. But YOU are already holding a set that you KNOW is 1/3, and you KNOW the other set is 2/3, despite any manipulated and pre-determined reveals. YOU are not choosing between a 1/2 set and a 1/2 set. You are choosing between a 1/3 set in the hand and a 2/3 set across the table. If you choose randomly, then after enough trials you should be around 50/50 in results. If you insist on holding the 1/3 set, you get hosed. If you insist on choosing the 2/3 set, you pull ahead. Posted by: Tully at March 29, 2006 10:57 AMDoesn't matter what you call the sets, as long as you keep them distinct. Posted by: Tully at March 29, 2006 10:58 AMThere is also a Wikipedia explanation of all this which is very detailed. But I'm very stubborn, so I haven't given up yet! Posted by: PatHMV at March 29, 2006 10:58 AMThe fractional math for long-term random picks: [1/2 * 1/3] + [1/2 * 2/3] = 1/6 + 2/6 = 1/2 The fractional math for one-shell set pick, always: 1/1 * 1/3 = 1/3 The fractional math for two-shell set pick, always: 1/1 * 2/3 = 2/3 The reveal doesn't change that. You're still picking set-on-set. Posted by: Tully at March 29, 2006 11:04 AMTully, would you very kindly do me a favor and address my X, Y, Z analysis I made earlier, as that is how I still fundamentally see this problem. It may appear that I am picking between 3 shells, A, B, or C, but in fact I can NEVER pick the reveal shell. Thus, my choice from the beginning is constrained not between A, B, and C, but only between X and Y, where Z is the shell which will be revealed. Thus from the very beginning my odds are 1 in 2, not 1 in 3, precisely BECAUSE I know from the beginning that one of the empty shells will be revealed. Posted by: PatHMV at March 29, 2006 11:06 AMJust to beat it completely to death and flatten it enough to call it roadkill, the reveal gives you no information that can be used to determine in which set the pea is located. None. So the initial odds do not change. Posted by: Tully at March 29, 2006 11:06 AMLOL. I agree that the reveal gives no new information. But it is because the reveal gives no new information that it seems to me that your initial choice is, in fact, between 2 shells, not 3. Posted by: PatHMV at March 29, 2006 11:17 AMOK, here goes: Your initial choice is between A, B, & C. Your choice A has a 1 in 3 chance of havng the pea. B & C form the other set, and the odds of BC having the pea are 2 in 3. AT LEAST one shell in the set BC MUST be empty. The person holding BC knows where the pea is. You know only that one of their shells must be empty. They show that one of their shells is empty. Since they know where the pea is, there is ZERO chance the reveal will show the pea. Since they always have at least one empty shell, they will always show an empty shell. If they have the pea, they will never reveal it. The reveal does not change the set odds. You have not learned anything new about set BC from the reveal. Their set remains set BC with a 2 in 3 chance of having the pea, your set A remains Set A with a 1 in 3 chance of having the pea. The XYZ analysis is superfluous, meaningless, becaue it's an analysis of conditions that do not exist. You are now given the choice of keeping Set A, or taking Set BC. To maximize your odds of getting the pea, which alternative should you choose? See above for the fractional analysis. IF you choose at RANDOM, you're 50/50. IF you hold Set A, you keep your original 1 in 3. IF you swap for BC, you claim their original 2 in 3. Does that make it clear? A hold is playing your original odds. A random pick (coin flip) is an improvement over your original odds, to even money. But a swap is a boost over random, by moving to the original odds of BC. Posted by: Tully at March 29, 2006 11:19 AMThe reveal does not discriminate on which shell will be turned from BC, only that one of the two will be. You're still choosing set vs. set. Posted by: Tully at March 29, 2006 11:20 AMYummy, yummy, crow for dinner and humble pie for dessert. What a meal! Ok, here's what convinced me. First rule of solving a computer bug, remove as many variables as possible. Forgot that for a bit. Distribution of the pea is random. To eliminate one variable, I decide that my pick will be non-random. I will always pick A. I will be right 1 out of 3 tries, and that just can't change no matter what. I went back and redid the Excel model to have the player always pick 1. Then I redid the "reveal" column by picking randomly between 2 and 3. Where that choice resulted in picking the pea, I had it pick the other number. Sure enough, out of 10,000 tries, I win with my original pick 3,347 times, and with the switch choice 6,653 times. I will be happy to take Brian out to dinner here in recompense for my slurs and insults. Tully, you get none because your first answer was the same as mine... ;) Posted by: PatHMV at March 29, 2006 12:17 PMWhat's really confusing is knowing that the long-term odds of a random pick produce a 50/50 result. But that's true for any opposed inverted set in a two-alternative scenario. That does NOT mean that the odds of any given set having the pea are 50/50. They're not. Knowing that a random pick is 50/50 is a trap, because if you think the odds of either set having the pea are 50/50, you have no reason to abandon your held set for the other set. But the odds are only 50/50 if you pick at random over a long run. In the short run, it's 1/3 to hold and 2/3 to switch. If you never switch, you get hosed. Posted by: Tully at March 29, 2006 12:19 PMHee hee. My first answer forced WHQ to reveal the assumptions. It was correct until the assumptions were made explicit. Posted by: Tully at March 29, 2006 12:22 PMI had no idea this would lead to a discussion of this magnitude. Joy. I'm not totally convinced that the person revealing the empty shell necessarily has to know which one the pea is under. It's just that the one revealed has to empty, whether by chance or otherwise. You would only consider those where no pea was revealed anyway in a long term analysis. If he shows you a pea, you either already lost because you can only switch from one without a pea to another without a pea OR, if you can pick the one revealed to have a pea, well... I think the human agency comes into play only when the revealer has the choice to reveal or not reveal. In that case, he may be malevolent and only reveal when he knows you picked correctly in the first place. Posted by: WHQ at March 29, 2006 12:55 PMAnd one thing that may clarify the heart attack/brother-taking-over scenario being 50/50 is that he is not choosing to switch or not switch. He is just choosing between two shells, not knowing which was picked the first time. And you're not really picking between two originally just because one will later be revealed. The one you pick cannot be the one revealed, so your 1/3 uninformed choice makes it different from just picking between two from the beginning. I know this has been covered, I just like to state it in my own terms in the hope that it might clarify something that may not have been clear for someone before. Posted by: WHQ at March 29, 2006 01:01 PMActually, I am convinced that the revealer's knowledge of the location of the pea matters. I started thinking about it as I was driving to a meeting. I'm sure there's a previous post from Tully explaining in detail what I thought of while driving. I read these numerous and, in some cases, long posts hastily as I was pressed for time. It's not very intuitive when considering a single instance of the shell game why it matters, but it does over time. For the 1/3 - 2/3 split to be fully maintained, you must know that the revealer knows where the pea is and has no choice but to show you an empty shell. If you know the revealer must lift a shell, but you don't know if he knows where the pea is, you should still switch. If he doesn't know, you haven't hurt yourself because your odds are the same either way. If he does, you're better off changing. If you know he knows where the pea is, but don't know if he has a choice, you can't really make a decision without some way of assessing the probabilities of his having a choice and of his possible intents. If you know he does have a choice, you need to be able to assess just the probabilities of his possible intents. Otherwise it's indeterminate. If he has a choice but no knowledge of the pea's location, it doesn't matter what you do. When he does exercise the option to reveal it's the same as if he had no choice but to do so when he lacks pea-location info. Posted by: WHQ at March 29, 2006 03:55 PMActually, I am convinced that the revealer's knowledge of the location of the pea matters. It does indeed. If you run a thousand trials of the situation as stated (the revealed shell is always empty) it's a safe bet the revealer always knows where the pea is...or at least where it isn't. :-) When I first ran into the problem I was a grad assistant teaching beginning econ, and it was the "Monty Hall" version. Monty always knew which door had what, so that certainly colors my analysis. To get the intuitively obvious 50/50 odds after an intentionally empty reveal, you still have to make the second choice between the two remaining shells randomly, not just reflexively hold your initial shell. The odds that you originally got the pea remain at 1 in 3. A random selection gives you 50/50 long-run results because the odds are direct odds inversions totalling unity, so any long-run trials will average out to 50% wins if the two alternatives are chosen randomly. Posted by: Tully at March 29, 2006 04:38 PMWhen the brother comes in, he is not picking 1 of 3, but only 1 of 2. One of the 3 options has already been foreclosed to him, an option that had not been foreclosed to the original player. There are not actually different results for the brother and the original player. Again, let's eliminate a variable by having the player always pick A. The brother makes a random pick between A and whichever of B and C is unrevealed. Over time, the brother will choose 50% A and 25% B and 25% C. (This is verified in my Excel worksheet). But in all scenarios, the pea is under each shell (A, B, and C) an equal number of times. The brother, who has no knowledge of the original picks or anything, still has the same 1 in 3 odds as the player BECAUSE his choice is constrained by the player's initial choice. He is choosing randomly between only the original pick and the unrevealed, unpicked shell. His choice is not completely random, therefore, because it is already constrained. There was always a 2/3 chance that the pea was not under the picked shell, and, as Tully and Brian kept telling me, removing one of the shells does not miraculously transform the odds of the player's #1 pick being correct. The brother does not have a 50-50 shot because the odds of the pea be under the picked shell were not 50-50. Posted by: PatHMV at March 29, 2006 05:56 PMThe first sentence of that last part doesn't make much sense. Please disregard. Posted by: PatHMV at March 29, 2006 05:57 PMAs I pointed out, if the original player makes that same RANDOM pick of the remaining two, they're actually getting even money (long-run) as well, a clear advantage over their original 1 in 3. It's only when they show a preference for one or the other because of the prior actions that they swing their odds either up (swap) or down (hold). That's the catch. If they assume it's a straight 50/50 shot on a single trial, then they will (psychologically) have a preference to hold, not swap. But it's only a 50/50 if they pick at random, and just knowing that your odds are improved by including the other choice should be telling you something! If my odds are 1 in 3 if I hold, and 50/50 if I pick at random, that means the swap gives me 2 in 3 odds. If the reveal is picked at random, and the swap choice is only offered when the reveal is negative, THEN you're getting single-trial 50/50 with knowledge of preceding events. It's when the reveal is non-random and assured negative that the original odds stay the same. Another Example: You have two horses in a race and no others. One of them is 1 in 10 to win, the other 9 in 10. You don't know which horse is which. If you randomly bet that same race over and over and over again, you'll average 50% wins. That doesn't mean that the horses themselves are evenly probable to win. This applies to ALL situations where you have an either/or choice of two possibilities whose combined probability equals 1. [1/2 * 1/10] + [1/2 * 9/10] = 1/20 + 9/20 = 10/20 = 50% Now, if you DO know the odds, and the payout is always 1 to 1, which horse do you bet? The 9 to 1 that should pay $9 but will only pay $1, or the 1 to 9 that should pay 11 cents, but will actually pay $1? Posted by: Tully at March 29, 2006 06:43 PMIf the reveal is picked at random, and the swap choice is only offered when the reveal is negative, THEN you're getting single-trial 50/50 with knowledge of preceding events. It's when the reveal is non-random and assured negative that the original odds stay the same. My initial reaction to the assertion that the revealer's knowledge mattered was that the information gained by the player was the same. You pick one. One of the other two is shown to be empty. What's the difference if the guy showing you the empty one knew or not? The end result is the same. My own assertion that you would only consider the scenario in which the lifted shell was empty in support of the position that the revealer's knowledge was irrelevent is, in fact, the very thing that makes it matter. Of the 2 out of 3 times that your initial pick was wrong, half of your potential successful switches would be carved out by revealer's random selection of the shell with the pea under it. Posted by: WHQ at March 29, 2006 08:06 PMIf the reveal has the pea, there is no swap to offer. Reformulate: The reveal IS random, but the swap offer is mandatory. How often would (should) you swap? Every time the reveal shows the pea, which would be 1 in 3, you lose. You are betting to trade for the remaining "down" shell. Hold odds, 0%. Swap odds, 0%. Swap, don't swap, still 0%. (Discard/disregard) trials. The remaining data set: When the reveal does NOT show the pea but IS random, what are the odds on hold/swap after the reveal? Posted by: Tully at March 29, 2006 10:47 PMAnd for all those who haven't had their fix of percentage wrangling and assumption-tweaking, here's more in a critical poll-debunking vein. Posted by: Tully at March 29, 2006 11:02 PMOf the 2 out of 3 times that your initial pick was wrong, half of your potential successful switches would be carved out by revealer's random selection of the shell with the pea under it. I think we're saying the same thing, Tully. Yup. I was hoping to get Pat to notice that the second scenario (random reveal, mandatory offer) was what he had actually analyzed the first time by spreadsheet. Posted by: Tully at March 30, 2006 09:54 AM |
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