18 Aug 2005 - 14:37

Doug Sundseth on the Monty Hall problem

Doug Sundseth (welcome, Doug!) posted an explanation of the Monty Hall problem, which I have never been able to understand:

It took me a long time to understand it, too.

The model that finally worked for me was something like this:

You have a 1/3 chance of being right to start with, and a 2/3 chance of being wrong. If you guessed wrong originally, Monte's pick will unambiguously determine the correct choice (he never picks the good door).

There are nine pairs of (your pick):(correct pick), A:A, A:B, A:C, B:A, B:B, B:C, C:A, C:B, and C:C. In three of those, you picked correctly, Monte's information isn't useful, and you shouldn't switch. In the other six, you picked incorrectly and Monte told you which of the other picks was correct; thus you should switch.

If you never switch, you have three chances in nine of being correct. If you always switch, you have six chances in nine of being correct and three chances in nine of switching off the correct choice.

Note that the latter possibility (choosing correctly at random then switching to an incorrect choice) may be more psychologically painful than just guessing wrong and not switching. This may have an undue effect on the choices of contestants.

Doug, thank you!

OK, I've just sat down and quickly thought this through.

[pause]

On my initial reading, I think it makes sense to me. What's particularly useful, for me, is the information that, yes, you could already have chosen the correct door, in which case, if you change your choice, you have moved to the incorrect door.

I think people who haven't studied probability get hung up on the 'what if I'm already right' issue.....and then, when math-savvy people try to explain Monty Hall without addressing, as Doug has, the issue foremost in their minds, the explanation doesn't 'take.'

metacognition again

I mentioned awhile back that metacognition is a huge issue amongst constructivists, both of the radical & the peer-reviewed , department of psychology cognitive science constructivists.

One of the main reasons for thinking about metacognition as you teach is that students may very well bring quite wrong ideas to the classroom, which they then 'build upon' as they acquire new knowledge. There's a lovely example of this in the National Research Council's book on learning. Many children, when told that the earth is round, picture it as a disk, not a sphere. (more t/k--I need to go take a look at these pages.)

In any case, Doug has addressed an aspect of metacognition that I haven't seen mentioned, which is to tell a student what it is they already know that's right, but incomplete.

I was having the same experience yesterday, puzzling through the 'false positives' problem. The objection both Ed and I were bumping our heads against--if it's 1 in 1000 and 50 in 1000, how can you ever have 1000???--was right; we just weren't seeing what to do about it.

I wonder how often it's the case that an incomplete right answer is the problem, as opposed to a Total Crackpot Misconception that has to be stomped out, obliterated, and disappeared without a trace before a person can learn Thing One about math? (And does this wording give you a feel for the challenge involved in attempting to re-learn elementary math in midlife?)

Or, as Steve H says, A little knowledge goes way too far.

a new question

This sentence confuses me:

You have a 1/3 chance of being right to start with, and a 2/3 chance of being wrong. If you guessed wrong originally, Monte's pick will unambiguously determine the correct choice (he never picks the good door).

[pause]

hmmm. Interesting. Reading this again, it makes sense.

I'm going to take a paper and pencil break, and see what I come up with working through Doug's explanation myself.

online Monty Hall simulation!

I love it!

Monty Hall dilemma

back again

OK, paper and pencil session complete.

I do understand this explanation, with one question: the funky, counterintuitive odds are created by the fact that Monty always opens the wrong door, correct?

That's why you shouldn't go with the 50-50 answer everyone automatically does go with--yes?

Carolyn was telling me the other day that a lot of Bayesian statistical results are counterintuitve (hey! just like the Bayesian proof of the existence of God!).

That's for sure.

other explanations

Here's a strictly mathematical explanation that will work for some people (and actually works OK for me....although frankly Doug's list helps move me a bit towards 'getting' the Monty Hall problem at a more intuitive level...):

After you pick but before you open any doors, there's a 1/3 chance that you've picked correctly, and a 2/3 chance that you've picked wrong. Assuming that the host can open doors, but can not move prizes, nothing that the host does will change the probabilities described above.

Now the host opens one of the doors, and there's nothing behind it. There's still a 1/3 chance that you've picked correctly, and a 2/3 chance that you've picked wrong. This means that the remaining door has a 2/3 chance of being correct.

This explanation helps me formulate exactly what it is that goes wrong for people: the chance to change your pick seems like a second event, with a second set of probabilities attached.

question: So how often does this happen in life?

How often do we perceive second events where we ought to perceive a continuation of the first?

update: an intuitive approach to Monty Hall that might work

I'm going to have to live with the Monty Hall problem for awhile....

But here's an interesting approach to rendering the answer intuitively correct:

It was a while ago that I accepted the idea that switching doors was the correct play every time because it improves your chances of winning, but I had trouble convincing my friends that it was the correct answer. However, a friend of mine just came up with this explanation that I think should really make it obvious.

Let's say that you choose your door (out of 3, of course). Then, without showing what's behind any of the doors, Monty says you can stick with your first choice or you can have both of the two other doors. I think most everyone would then take the two doors collectively.

Unfortunately, I don't think this works for me...

update: Keith Devlin's better version

OK, I think what the person above was trying to say was this:

...one last attempt at an explanation. Back to the three door version now. When Monty has opened one of the three doors and shown you there is no prize behind, and then offers you the opportunity to switch, he is in effect offering you a TWO-FOR-ONE switch. You originally picked door A. He is now saying "Would you like to swap door A for TWO doors, B and C ... Oh, and by the way, before you make this two-for-one swap I'll open one of those two doors for you (one without a prize behind it)."

I agree. Anyone told at the outset that he can pick one door or he can pick two doors would pick the two.

I give up

from Keith Devlin:

... suppose you are playing a seven door version of the game. You choose three doors. Monty now opens three of the remaining doors to show you that there is no prize behind it. He then says, "Would you like to stick with the three doors you have chosen, or would you prefer to swap them for the one other door I have not opened?" What do you do? Do you stick with your three doors or do you make the 3 for 1 swap he is offering?

OK, I'm switching doors.

But I'm doing so purely on the basis of 4/7 being greater than 3/7. Nothing common sense about it.

Of course, given that my family motto is no common sense-y, it's easy to dump my first pick and jump to Door Number Seven!


low birth weight paradox (& Monty Hall)
Monty Hall, part 2
Monty Hall, part 3
false positives
false positives, part 2
Doug Sundseth on Monty Hall
John Kay: We are likely to get probability wrong (subscription only)
Monty Hall diagram from Curious Incident
probability question from Saxon 8/7

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the intuition pump that works best for me:
think about what happens with a lot more doors.
or put it this way:
i deal you a down card from a standard deck.
you win $1 if it's the ace of spades.
i go through the rest of the deck and show you
50 up cards -- none of 'em is the A of S.
now i offer you a choice:
keep the card i originally dealt you
or switch to the one remaining card.
isn't it obvious that you should switch?

-- KtmGuest - 18 Aug 2005

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There's another nice explanation of Monty Hall in the book The Curious Incident of the Dog in the Nighttime by Mark Haddon. It's a really interesting book, very well written in the voice of an autistic boy, and it contains some fun math. I'd recommend it to anyone, but especially to those interested in autism/Asperger's.

-- GretaFrohbieter - 18 Aug 2005

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When explaining this to people, I always admit that they are thinking about it correctly. There really are only two choices--but those two choices are not one of two doors.

The two choices in this problem are your original door and the two other doors AS A GROUP.

You can stick with your door, or you can pick the other two as a group. If you switch, only one of the two doors has to be correct for you to win, but you get only one shot with your original door.

-- JdFisher - 18 Aug 2005

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Back to your metacognition theme:

A standard theme for comics is statistical reports of average (mean or median) family sizes. The joke usually takes the form, "The average family has 2.1 children. What, do they have two regular children and one that's really short?"

I think a subconcious variant of this might have been behind your quest for whole-number answers, and I think it's even more likely that such would be the case for kids. Understanding this sort of statistical abstraction about concrete objects that kids have extensive experience with is (I suspect) a stumbling block to understanding.

To get past this, I'd recommend dice. 8-)

I'm actually serious about this, not least because it is largely games that have caused me to be interested in probability theory. To be more specific, it's largely games with the worst press (Dungeons and Dragons, Magic: the Gathering, Poker ...) that have caused me to be interested in probability theory. D&D character generation will teach a pragmatic understanding of bell-curve distributions that even rather young children can understand.

And if you can generate a general solution to this problem:

What are the mean, median, mode, and standard deviation of a sum of N dice, each with M sides, discarding the lowest L dice before summing,

I'd appreciate it.

8-)

-- DougSundseth - 18 Aug 2005

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Title:	Doug Sundseth on the Monty Hall problem
TopicType:	WebLog
SubjectArea:	MathProblemHelpLine, StatisticsAbuse, StatisticsTeaching
LogDate:	200508180955