21 and the Monty Hall Paradox

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.

[more]

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:

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 on September 8, 2008 in Uncategorized, , , ,

Comments

  • Terry Smith says:

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

    • anonymous says:

      You are wrong, try doing some research before answering like that.
      the other reply to your comment explains well, if not you can use [http://en.wikipedia.org/wiki/Monty_Hall_problem][1]
      It’s called Monty Hall Paradox.

      [1]: http://en.wikipedia.org/wiki/Monty_Hall_problem

  • Anonymous says:

    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 says:

      You are wrong because your chances of you picking a goat is a 2/3 chance so if you stay with ur original choice there is a 1/3 chance you get the prize. So any time u pick a goat(which is a 2/3 chance) and swicth u win. To make a long story short just play the simulation.

    • Anonymous says:

      Try it. Your wrong. Anyone who thinks its 50/50. your wrong. simple as that.

  • Anonymous says:

    In rebuttal to
    [quote]How about you try thinking critically about this instead of blindly following the herd on this one…[/quote]

    [url]http://imgs.xkcd.com/comics/science.jpg[/url]

  • Anonymous says:

    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 says:

    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?

  • Ian says:

    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 says:

    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.

  • shiona says:

    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

  • Poker Forums says:

    Good to see some more regular updates!

    Keep it up ;)

  • Jordan says:

    Doors ABC. P(A) = 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 says:

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

  • Ens says:

    "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).

  • David Avraamides says:

    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.

  • Prospero says:

    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.

  • Ed K. says:

    Prospero that’s an excellent comment!

    Well put sir!!

  • DMonPoker says:

    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.

  • yq says:

    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 says:

    …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

  • ab says:

    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.

  • Atte says:

    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

  • JudgeD says:

    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.

  • noone says:

    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?

  • Anonymous says:

    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 says:

    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.

  • acb says:

    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.

  • Russell says:

    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.

  • Daniel says:

    "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.

  • ehsanul says:

    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 says:

    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.

  • Grey says:

    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%

  • ehsanul says:

    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.

  • hugo says:

    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 says:

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

  • Anonymous says:

    Thanks DMonPoker and Russell! My head can stop hurting.

  • jeremyp says:

    "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.

  • Anonymous says:

    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

  • davidm says:

    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%).

  • Anonymous says:

    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…

  • theMan says:

    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.

  • Anonymous says:

    "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 says:

    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.)

  • Frewfrux says:

    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.

  • Jeff says:

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

  • Nick says:

    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?

  • G says:

    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.

  • glyph says:

    "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.

  • Marco Maggi says:

    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?"

  • richard says:

    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.

  • Chad says:

    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.

  • ungood says:

    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.

  • acb says:

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

  • Thelonious says:

    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.

  • conrad says:

    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.

  • jkb973 says:

    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.

  • Vincent Toups says:

    "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.

  • Ens says:

    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.

  • Slider says:

    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.

  • Neil says:

    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.

  • Jon Hancock says:

    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.

  • ehsanul says:

    @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.

  • Anonymous says:

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

  • Ens says:

    @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.

  • Frank says:

    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.

  • Terry Smith says:

    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.

  • Jordan says:

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

  • Larry says:

    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 says:

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

  • Larry says:

    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.

  • wachtwoord says:

    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 :P

  • Anonymous says:

    Any chance of a update anytime soon?

  • Terry Smith says:

    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 says:

    Ok it’s up.

    Found it in Google Cache.

    http://www.google.ca/url?sa=t&source=web&ct=clnk&cd=1&url=http%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 says:

    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’

  • Anonymous says:

    "The original post has been deleted: long story"

  • Terry Smith says:

    Ok thanks. <sulks…>

  • Coding the Wheel says:

    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.

  • Steve says:

    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.

  • Joshua says:

    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.

  • Ron says:

    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!

  • iron says:

    @Steve

    Should be :

    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 A, you switch and get the other goat.
    4. You pick the Car at the beginning, Monty reveals a Goat B, you switch and get the other goat.

    That are ALL posible paths of choice.
    So I think it dosn’t metter if you switch or not => 50/50

    Cheers.

  • Robin Goodfellow says:

    I like the analogy made above about a million doors, though perhaps a million is excessive. If there were only 10 doors (9 goats, 1 car) and Monty opened 8 doors with goats behind them after your initial choice it would be obvious that the chance that your original door choice hid a goat was very much higher than the chance of the remaining door hiding a goat.

    It amuses me that so many people cherish their wrong opinions on this, even when it’s possible to scientifically verify the correct answer. Anyone who thinks the probability would be 50/50 should find out experimentally for themselves.

    Get a deck of cards, pick out two jacks and one ace, mix up the cards randomly then lay them out face down in a row. Pick one of the cards then see if it’s the ace. Do this a lot (i.e. 100 times), write down how many attempts you make and how many times you hit the ace with your first guess. Now, do the same except don’t look at the card you initially pick, instead look at the other two cards. Assume that "Monty" will show you one of the cards that’s a jack and you will pick the other card. If both of the cards are jacks, write down a loss. If one of the cards is the ace, write down a win (Monty would show you the jack you didn’t pick, and you would pick the other card, the ace). Do this as many times as you tried for the other method, figure out how often, as a percentage, you win with either method.

    Now, do the same as above, look at the 2 non-chosen cards, but choose one of the cards that isn’t your initial pick that is a jack and set it aside. Then, flip a coin to choose which of the remaining 2 cards to take (heads = card on the left, tails = card on the right), determine the winning percentage of this strategy as well. Note that 1/2 * 1/3 + 1/2 * 2/3 = 1/2, which I think is one of the primary sources of confusion in this.

  • iron says:

    I think anybody can agree that :
    Probability of winning = Count all choices that wins /(divides by) Count all possible choices.

    Now all possible choices when switching are:

    1. You pick Goat A at the beginning, Monty reveals Goat B, you switch and get the Car = WIN.
    2. You pick Goat B at the beginning, Monty reveals Goat A, you switch and get the Car = WIN
    3. You pick the Car at the beginning, Monty reveals a Goat A, you switch and get the other goat = LOOSE
    4. You pick the Car at the beginning, Monty reveals a Goat B, you switch and get the other goat = LOOSE

    2 wins + 2 looses = 4 possible choices.

    Probability of winning = 2 / 4 = 0.5

    That is so simple.
    Anyone could prove this wrong ?
    (maybe I am missing some possible paths, etc ?)

    Cheers again :)

  • Robin Goodfellow says:

    @iron, your above example is flawed, you have the probability of initially picking a goat or a car in the beginning as the same, and it’s not. The probability of initially picking goat a or b (items 1 and 2) is 2/3, the probability of picking the car at the beginning is 1/3. Again, I urge you to gather evidence, perform the experiment yourself and determine what the probability of the outcomes are. When the evidence disagrees with your reasoning, you should question your reasoning.

    My previous post got me thinking about how this problem is affected greatly by language.

    Consider this alternate way of stating the problem:

    You are on Monty Hall’s game show, you are presented with 3 closed doors, behind 2 of the doors is a goat, behind 1 of the doors is a car, which door hides which item is random.

    You choose one of the doors, however, Monty gives you the opportunity to switch your choice. You can choose BOTH of the other two doors and if EITHER of them hides a car, the car is yours.

    Obviously this is a better choice, because being able to choose 2 out of 3 doors is better than choosing only 1 out of 3 doors. However, this is exactly the same problem as the original Monty Hall problem, merely worded differently. Consider that in both this formulation and the original formulation you loose ONLY when your initial door choice hid the car and you win only and every time your initial door choice is not a car.

    Again I think this goes back to the problem that if you randomly choose between the original door and the other, non-goat door, the probability of winning is 50/50. However, this is not because the probability of winning is the same for each door, it’s merely because the probability of picking one of 2 doors randomly is 50/50! If the probability distribution was 0/100, you would still end up with a 50/50 chance of winning by choosing between the doors randomly.

  • Anonymous says:

    The women that wrote about the Monty Hall problem, is none other than Marilyn vos Savant. I believe that she holds the Guiness World Record for Highest IQ. Even she makes mistakes though. She wrote a column on "common sense", that instructed you to only use the information given in the question to answer them. One question was…

    You need a warm winter coat, and you’ve found two on sale. One is twice as warm as the other, but the other one looks much better. Which should you buy?

    a) The warmer one.
    b) The one that looks much better.

    She says that the correct answer is a) The warmer one. If you buy the one that looks much better, you will still need a warm winter coat.

    I disagree. The correct answer is b), because it has two benefits it is a "warm winter coat" and it "looks much better". In the question, both coats that were found were said to be warm winter coats {otherwise it would say that you found two coats}. This answer can be easily seen if you reword it as such.

    You need a car that seats 4 people, and you’ve found two on sale. One seats twice as many as the other, but the other one looks much better. Which should you buy?

    a) The one that seats twice as many.
    b) The one that looks much better.

  • astudent says:

    What will happen in first pick? There are only 3 outcomes possible

    OC1 G G [C] (~33%)
    OC2 C G [G] (~33%)
    OC3 G C [G] (~33%)

    So, the chance that we pick the car at first time is ~33%.
    After the first pick, the host will show a door with a goat

    Assume that we apply always switching door strategy (there are 2 doors left) :

    OC1(~33%)-> G G [C] (always lose, no matter which door shows)
    OC2(~33%)-> *G* C [G] (always win, because the goat would be revealed)
    OC3(~33%)-> C *G* [G] (always win, ..the same..)

    So we lose ~33%(OC1). And we win ~66%(OC2+OC3).

    Finally by always switching door after first pick, we have ~66% winning chance.

    P/s: for always holding the door of first pick, win odds will remain at ~33%.

    @iron:
    [quote]
    1. You pick Goat A at the beginning, Monty reveals Goat B, you switch and get the Car = WIN.
    2. You pick Goat B at the beginning, Monty reveals Goat A, you switch and get the Car = WIN
    3. You pick the Car at the beginning, Monty reveals a Goat A, you switch and get the other goat = LOOSE
    4. You pick the Car at the beginning, Monty reveals a Goat B, you switch and get the other goat = LOOSE
    [/quote]
    - Number 3and4 is one outcome (because there’s only 1/3 chance that you pick the car at beginning)

  • Anonymous says:

    I think the easiest way to understand this it to think in reverse…Your odds of picking wrong initially are 2/3. Therefore if you switch after the door is revealed, your odds of being wrong must be everything left, 1/3.

  • Anonymous says:

    Why don’t your odds of winning after a door is revealed increase regardless of weather or not you switch doors?

  • Anonymous says:

    Because you aren’t using the new information, it is as if you were never shown a door.

  • Coding the Wheel says:

    Great comments. Those interested in a similar puzzle should check out:

    http://www.codinghorror.com/blog/archives/001203.html

    if you haven’t already.

  • Dziorro says:

    The asumption that MONTY always choose the goats door is CRITICAL.
    Else its 50-50 situation
    Imagine the situation that we are in the game but we don’t know if monty always choose goats.
    So we choose one of the gate and then monty reveal other whic is goat.
    Assume that car is in gate number 1.
    all possible chooses are
    we 1 monty 2
    we 1 monty 3
    we 2 monty 1
    we 2 monty 3
    we 3 monty 1
    we 3 monty 2

    but we know that monty didnt choose 1 (couse the goose came up) so we are in situation
    1-2
    1-3
    2-3
    3-2

    so we have 50% of winning if we don’t switch couse the probability of each above situation is equol.

    The differenc between 33%-67% example is becouse we arbitrally rule out the posibility that monty choose car.

  • Real Money Poker says:

    If Monty always chooses either the correct door or the wrong door then this makes sense. I’m not sure how this makes sense unless he always gives the correct information.

  • Excelsior says:

    The answer is 50/50.

    Here’s why;
    First, the possible scenarios or realities as i like to call them can be described by the table below;

    Reality Door A Door B Door C
    I 1 0 0
    II 0 1 0
    III 0 0 1

    Where 1 = car and 0 = goat

    Now we examine the posibilities for Reality I.

    Possibilities for Reality I

    a) You choose door A. Host removes C and asks; -" Do you want to change?"
    If change = yes you open door B (=0) and loose
    If change = no you open door A (=1) and winn

    b) You choose door A. Host removes B and asks; -" Do you want to change?"
    If change = yes you open door C (=0) and loose
    If change = no you open door A (=1) and winn

    c) You choose door B. Host removes C and asks; -" Do you want to change?"
    If change = yes you open door A (=1) and winn
    If change = no you open door B (=0) and lose

    d) You choose door C. Host removes B and asks; -" Do you want to change?"
    If change = yes you open door A (=1) and winn
    If change = no you open door C (=0) and lose

    Conclusion;
    Reality I has four equally likely possibilities -> a), b), c), d)

    In a) and b) you winn if you don’t change your original choice.
    In c) and d) you winn if you change.

    The probability for winning if you change is thus =
    the probability for winning if you don’t change = 2/4 =1/2= 50%

    The possibilities for Reality II and III is shown in a similar manner.
    End of story.

  • Anonymous says:

    I’ve enjoyed reading all your articles and was especially excited about this one. I was the equivalent of a Engineering Statistics major in college and have seen this "paradox" many times. The best part of this post was anticipating how your readers would respond to it. Most all the responses are intelligent - even those who still refute the correct answer do so clearly and logically (albeit incorrectly). Your readers are not exactly average Joes.

  • Anonymous says:

    OK, Monty knows which door contains the car. If his algorithm is to only offer the deal if you have picked the door that contains the car, then you do yourself a disfavor by switching when you are offered a chance to switch (and you’ll never get the car, where the person who never switches will get it 1/3 of the time).

    The problem states that Monty knows which door contains the car, but doesn’t state that Monty will always make the offer. You know that in this particular instance Monty has made you the offer, but you do not know his algorithm or that he will always make the offer. In general, in puzzles like this, you’re not allowed to make up extra constraints like "he will always make the offer", so *as it’s worded*, the correct answer is "there is insufficient information to know whether it’s to my advantage to switch."

  • Anonymous says:

    ok so some others here have already pointed this line of reasoning, but i always found that it is easiest to look at the 3 possible decisions (for conveinience we will just look at the three possible events where the player switches doors)
    outcome 1: player picks donkey A, host reveals other donkey, player switches to—> CAR
    outcome 2: player picks donkey B, host reveals other donkey, player switches to—> CAR
    outcome 3: player picks car, host reveals one donkey, player switches to other—->DONKEY

    it couldnt be clearer…2 out of 3 outcomes ends in the player winning the car!
    some posters are getting hung up on the fact that there are the two sub-outcomes of outcome 3; namely that the host can pick either donkey, but these are not truly two different branches, because the player has already chosen their door, and choosing the car initially will always lead to failure, regardless of whether the host reveals donkey A or B. to further clarify this, imagine how absurd it would be to insert other sub-outcomes (monty opens the door with his left hand, for example.) the key thing to remember is that with the switching strategy, there are only 3 choices for the player (the doors), therefore, there are three possible outcomes, no more.

  • Anonymous says:

    @Excelsior - here is the flaw in your outcome tree (ill just put the corrcted version below):

    Possibilities for Reality I

    a) You choose door A. Host removes **A GUARANTEED DONKEY** and asks; -" Do you want to change?"
    If change = yes you open door B (=0) and loose
    If change = no you open door A (=1) and winn

    c) You choose door B. Host removes **A GUARANTEED DONKEY** and asks; -" Do you want to change?"
    If change = yes you open door A (=1) and winn
    If change = no you open door B (=0) and lose

    d) You choose door C. Host removes A GUARANTEED DONKEY and asks; -" Do you want to change?"
    If change = yes you open door A (=1) and winn
    If change = no you open door C (=0) and lose

    Conclusion;
    Reality I has three equally likely possibilities -> a), b), c)

    if you DONT change, the outcomes are win/lose/lose
    if you DO change, the outcomes are lose/win/win

    The possibilities for Reality II and III is shown in a similar manner.
    End of story

    see, the confusion lies in the fact that what monty does is remove a guaranteed donkey, and in two of the paths, that means he only has one possible door to open. but this is irrelevant; just remember, if you pick the car initially(1/3 chance), you will ALWAYS lose by switching, and if you pick a donkey initially(2/3 chance), you will ALWAYS win by switching! to put it yet ANOTHER way, you have a 2/3 chance of choosing a donkey up front, which ALWAYS leads to a win by switching!

  • Anonymous says:

    SO MANY of the above people seem to think that because there are 4 possible iterations of door openings, that they are all equally likely. but two of the four possiblilities start by the player picking the car according to these posters-thats all wrong, the chances of initially picking the car is NOT 50%! its 1/3!

    if you really really really really need to see EVERY outcome accounted for, try this:

    case 1: you pick car, monty reveals donkA, you switch to donkB == probability 1/6
    case 2: you pick car, monty reveals donkB, you switch to donkB == probabliity 1/6
    case 3: you pick donkA, monty reveals donkB, you switch to car == probability 1/3
    case 4: you pick donkB, monty reveals donkA, you switch to car == probability 1/3

    some people were obviously being thrown by the fact that half of these outcomes end in victory,
    BUT THE FIRST TWO OUTCOMES ONLY HAVE A 1/6 CHANCE OF OCCURRING! so combined, the first two outcomes have an equal chance of happening as the other two outcomes individually.

  • Rob says:

    Here’s what I did.

    I broke it down into a pick a number game.

    "I’m thinking of a number between 1 and 3"

    Pick a number at random.

    Whatever the number is reply with "It’s not" (one of the other 2 numbers it isn’t)

    Example

    I pick 1. Other person says "It’s not 3." If the answer was 1, I loose, if it was 2, I win.

    The ONLY way I can loose in this outcome is if I picked the number, which I have a 1/3 chance of doing.

    More examples

    Ans: 1 Pick: 3 Resp: "It’s not 2" - I pick 1 - Win
    Ans: 1 Pick: 2 Resp: "It’s not 3" - I pick 1 - Win
    Ans: 3 Pick: 2 Resp: "It’s not 1" - I pick 3 - Win
    Ans: 2 Pick: 2 Resp: "It’s not 1" - I pick 3 - Loose
    Ans: 2 Pick: 2 Resp: "It’s not 3" - I pick 1 - Loose
    Ans: 2 Pick: 1 Resp: "It’s not 3" - I pick 2 - Win

    Again, the ONLY way I loose is if I "win" by guessing correctly at a random probability of 1/3. The 2/3 times I pick incorrectly, I win by switching.

    Yeah me. Thanks to everyone before who helped me understand this.

  • James F says:

    @Jon: The relevance of Monty Hall to Blackjack is card counting.

    In the abstract if you get a Jack face up and a face down/unseen card your chances of having blackjack are 4 in 52 (four aces in a deck). However (and I admit I’ve never played blackjack for money), assuming cards dealt in previous hands are discarded, by keeping track of which cards have been discarded you have additional information about the probability that your facedown card is an ace, that is about your chances of having blackjack.

    By showing and discarding cards as Monty shows and discards a door it becomes possible to improve your probability of winning blackjack to better than 50% even though in general the game is designed to tip ever so slightly in the dealer’s favour (they win in the case of a tie).

  • Robert McDonnell says:

    I understand the logic behind this and the Million to One argument is compelling in the original context.

    Let’s take the Million to One scenario and flip it around a bit.

    You pick one out of a million, Monte then only shows you one goat. You now have the choice to switch with any of the other 999,998 choices. I like my original choice just as well as any other choices. The problem with the Monte Hall problem is that people only favor one side of the argument when they try to scale it. If Monte shows "all but one" then MHP holds true. If Monte shows "only one" then MHP starts to break down.

  • Anonymous says:

    Its a very simple problem: if the car is randomly placed behind one of the doors, then after 1 milion games, the car will be distributed 33.33% of the time behind door 1, 33.33% of te time behind door 2 and 33.33% of the time behind door 3. If you always switch, the only possibility to loose is when you choose the right door (with the car) from beginning . And that happens 33.33% of the time.

  • Ron Ford says:

    Most of you are making this more complex than it needs to be. It’s simple; starting out you are 66% likely to chose a goat, so you must assume you are going to chose a goat. So assuming you chose a goat, and Monty shows you shows you the other goat, then the remaining door must be the car. Everything is based on the fact that you are 66% to chose a goat from the get go.

  • Anonymous says:

    Here is yet another way of looking at it.

    First round, there are 3 and only 3 things that can happen.
    case 1) You can pick the car
    case 2) You can pick goat_a
    case 3) You can pick goat_b

    Now, look at the outcomes for switching in every case:
    case 1) You will lose (you had the car, and switched to goat_a or goat_b)
    case 2) you will win (you had goat_a, goat_b was revealed, and you switched to car)
    case 3) you will win (you had goat_b, goat_a was revealed, and you switched to car)

    Those are the only possibilities, and you will win 66% of the time.

    Next, look at NOT switching for the 3 cases:
    1) you will win (You picked the car, and stuck with it)
    2) you will lose (You picked goat_a, and stuck with it)
    3) you will lose (you picked goat_b, and stuck with it)

    Those are the only 3 possible outcomes, and you won only 33% of the time.

    Hope that is clear.

    Slingword.wordpress.com

  • Damian says:

    this is a problem I’ve been thinking about and have come to the conclusion that at the start of the game you have a 50/50 chance of winning, but because 1 of the three doors is discarded, by swapping you increase your chance of winning and not losing.

    imagine three doors, A, B and C, A is the winning door (it holds true for B and C if we compare permutations and combinations).

    what happens is as follows:-

    1. choose A and stick = 1/6 = win
    2. choose A and switch = 1/6 and lose
    3. choose not A and stick = 1/3 = lose
    4. choose not A and switch = 1/3 = win

    therefore you win and lose 50/50.

    the problem s this is a three door problem and not a two door problem. the expected value of a three door problem is win one game, lose one game and then what? win or lose, you can only have one and not both.

    therefore you have to pick a strategy that gives you best odds if winning, which is path 4 above. The odds of winning here are 1/3, whereas the odds of winning in path 1 is 1/6. combined they give 50% chance of winning, but path 4 has the greater chance of success ie switching).

    as I said the issue here is that you are expected to win 50% of the time but if you play three games what will the result be of the third game if you have won one and lost one?

    another approach to see it is 50/50 is as follows.

    the game is actually about choosing to play of game where you get to open one of two doors. you are shown doors A, B and C. A is the winning door.

    you have a 50/50 chance of playing game 1 with a and B and you have of playing game 2 with A and C, as follows:-

    I choose A and B is discarded leaving A and C: odds = 1/6
    I choose A and C is discarded leaving A and B: odds = 1/6
    I choose B and C is discarded leaving A and B: odds = 1/3
    I choose C and B is discarded leaving A and C: odds = 1/3

    so whether i play with A and B or A and C is 50/50

    now in each game I can either stick or twist, which is 50/50

    I choose A then stick: 1/2 * 1/2 * 1/2 = 1/8 = win
    I choose A then twist: 1/2 * 1/2 * 1/2 = 1/8 = lose

    and so on, you have a 50/50 choice at each stage and the win or lose outcome is equally likely.

    but because you are playing with three doors (three games needed to get expected result) you cannot have 1.5 wins and 1.5 losses. you have to have either 2 wins and 1 loss or 1 win and 2 losses.

    whereas you do not increase the odds of winning or losing, a switching strategy ensures you win more than you lose.

  • jayne from lancaster says:

    If you switched before any doors were opened you would have two doors, each with a 1/3 chance of hiding the car. The chance that the car is behind one of these doors sums to 2/3. If one of these is opened revealing a goat the chance that it hides the car falls to zero. So the entire (unchanged) 2/3 chance that the car is *not* behind the door you switched from moves to the remaining door.

  • Anonymous says:

    People that still dont get it, read.

    we have three door, A, B and C. one of it is a winning door, choose one and you have 1/3 chance of winning. One of the remaining two are opened and you know it is not a winning door. Then you are allowed to choose between the two remaining door, what will you do?

    1. you choose one of the two door: chance that you get the winning door is 1/2.

    2. what? you only choose once, you have to make a decision, stick with your choice or switch. what ever you do, the chance is still 1/2.

    —-> yes it is 1/2, now im convinced. continue reading please.

    what happened if you repeat the situation?

    1. you always choose one of the two door: chance that you get the winning door is 1/2.

    2. you always choose to stick with your first choice: your chance of winning is 1/3. why? you choose 1 of 3 door, and do nothing after that. your chance is still the same.

    3. you always choose too switch your first choice: your chance of winning is 2/3. why? you choose 1 of 3 door, and then switch to the other 2 of 3 door, your chance is 2/3. yes, by switching your first choice, you actually choose 2 of 3 door!

  • David from Bradford says:

    If Monty always offers a switch and always shows a goat.
    Easy way to see it.
    If you switch, you win every time you chose a goat (2/3) and lose every time you chose a car (1/3).

  • George Bush says:

    Stop posting idiots. Problem is solved!

  • Stephen Malinowski says:

    To anyone who thinks the odds are 50/50: read ehsanul’s post (on Tuesday, September 09, 2008). If, after reading that, you still think the odds are 50/50, let me know how many $100 rounds you’d like to play with me with you being Monty Hall and me being the contestant.

  • Mike says:

    There’s a psychology to this as well. Suppose for a moment, that the host has been trained to only reveal a door if your first selection was correct to begin with, knowing that due to the higher odds, you will likely change your selection. I wonder what the odds of that are from the perspective of the host?

  • Mike says:

    To clarify, hypothetically, if the host knows that your initial selection was correct and he knows that you will change your mind were he to reveal one of the wrong doors, but only does so if you were correct in the first place, then your chances drop to 0% because you will always be wrong. Since the host can’t possibly know that, is there a way of figuring his percentage of accuracy at deciding which contestants to reveal the door to?

  • Anonymous says:

    I know this is old.. but I have your answer right here… the game IS and WILL ALWAYS be 50%.. even with the 3 choices.. BECAUSE the show host will ALWAYS get rid of the goat.. period.. whether you chose the right answer OR not..if you can’t understand how HUMAN FACTOR plays into this equation and you go on BELIEVING that you had 33% chance in the first place YOU ARE WRONG.. sorry.. hope that helps.

  • Anonymous says:

    I am proving the MHP for a science fair project and TRUST ME, if you think the chances are 50/50, you are sadly mistaken. Try it for yourself.

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  • dave says:

    I don’t understand. Could someone explain this to me? There are two doors that are unopened. The percentage of the other door did not increase, since they are two seperate doors. How could the percentage of one door being correct increase, but not the other

  • Stevenh says:

    Minesweeper- you click a square and a big chunk of the game is revealed but hello… there is a suspect looking block in the middle of this now open area, and the square you initially clicked- which one has the mine 9 times out of 10? Same principle here with fewer squares. But, if you knew the suspect looking box in the middle was there by pure chance and not that the game avoided it, then your choice of which one is the mine is clearly the same, again same principle here with fewer boxes.

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  • Anonymous says:

    This is a age old post but I still have one question I think you could help clear up. If there were 2 players, each having chosen one of the non-opened doors, would they both be advised to change. If so, the chances of success for each of them would be 50%.

    Wanting to open old wounds… :)

    Thanks for any input on this - I’m flummoxed..! Barry.

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