Understanding the Monty Hall Problem

The Monty Hall problem is a counter-intuitive statistics puzzle:

This is a companion discussion topic for the original entry at http://betterexplained.com/articles/understanding-the-monty-hall-problem/
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You could explain it like this too:

If you stay with the door you picked initially you succeed if the initial door has a car, which has a chance of 1/3. If you’re strategy is to switch then you succeed if your initial pick is a goat, which has a chance of 2/3.

Welcome back!

I actually blogged about this a while back:

And one of my uncles was dead-sure that I was wrong about it, so I wrote up an application (source code included) to show the percentage differences between switching and not switching. It’s linked to from that blag post above.

I like your flash version though!

I never understood this before, but as I read your explanations, the following came to mind:

The doors can be divided into two groups; the one you picked, and the ones you didn’t pick. When you pick one out of 100 doors, chances are very high that the prize is in the group you didn’t pick, you just don’t know which of those 99 doors has the car.

Luckily for you, Monty narrows it down for you by opening all but one door. The odds of the selected door having the prize hasn’t changed at all. You just know more a lot about the doors you didn’t pick.

Great article - I especially like the visualization part with the green clouds. Sometimes I think the hardest part with mathematical concepts is being able to visualize them.

@uwe: Nice, I like seeing the win and lose probabilities next to each other like that!

@Aaron: Cool, I like the automated way to go through hundreds of trials :).

@Gary: Yes, that’s exactly it. I need to think of a good concise way to get that point across – something along the line of “getting rid of the weeds in the neighbor’s garden”, i.e. improving the choices in the items you didn’t pick. Great observation.

@CL: Thanks, and I totally agree. I often end up with some mental picture of what’s happening when I think about math, I wish we’d all share what we “really” think about when solving a problem!

[…] Understanding the Monty Hall Problem | BetterExplained betterexplained.com/articles/understanding-the-monty-hall-problem – view page – cached The Monty Hall problem:http://en.wikipedia.org/wiki/Monty_Hall_problem is counter-intuitive statistics puzzle: * There are 3 doors, behind which are two — From the page […]

Yes, great post. After I had time to think about this problem a bit, I thought about it similar to Aaron/uwe. But I use slightly different wording: 1st, I pick a door (1 out of 3). Then, I’m given the opportunity to either 1) keep my pick or 2) choose the other 2 doors (i.e., knowing that Monty would take one away for me later so to speak). Thanks for the articles, they are generally my favorite in the blogosphere.

Another great post. When I first encountered this problem in college it nearly drove me crazy for the first couple of minutes until my lecturer simply said… imagine it’s 1 million doors. It immediately became clear.
I like to use the Monty Hall problem when watching ‘Who wants to be a Millionaire?’ When a question is asked and I don’t know the answer, I pick one at random, and hope the actual contestant goes 50/50.
If they do, and my randomly chosen answer is still available, the correct answer more often that not is the other one.

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Funny, I came across this problem a couple of years ago and never intuitively understood it; I was all over the internet for a better explanation. Now, I just read the problem statement (not the entire reasoning) and I immediately understood (intuitively) that switching the door is the way to go, and I’m wondering why I got so confused back then.

The way I visualize this problem is with an adversarial scenario. First I imagine it’s 100 doors, then I think of it this way:

Imagine you’re playing AGAINST Monty Hall for a car. You each must pick a door. You go first. You are forced to pick your door at random.

Monty Hall, however, has the luxury of looking at every door other than yours, and not only can but MUST pick the car door if he finds it.

He has a wide grin on his face (99 out of 100 times) and is already celebrating his new acquisition when all of a sudden you are offered to trade places with Monty and get to pick his door! Do you switch then? I think so!

@lewikee: Awesome, I like that formulation! Monty is basically picking the best door he can out of the remaining choices. Very nice ;).

@ktr: Glad you liked it! Yes, I like thinking of it as “Pick 1 door or the best of the other 2”.

@Mike: Wow, that’s a really clever use of the paradox! I wasn’t sure if it had a ‘real-world’ use :).

Another way to think about it: If you pick a random answer (A B C D) you have a 3/4 choice of being wrong. If it’s still left after the 50-50 elimination, more often than not switching to the other choice will work out (since most of the time, your random guess is just there to be the wrong answer). Quite often your random guess will be eliminated, but if it remains it’s a good bet to go the other way. Neat!

@Srikanth: Cool, glad it clicked! Yes, sometimes it takes a second reading to have it snap into place.

Nicelly analysed and explained.

Ridiculously complicated. The probability the car is not behind ‘my’ door is 2/3. And now there’s only one door to pick. Change doors!

Interesting and crazy. I’ve read it twice and still don’t get it.

@3rojka: Thanks!

@Seldon: The 1-line explanation explains the “how” mechanics, but not the “why”. If Monty randomly revealed a door and it was a goat (i.e, he wasn’t trying to look at the other two doors and pick the best, he just randomly opened one for some reason), you’d still have an equal chance of being wrong and it wouldn’t give an advantage to switch. So, the secret is in the process Monty uses to reveal, not just the fact that he leaves you with 1 other door.

@geld: It is a tricky problem. You might try playing the game several times (in real life with coins under cups, 2 pennies [goats] and a quarter [car]). You’ll eventually see that your initial guess is right only 1/3 of the time.

I think this is overly wordy. What helped me is:

if you picked the door with a goat, the other doors have a car and a goat. Monty then CAN’T OPEN the door with the car because that would ruin the game.
the probability that you picked the right door out of 3 is 1/3. so you have a 2/3 of having picked the WRONG door, at which juncture, Monty opens the door with the goat. that means the other door has a 2/3 chance, and your door has a 1/3 chance.

@thegnu: Thanks for the comment. Yep, the goal of the article isn’t to just understand why switching works. As you mention, it can be done in a paragraph.

The more interesting principle is seeing the role of the filter itself, so you can handle alternative scenarios also (like your buddy playing, Monty mixing the doors again, Monty giving you 2 sets of doors to pick from).

And from there, we can see that these filters (new information that impacts earlier choices) exist all over the place in real life, from spam filters to analyzing experimental evidence. Hope this helps!

I still don’t get it and disagree. You have no extra information? It’s not a more informed decision. Maybe I’m wrong but this seems exactly like the situation of taking past occurences into an equation that have absolute no relevance. It’s a NEW decision/outcome. If 4 red’s in a row come up on the roulette table it’s no more likely to be black than 50/50 (negating zero) than it ever was. Same situation here no?