202. elliot803 - Feb. 25, 1999 - 2:39 PM PT
Rask:
No, the host opened C, and there was no prize behind it. The only doors still shut are A and B. The prize is behind one of those doors.
203. DanDillon - Feb. 25, 1999 - 2:41 PM PT
Kick the host in the nuts and smash down both doors.
204. Raskolnikov - Feb. 25, 1999 - 2:41 PM PT
Well, if said "A" and he opens "C", isn't it plausible that if choose "C" he will open "A" which has not yet been opened?
205. elliot803 - Feb. 25, 1999 - 2:41 PM PT
Rask:
Yes, I guess I should have been more explicit. When you picked door A, the host didn't open it. He opened C. Then he asked you if you wanted to switch your choice from A to B before he opened whichever door you end up choosing to see if you've won or not.
206. ChristinO - Feb. 25, 1999 - 2:42 PM PT
Oh, fucking duh.
Well.....no.....that doesn't make the sense I thought it did. The host is under no obligation to tell you that you chose correctly the first time around is he? Regardless of whether you've chosen correctly or not he's still going to offer you a chance to change your mind.
If he were obligated to let you know you'd chosen correctly then obviously the prize would be behind door B and you should change your choice......
Can we play Jeopardy instead?
207. DanDillon - Feb. 25, 1999 - 2:42 PM PT
If the host is female, kick down both doors and then go enjoy a Peanut Buster Parfait® at the local DQ. (This was your prize.)
208. elliot803 - Feb. 25, 1999 - 2:44 PM PT
Rask:
No. He just opened door C to give you a hint. Or possibly to fool you. Or maybe just to try and confuse you. Then he asks you to make your final decision (stick with A or switch to B), and he opens the door of your choice to see if you're a winner. Does that make it clear?
209. ChristinO - Feb. 25, 1999 - 2:45 PM PT
Gotta' go to lunch. I'll look when I get back. I'm exiting through door X.
210. DanDillon - Feb. 25, 1999 - 2:47 PM PT
This is why I mainly stick to Language.
211. elliot803 - Feb. 25, 1999 - 2:47 PM PT
Christin:
Gosh, you take late lunches there in La-La-Land.
212. ChristinO - Feb. 25, 1999 - 2:47 PM PT
Rask,
No matter which door you chose originally the Host is going to open one of the doors you did not choose in order to show you an empty space.
213. ChristinO - Feb. 25, 1999 - 2:48 PM PT
Elliot,
Well, I would've gone about 40 minutes ago, but I was playing. I don't even get to work until 9 so the idea of going to lunch and then coming back and having five hours left to work is depressing. I like to come back and have 3 or less.
214. elliot803 - Feb. 25, 1999 - 2:53 PM PT
Okay, I can't wait any longer. The correct answer is that you should switch to door B. Can anyone explain why, and say how much you increase your odds of winning by doing so?
BTW: I was amazed by this answer when I first heard it. I was convinced (incorrectly) that it made no difference whether you switched or not, because the chances of picking the winning door were 1-in-3, either way.
215. Raskolnikov - Feb. 25, 1999 - 2:54 PM PT
OK I think I understand the situation. I would lean toward staying with A. I don't know if the host is for me or against me. I don't know that if I had picked correctly first, I would have won. But I also don't know what would have happened if I had picked C. Would he have then instead opened the other empty door and left C closed, or would he have opened C as empty, making me lose. Since I don't know this, I will play the chance that incorrect choices are penalized immediately, and stick with A.
216. Raskolnikov - Feb. 25, 1999 - 2:55 PM PT
I give. state your explanation.
217. Raskolnikov - Feb. 25, 1999 - 2:59 PM PT
wait, I think I have it.
218. Raskolnikov - Feb. 25, 1999 - 3:04 PM PT
Imagine that the prize is behind door A. In which case, the host woudl have to choose door B or door C as an empty room.
Now, imagine the prize is behind door B. The host could only choose Door C as the empty room, since he doesn't want to open my current door and make me forego my choice of staying or moving.
So, under situation one, there is a 50% chance of him choosing door C. Under situation 2, there is a 100% chance of him choosing door C.
Knowing that he chose C, there is a greater likelihood that he did it because of situation 2. My hunch on the math is that you have a 75% of chance of being right when you choose B (but maybe its 67% I don't have time to think it out)
219. Raskolnikov - Feb. 25, 1999 - 3:10 PM PT
yeah, its 75% chance. 4 options, its behind A and he opens C, its behind A and opens B, it is behind B and he opens A, and it is behind B and he opens C. Since option 3 is precluded by your choice of A, and he choose C instead, you then get three outcomes where he chooses C and one where he chooses B to open. making it a 75% chance that the prize is behind door B.
220. elliot803 - Feb. 25, 1999 - 3:13 PM PT
Rask:
I think you've gotten sidetracked by your speculation about the motives of the host.
Switching to door B doubles your chances of winning, from 1/3 to 2/3. Here's why: When you chose door A, there was a 1/3 chance that you had picked the door with the prize, and a 2/3 chance that the prize was behind one of the other two doors. The host, who knew which door concealed the prize, opened one of those two doors (C), eliminating the possibility that it was the winning door. So there is now a 2/3 chance that the prize is behind door B. So you should switch to door B.
221. vonKreedon - Feb. 25, 1999 - 3:24 PM PT
Elliot - I don't understand. I choose A, a 1/3 chance of winning. That leaves B&C with a total of 2/3 chances of winning. The host then eliminates C leaving half of 2/3 chances of B winning: (2/3)/2=1/3 chance of winning, same as the chances of A. What am I missing? Note: math is not my strong suite.
222. ChristinO - Feb. 25, 1999 - 3:26 PM PT
Elliot,
So I got it right the last time. The thing is to remember that your chances are never 50/50 but always either 1:3 or 2:3. Cool.
223. elliot803 - Feb. 25, 1999 - 3:38 PM PT
VonK:
Where you go wrong is in assuming that the odds of the prize being behind either B or C change when the host opens C. There's still a 1/3 chance it's behind A and a 2/3 chance it's behind either B or C. When the host eliminates C, you know there's a 2/3 chance that it's behind B.
224. DanDillon - Feb. 25, 1999 - 3:39 PM PT
These types of mathematical brain teasers do nothing but remind me of of "Two trains leave Akron at 3:00. One train travels at a rate of 65 mph, the other at a rate of 45 mph...." Good grief.
225. vonKreedon - Feb. 25, 1999 - 3:48 PM PT
Elliot - Why don't the odds change when the situation changes? Please don't take this as argument, just confusion.
226. ChristinO - Feb. 25, 1999 - 3:52 PM PT
Hey Cos,
There's no math in Message #199 (well, not so you can tell anyway) but it amounts to the same thing as all the fractions.
227. elliot803 - Feb. 25, 1999 - 4:01 PM PT
VonK:
The odds do change when the situation changes. When the host opens C, the odds of the prize being behind that door fall to zero. But there's still a 1/3 chance it's behind A, and a 2/3 chance that it's behind either B or C. The important point--which is why I emphasized it--is that the host knows which door the prize is behind. He didn't choose C at random. (It wouldn't make much sense for a game show host to open the winning door and ruin the game).
Think of it this way. The odds of the prize being behind either A or B or C are one. Since there's a 1/3 chance it's behind the door you originally picked (A), and a zero chance of it being behind C, once the host has opened it, then the chances of it being behind B must be 2/3.
228. elliot803 - Feb. 25, 1999 - 4:05 PM PT
Yes, Christin, you got the basic idea in Message #199.
By the way, when Marilyn Vos Savant presented this problem and the solution in her column in Parade magazine, she got more mail telling her she was wrong than she had for any previous brainteaser, including indignant letters from college professors and other (supposedly) learned people.
229. elliot803 - Feb. 25, 1999 - 4:11 PM PT
And while I'm talking about Marilyn Vos Savant, three cheers to her for advising a young gay man to come out in her column last Sunday.
230. vonKreedon - Feb. 25, 1999 - 4:11 PM PT
I still don't understand why the odds haven't changed from 1 in 3 to 1 in 2. In the original instance we had no idea which of three doors contained the Lady...oops, wrong story... anyway, we had know idea which of three doors was the winning door. Then the host eliminates one of the doors. Now we, or at least I, have no idea which of two doors is the winning door. Why aren't the odds now the intuitive 1:2?
231. elliot803 - Feb. 25, 1999 - 4:14 PM PT
VonK:
Because you picked door A before the host had opened door C. If the host had opened C first, and then asked you to pick, it would be a 50-50 choice between A and B. But that's not what happened.
232. ChristinO - Feb. 25, 1999 - 4:21 PM PT
Elliot,
Does she print puzzles every week?
233. elliot803 - Feb. 25, 1999 - 4:23 PM PT
Christin:
Maybe not every week, but pretty often. And they're usually not very interesting. But she's had some great ones from time to time.
234. vonKreedon - Feb. 25, 1999 - 4:33 PM PT
But now I'm being given the choice once again, only now it is A or B. Why is this still 1/3 to 2/3 instead of 1/2 to 1/2? I don't know any more about what is behind doors A and B. Is this purely statistical or is there a reality that I am unable to grasp that now makes it more likely that the prize is behind B than A?
235. DanDillon - Feb. 25, 1999 - 5:01 PM PT
Message #231,
The fact that one has chosen door A *before* door C was opened has no bearing on any likelihood of which door the prize rests behind. The odds don't change.
236. ChristinO - Feb. 25, 1999 - 5:18 PM PT
Has she got a web site with these things?
237. resonance - Feb. 25, 1999 - 6:51 PM PT
I'm sorry, Elliot, but I'm going to play devil's advocate here and suggest that the odds do change, not merely relocate. The reason is that you are offered another choice -- the earlier odds do not matter at all. Believing that the odds carry over from one bet to the next in the manner described above is known as the 'gambler's fallacy'. I will give an example;
If you flip a quarter twenty-four times and by some statistical hiccup you get twenty-four heads in a row, what are the odds that you will get heads on the twenty-fifth flip? Still 1/2. Yet if you start off calculating the odds of tossing twenty-five heads in a row you will get a really low probability. The common tendency is to think that this probability will bear on the different probability that comes up each time you flip a coin. It doesn't. Each time you flip a coin and progress to the next flip you still go back to the same probability for the next flip -- the prediction beforehand is interesting, but ceases to bear upon the problem the instant the first heads comes up.
Similarly, the prize is definably in one location behind a door. The statistical probability that your guess is correct (that it's behind one particular door) is one divided by the number of remaining doors. Whenever you open a door and guess again, you are faced with brand-new odds.
Say, once the host opens a door, he were to ask a second contestant (who hasn't seen the previous stuff happening) to choose between the two closed doors. That contestant will have a 1/2 -- 1/2 option available to him. It doesn't really matter if the MC has opened one hundred doors in a row, out of one hundred and two. He won't know anything about the third door, or any others, originally being closed -- and it won't matter.
238. resonance - Feb. 25, 1999 - 6:51 PM PT
The important thing to keep in mind is that the guesses of either contestant in this second round have exactly the same chance of being right, for the same reason that that twenty-fifth flip of the quarter has a fifty-fifty chance of coming up heads. The past simply doesn't bear upon the new guess.
To put this another way: You can posit that there is a 1/3 chance that it is behind B and a 2/3 chance that it's behind either A or C. So once C is opened then there is, by this logic, then a 1/3 chance that it's behind B and a 2/3 chance that it's behind A. Yet there cannot at the same time be a 1/3 and a 2/3 chance that A is the correct door. The fact that the gameshow MC won't open door A as part of the odds doesn't mean that we remove door A from the calculations.
239. resonance - Feb. 25, 1999 - 7:16 PM PT
The thing to keep in mind is that 'odds' only reflect reality to the same degree that perception reflects reality, because odds measure nothing more than our own perceptions of a particular chance. In reality (unless you want to get all freakishly, wrongheadedly Heisenbergian) in this problem, the prize is behind one of the doors with probability one and behind all other doors at probability zero. Our odds are just a guess based upon what we can deduce; in this case, for any given door, one divided by the total number of doors. These 'odds' change every time new data are entered into our deductions -- whenever a door is opened, the quotient for each other door correspondingly rises. One an MC opens a door, then the guess odds drop to zero for that door and rise proportionately for all other doors -- the previous odds are no longer useful. But none of that changes the probability that the prize is where it is, and our probabilities don't even treat with this absolute quality. Therefore, they don't carry over from guess to guess.
240. uzmakk - Feb. 25, 1999 - 7:33 PM PT
Doesn't sound like philosophy, but it does sound dead.
241. Slackjaw - Feb. 25, 1999 - 7:41 PM PT
"Is this purely statistical or is there a reality that I am unable to grasp that now makes it more likely that the prize is behind B than A?" There is no difference.
Another great problem. Elliot is right, if you assume no ulterior motives on the part of the host. The key is, you as the contestant have to determine the right probability to try to figure out. Figuring it out is fairly easy if you know some probability. This is long but if you read it you will understand. I assume no knowledge of probability theory. Below, "pr(X)" is to be read as "the probability of X."
What you want to know is, *given* that the host (H) shows C, what is the probability that the prize is behind A?
Well, there's another key: only some of the possible realizations of this game (after you select A and before the host shows C) are relevant--others rule out the possibility that you would ever reach a 2nd stage where you are asked whether you want to change your mind. The realizations you need to worry about are these: (1) the prize is behind A and H shows B; (2) the prize is behind A and h shows C; (3) the prize is behind B and H shows C; (4) the prize is behind C and H shows B. (1) and (2) are equally likely. It is very important that you resist the urge to restrict attention to door C being shown at this point.
What do you need to know to figure the probability that the prize is behind A given that the host shows C? Well, think of the simpler problem of rolling a die. What is the probability of a 1? It's 1/6 of course, but that just the probability of a 1, divided by the probability of *something* happening: (1/6)/1 = 1/6. A little more complicated: what's the probability of a 1 showing up given that the outcome is odd? Simply the probability that the outcome is a 1 and it is odd, divided by the probability of an odd outcome: (1/6)/(1/2) = 2/6 = 1/3.
(cont.)
242. Slackjaw - Feb. 25, 1999 - 7:42 PM PT
The "and" in the previous sentence is quite important--you can't just divide the number of cases where a 1 occurs by the number of cases where an odd number occurs. Consider this: what is the probability of a 1 given that the outcome is even? Not 1/3! It's the probability that a 1 and an even number occurs, divided by the probability that an even number occurs, or 0/(1/2).
This is not different--in finding the probability of a prize behind A given that the door shown is C, you must first restrict attention to those cases where C is shown (*now* is the time to do that). So to compute the probability of a prize behind A given that the door shown is C, we want the probability that the prize is behind A *and* C is shown, divided by the probability door C is shown.
But what is the probability that door C is shown, once we have reached the point in the game where the contestant selects door A? One way to think of it is as the probability that C is shown and the prize is behind A, plus the pr. that C is shown and the prize is behind B, plus the pr. that C is shown and the prize is behind C.
But we know that the probability of X given Y (P(X|Y)) is just probability of X AND Y, divided by the probability of Y (see dice example). Or, multiplying both sides by pr. of Y, we can see that pr(X AND Y) is pr(Y) times pr(X GIVEN Y). Symbolically, P(X|Y) = P(X and Y)/P(Y), so P(X and Y) = P(X|Y)*P(Y). Now substitute "prize behind A" for "Y", and "door C shown" for "X"--you get: pr(C shown and prize behind A) = pr(C shown given prize behind A)*pr(prize behind A).
(cont.)
243. Slackjaw - Feb. 25, 1999 - 7:43 PM PT
(last one)
Applying this newfound rule above, we can see that:
pr(door C shown) = pr(C shown given prize behind A)*pr(prize behind A) + pr(C shown given prize behind B)*pr(prize behind B) + pr(C shown given prize behind C)*pr(prize behind C), or: (1/2)*(1/3) + 1*(1/3) + 0*(1/3) = 1/2. These calculations are correct if you assume: door A is chosen, the host has no ulterior motives, and the game actually reaches the point where the contestant is asked whether he wants to change his mind--all doors have an equal ex ante chance of having the prize; if A is selected and A has the prize, the host randomizes over B and C; if A is selected and B has the prize, C must be shown for sure; and if A is selected and C has the prize, C is not shown for sure.
Alright, that's the probability that C is chosen, having reached the point in the game where the contestant selected door A. The other thing we need is pr(prize behind A and door C shown). Obviously, though, this is just the same as pr(door C shown and prize behind A). But we calculated this in the previous paragraph! Remember, it is simply pr(door C shown given prize behind A)*pr(prize behind A)--we derived this two paragraphs up. Again, it is simply (1/2)*(1/3) = 1/6.
So, we are left with: pr(prize behind A given C shown) = pr(prize behind A and C shown)/pr(C shown) = (1/6)/(1/2) = 1/3.
But if pr(prize behind A given C shown) = 1/3, it must be that pr(prize behind B given C shown) = 2/3!! You should change your bet.
244. mintcar - Feb. 25, 1999 - 7:43 PM PT
Here's a way to see why you want to switch doors after the host opens a door.
Imagine that the game was slightly different, in that after you choose a door (say door A) then you are given a *pair* of choices: you could choose to guess (1) that the prize is behind door A, or (2) that the prize is behind door B or door C. In this format you have been given no extra information after your first choice, so the odds of it being behind any particular door are still 1/3. So choice (2) is clearly twice as good as choice (1).
The point is that this game is *equivalent* to the game in which one door is opened by the host. When the door is opened by the host, the host is effectively saying "you can choose both B and C at the same time, because if either one is right we won't let you pick the wrong one."
245. resonance - Feb. 25, 1999 - 7:58 PM PT
@#$%@#$*%$$.
246. Raskolnikov - Feb. 25, 1999 - 8:13 PM PT
Slack, why is it a given that if the prize is behind B, the host has to open C? Why can't he open A and show you that it is empty, forcing you to then pick between B and C?
I had assumed this was part of the rules of the game in my last guess, but in hindsight I don't see why I made that assumption. (however, if the assumption is correct, then my earlier reasoning stands, although I may have made a mistake in my math, as I was rushed)
Basically, why does my choosing of A matter? Why does it have any effect on the door the Host chooses to open? Why is the math any different than it would be if the host opened C *before* I chose A?
247. resonance - Feb. 25, 1999 - 8:14 PM PT
Okay. I see it now. The MC *must* show a door (something I overlooked).
This isn't philosophy. Nevertheless, more, please.
248. Slackjaw - Feb. 25, 1999 - 8:16 PM PT
Aha. I think it is implicit in the game that the host doesn't show you what's behind the door you actually opened--he will show you what's behind one of the other doors. This is certainly implicit in my calculations. It's like "let's make a deal."
249. Slackjaw - Feb. 25, 1999 - 8:23 PM PT
let's see...in the fray we've done the beauty contest, the lemons problem (both last summer), the dirty faces & the sage, game show, surprise hanging...
Landsburg did the Allais paradox from decision theory about a year ago in a column on rationality...
searching...
Trying to remember the one that makes people think there's some correlation in the gender of one's children, but it's not nearly as cool.
Can think of some good ones from game theory like improving your lot by burning a dollar & forward induction but they're not as cool either.
hmmm...
250. Slackjaw - Feb. 25, 1999 - 8:25 PM PT
I can think of lots of games with weird equilibria that will make you think game theorists don't know shit...centipede and travellers' dilemma come to mind.
251. Slackjaw - Feb. 25, 1999 - 8:29 PM PT
we've done Aces and 8s...
252. Slackjaw - Feb. 25, 1999 - 8:29 PM PT
obviously there are many, many more paradoxes out there...just trying to think of the ones I know very well
253. Raskolnikov - Feb. 25, 1999 - 8:36 PM PT
Slack: Ok, like I said, I had initially made the same assumption, and came to the same conclusion (through a different route), but my probability estimates were incorrect. I feel re-assured now.
Every once in awhile, Marilyn posts a doozy. She analyzed a new Vegas game a few weeks ago, showing where the vig was. I knew there had to be a vig, but couldn't see where it was for the life of me. I followed her explanation, but when I tried working it out just by calculating the probabilities involved, I kept getting different answers. It took two hours of thinking on a car trip, with my wife convinced I was mad at her because I was being so unnaturally quiet, before I figured out where my error was.
254. elliot803 - Feb. 25, 1999 - 9:15 PM PT
slackjaw:
I don't have the energy to digest #241 through #243. You're making a fairly simple (if subtle) problem in probability much more complicated than it really is.
Rask:
"Why is the math any different than it would be if the host opened C *before* I chose A?"
Because in that case you would be faced with two equally likely choices, A and B, so the odds for each would be 1/2. When you choose A before any other door has been opened, you are picking one of three equally likely choices, so the probability is 1/3. The subsequent opening of door C doesn't change the probability that the prize is behind a door other than A, it just eliminates one of those possible doors.
If you still find this hard to see, imagine there are 100 doors instead of just 3. You pick one, and then the host, WHO KNOWS WHERE THE PRIZE IS, opens 98 of the other 99 doors, leaving only two doors closed, the one you chose and the one he left closed out of the remaining 99. Isn't it obvious that in this variation the prize is much more likely to be behind other door he left closed than the one you chose?
255. Slackjaw - Feb. 25, 1999 - 9:23 PM PT
well elliot, then don't digest it. But I have yet to see a post from you that is as incontrovertible.
Incidentally, since you feel justified in using "simple" and "probability" in the same sentence, you should notice fairly quickly that maybe 2/3 of those posts can be skipped if you know Bayes' rule and the definition of conditional probability. I have not made it complicated at all--I have simply defined all the concepts I used in case someone has never seen them before.
256. elliot803 - Feb. 25, 1999 - 9:40 PM PT
slackjaw:
"well elliot, then don't digest it. But I have yet to see a post from you that is as incontrovertible."
Huh? The most incontrovertible explanation is the simplest one necessary to explain the solution. Your #241 through #243 are a long way from being the simplest explanation of the solution, even assuming they do not contain any errors.
"Incidentally, since you feel justified in using "simple" and "probability" in the same sentence, you should notice fairly quickly that maybe 2/3 of those posts can be skipped if you know Bayes' rule and the definition of conditional probability."
I don't know what Bayes' rule is. You don't need to know what it is to solve the problem. You just need to reason it out in the way I described.
257. Slackjaw - Feb. 25, 1999 - 9:50 PM PT
"The most incontrovertible explanation is the simplest one necessary to explain the solution." No shit. And none of your explanations, however simple, met the constraint of adequacy. I point you to the confusion following them.
"even assuming they do not contain any errors." I challenge you to find one.
Why, Bayes' rule is a very simple notion in probability. Trivial, really. You just need to reason it out as in my messages. Do you know how to multiply? Does it tire you to do it a few times in succession?
258. resonance - Feb. 25, 1999 - 9:53 PM PT
I thought that Bayes came up with 'hunch' theory.
259. Slackjaw - Feb. 25, 1999 - 9:58 PM PT
what in blazes are you talking about?
260. resonance - Feb. 25, 1999 - 10:02 PM PT
Why, hunch theory, Slackjaw. Isn't it obvious?
261. Slackjaw - Feb. 25, 1999 - 10:04 PM PT
no
262. Raskolnikov - Feb. 25, 1999 - 10:06 PM PT
elliot: I get it. I got it in my earlier solution to the problem. The explanation I gave is an alternative way of solving the problem, just as Slack and Marilyn explained it in different ways. The thing is, I questioned my solution because I thought I might be making an improper assumption. The key piece of information I was not sure on is that once A is chosen, it is eliminated from being opened. Without this, choosing A is no different from not choosing anything. This was not stated in your initial rules, and I guess I don't watch game shows enough to assume it without reservation.
263. resonance - Feb. 25, 1999 - 10:09 PM PT
If you have an unknown you want to explain, and you can't get susbstantial hard data, you poll 'experts' in the field, ask them to give their best guesses as to what happened, then you collate them according to some rigamarole that Bayes cooked up. I gather it's a way to utilize the unconscious competence of people. Maybe it's a different Bayes or a different name altogether, but it's kind of intriguing and seems to be helpful in the right applications -- the Navy used it to successfully locate a lost submarine in the middle of the Atlantic, for example.
264. Raskolnikov - Feb. 25, 1999 - 10:14 PM PT
My last sentence looks petty, which wasn't what I intended. It was an interesting problem elliot, but for some reason, whether it be lack of game show watching or whatever, I was unable to easily jump to one of the necessary assumptions. Slack was able to do it, so it certainly wasn't impossible. But if Marilyn didn't mention that part of the assumption as well, it is no surprise that a lot of people didn't understand her argument.
265. Slackjaw - Feb. 25, 1999 - 10:27 PM PT
aha. I know exactly what you are talking about, Resonance. It's the same Bayes. Bayes' rule is sometimes used to model that process. E.g., a plane has crashed, nobody is sure where, and you ask a bunch of experts who look at the evidence and name a region. If you know the probability that an expert names a region given that the plane crashed there and the probability of an expert naming a given region, and you have a few experts, you can find the probability that the plane crashed in a particular region given the experts claims.
Reminds me of the Condorcet jury theorem...a number of experts get a private signal about some state of the world and want to aggregate their information to make a prediction about it. This French fellow named Condorcet showed that if everyone is truthful, a unanimous jury verdict minimizes the probability that a false positive is obtained, for a jury of a given size.
However, these guys named Jeff Banks and David Austen-Smith showed a couple years ago that people would not want to be truthful about their private signal in this sparse representation of a unanimous jury. Think of two possible states of the world-- + or -. Each juror has a private signal. Suppose the jurors simultaneously cast their ballots for - or +. If all N say -, then the prediction is -. If anybody says +, the prediction is +. So suppose everyone else is being truthful, and your vote is actually going to matter. Given the rules, that must mean there are N-1 - recommendations. If your signal is -, great--vote -. But if your signal is +, then given that everyone else is truthful, N-1 signals say - and 1 says +. Well, the information in favor of - is much stronger--so it pays to sometimes say - even if you draw a + under unanimity. Truthful revelation is not an equilibrium.
266. Slackjaw - Feb. 25, 1999 - 10:42 PM PT
experts get a private signal about some state of the world LIKE + OR -.
The Austen-Smith and Banks argument is applicable in a multi-state world
267. AzureNW - Feb. 25, 1999 - 10:43 PM PT
Slackjaw, thanks for the great explanations of the three doors problem, Message #241. It reminds me of the shock of falling flat on my smug face at beginning of Probability 204.
268. Slackjaw - Feb. 25, 1999 - 10:45 PM PT
it's my pleasure
269. uzmakk - Feb. 26, 1999 - 5:36 AM PT
I had a woman call me last night seeking my support in her quest for a local magistrate's position. Before I sign her petition I shall ask her what her political philosophy is. Philosophy is not dead, my ducks, not by a long shot.
270. BunEBear - Feb. 26, 1999 - 6:58 AM PT
It looks like Slackjaw gave the definitive treatment of the game show problem, but in case anyone would like more, there is a web page, Marilyn Is Wrong - Gameshow which is interesting. The "is wrong" applies here only because she made the implicit assumption about the game show host always allowing a choice, independent of what the contestent chooses. The author of the web page acknowledges that, given this assumption, Marilyn's math is right.
A far bigger gaffe that he mentions on his page was a column in which she makes a silly claim that Wile's proof of Fermat's Last Theorem is incorrect because it relies upon non-euclidean geometry.
271. elliot803 - Feb. 26, 1999 - 8:19 AM PT
Rask:
"elliot: I get it. I got it in my earlier solution to the problem. The explanation I gave is an alternative way of solving the problem, just as Slack and Marilyn explained it in different ways."
With all due respect, if you are talking about the posts in which you concluded that the probability of the prize being behind door B was 75%, you obviously did not get it. It wasn't just your math that was wrong; it was your understanding of the problem at a conceptual level. You shouldn't feel bad about this. You're in good company.
"But if Marilyn didn't mention that part of the assumption as well, it is no surprise that a lot of people didn't understand her argument."
The description of the problem I gave was almost exactly the same as the one in Marilyn Vos Savant's column. I'm still not sure what incorrect assumption you were making.
272. elliot803 - Feb. 26, 1999 - 8:34 AM PT
BubEBear:
The creator of that website needs to get a life. And I think his claim about Marilyn's assumption is incorrect, anyway. In the problem as stated, the host does offer the contestant the choice to switch. If he didn't, that would be a different problem. The only unstated assumption in the problem as described, as far as I can tell (other than absurdly trivial ones such as the assumption that you know what a door is) is that the game show host does not open door C at random. And I think that condition is implicit in the statement that the host knows which door conceals the prize and the fact that it is a game show.
273. Raskolnikov - Feb. 26, 1999 - 8:35 AM PT
elliot: Evidently, Marilyn didn't specify the rules of the game, and the motivations of the Host sufficiently either, from the link that Bun E Bear posted.
Anyways, my logic was correct (although not the concluding math - which was rushed). If you don't think so, how about you tell me where my mistake was?
274. elliot803 - Feb. 26, 1999 - 8:37 AM PT
slackjaw:
"And none of your explanations, however simple, met the constraint of adequacy. I point you to the confusion following them."
I guarantee you that your epic explanation would confuse far more people than the standard one.
275. elliot803 - Feb. 26, 1999 - 8:42 AM PT
Rask:
"elliot: Evidently, Marilyn didn't specify the rules of the game, and the motivations of the Host sufficiently either, from the link that Bun E Bear posted."
Well, I just addressed that.
"Anyways, my logic was correct (although not the concluding math - which was rushed). If you don't think so, how about you tell me where my mistake was?"
The fact that you concluded there were four possible solutions, with the prize being behind door B in three of them. That is a conceptual error, not a mathematical one.
276. Raskolnikov - Feb. 26, 1999 - 8:59 AM PT
elliot: that section is the one that was rushed. It described it as a math problem since I didn't set up the problem right in the two minutes I had. My logic was in the previous post, which led to the conclusion that B was the proper choice. The logic is correct.
277. Raskolnikov - Feb. 26, 1999 - 9:04 AM PT
and anyway, I said that there are four options, one of which was precluded by the hosts inability to choose A. My math mistake was in not distributing the probability of the precluded choice evenly.
278. elliot803 - Feb. 26, 1999 - 9:43 AM PT
Rask:
Your list of possibilities in #219 is irrelevant to the probability calculation. In two of the three non-precluded possibilities you list, the prize was behind door A! That's not a math error.
You were on the right track in #218, but you didn't follow it through. Given the conditions that the contestant chooses A and that the host knows which door conceals the prize, and the assumption that the host is not going to open the winning door and ruin the game, the odds of the host choosing to open C are twice as great if the prize is behind B as they are if the prize is behind A; therefore, since the host did open C, the odds are 1/3 that the prize is behind A and 2/3 that it's behind B.
279. FreeToChoose - Feb. 26, 1999 - 10:09 AM PT
Raskolnikov
I checked out the Bunebear link.
Very interesting, but sounds like sour grapes from someone who got the wrong answer.
280. SeaSailor - Feb. 26, 1999 - 10:14 AM PT
Your discussion of the game show makes an excellent parody of some philosophy books. Much ado about nothing. Try to prove to each other that you exist or that you are not using two names on the same thread.
Would anyone argue that it is all taken on faith?
281. Raskolnikov - Feb. 26, 1999 - 10:35 AM PT
elliot: oops, I did screw that up in 219. Ok, 219 is a mess, from trying to get the answer out too quickly. But 218 has the right logic.
282. ChristinO - Feb. 26, 1999 - 11:29 AM PT
Seasailr,
Yeah, I had some idiot on the schoolbus in 10th grade try to explain to me how movement was impossible because you can't take 1/2 of .00000000005 of a single step.
He was actually trying to pick up on me. I get all the weirdos.
283. FreeToChoose - Feb. 26, 1999 - 11:33 AM PT
ChristinO
I've never heard Zeno's Paradox used as a pickup line.
Sounds like it didn't work, so I guess it isn't worth trying.
284. ChristinO - Feb. 26, 1999 - 11:40 AM PT
FTC,
A good 60% of the lines men have used to try and pick me up would never end up in a book of pick up lines. I don't know if I should be flattered that they think I'm intelligent enough to want to have a "serious" discussion or if I should switch to a perfume that doesn't attract psychos.
285. FreeToChoose - Feb. 26, 1999 - 12:27 PM PT
ChristinO
If you change your mind about the 10th grader, you can find responses here
286. ChristinO - Feb. 26, 1999 - 2:19 PM PT
Thanks for the link FTC. I think what I told him was that I didn't have a movement problem he had a perception problem but that it was all fine and good with me because I was moving to the other side of the bus and he could just watch me go while he tried to figure out how to move half an inch closer to me.
287. whhatlaw - Feb. 26, 1999 - 8:15 PM PT
I am sorry to be a late arrival, but I am and have concluded that the answers offered to the three-door game show are simply wrong. If the rules of the game are that you pick one of the three doors (A), then the host opens one of the other doors (C), then you are allowed to pick the other(B) or stay with your initial pick, there is in fact no reason to believe that you should change to door B.
The reason that the statistical explanations are wrong is that the game doesn't actually start until the host opens one of the doors. Therefore, it is always a 50-50 game. The first part, where you pick one of the three doors, is just foreplay. You know from the beginning that the host is going to eliminate one of the two incorrect choices, so you can take that one out of the equation from the git-go. (Mind you, it doesn't matter whether you personally actually subjectively know that he is going to do this; the fact that he does it makes it an objective part of the calculation.)
The game is really "you are going to be given a chance to choose between two doors, one of which hides the prize; since we have three doors on the stage, first you will pick the door that I *won't* open; make that choice now." Seen this way, it is clear that the odds are never 1/3 to 2/3. The first choice is merely part of the entertainment, and has no bearing on the odds of the game.
288. Slackjaw - Feb. 26, 1999 - 11:01 PM PT
ONCE AGAIN
under the assumption that the game is like "let's make a deal," the host DOES NOT randomly eliminate (or deterministically eliminate) one of the two incorrect choices in every state of the world. In some states, such as when you select A and the prize is behind B, the probability that the host shows, "eliminates" in your terminology, door C is 1. Other times there is 0 chance door C is shown, like when you choose A and the prize is behind C--then B is shown with probability 1.
This assumption was implicit in my calculations and evidently in Marilyn's as well. It was not explicitly stated by elliot but should have been.
289. FreeToChoose - Feb. 27, 1999 - 11:29 AM PT
whhatlaw
You are simply wrong. Try playing the game with a friend. If your sample data supports 50/50, then explain to me how you played the game, and I will explain to you which assumption you misunderstood.
If you follow the rules, your results will be consistent with 1/3, 2/3.
Hint: the key is that there is a non-explicit assumption, totally reasonable if you understand how the real game is played, and the motivations of the host. This assumption is either missed or ignored as irrelevant by virtually everyone who claims that the answer is 50/50.
290. FreeToChoose - Feb. 27, 1999 - 11:32 AM PT
elliot803
Thanks for posing this puzzle to the fray. It is one of my favorite puzzles. Many people have attempted to write down why the correct answer is 2/3. Your explanation in Message #254 is as good as almost any I have seen.
291. whhatlaw - Feb. 27, 1999 - 3:06 PM PT
Free to Choose and Slackjaw
OK, I'll take one more shot. My assumption was precisely that the host does not make a random choice of which door to open, but rather that he will always open a non-winning door. (If his choice of which door to open were random--excluding the door you picked first-- then in 1/6 of all games the second choice would never be offered, because he would open the door with the prize behind it.) If your first selection is of the winning door, then he can arbitrarily choose which of the other two to open; if your first selection is of a non-winning door, then he has no choice of which door to open -- he must open the other non-winning door.
This situation is indistinguishable from the complete elimination from the game of the first choice. That is, if he announced "Alright, there are three doors on the stage, but to make your chances better, I'm going to open one of the doors" then opens one of the non-winning doors and offers you a choice of the other two.
I believe the confusion arises here because there are too few doors in the puzzle as presented. Try it with one hundred doors. You select one. At this point you are either right (1%) or wrong (99%). Now the host opens 98 non-winning doors (selected either arbitrarily from the other 99, because you have chosen the winning door, or carefully selected by him, because he knows which of the 99 hides the prize), leaving only the one you selected and one other. Do you really believe that you now have 99 to 1 better odds of winning if you change your choice? Of course not!
It does not matter how many doors you start with. When you are ultimately offered the only choice that counts, it is between two doors, one of which wins and one of which loses, thus the odds are always 50-50. You try it with a friend. If you come out anything other than 50-50, you just haven't created a big enough sample.
292. whhatlaw - Feb. 27, 1999 - 3:17 PM PT
To whhatlaw from whhatlaw
I recant. Now that I've read what I wrote, I'm convinced. It was my problem that I was dealing with too few doors. When the number was upped, it becomes clear that the odds do favor changing the choice.
293. Slackjaw - Feb. 27, 1999 - 3:21 PM PT
unbelievable
294. Slackjaw - Feb. 27, 1999 - 3:23 PM PT
Poll:
what proportion of people in the Fray post with such authority yet without actually bothering to know what they're talking about?
295. whhatlaw - Feb. 27, 1999 - 3:41 PM PT
Poll:
what proportion of people in the Fray are both such obtuse writers (read message 288 and prior Slackjaw posts) and such gracious winners?
296. Slackjaw - Feb. 27, 1999 - 4:20 PM PT
"winners"? what have I won?
There's nothing obtuse in 288 or other posts on this subject, but you the reader may be asked to carry a few ideas around in your head for a minute. Some subjects require a little energy to understand. If you don't have it, I just can't help you.
297. FreeToChoose - Feb. 27, 1999 - 4:50 PM PT
whhatlaw
I'm glad I read ahead before posting. I started to write that the 100 door example is often given as a way to explain it. But I see that it worked.
I went over this puzzle with my daughter at dinner. She was insistent that the probability was 50/50, even after I covered some of the standard ways to explain it. But in the course of the discussion, I came up with a new way to explain it. I'll try to reduce it to writing soon.
298. BobaFett - Feb. 27, 1999 - 4:54 PM PT
I LOVE THIS QUESTION!
I mean, I hate it but I love it.
I've heard both sides of this, wrestled with it, and have come to the conclusion that FreetoChoose is (sorry big guy) all wet. I mean, your position is, at least.
299. FreeToChoose - Feb. 27, 1999 - 4:56 PM PT
For puzzle fans (since there seems to be little philosophy happening at the moment), here's another classic. Also mentioned in Marilyn's column recently.
A family moves into town and a nosy neighbor starts asking questions:
Q Any Children?
A Yes
Q How Many?
A Two
Q What ages?
A 7 and 10
Q Boys or girls?
At this point the person decides to give only part of an answer
A At least one of my children is a girl.
With that as setup, what is the probability that the other child is a girl?
(If you have seen this problem before, in Marilyn's column or elsewhere, please give others a chance to answer.)
300. FreeToChoose - Feb. 27, 1999 - 4:57 PM PT
BobaFett
I'll bet 100,000 real American dollars and give you 10-1 odds. Fair enough?