Monty Hall Problem Put to the Test
The following letter was sent to me in response to my column in Scientific American (which generated hundreds of letters in response, so I penned the following response) in which I discussed the now-infamous (and infuriatingly counter-intuitive) probability problem called the Monty Hall Problem, or the Three Door Problem, in which a contestant chooses one of three doors, behind one of which is a car and the other two goats. Monty then reveals what’s behind one of the other doors (only ever showing a goat and never showing you your own door pick), which is always a goat, then asks if you want to change doors. Most people say it doesn’t matter because now it’s 50/50, but the correct answer is that you should always switch, which will give you a two-thirds chance of winning. There are simulations of the game online, but my correspondent took it upon himself to test the game with his own computer program. Here are his very interesting results, which also nicely show the scientific method at work:
Mr. Shermer,
I am writing to thank you for your articles in Scientific American, specifically the one in the October 2008 issue discussing the ‘Monty Hall Problem’. Thanks to your essay, I think I finally understand the scientific method.
After reading about the ‘Monty Hall Problem’, I couldn’t shake the idea that switching doors shouldn’t make a difference. I knew that I must be wrong, but couldn’t get my head around the problem; I couldn’t get to sleep for a couple of hours that night either. So, instead of just believing that I was right or wrong and leaving it at that, I decided to see if I could find any objective data that would support one view or the other.
I wrote a little Visual Basic application within an Access database and ran 100,000 sessions where the contestant switched doors every time. The contestant was successful a little over 62% of the time. This seemed to lean to the conclusion that switching leads to a two-thirds success rate, but 62.2% seemed odd. I ran 1,000,000 sessions to see if the numbers be more definitive; they weren’t, still 62.2%. So, I looked through the database tables where I recorded the results to see what was going on. It was then that the true meaning of the scientific method became apparent to me. Looking through the data, I developed a new theory of the ‘Monty Hall Problem’ and why the strategy of switching doors should be successful two-thirds of the time. The new theory was elegant, the logic seemed clear, even obvious, and it seemed to agree with the data. The remaining problem was what was happening with the missing four-and-a-half percent. My suspicion was that this was caused by the random number generator I was using to pick the door with the car behind it and the door the contestant chose during each trial not being random enough. I rewrote the function choosing these doors, attempting to make them more random and ran 100,000 new trials and ended up with a success rate of 66.43%, close enough to satisfy me that the switching strategy is indeed the way to go.
As I mentioned, this little exercise opened my eyes to the true meaning and power of the scientific method. I was confronted with two competing and mutually exclusive theories explaining how something works. Instead of stubbornly standing by my own gut feeling, or believing another theory simply on faith, I ran an experiment to see if either theory would be supported or disproved. Examining the data led me to support the switching strategy and to develop a new theory explaining why this is so. I also developed a new theory to explain the remaining discrepancies in the data, ran a second, refined experiment, and gained further support for the theory behind the switching strategy.
I’ve read most of Stephen Jay Gould and Carl Sagan, I even subscribe to Scientific American. I always thought that I believed in the scientific method. However, it took your article, and its inspiring me to use the scientific method for myself to finally truly understand it.
Thank you,
Douglas Millar
Thank you Douglas Millar!
Michael Shermer
That was fantastic! I remember doing this kind of thing too when I was studying statistics, running little Matlab routines for a few minutes to check things I considered counterintuitive.
Cheers,
No description of the “new theory” is provided. Reminds me of Fermat’s last theorem.
The scientific method also involves explaining your new, elegant theories.
There’s an interesting variation on the Monty Hall problem, in which Monty doesn’t know which door hides the car. In that version, Monty guesses which door to open, and by chance reveals a goat. Should the contestant switch doors or not? It’s 50/50 in that scenario.
And, yes, I didn’t believe it was 50/50 when I first heard of this variation. Like Douglas Millar with the classic version, I had to write a computer simulation to convince myself.
… another important conclusion should be to “know your random number generator.” Poorly implemented or improperly used (pseudo-)random number generators are a never-ending source of computer bugs, security holes, and simulation failures (as demonstrated here). Good job to Mr. Miller for not just accepting his initial results but looking at the problem a bit more deeply.
(For a good discussion of a recent bug due to a failure to understand computer random number generators, see http://www.metasploit.com/users/hdm/tools/debian-openssl/)
The thing that convinced me of this solution was drawing out the truth table for it. The solution pops out imediately. By swapping your choice you win everytime you pick a goat on the first selection.
i.e.
C= car
2= goat
3= goat
D1 D2 D3 choice Val reveal change
C 2 3 D1 C 2or3 3or2
C 3 2 D1 C 3or2 2or3
2 C 3 D1 2 3 C
2 3 C D1 2 3 C
3 C 2 D1 3 2 C
3 2 C D1 3 2 C
C 2 3 D2 2 3 C
C 3 2 D2 3 2 C
2 C 3 D2 C 2or3 3or2
2 3 C D2 3 2 C
3 C 2 D2 C 3or2 2or3
3 2 C D2 2 3 C
C 2 3 D3 3 2 C
C 3 2 D3 2 3 C
2 C 3 D3 3 2 C
2 3 C D3 C 2or3 3or2
3 C 2 D3 2 3 C
3 2 C D3 C 3or2 2or3
ouch that looked better with the tabs.
Interesting… I remember studying a completely different thing using C programming. I wanted to see what the chance of choosing a set value out of a group of size n at random in n chances. It kept coming up with about 62.2%. I was a young programmer at the time, and I didn’t think about the default c random number generator not being truly random.
Come to think of it.. I was actually almost solving the Monty Hall problem.
Thor, thanks for that table (I tabbed it out myself…)
Now I finally get it! If you pick the car to start with, then by this strategy you always lose. If you pick a goat to start with, then by this strategy you always win.
=> 2 out of 3, sweet.
Performing a few sanity checks on a Monte Carlo simulation can expose bugs and save millions of dollars.
Bear in mind that when you generate a sequence of random numbers, it won’t have the exact statistics that you had in mind, so you might want to post-process it.
A friend of mine made a convincing argument that the stock market is doomed: If a $100 share goes up 1% and then down 1%, or vice-versa, it loses $0.01 either way.
By this logic, if a share goes up 100% and then down 100%, it hits zero. What’s missing here, is that the probabilities are not equal. The probability of going up 100% is closer to the probability of dropping 50% than dropping 100%. A simulation confirmed this.
I was *going* to write a short computer program to test it out because I had a hard time understanding the thing initially. However, as I thought about the algorithm and how to best implement it, it was immediately obvious that the result was p=2/3 if you switch – at that point there was little reason to go ahead with the programming.
As another reader of the original post pointed out: imagine you had two contestants, they both chose different doors, and Monty showed that the third door had a goat. Do both contestants now have a p=2/3 chance of winning if they take the other door? My brain hurts when I try to enumerate outcomes and calculate probabilities – so I left that one alone.
There’s not much to enumerate. A third of the time, the third door will have the car, so both contestants are screwed. If the third door has a goat, then there’s an equal chance that either contestant has the car, so switching won’t help.
Andrew, No. 8 above, puts it well.
I drew a probability tree (and I confess I don’t recall how Shermer originally articulated the solution.)
At first, p(car) = 1/3 and p(goat)=2/3. If car, then p(switch – – > car) = 0. If not car, then p(switch – – > car) = 100%. Overall, p(switch – – > car) = 1/3*0 + 2/3 * 100% = 2/3.
JRS81
You have a 1/3 chance of being right on your first guess, and a 2/3 chance of being wrong on your first guess.
If you always switch, you therefore have a 1/3 chance of losing and a 2/3 chance of winning.
RE: #3. Walter Brameld IV – There’s an interesting variation on the Monty Hall problem, in which Monty doesn’t know which door hides the car. In that version, Monty guesses which door to open, and by chance reveals a goat. Should the contestant switch doors or not? It’s 50/50 in that scenario.
In restricting it to the scenarios in which Monty by chance reveals a goat, you have eliminated all the scenarios in which Monty would have, by chance, revealed a car. Thus, since you are not looking at all possibilities, only a limited subset, it is no surprise that switching is then 50/50. There were 3 options, you pick 1 (33% chance), then out of the 2 doors remaining (66% chance), you eliminate the cases in which Monty picks the car (33% chance), leaving one door to switch to that has the same odds as your original pick (33% chance).
If you do it with a million doors it becomes very obvious.
1) Pick a door from among one million doors.
2) I open all doors but two: yours and another. I guarantee that there is a new car behind one or the other.
3) Now it would be foolish to stay with your one-in-a-million door, and not switch to the other door I offer. The only way a switch would make you lose is if by 1-in-1,000,000 odds you happened to get the car on your first guess.
I disagree with your 3)
Imagine a host who wants the house (the gameshow) to win.
If on your first guess you pick door #937, whether you picked the correct door or not, the host has at least 999,998 doors which he/she can open and show you to not have the car.
If they make the other unopened door to be door #1 or if they make the other unopened door to be door #974,532 probably has some psychological influence on what door you think to be correct, however, the rules of the game have changed.
The new game is defined as two doors, one car. In reality your first “choice” was no choice at all. The choice is made when you decide on a door and the host opens it. (post rule-change)
You have a 50/50 shot, even after the host opens 999,998 wrong doors for you.
:)
switch or don’t switch. It makes no difference… but in this new game your odds have increased from 1/1,000,000 to 1/2, so at least smile for the camera.
What. I’m sorry, but you’re not thinking properly.
Remember, none of the opened doors had the car. And, if you think about it, whenever you were to actually win the game you would have lost upon swap. However, isn’t the opposite true?
Think of it like this: Upon picking a door they are split apart into two sets: set A, the door you picked, has a 1/1.000.000 chance of winning. Set B, the doors you did not pick, have a 999.999/1.000.000 chance. Then, all but one of the doors in set B are shown to be wrong. Then, all those chances collapse into the one unopened door, which is now set B. The chances of the sets did not chance, but what set B was composed of was. All those chances are now set upon one, and only one door.
Do you understand now how this problem works?
It is pretty simple and doesn’t need a computer prog to understand.
3 doors, 2 goats, 1 car.
1. Open any door, what is “probably” behind it (2 out of 3 chance)? Ans a goat
2. Monty Hall reveals a goat, you “probably” have a goat in your box.
3. What is “probably” in the 3rd box – Ans a car.
Should you change? Only if you prefer cars to goats!
I first read this problem in Marilyn vos Savant’s column, and had a big argument with one of my brothers about it. I wrote a BASIC program to test it and came up with 66.5% odds of winning when you switch doors. My brother also wrote a program, but his came up with 50% odds. I asked him to send me his code so I could look at it, and found a line in it that forced the 50% result. He was making the classic error of assuming the result in formulating the test.
Logic puzzles like the Monty Hall problem are a lot like magic tricks — they fool you with misdirection. By way the problem is posed forces you to think of switching as a 50/50 chance. You have to get outside of that box to understand it.
Here’s one way:
After you choose a door, there are 3 possible scenarios for the other 2 doors. Goat/Goat (1/3 chance), Car/Goat (1/3 chance) or Goat/Car (1/3 chance). Monty Hall is always going to reveal the goat, so he effectively combines the second two options, making the chances 2 out of 3 that the door he didn’t open is hiding a car.
I don’t see why it would make a difference if Monty is choosing randomly. If he shows you the car, then you switch to that door. If he shows you a goat, there’s still a 2/3rds chance that the other door is hiding a car because only had a 1/3 chance of having picked the car originally.
What happens if the goat is a $1.5 million champion breed stock and the car is a ’73 Gremlin X?
While I understand the probabilities in the ‘correct’ answer I have always thought there was a little too much ambiguity in the problem to start with. Perhaps the price of the car comes out of Monty’s bonus check and he only shows you a goat and tempts you to change when you have correctly picked the car in the first place… They don’t really say that he will always offer you the choice to switch, just that he has.
Exactly. The human factor is nearly always at least as important as the purely statistical factor…
So, to get a better analysis of the situation, you’d want to watch Mr. Hall’s patterns for a while before trying your hand at the game.
The simple reason why this cool math works:
when you begin you have only a 1/3 chance of being correct.
by switching doors you are really just ‘betting’ that your first guess was wrong…thus changing your odds to 2/3.
Max’s comment yesterday made me pause a moment: “A friend of mine made a convincing argument that the stock market is doomed: If a $100 share goes up 1% and then down 1%, or vice-versa, it loses $0.01 either way.”
The way it is worded is not a true statement. My $100 share goes up 1% today, and I now have $101. However if it “then” goes down 1% the next day, my share is now worth only $99.99.
Statistics are all well and good, I love statistics, but YOU only get ONE shot at the car. It’s not like you get to play 100 times and want to maximize your take at @66 cars. How are you, the brilliant mathmetician, going to feel when having intelligently switched your choice, end up with the goat? Hah! Against the odds, you had picked the car on your first choice and just traded it away. And it was a nice car too. Your dream car. Manhood on four wheels. And you picked it and traded it away. Mental flagellation for years. :-(
No computer program is needed.
Door 1 has a goat
Door 2 had a goat
Door 3 has a car
Pick Door 1 > Shown Door 2 > Switching wins
Pick Door 2 > Shown Door 1 > Switching wins
Pick Door 3 > Shown Door 1 or 2 > Switching loses
I’m gonna switch.
Scott See
White Salmon, WA
RE: #21 Pete – I think you’ve misunderstood Max’s statement, which is true as stands. He says:
“A friend of mine made a convincing argument that the stock market is doomed: If a $100 share goes up 1% and then down 1%, or vice-versa, it loses $0.01 either way”
He’s positing 2 scenarios –
Scenario 1, the share goes up 1% one day (to $101.00), and then down 1% the next (to $101.00-$1.01 = $99.99, a 1 cent loss).
Scenario 2, the share goes down 1% one day (to $99.00), and then up 1% the next (to $99.00 + $0.99 = $99.99, again a 1 cent loss).
“Perhaps the price of the car comes out of Monty’s bonus check and he only shows you a goat and tempts you to change when you have correctly picked the car in the first place…”
When the issue of second-guessing an opponent’s motives comes into play it’s a whole different dynamic. There are champion ro-sham-bo players who have learned to cold read their opponents and outsmart them.
Marilyn Vos Savant published this in 1990, and it still generates so much interest that it is part of her website today. http://www.marilynvossavant.com/articles/gameshow.html
Far too many people who should have known better wrote in to tell Marilyn that she was wrong. Thanks Michael for reminding us to look at everything carefully.
Here’s another counterintuitive thing. If Monty offered to open a door with a goat BEFORE you choose one of the two remaining doors, doesn’t that sound like a good offer? Then, your chance of winning really would be 50/50.
The Monty Hall problem is interesting from a number of perspectives. In a sense the truth-table makes the logical solution almost trivial, but there also seem to be psychological or framing issues to the problem that make it difficult to explain and interesting to implement as a game show.
A logical framing of the three-door problem might go like this:
Condition 1:
a. On the first go-around of choosing, you choose a car.
b. It is then revealed that behind one of the remaining doors there is a goat.
c. What is going to be behind the other door? A goat – since you already choose the car.
Condition 2:
a. On the first go-around of choosing, you choose a goat.
b. It is then revealed that behind one of the remaining doors there is a goat.
c. What is going to be behind the other door? The car.
Since there are two goats and only one car, the odds that you chose a goat on the first go-around is 2 out of 3. In other words, Condition 2 will be the case 2 out of 3 times. In still other words, it is more likely that on the first go-around you made the “wrong” choice and that by switching, you are more likely to get the car.
Now comes the psychological part:
Consider a thought experiment:
A person with high self-esteem or a sense of self-efficacy probably feels or believes that his or her actions in the world will result in their intended consequences. I choose, for example, to go into computer programming because I feel or believe that this course of study will result in my happiness. If I believe that the outcomes of this world are under my control because I have applied reason or intuition, then if I am asked to reconsider my choice based on what seems to be no further evidence, I would be disinclined to do so. Generally, most of the time, when choice is deliberate and willed, I choose to make the “right” decision, not the “wrong” decision.
On the other hand, a person who believes that they are blown about by the winds of chance, who has low self-esteem or a feeling of low self-efficacy, who might feel that one choice is as good as another since they never make the right choice anyway, may have little commitment to any choice that they make.
Of these two people, who is more likely to stick with a decision that they have made?
The experiment would ask, Do people with a certain personality trait, as measured on a scale of self-esteem, or self-efficacy, or introversion/extroversion, be more likely to stick with their first choice, interpreting their unwillingness to switch as an affirmation of self (not being a “flip-flopper”, for example).
The hypothesis would be, people with high self-esteem are more likely NOT to switch on the second go-around in the Month Hall problem since doing so would disconfirm their decision and in a sense, themselves.
Since the logical problem embodied in the Monthy Hall problem is difficult to figure out in your head while being pressured by Monty Hall and a screaming audience, a person would tend under such conditions to go with their personality inclination: some would stick to their guns while others would allow the authority figure to sway them.
In other words, it may be our personality traits, or our culture of denigrating “flip-floppers”, that makes understanding the logical argument more difficult.
That’s the drama of the Monty Hall game.
As a framing of the problem, I like Joseph Francis’ example: So, as the probability of making the right choice the first time approaches zero, the probability of your making the right choice the second time by switching approaches certainty.
This reveals another aspect of the psychological problem: If Monty Hall had used 1000 doors and opened 998 of them to reveal goats, perhaps the odds of your making the wrong choice the first time becomes clearer. Confronted by 1000 effectively identical doors you may realize that any choice is as good as another and so any choice is random.
Instead of looking at this as a one-person game with a thousand choices, consider a two-person game between You and Monty Hall, where You get to make one choice and Monty gets to make 999 choices. The odds are certain that one of you has made the right choice, but who is more likely to have the right choice among their options? Regardless of when the right choice comes up for Monty, the odds are 999 in a 1000 that he will have the right choice among his 999 options. For Monty, revealing all his wrong choices first is only a matter of drama, the odds will always be in Monty’s favor. If Monty then asks You to come over to his side – the side that has 999 chances out of 1000 of getting the right choice – doesn’t it make more sense to switch?
With three initial choices which are then apparently whittled down to two, perhaps one might think that you now have two choices and so frame the problem as a fair-coin toss. But in fact, the initial number of choices remains three. It is just that certain outcomes have been revealed to you. Framing the second probability as a fair-coin toss may be caused by interpreting the problem as a one-person game. If you see the problem as a one-person game then you might think the issue is between you and the choices you have to make, and since opening doors seems to reduce the number of choices, you might conclude that once all doors but two (of say 1000 doors) have been opened, that you have only two choices. But opening doors provides no real information; you still don’t know what is behind the door you initially chose, and the odds that you chose the right one is still 1 in 1000.
So the Monty Hall problem is not only a logical problem, but one that also incorporates issues of seduction, personality traits and the ability to frame an issue.
Interesting psychological assessment. But you forget to take into account the psychology of Monty Hall. He’s more likely to offer the option to change doors (they were really curtains, by the way, not doors) if the contestant has picked the big prize initially. Thus, sticking to one’s guns might actually, statistically, be the best thing to do.
Doug Miller wrote, “As I mentioned, this little exercise opened my eyes to the true meaning and power of the scientific method.” I am not impressed. A true scientist would understand logic well enought that he would not have to conduct any such experiment. This is not a triumph of the scientific method – it is a failure of logic!
Last April, just ’cause I was bored, I wrote a sim that always changes from the original door picked, and ran it for 10,000,000,000 iterations, here are the results from the log:-
The time taken is in seconds
10 Iterations run of 1000000000 games. Time started 15:14:59
Won 666661364 Lost 333338636 Time taken = 8056.780762
Won 666659373 Lost 333340627 Time taken = 10846.781250
Won 666684099 Lost 333315901 Time taken = 10834.750000
Won 666665023 Lost 333334977 Time taken = 10874.563477
Won 666704475 Lost 333295525 Time taken = 10865.264648
Won 666666122 Lost 333333878 Time taken = 10298.046875
Won 666690047 Lost 333309953 Time taken = 6257.594238
Won 666692321 Lost 333307679 Time taken = 6523.046875
Won 666688961 Lost 333311039 Time taken = 7181.891113
Won 666669454 Lost 333330546 Time taken = 6774.219238
Won 6666781239 vs. Lost 3333218761
’nuff said :)
I think the most interesting thread of this is in the set up where Monty shows you a goat. In practice, I think he had other moves like any good magician who knows where the picked card is, offers a split of the cards and selectivly disgards the chunk without the chosen card only after asking the picker to choose a pile (‘Well, lets keep those then..’ and continues the trick or ‘okay, lets get rid of those’ etc.)
For instance, in the case of the initially picked door having a car behind it, skipping the revealed goat and offering a switch right away…
It becomes intuitive if your turn the problem on it’s side: Your first guess should not be to guess the door with the car. The first 2 turns (yours and Monty’s) should be made to eliminate the goats and therefore leave the car! We know Monty will reveal a goat. That means that you have a 2/3 chance of picking (eliminating) a goat on the first guess, and thereby leave the car behind the last door.
Look at it this way: After you choose door A, Hall gives you the option of keeping A or taking BOTH doors B and C. Clearly, having two doors gives you twice the chance of winning as having one, so you should switch. Then Hall opens B or C, obviously choosing one without the car behind it. But, assuming Hall is in the know, there is absolutely no difference between this scenario and the game as it is played! There’s a one-third chance the car’s behind A, and that you’re giving it up by switching, and a two-third chance it’s behind B or C, and that you’re winning it by switching. The timing of when a redundant, car-less B or C is removed from the game is irrelevant.
I just realised why people go with the 50/50 thing. It’s because they are thinking about what choices are left AFTER Monty has revealed a goat not what choices they start with. But after the goat reveal is not where the probability of winning is. It occurs at the beginning before anything is revealed. When the contestant is selecting their door at the start there are two chances they have selected a goat and only one they have selected the car so switching gives a higher chance of getting the car. It doesn’t matter what happens subsequently.
I was always convinced that switching was correct, but it was a while before I developed a good intuition as to why.
Basically, consider a similar problem:
There are 100 doors, only one of which has a car behind it. You pick one. Monty opens 98 of the other doors and then offers you to switch doors. In this case, it’s REALLY obvious: You should definitely switch!
Responding to #23, Scott See.
I think the more complete list of options would be:
Door 1 has a goat
Door 2 had a goat
Door 3 has a car
(Step 1) > (Step 2) > (Step 3) > (Step 4)
Pick Door 1 > Shown Door 2 > Don’t switch > Lose
Pick Door 2 > Shown Door 1 > Don’t switch > Lose
Pick Door 3 > Shown Door 1 > Don’t switch > Win
Pick Door 3 > Shown Door 2 > Don’t switch > Win
Pick Door 1 > Shown Door 2 > Switch > Win
Pick Door 2 > Shown Door 1 > Switch > Win
Pick Door 3 > Shown Door 1 > Switch > Lose
Pick Door 3 > Shown Door 2 > Switch > Lose
I admit, I still don’t “get it”. By the above logic, when you reach Step 3, switching or not switching leads to the same probability of winning or losing…
Sid, assign probabilities
1/3 chance: Pick Door 1 > Shown Door 2 > Switch > Win
1/3 chance: Pick Door 2 > Shown Door 1 > Switch > Win
1/6 chance: Pick Door 3 > Shown Door 1 > Switch > Lose
1/6 chance: Pick Door 3 > Shown Door 2 > Switch > Lose
That’s 2/3 chance of winning and 1/6+1/6=1/3 chance of losing.
This is a classic mathematical puzzle. We should have a contest on who can explain it in simplest terms so everybody can easily understand it. Here’s my take. I know my chances of being wrong in my first pick are two out of three (2/3). Monty shows me a goat and asks if I want to switch. I tell myself, I will only switch if I know my first pick was wrong. So what are the chances my first pick was wrong? Well, it’s 2/3 which means I’m more likely to be wrong than be right (1/3). So switch.
If there are 100 doors, my chances of being wrong in my first pick are 99 out of 100. Monty shows me 98 goats. I tell myself, I will only switch if I know my first pick was wrong. So what are the chances my first pick was wrong? Well, it’s 99/100. So switch!
I think half the trouble comes from regarding the goat as a LOSE! Consider the strategy in a society where the goat may be a coveted prize, and where a car is of no darn use at all..
— Pick a door, any door will do. They’re all twice as likely to carry a prize goat than a useless hunk of car, so it doesn’t matter which is chosen.
— What changes as the other goat is revealed, is that the odds have been *equalised*… not to 50/50, which is the mistake people make, but to 2:1 in favour on BOTH doors. Your door is STILL 2:1 in favour of a goat, same as it’s been all along, and the last door is also now ALSO 2:1, but in favour of a car!
— You maximise your chances of a nice goat to take back to the rainforest by staying exactly where are!! :-)
The correct response when Monty Hall offers you to switch doors is, “hey, what the hell’s going on here? This show hasn’t been on in like 20 years! This is a dream, right?”
Per #23-why switch?-what you listed was a 50/50 proposition if you split the last two options into two options instead of having them on the same line as one option.
If you pick door 3 with the car, and you indeed switch if they show either door 1 or door 2 each with a goat, that is 2 losing options. 2 losing options vs the 2 winning options that you listed- is 50/50.
Perhaps a computer is needed. :-)
Neil, see #37
This can be clarified by changing the problem slightly.
Just think how the dynamics change if there are two cars and one goat. Monty Hall always reveals a car (but you cannot have it).
Now by switching you would have a 2/3 chance of getting the goat.
By not switching you have a 2/3 chance of getting the car (ie there is a 2/3 probability you chose a door with a car on your first pic.
Max-question.
Why is-Pick Door 1 > Shown Door 2 > Switch > Win… a 1/3 chance.. but
Pick Door 3 > Shown Door 1 > Switch > Lose…. a 1/6 chance?
Why are the odds different?
This is kinda fun.
Hey Neil,
In answer to 44.
You have
Pick door 1 (1x) = 1/3
Pick door 2 (1x) = 1/3
Pick door 3 (2x) = 1/3 divided by 2 (or 1/6)
You are playing out FOUR separate scenarios when there are only actually three. If you pick door 3, he is going to reveal a door with a goat. It only happens once. It doesn’t matter which door he opens! It isn’t two separate scenarios where he opens door 1 in one instance or door 2 in another. It is one scenario where he opens either door 1 OR door 2. The specific goat he reveals is irrelevant. Switching is a loss if you pick door 3.
So:
Pick door 1 = switch = win
Pick door 2 = switch = win
Pick door 3 = switch = lose (again, it doesn’t matter which door he opens, it isn’t a separate scenario for each door).
Hope I helped instead of confusing the issue!
~Ben
Ah, thanks Ben. Sometimes I am a little dense, and obviously it is plain to see why I am not a mathematician!
Regards.
Thanks to Mark in #33:
“Look at it this way: After you choose door A, Hall gives you the option of keeping A or taking BOTH doors B and C.”
Lightbulb Moment!
There’s nothing wrong with splitting scenarios as long as the probabilities add up.
The scenario of picking a door has probability 1.
It’s split into three scenarios, each with probability 1/3.
Then, one of these three scenarios is further split into two scenarios, whose probabilities add up to 1/3.
By “scenario” I mean from start to finish.
Barrie and Anthony,
No matter how you set it up, people are afraid to switch the door because of loss aversion: they hate the possibility of selecting the right door only to give it up.
Concise General Statement and Solution of Monty Hall Problem
Two required math facts: For any integer n > 2, (n-1)/n > 1/n; The probability (Pr) of either event A or event B is the sum of Pr(A) and Pr(B) and equals 1, provided that Pr (A and B) = 0 and the two events exhaust the universe of events under consideration
Let there be n > 2 rooms, one which contains a car, all others being empty. If using a prescribed method of selection you pick the room with the car, you win the car. Here is the method: You pick one of the n rooms at random; i.e. with probability 1/n. Then (n-2) empty rooms are removed from the drawing and you are left with the option of selecting one of the two remaining rooms. Now, if you want to have the best chance of winning the car, do you keep your first selection or switch? That is the problem.
Solution: Since the car is certainly in one and only one of the rooms, the two probabilities are 1/n and 1-1/n = (n-1) /n. So our math facts say to switch and also that you increase your chances (n-1) times by doing so. The original Monty Hall problem was for n=3 and had several mentions of goats and doors.
So guys who has the simplest explanation? By simple, I mean the easiest to understand by anybody.
Look at my explanation #38. Isn’t it obvious? (self-serving statement) :-)
You don’t need a computer simulation to figure out the odds. There are nine alternatives when you switch (3 doors x 3 choices) and nine when you stay stay, for a total of 18 possibilities. Switching wins 12 times and staying wins 6.
You don’t need to run a C program 1,000,000 times to figure out the chances of getting any number on a six-sided die is one in six.
Yes, but Joseph Francis beat you to it.
Look at #13 and #15.
Let me take a crack:
1/3 of the time, you picked the car to start with; Monte has two goats, and if you switch, you lose.
The other 2/3, Monte has a car and a goat, and he tells you which one
is the goat. Switch and you win.
Again: 1/3 of the time you pick the car; 2/3 of the time Monte is telling you which one is the car.
I remain quite miffed by this underlying presumption that the goat is always considered to be the lesser prize.
Dr. Billy Nan Cudlow
American Goat Anti-Defamation League
Yeah, actually, I don’t want either. I hate cars, and goats are not something I think one can own, so I’d have to let it go free, it probably wouldn’t be very happy living in the Boston area.
A new commuter bike, with integrated generator lights and disk brakes, now that I’d think was worth doing the math for.
Please do not post if similar already posted.
The answer is that Monty Hall is not making a random choice. He knows the car/goat locations, so he only reveals goats. He is being nice to 2 out of 3 contestants. Probably enlightened self-interest, more happiness = more viewers.
Just watch the movie “21”. The hero explains it pretty well (he also tricks the bad guy and gets the girl – oops, spoilers).
That’s where instead of saying “two out of three”, he says “0.667” to sound fancy.
I thought of a simple explanation I don’t think has been commented upon yet…
The only way you lose by switching is if you got lucky and picked the car right out of the gate. There’s only a 1 out of 3 chance you did that, ergo there’s only a 1 out of 3 chance you’ll lose when you switch.
Suppose you are playing a seven door version of the game. You choose three doors. Monty now opens three of the remaining doors to show you that there is no prize behind it. He then says, “Would you like to stick with the three doors you have chosen, or would you prefer to swap them for the one other door I have not opened?” What do you do? Do you stick with your three doors or do you make the 3 for 1 swap he is offering?
I say you should swap. That gives you a 4/7 chance to win.
The interesting thing with these simulations is that the computer won’t generate a truly random pattern anyway :-)
RE: 61, Peter – Computers can easily use true randomness in their simulations. One example is the use of atmospheric noise in the RNG at http://www.random.org/
Re comment #3:”There’s an interesting variation on the Monty Hall problem, in which Monty doesn’t know which door hides the car. In that version, Monty guesses which door to open, and by chance reveals a goat. Should the contestant switch doors or not? It’s 50/50 in that scenario.
And, yes, I didn’t believe it was 50/50 when I first heard of this variation. Like Douglas Millar with the classic version, I had to write a computer simulation to convince myself.”
I believe this is in error for the following reason: On any given single trial, if Monty reveals a goat whether knowingly or by chance, the odds of your original pick is still 1/3 and the remaining door will be 2/3. Just because he guessed did not change your odds.
Re # 10 (MadScientist) “As another reader of the original post pointed out: imagine you had two contestants, they both chose different doors, and Monty showed that the third door had a goat. Do both contestants now have a p=2/3 chance of winning if they take the other door? My brain hurts when I try to enumerate outcomes and calculate probabilities – so I left that one alone.”
The puzzle is not valid for 2 parties, for the fact that it would require eliminating the scenario of both parties choosing the goats. If that happened Monty’s only option would be to reveal a car, which is not a condition of the puzzle. Removing that case every time it came up would change the statistical outcome to 50/50 ,but that is an apple not and orange, so to speak.
Thank-you Andrew #8 that is by far the easiest way to explain it to someone who doen’t get it. And for those who have a hard time accepting it, this explains it in a way that you can “feel” it’s the right answer! Simply put if you pick a car initially you lose, if you pick a goat you win- and there’s twice as many goats as cars!
This is meant to be a sceptic’s forum. The problem in today society is that too many people seem to be quite willing to abandoning genuine understanding for faith.
I cannot believe that Dr Shermer is promoting a computer simulation as a way to understand this, what is basically, an elementary statistical problem….
I will state the mathematically same problem in a different way. If people reading the problem cannot use their intuition to decide on the best outcome – then the problem lies in their statistical intuition. The majority find that they can be guided by common sense and intuition.
You are in the audience and Monty says come on down. There are 10 doors on stage and he explains that there is a car behind one of these doors and the remainder have goats.
Monty asks you to divide these doors up so there is one door on this side and nine doors on the other side of the stage.
He then asks you to pick a side which you think the car might be on.
You use your intuition to try to maximise your chances and choose the side with nine doors.
Monty says, now I want you to bring your wife down and look behind the doors on the side you have chosen and to eliminate eight goat doors .
His wife then removes eight doors which she shows are goats and discards them to the back of the stage.
Monty then says, Ok we now have two closed doors, one on either side of the stage. If the door contains the car then you will win it.
Would you like to swap sides to the single door or keep to the side of the stage you have chosen and chose the remaining door left by your wife?
I think you would have a very bad intuition for statistics if you decided to swap to the unopened door on the other side of the stage. (The side that did not undergo any elimination process by the contestants wife.)
Now this is mathematically the same problem. If you can understand the above problem, then you have also understood mathematically the problem as Dr Shermer posed it….
Why is this mathematically the same problem?
The fundamental of these types of problems is the ‘ELIMINATION PROCESS’ performed on a selected subset of potentials.
As Shermer stated the Monty Hall problem, Monty performs this ELIMINATION PROCESS on a subset of two doors (from a set of three potentials).
In the way I have stated the problem above it is the contestant’s wife that performs an ELIMINATION PROCESS (this time on a subset of 9 doors from a set of 10 potentials).
The increased door numbers was purely used to amplify this ELIMINATION PROCESS and therefore the disparate odds between the two choices.
There is a social issue that this touches on. We live in a society that wants to believe what it believes regardless of the truth.
A co-worker comes up and says, “He said, she said….” and then draws a conclusion. Yet they based that conclusion on their feelings and not facts. There are three variables. What was said, what was meant and what the person thought they were saying. (if you figure in a bad hair day or bad hair life, then it becomes the 10 door problem)
For instance, a co-worker was diagnosed with a glioma, a brain cancer. At 24, a new mom and just starting her career in health care, this was devastating to all around her.
A co-worker came up and said, “You must feel awful.”
On asking her how she felt she said, “That’s just it, I don’t feel anything. The part of my brain that it is affecting is the center of my emotions. I have no emotions.
We believe what we want to believe, sometimes failing to consider the facts, be they scientific or just plain asking.
Here’s another counterintuitive problem.
There are three closed boxes: one contains two gold coins, one contains two silver coins, and one contains a gold coin and a silver coin. You reach into a random box and pull out a random coin that turns out to be gold. What’s the probability that the remaining coin is silver?
In fact this is not an example of the “scientific method”.
I think it was Francis Bacon who first formally described the scientific method.
Computers can be used in mathematical problems and even some proofs like the ‘four colour problem’ where the computer was used to eliminate many of the mundane possible arrangements. The computer is used here as a tool. Theoretically this work could be done by humans.
The understanding comes from devising the proof (and possibly devising the code for programming of the computers so that it does the work you require).
So the use of computers does not apply to the advance of any real mathematical understanding.
As a society we have discovered a vast landscape of mathematics over the past 2500 years. It seems that our physical reality is a subset of this mathematical realm.
It is the scientist job to hypothesis what is the mathematics of our Physical World, then to prod nature to find out if it behaves as the hypothesised mathematics suggest.
Today the Standard Model and General Relativity describe the physical world to great accuracies. In fact there have not been experiments performed in the last 20 years or so that deviate from the mathematics of GR or the Standard Model of quantum mechanics.
Summing up the ‘scientific method’ is not relevant or applicable to computers churning through trials of a statistical problem and certainly no real understanding of the problem can come from such procedures.
I realize that this is a counter intuitive problem, and if I am wrong, I would really like someone to explain it further, but as Monty is going to reveal an incorrect door, and then ask you if you want to switch, doesn’t that reset the game? And as such, isn’t the choice now 50/50?
I agree that in the original choice, there as a 2/3 chance of picking a goat, but now the choice is again offered, and who will pick the revealed goat door?
I’d appreciate some illumination on this.
I propose an experiment that can be conducted on the web.
You’ll need a web programmer, because I’m sure as heck not going to do it.
Step 1: Randomize the placement of Goats/Car behind three doors (e.g. GGC, GCG, CGG).
Step 2: Participant chooses one of the doors.
Step 3: Show one of the two goats behind a door not chosen by the participant (not sure if this needs to be randomized).
Step 3: Participant is asked if he/she would like to stand or switch.
Step 4: The big reveal! Participant is informed of winning/loosing the game.
Step 5: Record the results in a huge spreadsheet and publish in shi-shi-poo-poo journal, noting me as co-author :-)
The key to understanding the Monty Hall Problem is random choice versus non-random choice in the 2 decisions made by the contestant.
The 1st choice is for a car-prize ‘randomly’ placed behind one of 3 doors. Random means that each door has an exactly ‘equal chance/probability’ of hiding the car– thus, a contestant can easily and correctly figure he has a 1-of-3 {.33} chance of picking the right door.
But the 2nd choice becomes ‘non-random’, after Monty Hall opens another door. There is NOT an ‘equal’ chance of the car being behind the 2 remaining doors… as most people assume. The instinctive probability calculation 1-of-2 {.50} chance is FALSE because it “requires” a random basis of choice.
Monty Hall personally interferes with the game probabilities, much like someone substituting a biased/trick coin into a previously ‘fair’ coin toss series… the other player doesn’t realize the game has changed significantly.
Since Monty will always reveal the “Goat-Door” before the contestant makes his second decision, Monty artificially doubles the original chance/probability (to .66) of the car being behind the remaining door not first chosen by the contestant.
My question is this. if there were 4 doors instead of 3, and monty opened only one door leaving you with an option of 3 doors instead of opening 2 doors and leaving you with an option of 2 doors. and he asked you 2 times if you would like to switch( because you always switch). does that mean you have a higher probability of winning in this situation?
Matt, no it would only give you an advantage to change from your original choice. After that it would be random.
i dont know if anyone has done this yet but…..
there is a car, a goat named bill, and a goat named jeff. bill and jeff are identical twins.
pick a car – show jeff – switch – win bill – GOAT
pick a car – show jeff – dont switch – CAR
pick a car – show bill – switch – win jeff – GOAT
pick a car – show bill – don’t switch – win car – CAR
pick bill – show jeff – don’t switch – win bill – GOAT
pick bill – show jeff – switch – win car – CAR
pick jeff – show bill – don’t switch – win jeff – GOAT
pick jeff – show bill – switch – win car – CAR
or
there are three people who each get to pick one of the three boxes.
two cannot pick the same box.
it is revealed one of them picked the goat.
he’s out.
the remaining two are both given the option to switch.
they can both pick the same box.
its not revealed until they have both decided.
2 people – 2 options – 2 boxes – 1 car – 1 goat.
a is right – switches – GOAT
a is right – doesn’t switch – CAR
b is right – switches – GOAT
b is right – doesn’t switch – CAR
a is wrong – doesn’t switch – GOAT
a is wrong – switches – CAR
b is wrong – doesn’t switch – GOAT
b is wrong – switches – CAR
the act of the contestant picking a door precludes the game ever being a 50/50 gamble in the 2nd round. the event determines the future probabilities
this game is designed to confuse people. it is nonsense like this that scares young people away from anything beyond basic math. if you want sleight of hand go to a magic show