Probability and the Monty Hall problem | Probability and combinatorics | Precalculus | Khan Academy
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- Опубликовано: 28 авг 2024
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Here we have a presentation and analysis of the famous thought experiment: the "Monty Hall" problem! This is fun.
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Thanks for the nice video. Especially the Intuitive way that "if you initially chose wrong you should always win by switching" - which is 2/3.
After a lot of confusion I really understand this now. The key point is that Monty opening a door is not a "random" variable.
Okay, let's go through every single possible outcome of this situation (Note: if you're too lazy to read all of this, skip to the end to read a summary!):
-Door 1 has a car, door 2 has a goat, and door 3 has a goat.
-You pick door 1.
-Door 3 is revealed to have a goat.
-You switch to door 2.
-You lose.
-You stay with door 1.
-You win.
-Door 2 is revealed to have a goat.
-You switch to door 3.
-You lose.
-You stay with door 1.
-You win.
-You pick door 2.
-Door 3 is revealed to have a goat.
-You switch to door 1.
-You win.
-You stay with door 2
-You lose.
-Door 3 is revealed to have a goat.
-You switch to door 1.
-You win.
-You stay with door 2
-You lose.
-You pick door 3.
-Door 2 is revealed to have a goat.
-You switch to door 1.
-You win.
-You stay with door 3.
-You lose.
-Door 2 is revealed to have a goat.
-You switch to door 1.
-You win.
-You stay with door 3.
-You lose.
-Door 1 has a goat, door 2 has a car, and door 3 has a goat.
-You pick door 1.
-Door 3 is revealed to have a goat.
-You switch to door 2.
-You win.
-You stay with door 1.
-You lose.
-Door 3 is revealed to have a goat.
-You switch to door 2.
-You win.
-You stay with door 1.
-You lose.
-You pick door 2.
-Door 1 is revealed to have a goat.
-You switch to door 3.
-You lose.
-You stay with door 2.
-You win.
-Door 3 is revealed to have a goat.
-You switch to door 1.
-You lose.
-You stay with door 2.
-You win.
-You pick door 3.
-Door 1 is revealed to have a goat.
-You switch to door 2.
-You win.
-You stay with door 3.
-You lose.
-Door 1 is revealed to have a goat.
-You switch to door 2.
-You win.
-You stay with door 3.
-You lose.
-Door 1 has a goat, door 2 has a goat, and door 3 has a car.
-You pick door 1.
-Door 2 is revealed to have a goat.
-You switch to door 3.
-You win.
-You stay with door 1.
-You lose.
-Door 2 is revealed to have a goat.
-You switch to door 3.
-You win.
-You stay with door 1.
-You lose.
-You pick door 2.
-Door 1 is revealed to have a goat.
-You switch to door 3.
-You win.
-You stay with door 2.
-You lose.
-Door 1 is revealed to have a goat.
-You switch to door 3.
-You win.
-You stay with door 2.
-You lose.
-You pick door 3.
-Door 1 is revealed to have a goat.
-You switch to door 2.
-You lose.
-You stay with door 3.
-You win.
-Door 2 is revealed to have a goat.
-You switch to door 1.
-You lose.
-You stay with door 3.
-You win.
Now let's look at the numbers:
-Number of times you played: 36
-Number of times you won: 18
-Number of times you lost: 18
-Number of times you switched: 18
-Number of times you stayed: 18
-Number of times switching made you win: 12
-Number of times switching made you lose: 6
-Number of times staying made you win: 6
-Number of times staying made you lose: 12
I'll let you come to your own conclusions about this data...
Best explanation yet. thanks
@@josephmorrow1881 since there are two doors and remaining....there is two possible choices......but one of the two door has a car in it.....so for the two possibilities he has to open the same door( the door with the goat) twice ....if he opens the other door then the door with the car will be revealed...
only last 12 part needed man
@@unknownera241 But they never show the car, only the goats.
You only have to consider one case: Door 1 has the car, door 2 and door 3 have a goat.
Cases:
Pick 1 and switch: L,
Pick 1 and stay: W,
Pick 2 and switch: W,
Pick 2 and stay: L,
Pick 3 and switch: W,
Pick 3 and stay: L.
Now you don't sum all the wins and losses, of course there will be 3 wins and 3 losses.
This data means that you win 2/3 of the time if you switch = lose 2/3 of the time if you stay.
Think about it.
I finally got it from this video, after about 2 hours of reading articles and watching videos.
It goes like this. If you stick with your original choice your odds are only one third all the way through. BUT... When you switch your odds DOUBLE because there is an EXTRA door! Monty HELPS you by telling you where the prize is not, thus increasing your odds by another one third, equaling 2/3. Lol
Correct.
Two doors always have double the probability of one door, and since one of these two doors is opened, we know that combined probability of these two doors must be behind that one unopened door.
@@max5250 needed this thanks
Maybe Monty can help me determine which side of the coin will come up 2/3rds of the time, because it seems to come up heads or tails evenly. More to your point, because Monty permits you to switch, the odds have to be recalculated. It was originally one out of three. Then Monty removed one. Now you're faced with one out of two. Switch or not, it makes no difference. And, incidentally, Monty CAN open the car door, he just chooses not to. That's just how Monty is.
I am a software programmer. I wrote a simulation with 2,000,000 contestants split into 2 control groups. Control group A stayed with their first choice. Control group B switched.
Control group A won the car 33% of the time.
Control group B won the car 66% of the time.
but what is your code. which probabiblity did you use
Care to open source it?
@@samp9053 I got ChatGPT to make it in JavaScript (you can actually run it in your browser right on this page by pressing Ctrl + Shift + J, pasting the code, and pressing enter) - the output said “Stay wins: 0.3335, Switch wins: 0.6665”:
function montyHallSimulation(iterations) {
let stayWins = 0;
let switchWins = 0;
for (let i = 0; i < iterations; i++) {
const prizeDoor = Math.floor(Math.random() * 3);
const contestantChoice = Math.floor(Math.random() * 3);
const remainingDoors = [0, 1, 2].filter(door => door !== contestantChoice && door !== prizeDoor);
const openedDoor = remainingDoors[Math.floor(Math.random() * remainingDoors.length)];
const switchChoice = [0, 1, 2].find(door => door !== contestantChoice && door !== openedDoor);
if (contestantChoice === prizeDoor) {
stayWins++;
} else if (switchChoice === prizeDoor) {
switchWins++;
}
}
const stayProbability = stayWins / iterations;
const switchProbability = switchWins / iterations;
console.log(`Stay wins: ${stayProbability.toFixed(4)}`);
console.log(`Switch wins: ${switchProbability.toFixed(4)}`);
}
montyHallSimulation(2000000);
6:59 - 7:18 made sense to me and helped me understand the problem. At the outset, you have a 1/3 of picking correctly and a 2/3 chance of picking incorrectly. Also at the outset, if you consider two doors together, you have a 2/3 chance of picking correctly. When one of the doors is eliminated, you essentially "capture" that 2/3 probability by switching doors from your initial selection. It's almost as if you got to pick two doors at the same time, but not really. More information - Monty revealing one of the doors that has a goat - improves your odds if you switch.
A two line solution:
1. By swapping you always get the "opposite" of your original choice.
2. Two thirds of the time, your original choice will be a goat.
That's it.
Very clever!!
Thanks for clearing this up!
Exactly! It really is that simple.
nah just simple logic. People always just assume
dudeeeee, that's hella smart! marry me
Thanks, Khan Academy, I finally understood the Monty Hall problem. You have by far the best explanation.
Khan you always explain things in order for viewers to easily understand. God bless.
If I am married, and have 2 mistresses, and my wife finds out, should I stay with mistress #1 or pick Mistress #2?
Mistress #2. If you had to get another mistress, she must be younger and better than the first one, you POS
@@josue_mejia Well then...
@Rentech In theory, my wife is that goat, so are you suggesting I stay with mistress #1?
@@TraceguyRune Considering the host knows everything I would ask him which one was better.
Threesomes are nice but I’d rather foursome!
Yes! Absolutely enjoyed every second of that! Especially the fact that that first explanation came from a student's point of view 👏🏻
Great explanation! I now understand the 2/3 chance of winning when switching! I mean I knew it's 2/3 but never really understood it.
forsenCD I understand why THIS explanation is right, but, with regards to the 50-50 explanation ONLY, can you please explain why THAT is wrong?
forsenCD I understand that, but my question is, since one of the doors is eliminated, why doesn’t the TOTAL, against which, the probability is calculated, change too? Like, since all of them have a 1/3 rd of a chance, when one of the doors is eliminated, why don’t we subtract that “1” door from the total of “3”? (I’m sorry if I sound dumb, but I genuinely cannot understand that part).
@@Raphael-fn9dr I am also thinking this??
FACTS:
1) Chance of a door hiding car = 1/3
2) Monty picks from 2 doors = 2/3 chance one hides the car.
3) Monty ALWAYS picks a goat...
4) ...leaving you CERTAIN to win a car in the 2 out of 3 cases Monty has one.
The *only* way you can lose by switching is if you correctly guessed the car originally (1/3 chance). Monty eliminates a goat in the other two scenarios, leaving only the car. Switching gives you a 2/3 chance of winning.
You're welcome!
The only video on this topic which made sense to me
The MH problem is great because it appears at first glance to be counter-intuitive and posing the problem to friends reveals just how stereotypical our decisions are when made under conditions of incomplete knowledge. This reliance on heuristics unveils the systematic biases in our decision-making yet problems like these can show us normative methods of reasoning like actually applying probability theory to problems involving uncertainty instead of "thinking with yer gut" in every situation.
I finally got it after suffering for three days. The clincher was: If you picked wrong and swich you'll always win. Thanks.
Thank you. That was great. I couldn't get my head around the fact that during the time that I have to decide betweening switching or not I didn't have 50-50 change but it was 66-33. The first explanation helped me understand that.
Ok have been watching the top videos for an hour and literally i just got it now. This is what i am looking for
***** I approve of your analysis - the host opens one of your two alternative doors for you - he shows you your own goat - what a nice man.
Then he makes the switch offer.
The revealed goat is not a pointer to the car - it's just one of two goats and can always be shown. The probabilities with the doors all closed are exactly the same when all the doors are opened - the door that actually hides the car has only a 1/3 chance probability of having the car.
Two doors are better than one.
+Richard Buxton
said _"The *probabilities* with the doors all closed are exactly the same when
all the doors are opened - the door that actually hides the car has only
a 1/3 chance *probability* of having the car."_
and also said: _"The Monty Hall Problem has _*_nothing whatsoever_*_ to do with _*_probability_*_"_
You couldn't make this up - yet somehow you did Richard!, LMAO
HumptyDumptyOakland People laughed at me when I said I wanted to be a comedian - they're not laughing at me now.
Bob Monkhouse.
1928 - 2003
Richard Buxton
No, they're all laughing at *you* now Richard
One of the best explanations I've come across 💯
Based on the way he is describing it Monty is showing you the goat curtain for no apparent reason. However, according to the laws of game show economics Monty cannot be giving out cars (which were always meh level but that's not important ) to every person who comes on the show. When the producers signed Monty and explained the routine they noted "Now look Monty forget about any fantasies about taking spokesmodel Carol Merrill to the Bahamas for a tryst if you let players win cars all the time. Because it's completely out of your bonus buddyboy. Monty thinks: I had better make sure they lose as much as possible. So I if you pick the goat curtain Monty's goto line is "You picked curtain number three, well lets see if you won that [Oldsmobile your grandmother always wanted]! At this point it gets all mathy which is I'd not really my thing.
monte didn't decide who wins
Very well-explained. I teach probability in college and threw in this Monty Hall problem. Only 10-20% of the class gets it on my first attempt at explanation. I'll send this video to them in future.
You should have known going in, or realized very quickly (because it stands to reason), that Monty Hall would show you a goat because that's the only way the game works, he can't show you the car or it's game over. So you have to look at it as if one door is already removed, you just don't know which one until he shows you. Your probability of picking car or goat initially is still 1/3 and 2/3 respectively. When he shows you a goat he's only giving you the illusion of new information, when really he's just revealing which door was the removed one. So it doesn't change the probabilities at all. Your choice isn't between 2 doors, it's between staying with your 1/3 odds that you picked car right away, or 2/3 odds that you picked goat and are switching to car. Smart money says take the 2/3.
If that didn't do it, try this. If you can agree that: 1) there are 3 events (your initial choice, monty's choice, and the choice to switch), 2) that YOU will ultimately only have 2 choices (to stay with your original pick and odds or switch your pick and odds), 3) that the sum of the probabilities of all options = 100%, and 4) that the only way to win is to pick a car. Then follow me and I'll show you the light. Event 1: your pick, 1 car, 3 doors means you have a 1/3 shot at picking the car right away or 33% (this is also what you're saying if you eventually decide to stay, and therefor also the same odds). Event 2: Monty's choice, which carries a 0% chance that he picks the car (because he knows where everything is). Now there is only 1 event left, which is the switch, and since all probabilities add up to 100, that means the remaining 66% chance of picking a car is behind the switch door. 33.3+0+66.7=100. That's the math of it, this is the WHAT is happening, everyone else is explaining the why/how and that can be confusing because there are so many ways and it's also counter-intuitive.
Thank god finally
By 3:19 the Monty Hall probabilities finally made logical sense to me. Thank you!
I agree fully. This is a 50/50 shot after the revealed curtain.
Yep the first rounds math is nothing but smoke and mirrors to over complicate something. No matter what door you choose. You end up choosing between two doors with a goat and a car.
Here's another way of thinking about it:
Say that, instead of 3 doors, the game involves 99 doors. After picking one door, the host removes every other door except the one you chose and one that he chose. He then tells you that either your door or his door has the car.
Picking the host's door is a no-brainer at that point; you have a 98.9% chance of being wrong the first time. The reason the Monty Hall problem is so tough is that the difference between 1/3 and 2/3 isn't THAT big. You could still have easily picked the right door the first time. And, if you lose, knowing that the odds were technically in your favor won't make you feel better.
You're right (like you need me to confirm that!). I looked into it a bit more last night, and found a site that emulated the scenario. I performed 100 passes with each strategy (always stay, always switch). 100 tries is a small sample, but it was enough to show me the probablities at work.It's just counter-intuitive to my thought process.
I've never understood this problem through adding probabilities, so THANK YOU SAL for sharing this problem a different way!
Impressive tutorial and excellently solving complex problems - wonderful
Are YOU KIDDING ME? What's the chance Khanacademy would pick something to explain that is already several places on RUclips?
Thanks Khan Academy. Finally I understood the Monty Hall problem.
I saw this on the movie '21', and wondered how it worked ever since, thank you very much!
Wow this video really cleared this up for me. Thanks
Fun fact: The Monty Hall problem actually only works with goats. Goats lead to counterintuitive probability problems.
The reason people are so confused is because they keep thinking 50% or 50-50 chance. Guess what? That's wrong and makes no sense. You begin with a 2/3 chance of losing and if you stay you still have 2/3 chance of losing. Simple as that. For example, you have 3 doors, door 1, 2, and 3. Let's say that the car is on door 2. Now if you choose door 1 or 3 initially and did NOT switch, you lose. However, if the host reveals a goat (and he always picks a goat) if you picked door 1 or 3 initially and switch, you win! You have a 2/3 chance of winning of you ALWAYS switch. If you stay, your odds remain the same 1/3 of winning. So stop thinking 50% chance because that makes no sense. Just remember switching when the goat is reveled always gives you a better odds of winning.
very well put. best explanation i have read so far
+cardog kitchen Here's the problem. You start with 2/3 of losing. One box goes away. Now the chances of this "new position" are 50-50 and not "1/3 and 2/3", becuse there arent' any more 3 boxes there, but just 2. It's like you step in one room, you have 3 boxes to chose, you have 1/3 of winning. But when they eliminate one box, I transsfer you to another room, forget the old one. You are in brand new room number 455, and there are only 2 boxes, you don't know if there were 3 or 4 or 50 boxes before that, one has a car one does not (50- 50 chance for you), and let's say somebody from your family chosed the box already. But you can switch. what does it matter if you switch? It's still 50-50.
+Erosis of common shuttles I wasn't convinced until I wrote a simulation and that finally convinced me. Also, the mythbusters ran an experiment that verified it too. Here's my simulation: github.com/pebreo/montyhallsimulation
+Erosis of common shuttles Just to answer: You pick your box and it’s removed to room 455 for later, it has ⅓ chance of a car. Two boxes left, the car has double the chance of being in one of the 2 boxes. You are asked do you want to stay in room one with the two boxes or room 455 with the one box. If you choose to stay with the 2 boxes you have already switched in essence. Because one will always be opened to reveal a goat, the other now has a second bite of the cherry, the ⅔ bite. Hope this helps.
So what you don't know, but picking random there's a 2/3 chance you randomly select a door with a goat. Then the host MUST open the only other door with a goat in it since he can't open the one with the car, you switch and you win. If you're REALLY sure the car is in door X, then you choose that door in stick with it, but you don't know so the chances are 1/3 if you don't switch.
I finally understand this MH problem completely. Thanks!
enlightenment at last, this one has bugged me for a while, thanks!!
Well explained ❤❤❤
Like always simple, short and the richest in information
Earthquakes may come 3 times at the same place: foreshock, mainshock, aftershock.
Also, pretty sure lightning isn’t random nor it is similar to the Monty Hall problem.
Lightning strikes for specific reasons and can strike at the same place, ask the empire state building.
If you have been struck once, stay there, keep everything equal, and you might get struck again as the best conductor on site.
This is the only video that made this make sense
Please help me solve :
Consider a game-show called \deal or no deal", where you have to choose between
three suitcases labelled (A, B, C), one of which contains Rs. 1 crore and the others are empty. On this show
contestants pick a suitcase (say suitcase A) and keep it temporarily without opening it. The host (who already
knows what's in the suitcases) then opens up one of the other two cases (say suitcase B) which is empty. The
contestant is now offered the option to keep the suitcase (A) or switch it with the remaining unopened suitcase
(C). Suppose that you are on this show, would you switch or keep your suitcase to get the prize? First make a
guess.
Now let us analyze this using likelihoods. Let Hi denote the hypothesis that the prize is in suitcase i where
i = A;B;C.
a) What are the a-priori probabilities of Hi before you pick a case?
b) Calculate the likelihood of each hypothesis Hi given that the host opened suitcase B after you picked suitcase
A.
c) Compute the a-posteriori probabilities of each Hi after the host has opened case B and shown that it was
empty.
d) To get the prize should you (i) keep A, (ii) switch to C, or (iii) does it not make a difference? Justify.
thank you
finally I understood the solution of the Monty Hall problem
That is really interesting, didn't expect the correct answer at all (I didn't think it would make a difference). I'm really enjoying these more informal lessons, they're fun and neat and the uber upper level calculus ones are still over my head x)
Nice explanation
Discussed in depth in "The Curious Incident of the Dog in the Nighttime," Great book, by the way -- especially the audio book. Look for it.
Another way you can think about it is this:
There are 3 doors to choose from. 2 doors have a goat and one has a car. The host asks you to choose a door. You have a 1/3 chance of picking the car and a 2/3 chance of picking a goat. Now, the host, without opening the door that you chose, opens another door. Of course, the door that he opens will have a goat behind it. The host asks if you want to switch your door with the one that is not open yet. Do you switch or not? Let's looks at the probabilities.
If you do not switch, you are staying with the original probability before the host opens the door of a goat. The probability of winning is still 1/3 and the probability of losing is still 2/3. You have more of a chance of closing that you do of winning.
If you do switch, think of it this way:
When the host opens the door (which has a goat), that eliminates that door from one you can choose. He asks you if you want to switch or not. This takes you back to the beginning, except instead of having to pick between 3 doors, you only have to pick between 2 doors: either the one you chose or the one you didn't in the first round of choosing. The third door has been eliminated because you now know what is behind it (the goat). So instead of having a 1/3 chance of winning, you now have a 1/2 chance of winning. If you do not switch, you are sticking with your original choice, where you had a 1/3 chance of winning and a 2/3 chance of losing. If you do switch, you have narrowed your chances down to 1/2. You can either keep the one you picked or switch. You should always switch because if you do, you have a 1/2 chance of winning instead of your original 1/3 chance of winning.
If you switch you have a 2/3 chance of winning, not 1/2.
If you were right, then 1/2 + 1/3 = 3/6 + 2/6 = 5/6. What is the remaining 1/6 chance?
By switching you are selecting going away from your initial 1/3 chance of winning, making that a 2/3 chance of winning.
Basically it is the group of 2 doors that you have not chosen that has a 2/3 chance of having a prize. Since every time you play at least one of those two doors will not have a prize, opening one door from the group that does not have prize is basically the same as saying: unless both doors did not initially have a prize (a 1/3 chance of that being the case from the beginning, since it would have been your door the one with the prize), then the other door in the group is the one with the winning prize.
-------
Just do the experiment, these are all the possible outcomes, and all of them have equal chances of happening:
Scenario 1: Door 1 = prize. Door 2 = empty. Door 3 = empty.
Scenario 2: Door 1 = empty. Door 2 = prize. Door 3 = empty.
Scenario 3: Door 1 = empty. Door 2 = empty. Door 3 = prize.
Let's say every time you play, you pick door 1. Every time you will be shown one of two doors, either door 2 or door 3, whichever is empty, or if both are (1/3 of the times), one of the two.
In scenario 1, you will be shown either door 2 or 3. If you change to the door that has not been opened you lose. However, in scenario 2 and 3, you will be shown doors 3 and 2 respectively. If you respectively switch to doors 2 and 3 you will win.
Since in the long run all scenarios would occur a similar amount of times, 2 out of 3 times you will win if you change doors instead of keeping your initial choice.
--------
Think of it this other way. Imagine you are given 100 doors to choose from, one has a prize. Let's say you pick door number 43, then the host opens all doors without a prize except one, let's say door 21 (all from the selection of doors you did not pick). You would be right to think that your initial door choice has a 1% chance of winning, but door 21 does not have a 50% chance of winning, it has a 99% chance. It is much more likely that the remaining door (21) has not been opened because the host won't open a winning door (he will just open losing doors), the other possibility is that you were right in your initial choice of picking door 43 and that one is the one with the prize.
Lope Bravo What I'm saying is that because you now know what is behind the third door, that door is eliminated. So it's 1/2: you're original choice or the other door that you can switch to. If you don't switch, you are sticking with your original probability of 1/3.
Since there is a winning prize in any of the two remaining doors, that means that there is a 100% chance that if you open both doors you'll find a prize, therefore the sum of the probabilities of having a winning prize behind each door have to be 100%. So if your first door has a 1/3 probability, the remaining has a 2/3 probability.
It is correct to say that by changing doors your probabilities change (they change to that of the door you switched to), but the probabilities of a specific door don't change based on your choice.
What changes the probability of a door is not what door you choose, it's the information the host is giving you by uncovering a non-winning door from the two doors that you did not initially choose.
From the beginning it's most likely that the winning door is one of the two doors that you did not choose, so that group together has a 2/3 chance of having the winning door. You switching to the remaining door is like switching to the whole group of two. Because in the group of two there will always be one door without a prize, you don't need that door, you just need the one remaining.
As if you said (this part is correct), sticking with your original door has a 1/3 chance of winning, the only remaining option is changing to the remaining door, so that has to be a 2/3 chance of winning. In total they have to be 3/3, since there is a prize in any of the two.
When switching you still only have 1/2 chance of winning ,saying 2/3 in incorrect.
wrong
So Monty Hall only operates on the rule of the game-in a scenario when he opened a the door random, then would it 50/50 for the player that switching or staying would be equal probabilities of getting prize?
sweet
Very good analysis of the problem, Sal. Could you do more computer science videos in another programming language?
Brilliant!!!
"had to account for variable change" -21
The explanation indeed only makes sense in this specific scenario where one door being eliminated is part of your strategy. If you would repeat this scenario often enough, you would find that you win 66% of the time by switching. Under any other circumstances it would basically boil down to 50%.
Yeah it was very counter intuitive, but you explained it well. Really interesting video. Thanks for making it. :).
First we have the option of chosing correctly at the first choice then staying with it. 1/3*1/2 so 1/6 chance of being correct.
second, we have the option of chosing incorrectly at the first choice then changing that choice at the second chance. So 2/3*1/2 so 1/3 chance of being correct.
So if we added up our chances of wining, it stays at 50%.
This is why I am confused with the video. It implies there is a singular choice, while there are two distintic choices for the person to make.
I finaly got it ,your right ,thank you !
fantastic explanation
Once Monty shows one of the incorrect door/curtain, then the probability of winning or losing changes to 50/50 as the cash prize will only be behind either door number 1 or door number 2 (in this example).
I know, what my odds are, but thanks for clarifying :)! But your last sentence is correct....
Is it correct to see it like this? It is ALWAYS one out of three doors. On your first selection (say door 1) your chances of success are one out of three: 33%.
Once the additional door (say door 3 ) is exposed, if you stick with door 1 you don't take advantage of the additional information, therefore it's still one out of three.
If you switch to door 2 it is STILL one out of three (because there are still three doors) but with a bonus (33%) because one of the doors has been exposed. That makes the chances now 33% + 33% = 66% if you switch.
Is that valid? It's almost as if you have to think of it as two completely separate selection processes, not just changing your mind. But as you stated "it's counter intuitive". But that's what makes it fun. Thanks
IMO your explanation doesn't quite spell it out. You must consider the difference between one of three being removed randomly, vs one of three being removed with knowledge.
In the first case, if Monty just opened a door blindly then he would have the chance to reveal the prize with this move. So in 1/3 of games this would happen and, one supposes, the game would be voided. That would leave the other 2/3 of the games for Monty to reveal a goat. In those remaining 2/3 of games, the location of the prize would be evenly split between the player's first choice door and the final door.
On the other hand, when Monty knows where the prize is, he will _never_ reveal the prize. He will always reveal a goat. The 1/3 of games that would have been voided in the preceding paragraph would now be played to conclusion, and in every case they would favor switching since, by definition, a game that could have been voided by Monty revealing the car are games in which the player has not selected the car, and so in switching he would have arrived at the car.
Nice analysis about Monty Hall Problem~~ Cool!
Classy representation
Why do I care about the third door with the goat he shows us if he shows us the goat every single time, doesn't it take 3 out of the equation, and narrows it to 2 every time?
the point is that it's more likely that you pick INCORRECTLY at first, leaving the host to have to reveal the other incorrect door to you. it's 2/3 that you pick WRONG first, so switching in that case will yield victory. the chances for picking RIGHT at first are 1/3, in which case you'll lose by switching, but not switching will also be a 1/3 chance of winning then, so switching is always better.
Not Legato, technically, if the host ALWAYS reveals one of the 2 goats regardless if you chose the car or the 2nd goat, you are left with switching or staying with the choice and that ultimately ends up with a 50% chance of winning. The only application where switching gives you the 2/3 advantage is when you are choosing 3 or more times (which is how you prove this to be statistically correct). But since the example is a game show with a win or lose situation with a single attempt, you always end up with 2 choices, staying or switching.
You either didnt understand half of what you read in my comment or you are a highly unintelligent being. I stated perfectly clear that a choice of 2 (same as your coin flip example) is 50%, no matter if you switch your choice while the coin is in mid air. Since the host is always removing a wrong choice, logically you are always choosing between 2, hence the explanation above. To prove the statistical advantage of switching is to have a best of 3 instance, which is irrelevant in this situation cause you get one pick only.
d1pz9, an absolutely ridiculous statement. There is a 66.7% chance to win the car by switching in any one game. If you flipped a coin one time, what are the chances it will land heads? 73/100?
Now you're being really silly. And quit giving a thumbs up to your own garbage.
If 100 contestants played one game each and they all switched they would win 67 cars, not 50 cars. Therefore each contestant playing one game has a 67% chance to win by switching.
Intuition is just a fancy word to describe the laziest part of the brain doing what it does best.
Since there are 2 goats and 1 car, you are 2 out of 3 times more likely to pick the goat on the very first try. So that means whatever curtain you pick you will more likely to have a goat so you should switch your answer when he reveals the other goat.
Fascinating.
When you remove the first option there are two remaining options. one has a 1/3 chance of winning, the other has a 1/2 chance of winning. Which would you rather have?
LOOK for RUclips user NIANSENX and play his "The Monty Hall Problem" several years later and with over a million views he used Goats and a Car too. Actually this is better done than Kahn's but I still give Kahn much respect for what he is doing. I only found it strange he would pick something so widely distributed already.
a year ago I had a hard time explaining the solution of this problem to one of my probability "Professors" :)
remember this from the movie 21.
er, what doesn't make sense to me is this:
The number of choices is actually higher then 3.
In order to see this, understand we have two sets of choices, not just one. So, we have a 1/3 chance of picking correctly the first time then on the second choice we have a 1/2 chance of picking correctly. The reason for the 1/2 is due to the show revealing one of the doors.
Now if we wish to count the number of wins: (I'll finish in the reply)
Perfect
They never taught this in undergrad econ or high school math. Interesting thought-experiment.
Class assignment was about this problem, we discussed the statistics, and everyone STILL chose to stay. What is it with this mentality? I think my class mates hate me.
Thank you!
When they show you a wrong box and they ask if you want to switch, they basicly change the game into: 'Here you have two boxes, one of the boxes contains a car while the other contains a goat... pick one'. In that case it doesn't matter what you pick. Or am I misrepresenting the situation now?
You totally get it. Round one is smoke and mirrors and has nothing to do with anything. The only choice that matters is your second round choice. Door A or B. With no information other then door C had a goat. Which you can't choose any more anyways. So switch doors or don't switch doors. It's all meaningless. Their is no point to the first round.
Wrong. And here's why: if we generalize this to 100 doors, and the host opens 98 doors, while KNOWING which one has the goats behind them, and being unwilling to open the door with the prize behind it, then it would be advantageous to switch. Switching is much more likely to get you the correct one, because your original guess had a lower probability than the UPDATED guess with the new information.
you were not really given the low priced cases by Howie, you were given a chance to open cases at random throughout the game, but once you opened it, the cash in that case was eliminated. So, there was a chance of eliminating the million throughout the game; however, if you had two cases, one with the million, then switch, because there is a 22/23 chance the other case is the million. The case you chose has only a 1/23 chance.
I admit that you start out with a 1in 3 chance them Monty shows you a door which gives you a 1 in 2 chance ,it was alway,s 1 in 2 ,not 2 in 3 ,but im not to smart and never met Monty so it matter,s very little to me .
Gens. Thank you soo much sal
In a nutshell you should always switch because when the goat is revealed that door has a 0/3 chance of a car. Your first pick, 1/3 chance of a car and the final door, 2/3 chance of a car.😂 Simple!
I understand this answer but why can't it be 1/2 if you ignore the fact that you chose anything before the first goat was revealed?
Yes, if you ignored that fact, like if you were another contestant who is going to do the second selection and didn't see the first step, it would be 1/2 likely to hit the car door, but not because the staying action is going to win 1/2 of the time, but because you couldn't know which is the staying door and which is the switching one, so the advantage that you could get by switching is compensated by the disadvantage that staying provides.
P(winning the car) =
P(selecting the switching door) * P(switching door has the car) + P(selecting the staying door) * P(staying door has the car)
= 1/2 * 2/3 + 1/2 * 1/3
= 2/6 + 1/6
= 3/6
= 1/2
It's like if you had a true/false question. There are two options, one incorrect and one correct. Does that mean a 1/2 chance to hit the correct? It depends on your knowledge. If you read something or someone told you something about the subject, you could get information and select an answer based on that information, being likely to be correct, but if you do a random choice like flipping a coin, there is a 1/2 chance for you to hit the right one.
give up
If you pick wrong and switch, you will always win because picking wrong means you picked a door that has a goat behind it and after Monty reveals the other door that has a goat behind it, switching means you switch to the only door that's left aka the one with the car
Thank you so much, I finally understand this :D
It's all about what you did in the beginning. The odds of picking a goat in the beginning was 66%, so most often you'll have a goat. When the other goat is revealed, the more likely case is that the car is behind the other door. Basically, when it's down to two options, it's more likely that your door is a goat door.
this is true it is a 2/3 chance switch
I wonder how this concept would have worked on "Deal or no Deal" all with briefcases with various amounts of cash prizes in them ranging from $1 to $1million. I think there were 23 or so briefcases. You picked one at the start, and then eliminated the others but you were always given a chance to swap out your chosen case with others. According to this, my guess is that it would have been wise to always switch there too.
The one difference is that on "Deal or No Deal", the reveals are completely random, and the host doesn't have any control over which briefcase gets revealed. So you don't get any more information that any insider knows when the reveals happen.
Had a lol moment. Played the game with my 2 boys separately. They both stuck with their initial choices and both won 3 out of 5 times each.
And I was hoping to teach them the math behind it.
So much for switching!
Yup your right!
If you don't switch, your odds of winning the car is 1/3.
If you switch, your odds of winning the car is 2/3 because your odds of picking a goat from the start was 2/3.
Probability is about the unknown. Using three doors or 100 does not matter if in the end you only have to choose between 2 doors. Because probability is concerned with unknowns. Kahn is either trolling you or forgot that is not how this thought exercise work.
I don't see what all the arguments are about, this problem is basic. If you choose a door with a goat in it to begin with (that's a 2/3 chance you do) the host has to open a door with the other goat, leaving only the car if you switch. Hence if you switch there is 2/3 that you win. If you stick to your door from start, no matter what, it's only 1/3 of a chance.
Nice sir
So basically, you have a higher chance of landing on a goat (2/3) than actually landing on the car (1/3) meaning that it would be a better idea of switching than staying on the same door because it would raise you odds if you switched theoretically.
OK, here we go, once and for all. You are playing Russian Roulette. The shooter has pulled the trigger 4 times, and the hammer has landed on an empty cylinder.
Do you think you have a 5/6 chance of dodging the bullet next time?
Mythbuster explained and did an experiment on it too .. !
It becomes really clear if you have 100 doors and the game-master opens 98 of them for you and you're allowed to pick another one each time. The likelyhood that there is a car behind the other door is much higher.
Boom! yeah i heard that
Learned this from movie 21. Movie learning ftw
0:12 I saw what you did with your moUSE!
🤣