21 and the Monty Hall Paradox
Monday, September 08, 2008   

Bringing Down the House by Ben Mezrich is, so far as I know, the only book which has ever succeeded in writing about the game of blackjack in an interesting way. 

No offense to blackjack authors/players, but blackjack suffers from the same problem that afflicts poker: it can be a lot of fun to play, but often not much fun to read about playing. As an old blackjack-pro-turned-poker-player once wrote:

Blackjack is a game of pure numbers and rote, algorithmic strategy, and the life of a professional blackjack player (as professional blackjack players will agree) can be an exceedingly dull grind. Why should anybody contend with huge variance relative to a measly 1-2% gain, hostile casino staff, and hours of never-ending boredom? Masochism? For this reason, I believe that inside every casino blackjack player is a poker player, waiting to get out. In poker, the edge is a fat 10%, 15%, 20% by some estimates. In poker, there is no hostile casino staff, only people who are glad you showed up to play. In poker, you can make more money in a year than many people will make in a decade, and you can do this even if you're not a world-class player. Compared to blackjack, the game of poker is like a breath of fresh air.

For this reason, I think Bringing Down the House is a work of real genius. It does the impossible: injects life back into the game of blackjack. Even if the story is a fraud, as a recent article in the Boston Globe questions: who cares. The story is what sells. And this story sold so well that it was turned into a major motion picture: 21.

21 tried to do for blackjack what Rounders did for poker. And okay, it failed. But from this forgettable jumble of poorly written, poorly acted scenes, we can extract one gem. It's the scene in which Kevin Spacey presents our hero with a puzzle:

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. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Welcome to the Monty Hall problem, which is admirably explained in layman's terms here, and demonstrated using Bayesian maths here.

The correct answer is: yes, by switching doors your chance of picking the "right door" goes from 33% to 66%, provided that the host knows which door contains the car. If the host doesn't know which door contains what and is just picking randomly, then the answer is no: your chances of picking the right door are 50-50.

And if that makes zero sense to you...welcome to the club. As Wikipedia notes:

When the problem and the solution appeared in Parade, approximately 10,000 readers, including nearly 1,000 with Ph.D.s, wrote to the magazine claiming the published solution was wrong.

This problem is still vigorously debated today. In Monty Hall Redux, Brian Hayes explains:

In the July-August issue of American Scientist I reviewed Paul J. Nahin's Digital Dice: Computational Solutions to Practical Probability Problems, which advocates computer simulation as an additional way of establishing truth in at least one domain, that of probability calculations. To introduce the theme, I revisited the famous Monty Hall affair of 1990, in which a number of mathematicians and other smart people took opposite sides in a dispute over probabilities in the television game show Let's Make a Deal. (The game-show situation is explained at the end of this essay.) When I chose this example, I thought the controversy had faded away years ago, and that I could focus on methodology rather than outcome. Adopting Nahin's approach, I wrote a simple computer simulation and got the results I expected, supporting the view that switching doors in the game yields a two-thirds chance of winning.

But the controversy is not over. To my surprise, several readers took issue with my conclusion. (You can read many of their comments in their entirety here.) For example, Bruce Sampsell of Chapel Hill, N.C., wrote: 

If you don't know who to believe, try an interactive version of the puzzle such as the one below by Shawn Olson:

I was just recently introduced to the Monty Hall Game paradox by my friend Andrew Penry. When he first proposed the game to me I thought it was simply absurd—and my intuitive thinking process would not allow me to accept the statistical conclusions that the game entails.

Sit down and play 50 or 100 games, and see whether you end up choosing the right door (or in this case, the right card) 66% of the time, give or take. Here's how my results looked after 60 games:

Shawn Olson's Monty Hall simulator

That's a highly unintuitive result from the perspective of anyone who's been trained to think that past results don't change the probability of independent future results. Flipping a coin and getting heads 10 times in a row doesn't change the probability of getting heads on the next flip. Your chances are 50-50, every single time. Similarly, just because the host has revealed 1 of the 3 doors doesn't change the fact that we're now presented with a choice between two doors. One door contains a car and one door contains a goat. Ergo, no matter which door you choose, and no matter what happened previously, your chances of picking the right door are now 50%.

That's rational, makes perfect sense, and yet, like many things in life which are rational and make perfect sense, it's 100% wrong.


Posted by James Devlin   79 comment(s)

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COMMENTS

Awesome that you chose this problem - it hurt my brain when I first stumbled over it!

Terry Smith on 9/8/2008 10:47:35 AM (73 days ago)

Lol.... biggest scam ever. THE CHANCES ARE 50 50. Listen, if you were an alien, and you interrupted the whole game show right at the point after the host guy opens one of the doors, you've got TWO DOORS.

TWO DOORS.

ONE DOOR has a GOAT.

ONE DOOR has a CAR.

FIFTY-FIFTY.

No matter how much verbal trickery you try to use, try as you might, you can't escape that. How about you try thinking critically about this instead of blindly following the herd on this one...

Anonymous on 9/8/2008 10:58:36 AM (73 days ago)

In rebuttal to
How about you try thinking critically about this instead of blindly following the herd on this one...

http://imgs.xkcd.com/comics/science.jpg

Anonymous on 9/8/2008 11:21:30 AM (73 days ago)

Hah. If you run the simulation, you'll see that the 66% chance of winning is unavoidably true. No amount of simple thinking can change that.

Anonymous on 9/8/2008 12:37:59 PM (73 days ago)

To the 50/50 person. How is this for a gross exaggeration to prove a point:

Imagine that there were a million doors. Also, after you have chosen your door; Monty opens all but one of the remaining doors, showing you that they are “losers.” It’s obvious that your first choice is wildly unlikely to have been right. And isn’t it obvious that of the other 999,999 doors that you didn’t choose, the one that he didn’t open is wildly likely to be the one with the prize... regardless of alien intervention?

Anonymous on 9/8/2008 12:58:38 PM (73 days ago)

monty.rb:

num_games = 10000 # Number of games to play
switch = true # Switch your guess?

wins = 0
num_games.times do
doors = [0, 0, 0] # Three doors!
doors[rand(3)] = 1 # One of them has a car!
guess = doors.delete_at(rand(3)) # We pick one of them!
doors.delete_at(doors[0] == 0 ? 0 : 1) # Take out one of the remaining doors that is not a car!
wins += switch ? doors[0] : guess
end

puts "You decided #{switch ? "" : "not "}to switch, and your win % is #{wins.to_f()/num_games}"

---

$ ruby monty.rb
You decided not to switch, and your win % is 0.3315
(edit code...)
$ ruby monty.rb
You decided to switch, and your win % is 0.6598

Ian on 9/8/2008 12:59:36 PM (73 days ago)

There's an assumption here that's not explicitly stated, and on which the whole chain of logic hinges: That Monty always opens a door after you've picked.

If you've seen this game played many times, and he always opens a door with a goat, then the logic holds. If you haven't, then you don't know. Maybe he opens a door sometimes, but not others. If he opens the door more often when you pick the door with the car, he could make the odds be whatever he wants.

In the original game show, Monty Hall did not always open a door.

Ian on 9/8/2008 1:18:41 PM (73 days ago)

Seems perfectly sensible now that I think about it.
First choosing a goat is 2/3, whichever of these two you choose, the computer / host is going to choose the other one. After that the only one left is the car.
Only if you hit the car on first try will changing your door get you end up with a goat.

Nice problem still

shiona on 9/8/2008 2:39:24 PM (73 days ago)

Good to see some more regular updates!

Keep it up ;)

Poker Forums on 9/8/2008 4:12:52 PM (73 days ago)

Doors ABC. PInnocent = 1/3. P(~A) = 2/3. Host has knowledge so those probabilities don't change. A or ~A yields true is a general axiom. After a door is removed, you get a chance to pick ~A instead of A. You should jump at this chance.

Jordan on 9/8/2008 5:10:36 PM (73 days ago)

lol, P( A ), not the angel thingy.

Jordan on 9/8/2008 5:11:12 PM (73 days ago)

"Past results don't change the probability of independent future results."

That's true. The problem is the future result isn't independent. Your choice of door influences Monty Hall's choice, which is 50/50 if you chose the car and deterministic the other times. The deterministic behaviour (which occurs 2/3 of the time) reveals the exact location of the car (aka...the other door). The alien intervening also knows that the first puny human influenced the second puny human's decision-making. Furthermore the alien can see through solid objects, and therefore chooses to stay, because aliens have advanced beyond cars but find goats to be delicious. Almost as good as puny humans.

Basically, it boils down to "chances are you're wrong on the first guess". That's how my brain settled upon it, and it's so obvious now. The version with 1 million doors is also helpful.

@Ian: usually the assumption is explicitly stated in the problem definition, but you're right; it isn't here. Therefore the proper answer to the question is "unable to determine probability". You don't have enough information to know whether Monty Hall is a trickster who only opens doors when you're right the first time (making switching a losing proposition 100% of the time); or maybe he's a fair trickster who opens a door even if you're wrong the first time, but only 2/3 of the time, thus balancing the equation back to 50-50.

I think we can say that, without that information, the odds of winning after switching are somewhere between 0% and 66.6666...% (but not necessarily evenly distributed within that range). It becomes more of a Poker game (but in the long run and assuming no psychic powers, the winning strategy for Monty Hall -- assuming he does not want to give up cars -- is to only open doors when the player selected the car).

Ens on 9/8/2008 5:44:29 PM (73 days ago)

I always thought a simple way of explaining it was "Would you trade your one door for the better of the prizes behind the other two doors?" If that was the question, everyone would trade up. And that's essentially what you are doing, it's just that Monty has opened one of the other two doors - the one with the goat - to make it obvious which of the other two doors is the right one.

David Avraamides on 9/8/2008 6:08:28 PM (73 days ago)

To the skeptic who says "No matter how much verbal trickery you try to use, try as you might, you can't escape that. How about you try thinking critically about this instead of blindly following the herd on this one..." - I just love playing poker with people like you. I learnt to figure odds when I was a kid from my dad, who was a greatly successful bettor of horses and player of card games. A few years ago I told him about the Monty Hall problem, and he got the right answer straight away. He was nearly 90 at the time, but still sharp as a knife.

I once explained the problem to two friends, both of whom are pretty smart. One got it, the other didn't, and he became increasingly indignant about it, insisting that the odds for the two doors are 50:50.

Originally you choose one of three doors which have equally randomised probabilities. You have on average a 1/3 chance of being right. Consequently, the chance that you are wrong is 2/3. If you stick with that original choice, then nothing that happens afterwards changes those probabilities. In other words, the chance of the winning door being ONE OF THE OTHER TWO is 2/3 - regardless of what happens next.

When Monty Hall shows you a losing door from these other two, he is helping you enormously. He is effectively saying, "Look, the chance of the prize being behind one of these two doors (of the three) is 2/3, and I'm going to help you by showing you one of them which ISN'T the winner." After he has done this, the probability of your original door being the right one doesn't suddenly jump to 50%, when it was only 33.3% beforehand. Why on earth should it? Your original choice was one out of three. If you played the game 100 times and stuck with the original choice you'd end up being right about 33 times, REGARDLESS of whether Monty Hall showed you what was behind any one, two or all of the doors. To repeat, you chose one of three doors originally, and if you stick with that choice you have a one in three chance of being right. End of story.

On the other hand, if you switch from a door which has only a 1/3 chance of being right to a different one which has a 2/3 chance - the one of the other two you have NOT seen - you will win 2/3 of the time - twice as often as if you stick with your original choice. Just like you will win more often if you bet on the better fighter, once you have been given the information. This is what people do not get - after Monty has eliminated one of the "wrong" doors, the two remaining ones no longer have equal chances of being correct. The same is true for a visiting alien, who arrives late in the game, and who unfortunately doesn't know that one of the choices is twice as likely to be right as the other one.

I explained it again to my disbelieving friend using three playing cards to demonstrate, and he still didn't get it. So I challenged him to play a simulation with the cards. Our other friend would play Monty Hall and reveal one of the cards after the original choice. I would switch every time, and he would stick with his original choice. Every time he won I'd give him a penny, and vice versa.

We started with 100 pennies each. After about half an hour and a few beers, I had twice as many pennies as he did. Then he twigged it. Later I wrote a simple computer program to simulate the situation and sent it to both of them. It's just one of those things - you either get it or you don't, but once you do, you kick yourself. Either way, the practical results cannot lie.

"Lol.... biggest scam ever. THE CHANCES ARE 50 50. Listen, if you were an alien, and you interrupted the whole game show right at the point after the host guy opens one of the doors, you've got TWO DOORS."

So what? In a boxing match you have two opponents. Do they always both have the same chance of winning? What about two poker players? Over a session will they always end up dividing the pot equally on average? Or is the better player likely to come out ahead? What makes one player better than the other? I'd say information and understanding.

In the present problem Monty Hall is giving you additional information about the distribution of probabilities by showing you a "losing" door. You can choose to ignore this information, or to act on it. I know what I would do.

Prospero on 9/8/2008 6:43:34 PM (73 days ago)

Prospero that's an excellent comment!

Well put sir!!

Ed K. on 9/8/2008 10:10:52 PM (73 days ago)

I loved this problem when I first saw it a few years back... although I must admit I didn't get it until I studied it a bit. And I loved that 1000+ mathematicians didn't get it and wrote in and told the author he/she was an idiot... until the author proved them wrong. With dependent probability, previous choices based on a foreknowledge of the correct answer... affects the outcome. If the actions taken by Monty were purely random then they wouldn't affect the outcome and this would be independent probability. But Monty's actions aren't purely random...

With 3 doors, each door starts with a 1/3 chance.
p(DoorA) = 33%, p(DoorB) = 33%, p(DoorC) = 33%.
Or put another way... If you pick A: p(DoorA) = 33%, p(DoorB or DoorC) = 67%.
Remove DoorB and the goat behind it and you're left with p(DoorA) = 33%, p(DoorC) = 67%.

With 10 doors, each door starts with a 10% chance.
p(d1) = 10%, p(d2) = 10%, p(d3) = 10%, ... p(d10) = 10%.
Or put another way... If you pick d1: p(d1) = 10%, p(d2 or d3 or d4 or d5 or d6 or d7 or d8 or d9 or d10) = 90%. or put another way: p(d1) = 10%, p(~d1) = 90%.
Remove d2 and the goat behind it and you're left with p(d1) = 10%, p(~d1) = 90%.
Remove d3 and the goat behind it and you're left with p(d1) = 10%, p(~d1) = 90%.
Remove d4 and the goat behind it and you're left with p(d1) = 10%, p(~d1) = 90%.
...
Remove d9 and the goat behind it and you're left with p(d1) = 10%, p(~d1) = 90%.
Now there are only two doors but you started with a 90% chance of having picked the wrong door, and you still have a 90% chance of having picked the wrong door.

That's why the million door analogy is so helpful in clarifying. You started with a freakin huge probability that you picked the wrong door And only a 1/1,000,000 chance that you picked the right door. If Monty then removes doors with goats behind them until their are only two doors left, the one you originally picked and one other (the other being the one door he's been avoiding picking). The odds that you originally picked the right door are still 1 in a million, the odds that the one other door is the correct door are 999,999/1,000,000. So given those odds you should probably switch doors when given the opportunity. I certainly would.

DMonPoker on 9/9/2008 2:57:28 AM (72 days ago)

what if alien comes after player choose door and before monty hall shows one loosing door, and choose different door. monty showing third door as empty. both player and alien should switch ? both have 2/3, 1/3, 1/2 chance for win?

yq on 9/9/2008 3:32:23 AM (72 days ago)

...and player doesnt now about invisible alien. so for him its the same situation as befor. and alien doesnt know what happens before he comes - so he is in situation of player

yq on 9/9/2008 3:37:20 AM (72 days ago)

Thanks for keeping my brains up with this clever discussion!

It is so difficult to admit that your first guess, driven only by feeling, can be proved wrong by pure logical deduction.

ab on 9/9/2008 4:36:21 AM (72 days ago)

Three doors: G for goat and C for car, brackets is the players choice,
plus-sign is the door Monty shows containing a goat after player's pick:

1 2 3
-----
G G C

At this moment, you have 1/3 chance hitting the car. Now, the only case
you don't want to change the door is when you've picked the car straight:

1 2(3)
+----
G G C

or

1 2(3)
--+--
G G C

So, in this scenario, if you change the door, you'll loose.

All other (two remaining) choices requires you to change the door (giving 2/3 chance):

(1)2 3
---+--
G G C

1(2)3
+----
G G C

Atte on 9/9/2008 5:17:53 AM (72 days ago)

Always knew this was true but found it hard to understand intuitively.

Finally got it (mathematically and intuitively) after reading DMonPokers simple explanation. Thanks...

Great blog by the way, love that poker bot stuff.



JudgeD on 9/9/2008 5:23:45 AM (72 days ago)

What about if we have two players both choosing different doors. Then Monty opens third door that neither of players has chosen. (let's assume that one of the players chose the car and other one goat.) Now, should both players switch doors?

noone on 9/9/2008 6:30:30 AM (72 days ago)

In fairness to the second poster (skeptic) - his argument is partially correct - if an alien walked in when there were two doors left, it would have a 50% chance.

The problem with his argument is that THERE IS MORE INFORMATION AVAILABLE!

Sure, if you don't use that info, then there is only a 50% chance, like if you used a coin flip to bet on a sports game rather than analysis of players, conditions, etc... But just because you can ignore the information and make it a 50-50 problem doesn't mean that's all it is.

Anonymous on 9/9/2008 8:58:12 AM (72 days ago)

I always think I 'get it' after seeing the explanation - but deep down I still don't 'get it'. I never had the AHA feeling about this problem.

Anonymous on 9/9/2008 10:07:00 AM (72 days ago)

Another way to look at it skeptically.

We have 2 doors - 1, 2 - one with a Goat and one with a Car . You are asked to pick one?
- there is a 50-50 chance so I picked 1.

Now one more door(#3) gets added, which has a goat in it - will switching increase my odds?

Is this version different from the actual problem? If so what am I missing?
The only reasoning I can think of is in the actual problem there was a chance that I picked #3, but it beats me because at the end I always have to pick between 2 no matter how many doors I start with.

acb on 9/9/2008 10:19:26 AM (72 days ago)

Like mentioned above, the Monte Hall paradox is really not intuitive when discussing probability. The key things to realize is not just that Monte Hall reveals a door after you have chosen one first, but also that he knows where the car is and always picks a door that is NOT the car. To understand why it is not 50% odds when you have another choice between switching or staying, think of it like this: If you initially pick the door that is hiding the car first (which you exactly have a 1/3 chance of doing), if you switch after Monte opens one that is not the car, you will lose, and if you stay you will win. So with that 1/3 chance you lose with switching and win if you stay. But, if you initially pick a door that is hiding the donkey (which you have exactly a 2/3 chance of doing), if you switch after Monte opens one that is not the car, you will win, but if you stay you will lose – so 2/3 times you win if you switch and lose with staying. Hope this helps along with the other explanations above.

Russell on 9/9/2008 10:26:03 AM (72 days ago)

"The correct answer is: yes, by switching doors your chance of picking the "right door" goes from 33% to 66%, provided that the host knows which door contains the car. If the host doesn't know which door contains what and is just picking randomly, then the answer is no: your chances of picking the right door are 50-50."

That is not correct at all. The author of this article should not have written this paragraph. The probabilities of switching vs. not switching have no bearing on whether or not the host knows where the car is. The math works out the same. Once a door is picked, it's ALWAYS in the contestant's favor to switch. Maybe what the writer meant to say is that a door is opened before the contestant picks a door. In that case the choice would be a 50/50.

Daniel on 9/9/2008 11:01:46 AM (72 days ago)

Haha, it's really hilarious to read everything people say, having not been convinced by the logic. Not to say that I was instantly convinced, that would be foolish. But after reading enough logical babble over this, and especially after listening to the case with one million doors (which makes the problem much more intuitive), and then even finding a computer simulation that finds the answer out empiricallly, I don't see how one can still so stubbornly stick with their 50-50 answer.

ehsanul on 9/9/2008 11:09:26 AM (72 days ago)

acb, your example is extremely different. You're right, it lies in the fact that in the start you start with a 1/3 chance of getting the prize, when you make an initial choice. The fact that the initial choice has a 1/3 chance of being correct is extremely important.

It makes much more sense if you consider the case if there were a million doors with 1 prize behind one of them. When you make an initial choice, you had a 1 in a million chance that you had the door with a prize behind it. Then Monty comes along and reveals 999,998 doors with goats behind them, and he leaves your choice, and another mystery door behind. So, you have to see that since Monty knew which door had the prize, he couldn't reveal that one. Now you have a choice between your initial choice, which had a 1 in a million chance of being correct, and the one Monty left behind. It should be obvious now that switching doors makes a lot of sense, atleast in this exaggerated case. It's the same with 3 doors, just a little less intuitive.

Now, acb, if we apply your proposed change to the million doors version, then yes, having two doors initially to choose from and then adding in 999,998 doors with goats behind them doesn't change the probability of picking the right door from the two you were presented with. But this is a different problem, it's not the Monty Hall problem. It should be apparent here that it's extremely important that you start out with all the possible choices, and end up with a smaller chance of picking the right door initially. I hope that made some sense.

ehsanul on 9/9/2008 11:22:31 AM (72 days ago)

can someone explain to me the 66%? I understand that you're original choice of 1 out of 3 is a 33%, and that once the host reveals one of the "loser" doors, your chances are better to switch, but it seems to me that your improvement in chances would be to 50%. Originally you choose a card and its 1/3, 33% - simple. As soon as the host eliminates one, it is in your interest to switch because while your inital pick had a 33% chance, the remaining card has a 1/2 chance of being correct... or 50%.

I guess I understand why chances improve, but not why they improve to 66%

Grey on 9/9/2008 11:23:12 AM (72 days ago)

Daniel, you said:
"The probabilities of switching vs. not switching have no bearing on whether or not the host knows where the car is. The math works out the same. Once a door is picked, it's ALWAYS in the contestant's favor to switch. Maybe what the writer meant to say is that a door is opened before the contestant picks a door. In that case the choice would be a 50/50."

Actually, I believe the point was that if the host were choosing randomly, it would affect the case overall because sometimes the host would end up picking the door with the car behind it, in which case the player loses whether he switches or not. The fact that Monty knows which door has a car behind it and consciously avoids it is important.

ehsanul on 9/9/2008 11:26:11 AM (72 days ago)

Some people are resistant to reality and refuse to accept logic, albeit difficult at first. It does ABSOLUTELY matter that Monty picks a door with a donkey behind it with prior knowledge of where the car is after you pick one first!! It is not statistically 50:50. This is proven, just research it or read some good explanations above like Russell’s – thanks for making it simple dude.

hugo on 9/9/2008 11:35:10 AM (72 days ago)

It is not a coin flip - just read about it.

hugo on 9/9/2008 11:36:04 AM (72 days ago)

Thanks DMonPoker and Russell! My head can stop hurting.

Anonymous on 9/9/2008 11:56:20 AM (72 days ago)

"Listen, if you were an alien, and you interrupted the whole game show right at the point after the host guy opens one of the doors, you've got TWO DOORS."

Aliens have x-ray vision so they always win.

jeremyp on 9/9/2008 12:06:10 PM (72 days ago)

Another GREAT blackjack book is "Ken Uston on Blackjack". It chronicles the late Ken Uston and his ground breaking techniques for beating the casinos at 21 in the 80's. It also tracks his legal battle with the Nevada Gaming Commission to decriminalize card counting. An interesting read even for non-players

Anonymous on 9/9/2008 12:42:17 PM (72 days ago)

When I was finally convinced of this, it required a friend with a PhD in Math to do the convincing.

I bought into the alien 50-50 argument. Here's the killer:

Show the audience what's behind two doors - one car, one donkey. Close the doors.

Introduce alien. His choice isn't 50-50, it's 100-0, because the events before that point can't be separated from the choice. In the usual problem, Monty always has a junk door to give away - either one of the two you didn't pick, or both of them, so revealing junk doesn't tell us anything new BUT, you first picked one of 3 doors (33%), and you now have the option to swap for what is effectively the contents of both of the remaining doors (66%).

davidm on 9/9/2008 12:54:14 PM (72 days ago)

This paradox is best explained by increasing the number of doors. Say doing the test with a deck of cards. Your first choice has an odd of 1/52 of being right. When the game show host throws out the remaining 50 cards, it's obvious that you've "compressed" the odds highly. The odds that the card was in the remaining set is 51/52. Hence, now, you have the choice to stay with your initial choice (odds of 1/52) or switch to the other card (odds of 51/52). This is the same thing reduced to three cards (odds of 1/3 versus 2/3).

When looked at this way, this paradox is so trivial, people understand it intuitively. And yet...

Anonymous on 9/9/2008 1:00:56 PM (72 days ago)

I just understood this. When you first pick, you have a 2/3 chance of getting the goat. This means there is a 66% chance you chose the losing card the first time.

theMan on 9/9/2008 1:42:26 PM (72 days ago)

"What about if we have two players both choosing different doors. Then Monty opens third door that neither of players has chosen. (let's assume that one of the players chose the car and other one goat.) Now, should both players switch doors?"

In the normal Monty Hall problem, I know that if the third door contains the car then Monty will choose the second door and vice versa (assuming I pick the first door). In your new version, you don't say what happens. So the answer is "it doesn't matter if you switch or not".

It's crucial that Monty Hall knows where the car is _and_ you know that he will always pick the door with the goat. Ironically, most people state the problem incorrectly. For example, since Kevin Spacey in the above example never states that Monty will always pick the door with the goat, he is actually incorrect and it doesn't matter if you switch. But since it's only a movie, we will let him off.

Anonymous on 9/9/2008 2:03:06 PM (72 days ago)

The way Kevin Spacey phrased the problem (from your quote), it is not necessarily advantageous to switch. He's missing a key piece of information: that the host *always* will open a door with a goat behind it.

To see why this is important, consider the host is following this set of rules:
- if you choose the door with the car, the host will open a door with a goat and ask you to switch.
- if you choose a door with a goat, the host will immediately inform you that you've lost.

In this situation, it's to your advantage *not* to switch. Since you don't know what set of rules the host is following, you don't know one way or the other whether switching benefits you.

(Though it's true that if the host always shows you a goat and offers a chance to switch, it's better to switch.)

Anonymous on 9/9/2008 2:06:52 PM (72 days ago)

I actually wrote a simulation to run through millions of test cases for this problem. The end result, after more then 10,000,000 "games" was that if you switched doors you won 67% of the time, and if you stuck with the door you initially picked you won only 33% of the time.

I can't explain the logic, but that's the reality.

Frewfrux on 9/9/2008 2:13:43 PM (72 days ago)

To anonymous commenter #2: brilliant troll! The alien freezing thing as an especially nice touch.

Jeff on 9/9/2008 2:14:10 PM (72 days ago)

Love this problem!
Here's how I think about it. From the offset there are three randomly distributed possible consequences: you get the good prize (outcome A), you get Fred the goat (outcome B) or you get Sally the goat (outcome C).
pA = 33 1/3 %
pB = 33 1/3 %
pC = 33 1/3 %

So, if you know that every time you pick, Monty is going to show you a goat, and you're going to switch your choice every time, here are your results:
-You pick A. Monty Shows you Fred(B), or Sally(C), and you pick the other goat. You lose.
-You pick B. Monty shows you Sally(C), and you pick the good prize, A. YOU WIN.
-You pick C. Monty shows you Fred(B), and you pick the good prize, A. YOU WIN.

Two out of three. 66 2/3 % chance of getting the money (or the jet-ski or the vacation or the BRAND NEW CAR) if you switch every time.
I'm sure it's got a more complicated element to it, but that's how I see it. Anyone care to correct me?

Nick on 9/9/2008 2:19:31 PM (72 days ago)

Just to be clear, the alien question does make sense, if he stepped in after the door opening and knew nothing of what was going on before. It is 50/50 then, without prior knowledge.

G on 9/9/2008 2:37:09 PM (72 days ago)

"provided that the host knows which door contains the car"

Yeah, no. It doesn't matter if the host knows what door is what. He asks you if you want to switch AFTER he opens the door. So if he happened to open the wrong door (revealing the car) he would just say "oops, oh well, let's play a different game now". It's ALWAYS 2/3 good to switch doors AFTER he shows you the open door.

glyph on 9/9/2008 2:38:40 PM (72 days ago)

Gee!! Thanks for all these comments! Now I think that I get it. I see it like this: the correctly stated problem is equivalent to the following:

1) there are 3 closed doors, you pick one;
2) without opening any door, the host asks you "do you want to stick with the door you chose or do you want to switch to BOTH the other doors?"

Marco Maggi on 9/9/2008 2:46:01 PM (72 days ago)

Think of it this way. Instead of 3 doors there are 100 doors (99 goats and 1 car). Monty opens 98 doors showing goats and asks you if you want to switch. When you first picked your door you had a 1 in 100 chance of getting the car. If you switch you have a 99 in 100 chance of getting the car. Same logic for 3 it's just not as obvious.

richard on 9/9/2008 3:03:31 PM (72 days ago)

Lots of good comments here. The one that works best for me seems to be related to a couple of the comments:

Re-frame the game as Monty asking "Either pick 1 door or 2 doors. If you pick 2 doors, I'll give you the better prize of the two." (I think somebody else here said something similar.) Who would turn down that offer? You get 2 choice instead of 1 and you'll get the better of those 2.

You can do this in the real game. The way to pick two doors is to pick them in your head, but tell Monty you picked the third one. You then flip from that one to the two you actually wanted by telling him to switch. By opening the door all he's really done is throw away the lesser of the two prizes of the two doors you actually wanted, leaving you with the better of the two.

Chad on 9/9/2008 3:06:52 PM (72 days ago)

To those of you who did/do not get it easily, don't feel too bad. Paul Erdos ended up being on the wrong side of a bet involving the Monty Hall problem... and he was one of the greatest mathematicians of our time.

ungood on 9/9/2008 3:19:30 PM (72 days ago)

@ehsanul, thanks for clarifying that the variation was indeed a very different problem (adding more dummy doors after presenting only two initially).

acb on 9/9/2008 4:12:32 PM (72 days ago)

I thought that David Avraamides' comment offered a great way to think about this (idea since repeated by Marco Maggi and others).

For me, the light came on when I sat down and played 3-card Monty - it becomes completely obvious that switching doors wins if and only if you picked a loser on your first try - which happens 2/3 of the time.

Thelonious on 9/9/2008 5:21:32 PM (72 days ago)

The top level post is the first time I've ever heard the answer for the Monty Hall problem explained correctly, with regards to the foresight of the host playing an important role.

The reason people argue about this problem, I now realize, is because the problem statement is purposely confusing on that point. That's why all those "PhDs" write letters, not because they don't understand the answer.

conrad on 9/9/2008 5:30:10 PM (72 days ago)

Here's what did it for me so many years ago -- and it doesn't have any numbers!

- When you picked your original door, you probably* picked the wrong one.
- So, given the opportunity, you are no worse off picking another door.
- And if one door is eliminated, you're actually better off switching. After all, your first pick was probably* wrong. With only one other door left, that door is probably* right.

*In the long run.

Easy peasy.



jkb973 on 9/9/2008 5:44:18 PM (72 days ago)

"That's rational, makes perfect sense, and yet, like many things in life which are rational and make perfect sense, it's 100% wrong."

Please don't say things like this. The analysis you present does not apply to the Monty Hall problem, therefore it is NOT logical and it DOES NOT make perfect sense. Logic is a real, formal thing. It gets the right answer when you use it correctly, and if you AREN'T using it correctly, IT ISN'T LOGIC.

The whole problem with the Monty Hall paradox is that people are not trained to TRUST the formal application of logic. The formal proof the outcome is so simple it is practically trivial, and a person comfortable with mathematical reasoning can easily be convinced. The rest of the world, because they hear things like "That's rational, makes perfect sense, and yet, like many things in life which are rational and make perfect sense, it's 100% wrong." persist in believing that the logical proof is "an academic trick" of some kind, and are difficult to convince exactly because they still trust their intuition better than the principles of formality and rigor.

Vincent Toups on 9/9/2008 7:59:54 PM (72 days ago)

I agree with Vincent. However, we have to recognize that some people, for whatever reason, seem to be incapable of correctly applying formal logic so such mistrust isn't necessarily bad as long as they aren't going to improve their own formal logic skills. Thus it is useful and interesting, where applicable, to provide an intuitive rationale for why things happen. And, overall, I think such intuitive breakthroughs increase people's skill and trust for logic. To that end, I like the explanations in David's "it's like choosing 2 doors instead of 1" or the "you're probably wrong the first time" one that a few of us have proposed (myself included).

A couple of places where intuition is not currently applicable is in what-if hypothesizing about "faster-than-light" travel, the behaviour inside black holes, behaviour at the Universe's age << 1 second, and most of Quantum Mechanics. Mind-you, I find the many-worlds hypothesis makes QM -much- easier to grasp, even if I accept that many-worlds is probably not literally true in any sense. The biggest problem with many-worlds is the way people will once again abandon logic in analysing the idea. Sci-fi writers absolutely LOVE the idea of many-worlds and use it with some treknobabble to make alternate universe plotlines. New agers justify whole mythological belief systems with it. I think that's the far more pressing problem than people just not trusting logic -- it's when they trust logic so much they'll swallow illogic down with it. Not that I really hate sci-fi or anything. Just that I hate it when people take it too seriously.

Ens on 9/9/2008 8:41:00 PM (72 days ago)

Some answers are confusing the problem more. I like it simple like Russell's, breakdown or with the increased numbers analogy like Richard's and an Anonymous one above. It just doesnt make sense at first, but then it comes together.

Slider on 9/9/2008 10:06:53 PM (72 days ago)

Grey, let me explain why the odds are 67%, via illustrating why they aren't 50%.

The 50% camp believes that if you pick A (from A-C), you have a 33% chance of it being X; and that if you then take away C, then you have a 50% chance of having X. (Forget for a moment about opening the door; that's irrelevent. Just remove the door.)

If you follow that to its logical extreme, then taking away B and C would mean that A had 100% chance of A being X, regardless of what it originally was.

If you follow that; whatever door you choose, if I take away the other two doors, you don't automatically win. The truth is that THE ODDS DON'T CHANGE for your original selection. When you pick a door, it has a 33% chance of being the prize. No matter what I do to the other doors, those odds don't change. That door *always* has a 1/3 chance of being the prize.

So, if we take away door C (assuming we do so with knowledge, ie, opening the door to see the goat), then door B's odds are the remainder: 100 - 33 = 67.

If that hasn't cleared it up, then think about it another way; if you pick door A, you have a 33% chance of it being a car. If I then go change doors B and C so they both have cars behind them, what are the odds that door A has a car? It doesn't change. It still has a 33% chance of being a car.

To get a bit more complicated, the same thing would apply to all of the 'million door' variations. If you take a million doors, and choose one, its odds of being the prize are 1/million. If you take away all but nine of the remaining doors, then door A still has a 1/million chance, and doors B-J have the remainder (999,999) between them, or 1/111,111 each.

Hope that's helpful.

Neil on 9/9/2008 11:36:10 PM (71 days ago)

This "math" problem is well explained. Thank you. What is missing in relation to the movie 21 is that this paradox has no meaning in the context of blackjack. Its just a cool sounding math puzzle thrown into the movie to make the professor and student look smart. But is has nothing to do with winning at blackjack.

Jon Hancock on 9/10/2008 4:58:54 AM (71 days ago)

@Jon

Perhaps it's just that consideration of the Monty Hall problem makes you a more logical thinker, or a more careful analyzer of problems relating to probability, or more appropriately skeptical of your intuition? And perhaps in this indirect way, it makes you a better blackjack/poker/any_game_involving_probabilities player? I mean, it can be reasonably demonstrated that humans on average are very inept in dealing with probabilities in an intuitive manner. And the Monty Hall problem is a good example of the type of situation where humans don't have good intuition.

ehsanul on 9/10/2008 5:33:22 AM (71 days ago)

I never got it until I read Prospero's comment. It was driving me nuts. Thanks a lot, it now clicks.

Anonymous on 9/10/2008 4:45:01 PM (71 days ago)

@Jon

The point of the Monty Hall trap is to learn to properly interpret probability. Blackjack is about playing probability. Ergo, the example is appropriate.

Ens on 9/11/2008 2:29:51 PM (70 days ago)

I find the easiest way to explain this to the people that argue is simply to say that it isn't two new doors, it is not a new situation, it is just more information given in the same situation. If you make a completely new decision (a 50-50 chance of winning) then you are ignoring the previous information you have - that the door you originally picked only had a 33% chance of being correct, and hence the chance is 66% that you originally chose the wrong door, giving the now only remaining door a 66% chance of being correct.

Frank on 9/12/2008 12:49:45 PM (69 days ago)

Yes - there's a two thirds chance that the other two doors are the proper ones, then Monty goes and eliminates one of those doors for you. STILL two third chance that the now-one-door is the proper one.

Terry Smith on 9/13/2008 8:13:03 AM (68 days ago)

You should trade because your first pick is only a winner 1/3 times. Bottom line.

Jordan on 9/15/2008 9:20:00 PM (66 days ago)

Let's re-phrase the problem just a bit: Instead of Monty knowing which door has the car, let's just have him open one of the doors thazt you didn't choose. Now, if the rule is either #1. even after looking at the result you must keep your first choice, or #2. that after he opens a door you could choose either of the two you didn't pick first but can't keep the one you chose first. You MUST pick rule #1 or rule #2 to go with before he opens a door, it really doesn't matter whether this is before or after you make your first choice. Which rule do you like? When it comes down to it, this is what you are being offered (your choice of one door, or the best of all the rest)!

Larry on 9/18/2008 11:51:47 PM (62 days ago)

Also Neil must have meant 111,111/1Million, rather than 1/111,111. above.

Larry on 9/18/2008 11:56:28 PM (62 days ago)

But, just like poker, someone can be wrong and still win - just not as often.
Rule#1 above doesn't ALWAYS lose, just 2/3 of the time.

Larry on 9/18/2008 11:58:48 PM (62 days ago)

What struck me most (besides the total inaccurateness of my intuition) when I heard it in class was the mockery in the scientific community. The woman that originally posted this was mocked, until people found out she was right Tong

wachtwoord on 9/21/2008 8:47:19 AM (60 days ago)

Any chance of a update anytime soon?

Anonymous on 9/24/2008 1:07:07 AM (57 days ago)

This may be weird, but why does your rss feed say the current article is 'The 10 Most FREAKING AMAZING Pieces of Software in the World' - and it isn't?

On topic, my most FA software experience was the PowerLAN Server. It was written in Assembly (a lost art) and sat like a rock on a 16mb 386 I had and served my old BBS perfectly. Never a problem at all.

You booted it up in DOS and it took total control of the computer, keyboard, disk IO, screen, everything the BIOS hadn't already grabbed.

On the enterprise side, I was part of a firm that set one up on a 486 to serve a 50 person office. Its served fiels and unspooled rthe printer.

You could configure it to wait a set period of time of idleness before it did disk commits - we used 0.5 seconds. Why can't you do that on todays servers?

Never a problem until the day the disk died and started making noises like a cement mixer, and the other day the motherboard died and insisted it was 4700 AD.

The PowerLAN people eventually gave it away and went into Internet apps in the face of - who else - Bill Gates and MS networking. This was before Samba.

Here's to PowerLAN Server, one of the most freaking amazing pieces of software in the world.

Terry Smith on 9/27/2008 9:45:40 AM (54 days ago)

Ok it's up.

Found it in Google Cache.

www.google.ca/url%3A%2F%2F72.14.205.104%2Fsearch%3Fq%3Dcache%3Ast9JGlxNZ3gJ%3Awww.codingthewheel.com%2Farchives%2F10-most-amazing-software-world%2BThe%2B10%2BMost%2BFREAKING%2BAMAZING%2BPieces%2Bof%2BSoftware%2Bin%2Bthe%2BWorld%26hl%3Den%26ct%3Dclnk%26cd%3D1%26gl%3Dca&ei=Fe7eSOWsNpnAgwK9sZDwAw&usg=AFQjCNFQj7KFl0qJXHcFQsyoCndLQX67ew&sig2=59Ude7T07wxgTthlEJuAbQ

Terry Smith on 9/27/2008 10:49:47 PM (53 days ago)

It works, I post it, I test it and it doesn't work.

Just enter 'The 10 Most FREAKING AMAZING Pieces of Software in the World'
in Google and hit 'cache'

Terry Smith on 9/27/2008 10:53:09 PM (53 days ago)

"The original post has been deleted: long story"

Anonymous on 10/1/2008 3:03:08 PM (50 days ago)

Ok thanks. <sulks...>

Terry Smith on 10/2/2008 12:09:36 AM (49 days ago)

Terry, I went ahead and posted even though this time you did NOT mention Harry Potter. Sorry for breaking form, but we've got a bunch of content to get through if we're all to have (mostly) working poker bots in time for Christmas.

James Devlin on 10/3/2008 7:18:00 AM (48 days ago)

I find the simplest way to see the real solution is to simply look at the possible outcomes.

If you DO NOT SWITCH:
1. You pick Goat A at the beginning, that's what you get.
2. You pick Goat B at the beginning, that's what you get.
3. You pick the Car at the beginning, that's what you get.

You win the car 1/3 of the time (assuming your choice is completely random, which it will be if you have no idea which item is behind which door).

Now, if you DO SWITCH:
1. You pick Goat A at the beginning, Monty reveals Goat B, you switch and get the Car.
2. You pick Goat B at the beginning, Monty reveals Goat A, you switch and get the Car.
3. You pick the Car at the beginning, Monty reveals a goat, you switch and get the other goat.

You win the car 2/3 (66%) of the time.

Steve on 10/15/2008 9:04:10 PM (36 days ago)

To me the easiest way to look at this, isn't that him picking a goat is important, its that he didn't pick a door. 3 Doors ABC, you pick A and he reveals B as a goat. Is that Monty DIDN'T pick C which makes it important. You had a 1/3 chance of being right. Then he takes he basically gives you 1/3 by revealing a door, which you can only take advantage of by switching doors. If you don't, even though it looks like 50%, you are still working your original guess you made when you had 3 doors.

To explain it slightly differently. If your original choice was 1/3 and not changing it doesn't change your chance of being right, which should be obvious. Then the other door cant be 1/3 or 1/2 because that leaves some % hanging out there. So its obviously 2/3 to complete the math.

Joshua on 10/25/2008 1:31:30 PM (26 days ago)

The point, three doors, one is removed by the game show host opening it revealing nothing, leaving you with a 50/50 decision. I think this is the point everybody is missing, if the host 'knows' which door the car is behind and he is offering you a choice of changing your answer, he is literally 'telling you' that the car is behind the door you did not select. I think that this is a psychological nuance common to game shows, once you have got to the final stage of a show they really don't want you to loose, it lowers ratings and makes 'them' (the tv company) look like smucks. Let’s face it; they make much more money than the value of the car from advertising etc. However, if the show host does not know, then you are back to a coin flip decision.

Or you could look at it this way: two doors remain (A,B) giving you 50/50 chance to win. However, the host is suggesting that you change your choice, this is the crux, he may know which door the car is behind and is suggesting that you change (if he knew you already had the right door would he give you a choice?). This addition piece of information, the choice, has just doubled your chances, so now you have a 66/33 (or 2/1) chance that the car is behind the other door, so change your choice!

Ron on 11/20/2008 5:29:54 PM (5 hours ago)

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