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Brainteaser

11:38 / 26.05.02

Ok, this is an old one. But I like it....

You are playing a quiz show and you have to choose your prize. Thing is, there are three closed doors and your prize is behind one of them. Behind the other two are goats (this topic is not for goat "lovers"). You then choose a door.

Your irritating gameshow grins broadly and tells you that, since you've been a good sport, he'll help you. He points to one of the doors you didn't pick and tells you that there is definitely a goat behind that one. So the prize is either behind the door you picked or the remaining door. Here comes the crunch - you get to decide again.

Should you stick with your initial choice, or change your mind to that third door (ie the one you didn't pick and the host didn't indicate). Or does it make no difference?

Elijah, Freelance Rabbi

11:53 / 26.05.02

this reminds me of the "1 of us always lies, the other always tells the truth" there really isnt an easy way to go solve it.

Theoretically, if he chos door 3 as the goat door, and you have picked door 2, than whats happening is he is trying to influence your thinking, since he wouldnt tell you the door you picked was wrong, he picks the pther wrong one. However if both remaining doors are wrong, he may have just picked the one closest to him.

I would hit him over the head and make him tell me which door had my new car.

12:11 / 26.05.02

ah, the Monty Hall dilemma... it's very counter-intuitive, this one. I find the only way I can actually believe the result is if I draw out a decision tree of some sort.

12:18 / 26.05.02

Oh hey, this is simple.

You grab the host's microphone, beat him over the head with it until he's unconcious, and take the stuff behind all three doors.

12:25 / 26.05.02

Thats right, fridgemagnet. The first time I heard it, it took me a good half hour to believe the result. Even when I drew the tree it seemed wrong. I think the counter intuitive nature of the problem may offer some insight into how we think. But it might be little more than the fact that we are crap at assessing risk.

Solitaire Rose as Tom Servo

12:40 / 26.05.02

You should switch. And Car Talk spent a good month on this puzzler.

12:44 / 26.05.02

ok, no one's said this explicately yet... yes it is counter intuitive. the answer is that you always CHANGE YOUR GUESS in this situation. you would think that the two remaining choices have equal 50/50 probability -- but this is not the case.

when the contest begins, each door has a 33.3% chance of being correct. this is unrefutable. so you choose door A. without question, there is a 33.3% chance that your prize is behind that door. this much is certain. and NOTHING CAN CHANGE THAT.

so when monty hall shows you that behind door B is a goat, what does this mean?

as we said, door A has a 33.3.% chance. And nothing that doesn't directly change its contents affects this. DOOR A STILL at this point has a 33.3.% chance of having your prize. And door B, it has been revealed, has a 0% chance. This leaves the braintwisting and unrefutable truth that DOOR C has a 66.6% chance of having your prize. So you switch your guess.

It does hurt the brain. because we have the impulse to think -- but what if i chose door c from the beginning. yes, what indeed.

i would like to point out before anyone jumps to conclusions that although it seems similar, this problem has nothing to do with probability vectors collapsing from perception. that is, this is strict classical probability without anything quantum about it.

19:59 / 26.05.02

From the "things Spiro Agnew could've said, but didn't" file:

"Don't move your choice over to the other door. Stay with the origional choice. Why? Because obstinance will get you everywhere in life."

00:40 / 27.05.02

when the contest begins, each door has a 33.3% chance of being correct. this is unrefutable. so you choose door A. without question, there is a 33.3% chance that your prize is behind that door. this much is certain. and NOTHING CAN CHANGE THAT.

so when monty hall shows you that behind door B is a goat, what does this mean?

as we said, door A has a 33.3.% chance. And nothing that doesn't directly change its contents affects this. DOOR A STILL at this point has a 33.3.% chance of having your prize. And door B, it has been revealed, has a 0% chance. This leaves the braintwisting and unrefutable truth that DOOR C has a 66.6% chance of having your prize. So you switch your guess.

I don't buy it, Gypt. Once Door B is proven to be prizeless (or goat-full, if you're an optimist), Door A and Door C each assume half of Door B's 33.3% chance of winning.

This is a slightly more confusing variation on the old puzzle that asks, "If you flip an ordinary quarter nine times and get nine heads in a row, what are the odds of it coming up heads again?" The answer: fifty-fifty. Why? Because the previous nine flips don't have any impact on the tenth flip.

Likewise, Door B has no impact on the odds of wheter the prize is more likely to be behind Door A than Door C. Whether the game show host shows you what is behind Door B or not, the odds of the prize being behind Door A or Door C remain equally probable: 50% for both doors the first situation, 33.3% for both doors in the latter.

Switch doors or not; it makes no difference to your bookie. It's all in the psychology, not in the math.

Solitaire Rose as Tom Servo

01:06 / 27.05.02

Be enlightened:

http://cartalk.cars.com/About/Monty/proof.html

01:37 / 27.05.02

Duly noted, and still utterly unconvinced. If I'm wrong, it'll take a more articulate argument than anything on CarTalk to convince me.

The central fallacy of this argument is that the odds of Door A hiding the prize remain constant from the original round of choices, but that Door C's odds change.

I'm going to test my hypothesis; I expect the results will back me up and will report thusly if they do not. I will stick a dollar bill in one of three books at random with my eyes closed. Next, I'll select one of the three at random. Then, I'll pretend I'm Monty Hall and take one of the two remaining books and open it up. If there's not a dollar bill in it, then I will switch to the final book. (If there is a dollar in the book I picked as Monty, then I will discount that game on the grounds that I was a lousy Monty Hall and ruined the experiment.) After an extended run, I expect to find that I have won roughly fifty percent of the time, not sixty-six percent of the time.

Any complaints about this test? Any wagers on what happens?

01:57 / 27.05.02

More a matter of the wording of the question.
If you disassociate the second choice from the first it is 50/50, but the question means the second choice includes the parameters of the first.
Shitty maths word questions, always meant to confuse.

03:11 / 27.05.02

You know, for a minute there, I thought I might actually prove myself wrong. Silly me.

Okay, according to the parameters given, if I follow Gypt's "switch to the other door" gameplan, he/she says I should win 66% of the time, while I say I'll win a mere 50% of the time.

Shuffling the Elric books now...

Here's the results of the first ten times; "L" is for a loss, "W" is for a win, "X" is for an invalid game wherein Monty accidentally opens the door hiding the prize: L-X-L-L-L-X-W-L-L-L. Gee, one win out of eight valid games, for a 12.5% success rate. Let's keep playing.

Second set of ten: W-L-X-W-X-X-X-W-W-L. Four out of six, for 66.6%. Of course, that's only for this set; overall, my sucess rate is only a measly 35.7%.

Third set of ten: L-X-L-L-W-L-X-L-X-X. One out of six. 16.6% success for this set, 30% overall.

Fourth set of ten: L-W-W-X-X-X-X-X-X-X. Damn, that's a lot of invalids in a row. Interesting statistical anomaly. Two out of three, for 66.6% for this set, 34.7% overall.

Fifth set of ten: X-L-W-X-X-X-X-L-W-X. Two out of four. 50% for this set, 37% success overall.

Okay, so we finished well below even my meager prediction, and far below Gypt's enthusiastic claim. What does this prove? First of all, a bad run of luck early on will take a while to overcome. Second of all, this is an even contest, since you can clearly see that the overall rating is approaching equilibrium at 50%, though not through a smooth progression.

I'm sorry to disillusion anyone who buys into this nonsense, but some people know just enough math to pull the wool over your own eyes. Reality doesn't back up these mathematically spurious claims about Door C inheriting all of Door B's odds while Door A gets none. Feel free to insist that my experiment wasn't scientific or that I'm lying about the results; I'll be reading the gullibility thread and chuckling.

04:51 / 27.05.02

you test doesn't work because we're arguing logic. if i sat here and told you i just flipped a coin 5 times and got heads five times it wouldn't refute anyone, either.

i think the theoretical main fallacy in your experiment is that you discount when monty makes a bad call. he knows the answer 100% of the time, so yr throwing away 2/3 of your statistics. in real life, yuo're just not doing it enough.

what you're arguing is the intuitive reasoning, that which we all know -- that flipping 9 times a head does not affect the 10th. and that's true. and it is by that logic that this puzzle is true. try to aruge against the logic rather than through dodgy experimental data, and we'll see. or else run your experiment a hundred times with a real monty hall, but still. the point is to help us understand the maths, not the experimental method, right?

05:16 / 27.05.02

Right.

My fault for being educated-stupid.

Monty Hall disproves 1-day earth rotation.

Timecube proves what is behind Door B: the absolute unrefutable proof of 4 days simultaneously in a single rotation of the Cubic Earth.

16:37 / 28.05.02

I'll try to explain this as best I can. Your first choice was made with three doors, giving you a 1/3 chance of being right. Then, you were shown that Door B was an incorrect choice. If you switch, you have made a decision based on two doors, and have a 1/2 chance of being correct. Revealing the contents of Door B allows you to make a more accurate decision.

We actually spent a good portion of a week on this problem back in my Modeling and Simulation class. One of our homework assignments involved writing a computer program to test this problem. Switching is the more accurate method.

Feel free to insist that my experiment wasn't scientific or that I'm lying about the results; I'll be reading the gullibility thread and chuckling.

I think you need a larger sample size, and a more random method of shuffling the dollar around. Using CD cases, and having someone else shuffle the dollar while you were outside of the room would have made for a more accurate experiment. With these changes, I think you'll find that the final probability is 1/2, and not 1/3 or 2/3.

16:50 / 28.05.02

so we've got a couple different fully insisted-upon answers here... anyone have a reference to a definitive mathematics or statistics article that might settle it? the 2/3 vs 1/2 argument is particularly bothering me.

17:22 / 28.05.02

errrmm. just in case it wasn't clear, IMO (and every other mathematician I know) you should switch. There is a 2/3 chance of winning if you switch. Mystery Gypt has explained it well and you should trust hir. I'm not sure that you are going to find a better explanation out there.

Isn't it great to argue about stuff like this? OK, maybe not that great...

18:53 / 28.05.02

Lurid is a math graduate student, so I'll defer to his judgement. I was fairly certain it was a 1/2 chance though.

pointless and uncalled for

19:09 / 28.05.02

The fairly simple answer to this. Before going on this kind of game show you should take an class in identifying tells.

Then during the selection procedure you will be able to tell if the Montyesque one has any knowledge of which door the prize is behind and if he is trying to help or hinder.

If he doesn't then you'll probably have more luck if you toss a coin to see whether or not you should change your decision.

19:54 / 28.05.02

This caused a big flap in Maria Vos Savant's games column. She originally got the answer wrong, despite an IQ of 200+.

The correct answer is: if you've got two doors, and the prize is behind one of them, then you've got a 1 in 2 chance of picking the right door. It is not affected by how many doors were involved in previous or subsequent games.

Informal Proof #1:

Originally, we have three doors, one of which has the correct prize behind it: Door A, Door B, and Door C.

Each has one chance of being the correct door: Door A: 1 chance, Door B: 1 chance, Door C: 1 chance. 1+1+1=3 total chances. Each door has a 1 in 3 chance.

If you eliminate one door, one of those chances goes away. We have two remaining doors to choose from, say Door B: 1 chance, and Door C: 1 chance. 1+1=2 chances. In the second round, we've got 1+1=2 total chances, for a 1 in 2 chance. It's a smaller pool, a smaller denominator.

Only if you got to pick two of the doors from the start would we have a 2 in 3 chance, because we're getting 2 choices from a pool of 3.

Informal Proof #2:
Still don't believe it? Start with two doors, A and B. The prize is behind one of these two doors. Now, add all other doors in the universe. You may look behind any door you wish, but you still know the prize is only behind A or B. Now: your chance of finding the prize behind door A? 1 in 2. Your chance of finding the prize behind door B: 1 in 2. Your chance of finding the prize behind any other door: zero.

.

19:54 / 28.05.02

Sorry, maybe I'm being stupid or something, but I just don't get it... When it is revealed that there is a goat behind door B, the probability of there being a goat behind A is 50%... There are two doors, a goat is behind one, and not behind the other. Hence a 50/50 split... Surely? Surely? Why not?

.

19:56 / 28.05.02

Saw Logos' post after I replied, and I have to say, it makes sense like that to me.

20:59 / 28.05.02

Thiazi, I'm just a smidge older than that, but you shouldn't believe me just because I'm a maths nerd and you shouldn't defer to my judgement. No matter how many letters come after my name. Really, the way to do it is to follow fridgemagnet's suggestion. Draw yourself a tree. Or...

If you chose the prize to begin with then you will always lose by switching. If you chose a goat to begin with then you will always win by switching. So the key question, if you go for the switching strategy, is whether or not you chose the prize or the goat to begin with. You choose the prize 1 in 3 times and choose the goat 2 in 3 times.

Its another way to look at it, almost. I liked Mystery Gypt's explanation, though.

But I'm not sure that explanation would have convinced me at the start. It is so excellently counter intuitive.

23:26 / 28.05.02

thanks lurid! ... it does seem like something is missing in these explanations though, something that would make it seem obvious and necessary even to people that aren't inclined to believe it. isn't that what a good proof should be?

you're explanation is totally clear, a completely different way of thinking about it for me also... probably will be for anyone else, too, since my description does make it appear as though i'm suggesting the probablity changes, which i'm not. cheers for this thread --

a question i have is how can we use the truth of this counter intuitive puzzle to affect decision making / game playing policy in other contexts? what are other ways of envisioning the goat / prize / door / monty hall symbol in ways that have to do with, say, politics, negotiation, or personal decisions?

16:58 / 29.05.02

Thiazi, I'm just a smidge older than that, but you shouldn't believe me just because I'm a maths nerd and you shouldn't defer to my judgement.

I know, but it's easier to do that than to rewrite the computer program we made back in my class from scratch.

22:01 / 29.05.02

Either way, you still get a door right?

22:03 / 29.05.02

I'm so sorry, I couldn't resist.
I too, will go for the 50/50 theory, there being only 2 doors to contend with.

23:07 / 29.05.02

Lurid Says:

If you chose the prize to begin with then you will always lose by switching. If you chose a goat to begin with then you will always win by switching. So the key question, if you go for the switching strategy, is whether or not you chose the prize or the goat to begin with. You choose the prize 1 in 3 times and choose the goat 2 in 3 times.

Endquote.

I'm sorry, Lurid. You've lost me there.

As I understood the initial problem, you start by picking a door. The door isn't opened, however; instead, the host removes one of the other doors. We still have a choice, but now it's down to two doors.

Schematically (where x=unknown t=prize f=goat P=probability, P=1=100% chance):

X X X
1/3 1/3 1/3=1=P

Host removes a door, proving it false.

F X X
0 1/3 1/3=2/3=P.

Now, correct P to 1 by multiplying everything by 3/2.

F X X
0 1/2 1/2=1=P

Yes?

A

06:40 / 30.05.02

I've been caught out by this one before. It seems so obvious that it would be a 50/50 chance, because there are only two doors, right?

BUT, the fact that the game show host knows where the prize is changes things.

Look at it this way:

First- you make your choice. Clearly, there is a 1/3 chance of the prize being behind your door, and a 2/3 chance of it being behind another door.

Now, the host removes one of the other doors, but because he KNOWS where everything is, he removes one with a goat behind it, not a prize.

The odds don't change. There is still a 1/3 chance of the prize being behind your door, and a 2/3 chance of it being behind another door, it's just that there is only one other door now.

I know it seems odd, but i'm positive that that is how it works.

08:37 / 30.05.02

Looking at it from Monty's perspective, in a sense the contestant can only do one of TWO things (even though there are three doors):

1) Pick the prize door. Monty can open whatever other door he wants to reveal a goat. (1 in 3 chance, 'cause there's only 1 prize door.)

2) Pick a goat door. There is only one door that Monty can open without showing where the prize is. (2 in 3 chance, 'cause there are 2 goat doors.)

In case 1, switching will get you a goat. But case 1 only happens 1/3 of the time.

In case 2, Monty "tells" you where the prize is--it's whatever he didn't open. Case 2 happens 2/3 of the time.

Therefore, switching gets you the prize 2/3 of the time.

(Hey, this is the first time I've *really* believed the answer to this one--insomnia rules!)

.

08:54 / 30.05.02

Ok, I'm almost being convinced now... Thanks to Count Adam and Lurid... but not quite... Let me see if I've got this right- the trick is that one door is removed, but the probabilities don't reset to 50/50 because the one door removed is always a wrong one. But in my mind it still looks like this:

1/3 unknown goat 1/3 unknown goat 1/3 unknown prize

Then one wrong door removed:

1/3 known goat 1/3 unknown goat 1/3 unknown prize

The odds on the unknown doors are still split 1/3 1/3 , or 50/50.

We're talking about two things hidden behind two doors. Epistemically I don't see how knowing where one goat is can possibly help you know where the other is.