I am taking Prob & Stats this semester, and this article does as good a job as my teacher in explaining this concept.
Very well done!
The java script game is fun and helps explain the concept. Not that it matters but when there are more than 3 doors I can almost always predict which door has the car due to how your code is written. Letâs say for 10 doors, if I always pick the first door then the other door remaining will be the car. Unless the first door is the car in which case the remaining door is the last in the series of doors. So in the scenario where I always pick the first door and the first door is the car then I know it will be a car because the last door in the series is not revealed as goat (the other scenario is the last door is the car). You should randomize which door is left unrevealed when the user picks the car on the first choice.
i still donât get itâŚ
Suppose we have 4 doors.
suppose the player choose door A. Considering the 4 possible cases:
A,B,C,D
i) 1,0,0,0
ii) 0,1,0,0
iii) 0,0,1,0
iv) 0,0,0,1
1 mean that car is behind the doorâŚThe solution suppose that if we donât switch, we win for 1st case only, so 1/4. If we switch, we have 3 case to win, so 3/4âŚ
But why case ii, iii and iv are different? I still believe that we have only 2 cases (i and ii or iii or iv) , because they will alway open the door without carâŚ
@vichet: Great question. Youâre right that situation i) has one outcome (you win), and situation ii) iii) and iv) have a different one (you lose).
However, since each situation is equally likely, itâs much more likely that youâll lose. Imagine a dice with 4 sides â if you roll a 1 you win, if you roll a 2 3 or 4 you lose. Even though itâs only 2 cases (1 or 2/3/4), you wouldnât say winning and losing are equally likely, right?
Thatâs one of the tricky things â separating the 2 possible outcomes (win or lose) from the number of ways to get those outcomes (1 way vs 3 in your case). Hope this helps!
Suppose there are three doors, A,B and C and you originally chose door A. If you stay with your original door, then the only way that you win is if originally the prize was behind that door A, which has a chance of 1 in 3. If the prize was originally behind door B on the other hand (which has a chance of 1 in 3), then when you pick door A, door C will be removed. Hence, if you switch you will be switching to door B, and therefore you will win. Finally, if the prize was originally behind door C (which again has a chance of 1 in 3) then door B will be removed, and if you switch you will be switching to door C and therefore will win. Hence, if you stay with your original door, you win if and only if the prize was originally behind door A. If you switch though, you win if it was originally behind either door B or door C. Since the chance the prize being behind door A from the get go is 1 in 3, whereas the chance of it being behind either B or C from the get go is 2 in 3, you are better off switching!
In a way, itâs still a 50/50. âMontyâ may have eliminated a door, but youâre then left with two doors. Just because he eliminated the third doesnât mean itâd be better to switch.
By the way, from switching I won 3 times, lost 7.
Let me post 2 cases to challenge those that have assumed 2 things.
- Initial probability of the chosen door remains unchanged no matter how much info is revealed by the host on the remaining doors.
- Given the choice to switch, always better to switch to higher probability on the accumulated remaining choices.
My case assumptions: 4 doors, one with prize, Door1 was chosen, Host will reveal 0, 1, 2 or 3 doors ; Choice will be given to switch/stay.
(Bear with me on the slight differences to the original example)
Case A: Host decides to reveal 0 doors
: P ( Door1 = Prize ) = 25%
: P ( Door2/Door3/Door4 = Prize ) = 75%
Ambiguity: It appears that switching your choice will increase your chance by 3 times ( 25% -> 75% ). However it doesnât make sense since no extra info is revealed to you. It be the same as choosing a different door with 25% chance of hitting jackpot.
Case B: Host decides to reveal ALL of the remaining doors.
: P ( Door1 = Prize ) = 25%?
in my opinion,
: P ( Door1 = Prize ) = 0% or 100% (base on what is revealed)
Ambiguity: Many have stated that the initial probability of the Door choice is to remain unchanged despite additional info provided by the host. In this case, 2 possibilities can happen, host reveals the prize by opening all the remaining doors or reveals all the goats. Base on what is revealed, probability of Door1 being prize or not is DIRECTLY AFFECTED.
Questions:
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Why do many claim that the probability of the initial door choice should remain unchanged when more info is given? I have given a counter case that suggests otherwise.
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Why do many claim switching your choice to a larger portion of aggregate probabilities always earns you a higher chance? I believe in some cases, it is not apparent that you should switch like in Case A, switching or not makes no difference if no information is revealed. Even though it would be an act of switching to a higher probability.
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Is it right to say probability is just a measure of how confident one is of hitting the jackpot, it does not actually mean a higher number corresponds to higher occurrence in every situation?
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Probability is indirectly proportional to the total number of possibilities unless i am mistaken. When a host reveals the location of some of the goats, the total number of possibilities of the actual location of the prize is directly reduced, how can the probability if Door 1 is a prize be unaffected?
Will the following measurement make more sense?
Case C: Host decides to reveal 2 out of 3 doors
(assuming Door2 / Door3 is revealed to be goats )
: P ( Door1 = Prize ) = 25% + Adjustment
: P ( Door4 = Prize ) = 25% + Adjustment
Adjustment = Total% from revealed Doors / # of Doors remaining.
It just makes sense that ALL the unrevealed doors should have higher probability of being the prize including the initial choice that was chosen with a 25% confidence.
Sorry for the presentation if it confuses anyone, I am rather unconvinced at what the majority is accepting which motivates me to find out if I am wrong in my understanding.
Do advise if you understand what I am trying to say in my post.
only problem with your implementation is that it always picks the leftmost door that isnât the car so if you always choose the rightmost door and it shows the middle door as a goat then the leftmost door will always be a car
position of the door do not matter.
whichever chosen in the beginning can be assigned as door1 for reference. The remaining doors regardless of position will be considered as per stated in my previous example.
my main point is that once additional information on the remaining doors are revealed, the effect is propagated to the probability of the first chosen door. So eventually, it always end in a 50/50 chance whether your first door contains the prize. Which in my opinion is a rather true reflection of real life choices.
Much has argued that remaining doors collectively contain a higher chance, then follow through their calculations while assuming the probability calculated on the first chosen door should remain the same throughout regardless of additional information revealed on the remaining doors, which i find rather weird.
I think youâre wrong. In the game, whatever your choices are at the start, in the end you will have two doors. You shouldnât consider the probability of 1/3 from the first stage at all because it doesnât mean a thing; it doesnât help; but unfortunately you do compare it to the second stage probability of 1/2, the actual game probability.
By eliminating the rest of the bad doors you are not helped at all, in the end there will always be just 2 doors with equal probabilities.
Itâs like you always start the game with 2 doors: One with the car ( which might be your first choice ) and one with the goat.
If youâve picked the car(but donât know it) and switch, your chance of winning are 50% (because the other door, that remained after elimination can have your car or a goat - you donât know what is behind your picked door, i.e a goat or the car).
If you stick with your first choice you have the same probability of winning the car -50%.
I think youâre letting yourself be confused by the initial probability of winning equal to 1/3 (in case of 3 doors) that is less than the actual probability (1/2) at which you play the game in the final stage.
The first stage and the last stage of the game are not interconnected, so in the final stage youâll not be helped by the decision you made initially.
Youâll always start from scratch and have to pick between a car and a goat, hidden from you. Revealing the other doors, doesnât mean that you picked a goat, it doesnât give you any info about the quality of you pick, it just narrows the the possibilities from many to just 2.
That is what I i understood from reading your article, my opinion is different from yours, i believe am right, but thatâs just me.
I love when this question comes up, sooooo many people never get it. The way I get most people to see it is by looking at Deal or No Deal. Say you have selected your case and there are 24(or is it 25? no matter) you have a 1/24 chance of having the million in your case, most people can agree with that. If that is true, then the horde of models have a 23/24 chance of having the million.
When it gets to the final case, and you can either keep your case or switch, you do not have a 50% shot as seems intuitive (2 options) you still have 1/24, they still have 23/24.
If you still have a problem, would you agree that your odds are 1/24 if they were all opened at the same time? Then what is the difference if you open them dramatically.
Not the same exactly as the Monty Hall, but sometimes helps people figure out what is going on.
Thanks for the great site!
@John: Thatâs a great example. Yes, intuitively we think about 2 options as being 50-50, even though one went through a screening process and the other didnât! Itâs hard to fight our first gut reaction sometimes ;).
My wife was not convinced of it until we did the experiment with 10 doors, haha.
Frank is right (comment #18). Sorry Kalid, but on this one youâve misconstrued the model as a conditional probability issue. Monty is always going to show you a door with a goat, no matter which door youâve picked initially. Therefore, your initial pick has no relevance to what he is going to do and is a completely independent event. There is a a 100% chance of that he will show you a goat and your pick has nothing to do with that. Of the remaining two doors, exactly one contains a car. The chance that your prior pick contains a car is exactly 50%. If you change your mind after he shows you what he is always going to show you, that does not influence the probability that youâll get a car, therefore the ânew informationâ isnât helpful and it really isnât âinformationâ at all. All Monty has changed is the setting of the âgreat revealâ â heâs moved it from a studio without a picture of a goat to a studio with a picture of a goat. Monty is essentially asking you to flip a fair coin and then showing you a picture of a goat. The intuitive answer is correct â your probability of getting a car is 50% and the âthird doorâ is a canard.
Reading through the comments, the key mistake is assigning the initial probability of getting a car at 1/3 just because three doors appear before you. Monty is always going to show you a goat and therefore the probability of getting the prize is 1/2 before he shows you a goat and 1/2 after he shows you a goat. Ten trials are not enough to get past the randomness â do the experiment 1000 times and almost 80% of results will be within a range of 480 and 520 âwinsâ out of 1000 trials.
John B (#49): In âDeal or no Dealâ the probability of winning the $million does change throughout the game. That game is a demonstration of conditional probability. If the last step is to pick between 2 cases, one worth $1 million and one worth $.01, then at that point you have a .50 chance of winning 1 million.
Smith Jr. (#48) hits it spot on when he stated âThe first stage and the last stage of the game are not interconnected, so in the final stage youâll not be helped by the decision you made initially. Youâll always start from scratch and have to pick between a car and a goat, hidden from you. Revealing the other doors, doesnât mean that you picked a goat, it doesnât give you any info about the quality of your pickâŚâ The rest of his statement about ânarrowing the possibilitiesâ appears to be a misstatement.
Scott (#43) stated âSuppose there are three doors, A,B and C and you originally chose door A. If you stay with your original door, then the only way that you win is if originally the prize was behind that door A, which has a chance of 1 in 3.â That is not correct, because the choice is not among three doors, itâs between two doors. The elephant (goat?) in the living room is that Monty is always going to show you a goat no matter which door you pick. You already know this. So the presence of the third door is an ILLUSION⌠itâs an empty shell 100% of the time that has zero bearing on your chance of picking the door with the car. The âthird doorâ might as well be a picture of a goat, or a picture of a car or a picture of the moon. Monty is always going to leave you with the same result NO MATTER WHAT. That is the very definition of an independent event. Come on, people. Switching doors canât possibly help you because Monty didnât base his decision to show you a goat on the fact that you made a choice. He doesnât care which door you chose initially. His instructions to the director say, âShow the goat.â Itâs irrelevant.
I choose door #1: Monty says show the goat. I am REALLY choosing between two doors.
I choose door #2: Monty says show the goat. I am REALLY choosing between two doors.
I choose door #3: Monty says show the goat. I am REALLY choosing between two doors.
1/3 divided by 2/3 is 1/2. The choices are constrained the choices to two doors before youâve even started, because âdoor number threeâ isnât the door labelled âDOOR 3ââŚitâs whichever door Monty wants it to beâŚwhich is one of the two with a goat. The probability of winning is 50% before the event of seeing a door with a goat and after seeing a door with a goat because all youâre doing is waiting to SEE the results of your coin flip until after Monty has shown you a door with a goat and Monty is giving you a chance to change your call of âheadsâ to a call of âtails.â
Have a great day and thanks for an awesome site. Iâm supposed to be working on my precalculus so I better get back to itâŚ
Typo fix: 2nd to last para. above: First part should read â1/3 divided by 2/3 is 1/2. The choice is constrained to only two doors before youâve even started, because âdoor number threeâ isnât the door LABELLED âDoor 3.â Itâs whichever door Monty wants it to beâŚwhich is one of the two with the goatâŚâ
Point being: If Monty is always going to remove from the âuniverseâ 1 of the 2 goats, then that itself defines the condition under which the game is played. Thatâs your sandbox and you canât get out of it.
Kalid has caused me to spend the day in primal scream therapy with this Monty Hall problem. Fortunately my head did not explode notwithstanding the impassioned efforts of several posters here to make that happen.
Monty is a showman, yes, with the costumes and goats and tricky switchbacks. But the key to this (and I didnât see it until the third primal scream ) is that after our contestant has picked her door, heâs showing her that one of the remaining doors is a dud. That doesnât mean sheâs guaranteed to win if she switches, it just improves her chances. Hereâs why.
The contestant is still going to lose by switching if sheâs picked the car to begin with, and remember picking a car in the first place has a probability of 1/3. The other 2/3 of the time, sheâs going to win by switching because Monty has, by revealing one of the goats, provided VERY useful information about the door he just showed her (It has a goat) AND the one he didnât show her, too.
Thatâs because heâs NOT showing her the door with the car in those 2/3 of cases where she indeed picked the goat to begin with. If heâs not showing her a car 2/3 of the time, THATâS WHERE THE CAR IS. The other 1/3 of the time, sheâd be switching from a car to a goat.
If 2/3 of the time she would win by switching, and 1/3 of the time she would win by staying with her initial pick, SWITCH IT UP, GIRL!
BTW: I read out there in cyberspace that presented with this problem, Monty himself said that yes, if the host is always going to show a goat and give the contestant a chance to switch, then switching makes sense. And we know that Monty wouldnât mess with our heads in cyberspace!
Let me try this approach:
G = Goat
X = Car
Before you make your choice, here are the possibilities with the probability after each.
G G X = 1/3
G X G = 1/3
X G G = 1/3
Letâs say you pick the first door. Here are the possibilities of the other 2 doors with the probability after each. Note that all I did was remove the first column to concentrate on our next choice:
G X = 1/3
X G = 1/3
G G = 1/3
Letâs go deeper into each possibility.
Letâs first look at G X = 1/3. Monty MUST open a door. That door CANNOT be a car. Under this possibility, the fact that Monty opens a door at all means the unopened one MUST HAVE A CAR. Awesome, we just figured out that the first possibility (with a likelihood of 1/3) means switch = car!
Letâs look at the second possibility, X G = 1/3. Monty MUST open a door. That door CANNOT be a car. Under this possibility, the fact that Monty opens a door at all means the unopened one MUST HAVE A CAR. Awesome, we just figured out that the second possibility (with a likelihood of 1/3) means switch = car!
So far, looks like 2/3 of the time, Monty opening a door at all (which he is compelled to do) means switch = car. This is great!
Letâs look at the third line. G G = 1/3. Here Monty picks a door randomly. So half the time (UNDER THIS POSSIBILITY which is already only 1/3 of the time) heâll open one door, and half the time (again, UNDER THIS POSSIBILITY which is already only 1/3 of the time) heâll pick the other. It doesnât matter which though, since when you add each half probability, you are still left with a whole 1/3 of the time where you switching = no car.
So letâs look back. 2/3 of the time switching = car. 1/3 of the time, switching = no car.
Switch!
to Groove: I also initially refused to accept that switching improves the odds (though perhaps not as vociferously as you). Every argument (including the one I just made) felt completely wrong to me.
The only thing that convinced me was me writing a program simulating the game myself, and running it a million times. Switching gave me the car 66% of the time.
Since I was where you are now, I offer no argument to you, as I know arguments led me nowhere. I only ask (for your own curiosity) that you take some time to write the program yourself with the game exactly as it is stated, and to run it a ridiculous number of times.
And hey, itâs sort of scientific too, right? You disagree with the results of other scientists, but thanks to the reproducibility property of the scientific method, you get to check for yourself!