Monty Hall problem - What do you think?

[matthatter]

I am pasting the below from Wikipedia:

The Monty Hall problem is a probability puzzle based on the American television game show Let’s Make a Deal. The name comes from the show’s host, Monty Hall. The problem is also called the Monty Hall paradox, as it is a veridical paradox in that the result appears absurd but is demonstrably true.

The problem can be unambiguously stated as follows:Suppose you’re on a game show and you’re given the choice of three doors. Behind one door is a car; behind the others, goats [that is, booby prizes]. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you “Do you want to switch to Door Number 2?” Is it to your advantage to change your choice? (Krauss and Wang 2003:10)

As the player cannot be certain which of the two remaining unopened doors is the winning door, most people assume that each of these doors has an equal probability and conclude that switching does not matter. In fact, the player should switch—doing so doubles the probability of winning the car from 1/3 to 2/3.

I don’t get how people actually manage to convince themselves the answer is 2/3.

The question concerns whether it is to your advantage to switch from one box to another. That situation only involves two boxes, one with a goat in it and one without. A third box is irrelevant.

“Yeah well before there were three at the beginning and your initial choice had a probability of 1/3 which was 2/3 wrong; those other two boxes you did not pick had a total probability of 2/3, so it is in your advantage to choose that one”

Only if you think human probability somehow magically alters future situations in order to preserve the probability of now non existing situations.

“We are confident that the new cell phone design will be well received. Statistics show that blah blah blah”

“Makes sense. What are your sources?”

“A, B, C, D and E”

“What about previous statistics? Did you take those into account?”

“I’m sorry, what previous statistics?”

“You know, past chances of making a huge product on new cell phones”

“Oh yes of course we took into account the iphone, blackberries, and also an upcoming release that is combination of touch screen technology and–”

“Yes yes, cell phones that are in the market now, I understand that.”

“Well… what else is there?”

“All statistics concerning all market value for past cell phone releases”

“All of them?”

“All of them”

“All statistics concerning probabilities of success for cell phones nobody can even buy, in markets that don’t even apply to the one we’re now in?”

“Absolutely, all of them.”

“Well… if you take the probability of all the cell phones being successful in their respective situations I guess that gives a total chance of, I dunno, at least 10/1…”

“What does that mean?”

“It means 10 x absolute certainty that the sum of the past chances of past cell phones in their past markets would be a profit”

“That’s incredible! I wish I was in this business earlier, HAHA!”

“Well, I mean those statistics really mean anything… I am actually just calling success ‘1/1’ and then, for the real past money-makers, giving them values like ‘3/1’, because they made three times more than what would have been a success… you see I am judging their past probability based off what happened after the fact”

“But it’s statistics, right?”

“Well… I mean, I guess…”

“Great work! Is that all?”

“All what? I mean, I didn’t include most failures because I just don’t remember them.”

“Oh no no, I don’t care about cell phones, I care about the past statistics concerning them”

“Oh, yeah, that’s all”

“So how does this affect our chances?”

“Ohh… well since we want to make a profit, and our cell phone is not any of those cell phones, I guess our chances are 10 times absolute failure”

“THEN WHY ARE WE WORKING ON THEM?!”

“Sir I think you’ve mis–”

“STOP WASTING OUR MONEY!”

“But SIR, our current numbers tell us that we are very likely to succeed here! Not only would we lose an opportunity but all the money that we have already put into this would b–”

“What are our losses?”

“Well… let me see… about 8 times that which we needed to make in order to make a decent enough profit to label the project a success.”

“8? Are you sure it’s not 10?”

“Yes I’m sure”

“Which is why you’re fired! I need someone who knows his statistics”

[Only_Humean]

@matthatter

I am pasting the below from Wikipedia:

As the player cannot be certain which of the two remaining unopened doors is the winning door, most people assume that each of these doors has an equal probability and conclude that switching does not matter. In fact, the player should switch—doing so doubles the probability of winning the car from 1/3 to 2/3.

I don’t get how people actually manage to convince themselves the answer is 2/3.

… because it is. I do get how people fail to realise it, though.

“Yeah well before there were three at the beginning and your initial choice had a probability of 1/3 which was 2/3 wrong; those other two boxes you did not pick had a total probability of 2/3, so it is in your advantage to choose that one”

Only if you think human probability somehow magically alters future situations in order to preserve the probability of now non existing situations.

Not probability, knowledge. Monty himself knows for a fact which door contains the car, and will not open the door containing it. You have an additional system constraint; you can think of it as an additional determinate step, an injection of information.

Your 50/50 idea only holds true if he picks a box at random. Then it doesn’t matter if he filters out a third box before or after you make your selection.

Think of it with twenty boxes, one of which has the car. You’d agree you have a 5% chance of picking the right one at random? If Mr Hall then opens eighteen other boxes, leaving only one over, is there still a 50/50 chance that that one he didn’t pick out, knowing where the car is, is the lucky box?

[The_Last_Man]

I get that part, but the problem is this: after he has opened one of the three doors, and now offers you the choice to switch your guess, you know for a fact that the car is behind one of the two remaining doors, one of which you initially chose. At this point, it does not matter if you switch your guess, because whether you keep the original door or switch to the other one, you still have a 50% chance either way. . . what Im saying is that whether you switch or not does not affect the odds of your picking the door with the car.

In your example of 20 doors, this highlights the fact that you have narrowed the odds from 1/20 to 1/2. This means that you have a 50% chance of picking the door with the car behind it now that the choice is offered to you. But if you switch your guess to the other door that you did not initially pick, you have a 50% chance, AND if you keep your guess on the door you initially chose you still have a 50% chance. It doesnt change anything.

If you decide to stay with your initial choice, you are in effect picking that door out of two possibilities, thus giving you 50% odds. Likewise if you choose to switch, you are picking the second door of two doors, thus giving yourself 50% odds still. It is the same either way.

I think this “paradox” and the fact that people seem to think it matters whether you switch or not results from a misunderstanding of how probabilities work. Its like people who think that by recording all the past lottery powerball numbers you will eventually narrow down the odds of picking the correct one, because it is unlikely that the same combinations will come up more than once; but it doesnt matter at all even if you know every single past combination, because the selection of THIS combination of winning numbers is still random, thus, any such past information has no bearing whatsoever on which numbers are picked. Its a similar misconception there as it is for the Monty Hall.

[matty]

You’re focusing on the decision to switch rather than the original decision to pick one of the three doors. In the original decision you had a two-thirds chance of being wrong, and that’s the point. Even after the goat has been revealed it’s still a two-thirds chance you were wrong, so on balance of probability you’re better off switching (although of course very far from guaranteed to win in a one-off scenario).

Plus, it’s empirically verifiable (try it at home!):

[youtube]http://www.youtube.com/watch?v=axqT323GHVc[/youtube]

Also, the contrast with the lottery is a false one because there is no narrowing down of choices from one lottery draw to the next - next week is a new draw and the chances of it being any particular set of numbers remains the same.

[Only_Humean]

@The Last Man

I get that part, but the problem is this: after he has opened one of the three doors, and now offers you the choice to switch your guess, you know for a fact that the car is behind one of the two remaining doors, one of which you initially chose. At this point, it does not matter if you switch your guess, because whether you keep the original door or switch to the other one, you still have a 50% chance either way. . . what Im saying is that whether you switch or not does not affect the odds of your picking the door with the car.

Yes, it really does. It’s a practical problem, you can try it out with a friend. Use cards, or coins under shells or something; cars and goats make for a lot of work. It has a real effect in the real world.
upload.wikimedia.org/wikipedia/c … _carlo.svg

In your example of 20 doors, this highlights the fact that you have narrowed the odds from 1/20 to 1/2. This means that you have a 50% chance of picking the door with the car behind it now that the choice is offered to you. But if you switch your guess to the other door that you did not initially pick, you have a 50% chance, AND if you keep your guess on the door you initially chose you still have a 50% chance. It doesnt change anything.

No, for the reason I gave above; there is information in a sequential process. You haven’t narrowed any odds at all, you’re either ignoring information outright or treating non-random information as if it were random.

If you decide to stay with your initial choice, you are in effect picking that door out of two possibilities, thus giving you 50% odds. Likewise if you choose to switch, you are picking the second door of two doors, thus giving yourself 50% odds still. It is the same either way.

You’re ignoring information. Probability is a construct, a measure of confidence, not a Real Thing, although it has real outcomes that can be measured post hoc.

Take twenty people doing the 20-door problem. On average, 19 out of the 20 people will pick a door with a goat. The doors are opened, leaving everyone with a choice to switch. Those 19 people who’ve picked a goat profit by switching, the one who had the car all along makes a loss. There is no 50/50 involved, if you are one of those people you’re 19 times more likely to go goat → car than car-> goat.

I think this “paradox” and the fact that people seem to think it matters whether you switch or not results from a misunderstanding of how probabilities work. Its like people who think that by recording all the past lottery powerball numbers you will eventually narrow down the odds of picking the correct one, because it is unlikely that the same combinations will come up more than once; but it doesnt matter at all even if you know every single past combination, because the selection of THIS combination of winning numbers is still random, thus, any such past information has no bearing whatsoever on which numbers are picked. Its a similar misconception there as it is for the Monty Hall.

The misconception is yours. Lotteries and coin tosses are random information; there’s no asymmetric knowledge involved. You’ve actually fallen into a reverse-lottery fallacy, treating non-random information as if it were random rather than the other way around.

Seriously, try this yourself, it’s easy. Get a friend to spread out all the picture cards face down, let him check which one is the ace of spades without you knowing. Now you guess, he takes away all but one, and you switch choices. Do it 16 times, see if you’re right 50% of the time or 94% of the time. Do it 160 times, see if you get closer to 80 or 150 right.

[The_Last_Man]

Yes, the effectiveness lies in the elimination of choices which can ONLY BE WRONG. I see that now, thanks for the clarification. After my initial post here it got me thinking, and I ended up realising the importance of the fact that of those choices which are removed, only the incorrect ones will be selected, and furthermore that the initial choice which you have made cannot be selected for removal - so with the 20 doors example, a wrong choice 19 out of 20 times is locked in by your having picked it, and must survive through to the final two choices, after elimination of 18 other doors. Whereas the remaining unknown door, because all the other 18 doors are KNOWN to be INCORRECT (have a goat), therefore gains huge odds because either that new door or the original one you chose MUST have a car behind it, but we know that originally you only had a 5% chance of picking the door with the car. I actually came back to this thread just now to post all that, but I see that you have effectively explained the misconception already I admit though, that until I thought about it from the perspective of the probabilities behind the ELIMINATED CHOICES, I could not seem to understand it. It was tricky to grasp, but once that perspective is adopted its surprisingly obvious. . . strange how the mind works. Simple changing a frame of reference or a slightly new perspective can illuminate things in a whole new way. Amazing.

[Only_Humean]

@The Last Man

After my initial post here it got me thinking, and I ended up realising the importance of the fact that of those choices which are removed, only the incorrect ones will be selected, and furthermore that the initial choice which you have made cannot be selected for removal - so with the 20 doors example, a wrong choice 19 out of 20 times is locked in by your having picked it, and must survive through to the final two choices, after elimination of 18 other doors.

That “locking in” is a good way of clarifying it, I hadn’t considered it that way. Another insight.

You’re in good company, anyway - apparently even Paul Erdos got it wrong the first time he looked at it.

[Impenitent]

if you aren’t wearing a bad costume, you don’t get a choice

-Imp

[matthatter]

… because it is. I do get how people fail to realise it, though.

Man oh man, considering my OP I so badly wish I could’ve came up with some counter argument (and even after I had pretty much accepted the idea that it was definitely 2/3s, I was still hoping I could find some kind of verbal loophole).

But the simple fact that the first choice is either a goat, another goat or a car, for which switching is either car, (“another”) car or goat, has made me unable to do so.

I admit my error, as well as the overconfidence of my assertion.

apparently even Paul Erdos got it wrong the first time he looked at it.

That softens the landing a little.