I hope the movie didn't have him actually end up picking the right door, which would just make his smugness even more insufferable. Yeah, he has a 2 in 3 shot, but it's not 3 in 3.
I dont understand, you know behind 1 door is a goat. then there are 2 doors left. 2 door to gamble, why is there not a 50% to get the car? door number 1 + 2 is a big gamble isnt it? if he opens door number 3 you still cant know if 1 or 2 is the car.
@HappyBrotherBeat You choose door number 1. You have a 33.3% chance of your door being the car and a 66.7% chance of it being a goat. It's more likely you will pick a goat. Now the host opens a goat door. Now that only a goat and a car remain you can stay, hoping your pick was a car or switch hoping the pick was a goat. Since your pick was most likely a goat switching means it is more likely to win a car. Does that make sense?
@GonnaBeALongLongTime If i pick door number 1 (which is the car) and he shows me door number three. So i now have door number 1 and 2 to choose from. Wouldnt it be a 50 50 chance? Why should i change if the odds are the same? im so confused at how this works, its all about luck, what if everytime i somehow manage to guess the right one first, if i went by this i would always lose.. dont see the logic in it...
It would only be 50/50 if from the start there were only 2 doors to pick from. since there are 3 doors, 2 of which do not contain the prize you are more likely to pick them over the prize.You WILL pick a goat 66.6% of the time, that's called statistics, you saying, well maybe i will get lucky and pick the car more often is ridiculous. The odds do not change... just because I may flip a coin and 3/4 times it's tails doesn't mean every time it's 75% that the coin will b tails...
@sch3ffs Monty always opens a goat door. If you pick a goat door at the start then switch you always get a car. The odds of picking one of the two goats at the start is more likely than picking the car.
@HappyBrotherBeat no, the events are not independent of each other, means that if one event occours (opening of the first door) it changes the nature of the other events as well, its called conditional probability and dependent events, its an application of the famous bayes theorm, search for bayes theorm of probability and read a lil about conditional probability :)
The teacher bungled the problem by suggesting that the host might be playing mind games. This totally changes the nature of the problem. It is supposed to be clear that the host will always open a door with a goat after you make your original choice.
I think the movie relied on the fact that him knowing the 33.33.. and 66.66.. percentage chance would be considered impressive by the average American.
He should have made clear that the host always opens the door with the goat. He said the host knows what is behind the doors and opens another one. It only works if the host always reveals a goat. :)
@Eudaletism Because the problem is a problem and nothing more, there shouldn't be any "maybe". The host DOES know where the car is, always. The problem is, "You choose a door. The host reveals a goat. The host asks you if you want to switch, no ifs, ands, or buts. Do you stick or change?" Remember, this is a sort of a riddle. You can't say, "what if." If the host didn't know where the car was, and if he only asks to switch when you choose the car, then it is no longer the Monty Hall Problem.
yes there is variable change but once door number 2 is gone his odds in crease that door number 1 is still the right door. either way its a 50 50 chance in the end regardless which door he chooses.
@fliptthescript the door you choses starts with a 1/3 chance, then he opens door number 3, and there is a goat. He offers you to change to door number 2, and you should because door 2 and 3 together has a 2/3 winning chance, and door 1 has a 1/3 winning chance
When you choose the first door, you have a 1/3 chance of winning. If the game show host gave you a "two for one" deal, permitting you to choose the two other doors, not just one door, then you would have a 2/3 chance of winning. So by switching doors, you're essentially getting the benefit of the "two for one" deal. It's as if you had picked the "two other doors" from the very beginning.
@Akilles047 No, he is correct. There is no explanation given here. The kid merely states the correct probabilities. That's hardly an explanation of WHY those probabilities are correct. Why did the other door get a 1/3 boost? Because your confidence in that door is allowed to increase since the host COULD have legally opened that door and DIDN'T. Your door, however, he cannot legally open at all. So there's no reason to gain confidence in it when you see him dodge it.
@DMK000 the door you choses starts with a 1/3 chance, then he opens door number 3, and there is a goat. He offers you to change to door number 2, and you should because door 2 and 3 together has a 2/3 winning chance, and door 1 has a 1/3 winning chance
Think of it this way. YOU'RE ON THE ASSUMPTION THAT YOU INITALLY PICKED A GOAT. Why? Cause there's 2 goats and 1 car. Therefore there's a higher chance (66.7%) of picking a goat. So when the gameshow host reveals a GOAT (can't reveal car) there's, now only a 33% chance of the other door being a goat and 66% chance of it being a car. Get it? Cause the 1 you picked (most likely goat) + goat shown = SWITCHING.
it's 66 percent that you will initially choose a goat. therefore, there's a 66 percent chance that when you switch, you will switch to the car. since of course, you most likely (66 percent) picked a door with a goat behind it.
EXPLANATION FOR IDIOTS: if you pick one door from 1000 doors. the odds are you did not pick the door with the car. if every door expect one is opened you might think you have a 50/50 chance if you switch or don't. but your still stuck with your original door which is most likely wrong. so switch to the single closed door left and you will increase your chances by 999%.
Actually I'm wrong. You increase your chances from 0.001 to 0.999 which is 999 times greater and thus an increase of 99800%
The original door doesn't have a 0.5 chance of being correct, it has it's original 0.001 chance. However in the other 0.999 of the time the car will be the only other unopened door so switching to it increases your chances of winning massively.
The teacher never said that the host will always open a door with a goat. He just said "the host opens another door, and behind it is a goat". That is completely different from "the host then opens another door, which will always contain a goat".
While the main stats have been discussed to death, this has not:
If a person who knows the proper stats was to play against a person who knew the proper stats (who would sometimes randomly ask to change, other times will not), statistically speaking, each person would win 50% of the time. This is because on average the player will call the bluff 50% of the time accurately (2/3 chance of winning), and 50% of the time the player will not call the bluff accurately (1/3 chance of winning).
This means that over multiple playthroughs if the host sometimes asks you if you want to switch and sometimes doesn't (essentially randomly), it doesn't matter if you switch or don't switch per se, as long as you keep it mixed up (if the player never switches or always switches, then the host could abuse that)
When you start, you have a 1/3 chance of picking the car, and there is a 2/3 chance that the car is in one of the other two doors. Assume you pick door number
1. There is a 1/3 chance that this door contains the car, and a 2/3 chance that the car is in door number 2 or 3. Then, assume the host eliminates door 3. The same 2/3 chance from before now applies to this door only. In other words, there is a 2/3 chance that the car is behind door number two and a 1/3 chance that the car is behind door
guys I have a question. so if you have 3 doors (or however many doors) and the game show host always offers you a choice to switch, no matter what, when should you stop switching? like do you switch once and stick with it or what?
@reallyboredist May I? :) First of, you can't always be right! Imagine how the world would be like if we could. "There's 66.6% chance of picking a goat.": 2/3 times you'll pick a goat, and you rely on that - it is your best shot, you'll never be 100% sure. Assuming you have picked a goat, the host has no other goat to reveal but the second one. In your mind both goats are revealed, hence you switch.
Assume that you and I play the same game 3 times (without changing the door objects inbetween) and you choose a different door in each game. If we follow the patern of 1. You choosing a door 2. Me revealing the object behind another door, and 3. You switchin', 2/3 times you win.
The man obviously doesn't know for sure he'll win! It is just that probability dictates that he will win 2/3 times. Though he might lose too, if he switches and is unlucky!!
I understand peoples misconceptions about this problem and 50% would be completly right IF, IF the host chose at random. Then based on the fact that odds have no memory it would be 50%. But because the host knows which door has the car your odds increase. But this clip is bad because your odds only increase if the host always reveals a goat and asks if you wanna switch. However he says how do you know hes not using reverse psychology which makes me think its more conditional probability.
just because there's two options at the end and one answer it doesn't mean they have equal chances. Those two doors have been through different circumstances and undergone different selection processes before they've been presented for your final choice, so they aren't necessarily the same.
Anyway, the way it's explained in the clip it isn't 1/3 and 2/3, the probablity is basically unknown and depends on how much you trust the game show host. If you don't switch it's 1/3, if u do it's unknown
Is everybody here a drug addict, or do people realize that this scene from a fairly good movie is rubbish? you're chance to begin with is 33.33 % as one correct answer out of three possibles. Once one is eliminated it becomes 2 possible answers with one correct, hence fifty perce..... AGGGHHH Screw it, if you can't see it just don't quit your day job as a painter, and forget card counting. Kevin Spacey was given a burn here as the scene is just wrong, simple, my sister even figured that out.
@pacmandem I don't know I would say his sister is very smart I didn't read the problem, but I read your post, We will call the car C and the goats G, C G G,
you pick C(1/3 ev), he reveals a G(reverse psychology) and you switch, you get a G,
You pick G(2/3 ev), he opens the door you picked and you get a G.
If we did this with a million goats and one car, pick car (1/million chance) he opens one G, you switch get G, you pick G, he opens door you chose, you get G. Therefore never switch.
@diablovt108vt109 bud once he gives you the option of switching it becomes 66.6% because you realize that the door you didn't pick had the goat and now there is only 2 doors.
@diablovt108vt109 Oh dear. This is a well known problem in Mathematics/Statistics. Given this perhaps you should show some humility and read up on it, before your sister and you start typing. Just some friendly advice. Bayles theory might help you.
um. they make this idea so difficult but he's just stating the obvious. Of course there are 3 doors. he's just giving the numbers to make it sound difficult.
The 2/3rd probability he is talking about isn't the probability of choosing the car. It's the probability of choosing the goat your FIRST time. That's a 2/3rd chance. If this scenario happens, the host will then reveal the LAST remaining goat door. So if you switch, you will win the car 2/3rd of the time. It's not based on your second decision, but rather your first decision IF you have chosen a goat door first.
@Azinoskateteam No, because at first, he had a 1/3 (33.3%) chance of choosing right. After one door is opened, changing his choice gives him a 2/3 (66.7%) chance because each door gives 1/3 of a chance. This is all of course assuming the host chose to open a door with a goat in it.
This scene gets the Monty Hall problem wrong. The strategy of switching only works if we assume that the game show host always gives the contestant a choice of switching doors, whether he's chosen a goat or not. In that case, as argued, switching doors will win 2/3 of the time.
But if the host follows a strategy of only offering a switch when the player has selected a car for example, to attempt to fool him (as suggested in the video), then switching when offered a choice is a losing strategy.
@MistyGothis This is what I don't understand about The Straight Dope's argument. You're adding variables unnecessarily.
There is no "strategy" in this problem, no assumption about whether the host always/never/sometimes offers the switch. The problem is simply as stated, Monty gives you the choice and in this problem, the math is correct.
Suppose that Monty Hall only offers the contestant the choice to choose another door when he or she has already chosen the correct door; when the contestant chooses a door with the goat, he immediately awards him the goat. Then switching doors will always lose. So the reasoning behind the usual solution does depend on an unstated assumption that Monty Hall is not following such a malicious strategy.
@MistyGothis mind games mean nothing here, it is all about statistics. it is a 1/3 chance of choosing the car right away, where there is a 2/3 chance of choosing one of the goats. Now if there is no option to chose the second door than there is no point of the show, a person chooses the right door immediately or doesn't and that is t he end.
@MistyGothis The scene actually doesnt get the problem wrong. You have an increase of 33.3% of earning a car instead of a goat as the goat was revealed. Revealing one goat gives you only one change of picking a goat 33.3% Switching doesnt guarantee you the car but it increases the chances of you earning it. Like he said. It is based on the statistics. Which sounds more appealing to go with 33% of earning the car by staying with your choice or 66.7% by switching?
@MistyGothis then you don't truly understand the problem. also he's only offering him a chance to switch not suggesting him to switch this is not slum dog millionaire.
I do understand the problem. The conclusion given in the film is correct, provided one makes the assumption that the host will open a door with a goat behind it regardless of whether the contestant chooses a door with a car or not. I'm just pointing out that assumption is necessary to the traditional analysis of the problem. It's not stated in the film, and the teacher's suggestion the host might be trying to "trick" the contestant is actually contrary to it.
@MistyGothis then you still don't understand the problem it doesn't matter if the host is trying to trick him or not because the odds will always be better for him to switch. the fact that the host knows exactly where the car is the reason why switching increases your odds. the point is no matter what. you're trying to add reverse psychology and intuition to a mathematical problem.
What you're saying is correct provided we assume the game show host reveals a goat regardless of whether the contestant chooses a door with a car or a goat. If, for example, the game show host only reveals anohther door when the contestant picks a car, and if when the contestant chooses a door with a goat he awards them the goat without giving them a chance to switch, then switching doors when offered the chance is a losing strategy. That's the only point I'm making.
@MistyGothis Doesn't get it wrong at all. The motives of the host are not part of the equation, that is a human element not considered in the equation as it can't be measured by numbers. The scene is correct, the argument you throw up belongs in a psychology class, not a maths class.
@rhinos111 I think MistyGothis disliked the fact that the professor himself introduced the "possibility" that the game show host, who knows everything, is revealing the other door and offering the contestant the chance to change in order to "trick" him into changing his answer and choosing the wrong door. But the professor should know better because--as you said--he's teaching math, not psychology.
@MistyGothis Notice how he says, "I DON'T CARE", implying that he doesn't care about any psychological bullshit the host is trying to use. The post before me has it right, the mathematical model is correct, and the main character agree's with me.
@Dillingerman1 He SHOULD care about the "psychological bullshit." Numbers are predictable. But humans? Not so much. That's probably why you consider psychology to be bullshit to begin with. It's a difficult science. For the mathematical answer to have practical significance (and answer the "what would you personally do to win" question), we need to assume that the host has no ill motives, but is acting methodically. That is, he will ALWAYS reveal the door with "the other goat" in every scenario.
There is no longer so clear of an advantage once we start questioning the motives of the game show host. That's when we leave the ordered realm of mathematics to play around in the monkey house that is psychology.
There's still some mathematics that can be done, even if the game show host is being tricky. Suppose the probability the game show host offers the contestant the opportunity to pick another is p if the contestant selects the correct door originally, and q if he picks a goat originally. Then the probability the contestant wins by switching doors is 2q/(p+2q). This reduces to 2/3 when p=q=1 (ie when Monty always offers a choice, as is usually assumed)
I think the advantage is largely irrelevant if we consider the host's motives. For instance, if a host is aware of the "advantage" of switching and believes contestants will likely switch for this reason, then perhaps he will offer "the choice" if and only if the correct door was first selected. Thus, the person switching may be more likely to lose, depending on the psychology of the average host. My point: once we consider the "human factor," mathematics is largely lost in the fog.
That would correspond to p=1 and q=0. Switching when given the opportunity would then have a probability of 2*0(1+2*0)=0 chance of winning. If the host always offered the contestant the opportunity to switch when he chose a goat, but only half the time when he chose the car, then switching would win half the time. As you say, we can't predict what the host will do, but we can still compute probabilities of winning under various assumptions of what the host will do.
This only works if the game show host asks every time. Otherwise, you wouldn't know if he's just trying to trick you. If his strategy was only to show you a door if you picked the car (and you didn't know that), then staying would be a better strategy.
So really, there's an element of chance in this strategy itself, because you don't know that the host isn't just trying to trick you, but since in the game show Monty Hall they did this every time, it is correct.
@mathijs91 With all 3 doors closed: 33% chance of picking the car & 66% chance of picking a goat (any of two goats)
This means that on your first pick, you have a 66% chance of picking a goat. When he opens the door to reveal a goat, the only options left are one goat and one car. Then you switch doors because you had 66% chance of picking a goat the first time, and after the reveal there's 66% chance that Door #2 is the car. If you don't switch your car chances stay 33%
the way they explained it is totally wrong. It most definitely matters if the host knows what's behind the doors and he's doing his best to cheat you. With these dumbasses playing the game the host will never have to give away one car. If they pick a goat with their first pick then the host says 'game over you lose', if they pick the car the host reveals a goat and asks if they want to switch to the remaining goat and they switch everytime. What a bunch of suckers.
I just visited stayorswitch (dot) com and you can play the actual game that they're talking about and see animated explanations. Kevin Spacey would be proud
The problem is not explained correctly. He doesnt tell you that the host has to give you the opportunity to switch no matter what you initially pick. It makes it seem like the host could trick you and only give you the chance to switch if your first pick was the car.
the dude knew that he had an equal chance of getting the door no matter what door he chose. so he picked door number one. the host knew it was wrong so he pulled down door number three to reveal nothing. now, why did he decide to pull down door number three? becasue he knew the car was behind door umber 2 so he couldnt pull that one down, and he couldnt pull door number one down since the dude already picked it and it would have revealed the answer(that being he was wrong) so he chose door 2
a good way of thinking of this is by imagining the same scenario but with 100 doors. You pick one and the game show host opens 98 of them leaving behind your door and one other door behind. Now what is the chance you picked the right door to start off with, it is 1/100. The chance the other door is the right one is much more likely
If you're still stuck on 50/50 as an answer, take a look at my video response "Proving The Monty Hall Problem." It'll give you a way to prove to *yourself* just what the correct answer is.
This is as tricky as flipping 5 times a coin and getting all 5 times heads, and for a 6th toss you think the universe is "due to itself" to get a tails. But as every coin toss is an independent event, the probability always will be 1/2 for each side. The probability tends to its expected value (1/2) after 'n' coin flips when n tends to infinity (Law of large numbers)
The difference between the host problem and the coin toss is that one is dependent of previous events while the other isn't.
First, you have 1/3 of picking right, but the other 2 doors contain the other 2/3. The important part of the problem is seeing that the problem stays but the variable is the only thing that changes. When the host opens the 3rd door, the 2nd and 3rd door STATISTICALLY still have 2/3 of being correct, but you already know that 3rd one is a goat; from there you conclude the 2nd one gets all the odds from both.
Statistics are not always intuitive as they seem. BTW, the real Ben was asian: Racism!
You get one door, the "house" gets two. At least one of the house's doors is always going to contain a goat, that is a given. So revealing one of the goat doors makes no difference either way; it's just smoke and mirrors. The advantage in switching is simply this: you are trading your ONE chance for the house's TWO chances. Just like buying two lottery tickets instead of only one doubles your chances of winning. It's no more complicated than that so you all get some sleep!! ~Mommy~ :)
actually. The problem would be much harder if the host did not open the door every time but only did so a certain percentage of the time. that would mean that you would have to change your chose a certain percentage of the time.
That's exactly correct. I wonder why this "problem" is still considered a problem. Once one of the door is open, your chance will stay at 50/50. Switching or not switching will give you equal chance of winning the car.
@ahlpa06 I thought so too for the longest time, and it seemed simple enough. But it apparently isn't the case. If you chose a goat (66% chance), and the host reveals another goat and you switch, you get a car. If you chose a car (33% chance), and host reveals a goat and you switch, you get a goat. 66% car, 33% goat. If you do not switch in both scenarios, then you have 66% chance of choosing a goat and 33% chance of choosing the car. The misconception is that the events don't affect each other!
@pixelsage Doors 1 2 and3. You have 33.33% chance of getting the car. why? because there are 3 options. simple enough right? right, we both agree on that. Now, the host opens a door that reveals a goat, how many doors are left? TWO! At this moment, the opened door doesn't matter anymore BECAUSE IT IS ALREADY OUT OF THE GAME. There are 2 doors left. Only 2. Not 3. And with 2 options, i'll say the chances are 50/50.
@ahlpa06 but it does not matter at that point, what matters is the initial choice. imagine if you had to choose betweeen 1000 doors. you have 1/1000 chances to get it right. now they open the wrong ones all the way down until you only have 2 remaining. whould you change or not? well you had only 1/1000 to hit it right at the first place, so you were wrong most likely. so you better change it. the probability is "of having it right at the first place".
@ahlpa06 it took me a while as well... the caviat though is that stats do not take in consideration emotions. most likelly the suicide rate is higher in people who decided to switch and lost because they did hit it in the first place... so the choice is do you want to increase your chances to win or your chances to commit suicide if you lose :p because the impact is much higher if you know you had it right and lost it by switching.
If you don't get it from this explanation, don't worry. It's not because it's incorrect, (it's not,) it's because it's simplified. Just look it up somewhere else for a more in-depth explanation.
@jimmayl1 Not exactly; probability and statistics are intertwined and studied together. We can't bash him for saying that it's statistics, where probability distributions are very often used.
@MSJjoshua Although the two are very closely related, I'd still maintain that the reasoning given is from a probabilistic standpoint. If it were from a statistical standpoint, the answer would've been along the lines of "In experiment, it seems to happen about 2 thirds of the time." I see your point though :)
@bakedbean1232 Depending on how you read it, that does not necessarily imply that the game show host *ensures* that he opens a door behind which sits a goat. :)
Considering the fact that door number three was a goat and is now out of the equation that increases the chances of both doors 1&2 being the car. wouldnt switching doors not make a difference because its now 50/50.
@TheMegaman321 no, because you know that in 2 out of three cases you had initially picked the wrong one, now when one wrong one is taken out of the game, your chances that you are on the right one is still 1/3, by switching you increase your chances to 2/3. Incidentally, it doesnt even matter if the host knows what is behind what door as long as you find yourself in the situation that one door is opened and there is a goat behind it.
@TheMegaman321 if you chose a car first and then switched, you would get a goat. If you chose a goat first and then switched, you would get the car. Since there's a higher probability of you having chosen a goat at first, it is in your interest to switch.
@Dathinkingman The professor is playing Devil's Advocate. Of course he knows how the Monty Hall problem works in great detail. By acting surprised, he forces the student to defend his stance with conviction, whereas a less confident student would simply assume he or she was wrong. It wasn't just a test to see if he could guess right, it was a test to see if he knew he was right beyond a doubt, and could explain why.
I hope the movie didn't have him actually end up picking the right door, which would just make his smugness even more insufferable. Yeah, he has a 2 in 3 shot, but it's not 3 in 3.
km9OOO 4 days ago
jUST ALL YOU BE QUIET, and enjoy the movie :)
dbknottyboby 6 days ago
Comment removed
DMK000 6 days ago
most clear explaination: /watch?v=mhlc7peGlGg
HexerPsy 1 week ago
I dont understand, you know behind 1 door is a goat. then there are 2 doors left. 2 door to gamble, why is there not a 50% to get the car? door number 1 + 2 is a big gamble isnt it? if he opens door number 3 you still cant know if 1 or 2 is the car.
Can someone msg me plz?
HappyBrotherBeat 2 weeks ago
@HappyBrotherBeat You choose door number 1. You have a 33.3% chance of your door being the car and a 66.7% chance of it being a goat. It's more likely you will pick a goat. Now the host opens a goat door. Now that only a goat and a car remain you can stay, hoping your pick was a car or switch hoping the pick was a goat. Since your pick was most likely a goat switching means it is more likely to win a car. Does that make sense?
GonnaBeALongLongTime 2 weeks ago
@GonnaBeALongLongTime If i pick door number 1 (which is the car) and he shows me door number three. So i now have door number 1 and 2 to choose from. Wouldnt it be a 50 50 chance? Why should i change if the odds are the same? im so confused at how this works, its all about luck, what if everytime i somehow manage to guess the right one first, if i went by this i would always lose.. dont see the logic in it...
sch3ffs 2 weeks ago
sch3ffs
It would only be 50/50 if from the start there were only 2 doors to pick from. since there are 3 doors, 2 of which do not contain the prize you are more likely to pick them over the prize.You WILL pick a goat 66.6% of the time, that's called statistics, you saying, well maybe i will get lucky and pick the car more often is ridiculous. The odds do not change... just because I may flip a coin and 3/4 times it's tails doesn't mean every time it's 75% that the coin will b tails...
igotsachubby 1 week ago
@sch3ffs Monty always opens a goat door. If you pick a goat door at the start then switch you always get a car. The odds of picking one of the two goats at the start is more likely than picking the car.
GonnaBeALongLongTime 1 week ago
@HappyBrotherBeat no, the events are not independent of each other, means that if one event occours (opening of the first door) it changes the nature of the other events as well, its called conditional probability and dependent events, its an application of the famous bayes theorm, search for bayes theorm of probability and read a lil about conditional probability :)
P(A|B).P(B) = P(B|A).P(A)
HAPPY LEARNING! :)
nextblain 1 week ago
The teacher bungled the problem by suggesting that the host might be playing mind games. This totally changes the nature of the problem. It is supposed to be clear that the host will always open a door with a goat after you make your original choice.
abuchanan821 3 weeks ago 11
@abuchanan821 It doesnt, since you cannot ever know if he is or not, and then the probability stay the same.
Ko252 1 week ago
I think the movie relied on the fact that him knowing the 33.33.. and 66.66.. percentage chance would be considered impressive by the average American.
PodzniakovisElite 3 weeks ago
He should have made clear that the host always opens the door with the goat. He said the host knows what is behind the doors and opens another one. It only works if the host always reveals a goat. :)
gabspen 3 weeks ago 4
SHOW OFF, LITTLE SMARTASS, PRICK, LITTLE SHOW OFF.... DUMB SMARTASS
Zee96969696 1 month ago
Reverse psychology really does come into play. Maybe the host only asks you to switch when he knows you chose the car. It matters.
Also, on the off chance that the host doesn't know where the car is, and chose a goat by accident, the two doors are 50-50.
Eudaletism 1 month ago
@Eudaletism Because the problem is a problem and nothing more, there shouldn't be any "maybe". The host DOES know where the car is, always. The problem is, "You choose a door. The host reveals a goat. The host asks you if you want to switch, no ifs, ands, or buts. Do you stick or change?" Remember, this is a sort of a riddle. You can't say, "what if." If the host didn't know where the car was, and if he only asks to switch when you choose the car, then it is no longer the Monty Hall Problem.
PinoyYoung 1 month ago
Is it true he introduces himself as SALVADOR SANCHEZ TO A GIRL IN THIS MOVIE!!! ONLY THE GREATEST BOXER EVER HAHAHA
TheFUTUREZHizTORY 1 month ago
Yes it is, thanks
MisterBamBam89 1 month ago
MYTH CONFIRMED!
glenjamz 1 month ago
Notice you win 66%. Doesnt matter if you play 9 times or 900.
MisterBamBam89 1 month ago
Saw this somewhere. Assume you play this game 3 times and always pick door 1 and switch. C=car G=goat
#1. cgg. You pick car and win goat
#2. gcg. You pick goat and win car
#3. ggc. You pick goat and win car
MisterBamBam89 1 month ago
yes there is variable change but once door number 2 is gone his odds in crease that door number 1 is still the right door. either way its a 50 50 chance in the end regardless which door he chooses.
fliptthescript 1 month ago
@fliptthescript the door you choses starts with a 1/3 chance, then he opens door number 3, and there is a goat. He offers you to change to door number 2, and you should because door 2 and 3 together has a 2/3 winning chance, and door 1 has a 1/3 winning chance
Akilles047 1 month ago
When you choose the first door, you have a 1/3 chance of winning. If the game show host gave you a "two for one" deal, permitting you to choose the two other doors, not just one door, then you would have a 2/3 chance of winning. So by switching doors, you're essentially getting the benefit of the "two for one" deal. It's as if you had picked the "two other doors" from the very beginning.
flavoredwallpaper 1 month ago
hungrybox
hussein335 1 month ago
It's easier to understand if looked from the car perspective (the prize u want):
1) picking a car = 33.3% (one door)
2) picking a goat = 66.6% (two doors)
If you stick to your 1st choice, your best is 33.3%,
but by switching you're trading your 33.3% for the remainder--the 66.6% (in effect, you are choosing TWO doors vs. one door).
So, staying = choosing one door = 33.3%.
switching = choosing two doors (opened door and switched door) = 66.6%.
Basically, opening two doors is better.
R2Mintus 1 month ago
can someone tell me what this movie is called becos i only know it as ''21''
sweetzez 2 months ago
Thumbs up if you watched Mythbusters and then searched for this :)
efenili 2 months ago
Why is this so special? Easy... logical... still... in the end... it is pure luck
ValmisFilm 2 months ago
Goats are way better than a fucking car. I actually really like Goats.
yesiamawizardjonny 2 months ago
he gave the answer. but he didn't explain how he got to that answer. so this scene is a failure.
blklightning330 2 months ago
HAHA I get this. This is such a cool math problem
technologycrazy 2 months ago
That was a terrible explanation.
DMK000 2 months ago 60
@DMK000 thats probably because your matematical understanding is not high enough
Akilles047 1 month ago
@Akilles047 Haha, is it a hobby of yours to troll on YouTube? He simply states facts in this video. He doesn't "explain" anything.
DMK000 1 month ago
@Akilles047 No, he is correct. There is no explanation given here. The kid merely states the correct probabilities. That's hardly an explanation of WHY those probabilities are correct. Why did the other door get a 1/3 boost? Because your confidence in that door is allowed to increase since the host COULD have legally opened that door and DIDN'T. Your door, however, he cannot legally open at all. So there's no reason to gain confidence in it when you see him dodge it.
SandBarAndYou 1 month ago
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@DMK000 the door you choses starts with a 1/3 chance, then he opens door number 3, and there is a goat. He offers you to change to door number 2, and you should because door 2 and 3 together has a 2/3 winning chance, and door 1 has a 1/3 winning chance
Akilles047 1 month ago
@DMK000 Yes it was. In fact it was not an explanation of the problem. It was merely a statement of the problem. Another false YouTube video title.
SandBarAndYou 1 month ago
Think of it this way. YOU'RE ON THE ASSUMPTION THAT YOU INITALLY PICKED A GOAT. Why? Cause there's 2 goats and 1 car. Therefore there's a higher chance (66.7%) of picking a goat. So when the gameshow host reveals a GOAT (can't reveal car) there's, now only a 33% chance of the other door being a goat and 66% chance of it being a car. Get it? Cause the 1 you picked (most likely goat) + goat shown = SWITCHING.
It's simple logical statistics.
VincentiusLeonidas 2 months ago
it's 66 percent that you will initially choose a goat. therefore, there's a 66 percent chance that when you switch, you will switch to the car. since of course, you most likely (66 percent) picked a door with a goat behind it.
blklightning330 2 months ago
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Shocked! ! The most incredible secrets of blackjack sites:bkk.nn.cx
0737sa 2 months ago
that sounded easy
TheJasonDR 2 months ago
hungrybox
hussein335 2 months ago
EXPLANATION FOR IDIOTS: if you pick one door from 1000 doors. the odds are you did not pick the door with the car. if every door expect one is opened you might think you have a 50/50 chance if you switch or don't. but your still stuck with your original door which is most likely wrong. so switch to the single closed door left and you will increase your chances by 999%.
EPICacceptance 2 months ago
@EPICacceptance
You actually increase your chances from 0.001 to 0.5 which is 500 times greater and thus an increase of %49900... just saying
Gastrophetes 2 months ago
@Gastrophetes
your correct, i was simplifying it for others.
EPICacceptance 2 months ago
@EPICacceptance
fair enough
Gastrophetes 2 months ago
@EPICacceptance
Actually I'm wrong. You increase your chances from 0.001 to 0.999 which is 999 times greater and thus an increase of 99800%
The original door doesn't have a 0.5 chance of being correct, it has it's original 0.001 chance. However in the other 0.999 of the time the car will be the only other unopened door so switching to it increases your chances of winning massively.
Gastrophetes 2 months ago
The teacher never said that the host will always open a door with a goat. He just said "the host opens another door, and behind it is a goat". That is completely different from "the host then opens another door, which will always contain a goat".
lazylink 2 months ago
to win without switching you have to choose the car to begin with (1/3)
to win with switching you have to choose a goat to begin with (2/3)
springeeljim 2 months ago
33.3+33.3=66.6 not 66.7 -_-
rusty8785 2 months ago
@rusty8785 the statictics is actually 33.333% 33.333 + 33.333 = 66.666 which is rounded to 66.7
JesseAesthetic21 2 months ago
The sound is out of phase.
Danderson07 2 months ago
I Will Choose Door Number four, and go out from that class..
Kinokunii 3 months ago 2
While the main stats have been discussed to death, this has not:
If a person who knows the proper stats was to play against a person who knew the proper stats (who would sometimes randomly ask to change, other times will not), statistically speaking, each person would win 50% of the time. This is because on average the player will call the bluff 50% of the time accurately (2/3 chance of winning), and 50% of the time the player will not call the bluff accurately (1/3 chance of winning).
MsHojat 3 months ago
This means that over multiple playthroughs if the host sometimes asks you if you want to switch and sometimes doesn't (essentially randomly), it doesn't matter if you switch or don't switch per se, as long as you keep it mixed up (if the player never switches or always switches, then the host could abuse that)
MsHojat 3 months ago
so did the dude get the car or the goat thats what i want to know!
peippoRebekks 3 months ago 2
When you start, you have a 1/3 chance of picking the car, and there is a 2/3 chance that the car is in one of the other two doors. Assume you pick door number
1. There is a 1/3 chance that this door contains the car, and a 2/3 chance that the car is in door number 2 or 3. Then, assume the host eliminates door 3. The same 2/3 chance from before now applies to this door only. In other words, there is a 2/3 chance that the car is behind door number two and a 1/3 chance that the car is behind door
matthew8333 3 months ago
guys I have a question. so if you have 3 doors (or however many doors) and the game show host always offers you a choice to switch, no matter what, when should you stop switching? like do you switch once and stick with it or what?
reallyboredist 3 months ago
@reallyboredist May I? :) First of, you can't always be right! Imagine how the world would be like if we could. "There's 66.6% chance of picking a goat.": 2/3 times you'll pick a goat, and you rely on that - it is your best shot, you'll never be 100% sure. Assuming you have picked a goat, the host has no other goat to reveal but the second one. In your mind both goats are revealed, hence you switch.
failingwithin 3 months ago
@reallyboredist Otherwise:
Assume that you and I play the same game 3 times (without changing the door objects inbetween) and you choose a different door in each game. If we follow the patern of 1. You choosing a door 2. Me revealing the object behind another door, and 3. You switchin', 2/3 times you win.
The man obviously doesn't know for sure he'll win! It is just that probability dictates that he will win 2/3 times. Though he might lose too, if he switches and is unlucky!!
Hope this helped!
failingwithin 3 months ago
THE JOKES ON THE GAME SHOW...I WANTED A GOAT!
SecretSocietySK8kru 3 months ago 89
Is this game theory, math, stats, or what?
ICarnag3I 4 months ago
(O_o)
70rm3n70r 4 months ago
This thing is pretty straightforward. Don't know why some people don't get it.
ultimokatiting 4 months ago
I understand peoples misconceptions about this problem and 50% would be completly right IF, IF the host chose at random. Then based on the fact that odds have no memory it would be 50%. But because the host knows which door has the car your odds increase. But this clip is bad because your odds only increase if the host always reveals a goat and asks if you wanna switch. However he says how do you know hes not using reverse psychology which makes me think its more conditional probability.
InsanePorcupine 4 months ago 3
28 people chose the door with the goat
MrBinet1976 4 months ago
Why are people disagreeing with the answer to this problem? The answer is supposed to be counter-intuitive.......The movie is right.
PapaMooCow 4 months ago
just because there's two options at the end and one answer it doesn't mean they have equal chances. Those two doors have been through different circumstances and undergone different selection processes before they've been presented for your final choice, so they aren't necessarily the same.
Anyway, the way it's explained in the clip it isn't 1/3 and 2/3, the probablity is basically unknown and depends on how much you trust the game show host. If you don't switch it's 1/3, if u do it's unknown
JohnnyGeologist 4 months ago
Is everybody here a drug addict, or do people realize that this scene from a fairly good movie is rubbish? you're chance to begin with is 33.33 % as one correct answer out of three possibles. Once one is eliminated it becomes 2 possible answers with one correct, hence fifty perce..... AGGGHHH Screw it, if you can't see it just don't quit your day job as a painter, and forget card counting. Kevin Spacey was given a burn here as the scene is just wrong, simple, my sister even figured that out.
diablovt108vt109 4 months ago
@diablovt108vt109 lol what is so difficult to get?! RUN THROUGH IT. We'll call the goats "A" & "B" and the car "C".
You pick A, he reveals B, you switch to C.
You pick B, he reveals A, you switch to C.
You pick C, he reveals A or B, you switch.
To begin with you're more likely to pick a goat. Therefore 66.6% of the time it is favourable to switch.
Btw, your sister is a retard. I wouldn't go to her for any maths related problems in the future, if I were you.
pacmandem 4 months ago 3
@pacmandem I don't know I would say his sister is very smart I didn't read the problem, but I read your post, We will call the car C and the goats G, C G G,
you pick C(1/3 ev), he reveals a G(reverse psychology) and you switch, you get a G,
You pick G(2/3 ev), he opens the door you picked and you get a G.
If we did this with a million goats and one car, pick car (1/million chance) he opens one G, you switch get G, you pick G, he opens door you chose, you get G. Therefore never switch.
Kneecap22 2 months ago
@diablovt108vt109 bud once he gives you the option of switching it becomes 66.6% because you realize that the door you didn't pick had the goat and now there is only 2 doors.
mmarano869 4 months ago
@diablovt108vt109 Oh dear. This is a well known problem in Mathematics/Statistics. Given this perhaps you should show some humility and read up on it, before your sister and you start typing. Just some friendly advice. Bayles theory might help you.
bakersteven3 4 months ago
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@diablovt108vt109 you are a dumbass .
nmc400 4 months ago
actually he should thank him only for 16.67 pct. not 33.3pct because if he stayed he still had a 50pct chance. SO....
mmarano869 4 months ago
He's talking about the start when he was at 1/3 odds then getting to 2/3 after removing one of the doors, so it is 33.3 %
Jem93rocks 4 months ago
@Jem93rocks yea if he's talking about the start it would be, but its a little ambiguous. /closed :D
mmarano869 4 months ago
um. they make this idea so difficult but he's just stating the obvious. Of course there are 3 doors. he's just giving the numbers to make it sound difficult.
bloosea123 4 months ago
what if you shot the gameshow host and took the three goats and the car
gunny0488 5 months ago
The 2/3rd probability he is talking about isn't the probability of choosing the car. It's the probability of choosing the goat your FIRST time. That's a 2/3rd chance. If this scenario happens, the host will then reveal the LAST remaining goat door. So if you switch, you will win the car 2/3rd of the time. It's not based on your second decision, but rather your first decision IF you have chosen a goat door first.
MrSpeedyAce 5 months ago 2
@Azinoskateteam No, because at first, he had a 1/3 (33.3%) chance of choosing right. After one door is opened, changing his choice gives him a 2/3 (66.7%) chance because each door gives 1/3 of a chance. This is all of course assuming the host chose to open a door with a goat in it.
Kivexor 5 months ago
its not 67% its fuckin 50% what is he on crack?
Azinoskateteam 5 months ago
And I see Evilerjolly already made that point.
MistyGothis 5 months ago
This scene gets the Monty Hall problem wrong. The strategy of switching only works if we assume that the game show host always gives the contestant a choice of switching doors, whether he's chosen a goat or not. In that case, as argued, switching doors will win 2/3 of the time.
But if the host follows a strategy of only offering a switch when the player has selected a car for example, to attempt to fool him (as suggested in the video), then switching when offered a choice is a losing strategy.
MistyGothis 5 months ago 30
@MistyGothis This is what I don't understand about The Straight Dope's argument. You're adding variables unnecessarily.
There is no "strategy" in this problem, no assumption about whether the host always/never/sometimes offers the switch. The problem is simply as stated, Monty gives you the choice and in this problem, the math is correct.
Pachuho 3 months ago
@Pachuho
Suppose that Monty Hall only offers the contestant the choice to choose another door when he or she has already chosen the correct door; when the contestant chooses a door with the goat, he immediately awards him the goat. Then switching doors will always lose. So the reasoning behind the usual solution does depend on an unstated assumption that Monty Hall is not following such a malicious strategy.
MistyGothis 3 months ago
@MistyGothis The problem is about a game in wicth the host always ask to switch, so accepting is always the best choice
Kibofro 3 months ago
@MistyGothis mind games mean nothing here, it is all about statistics. it is a 1/3 chance of choosing the car right away, where there is a 2/3 chance of choosing one of the goats. Now if there is no option to chose the second door than there is no point of the show, a person chooses the right door immediately or doesn't and that is t he end.
Greginda11 3 months ago
@MistyGothis The scene actually doesnt get the problem wrong. You have an increase of 33.3% of earning a car instead of a goat as the goat was revealed. Revealing one goat gives you only one change of picking a goat 33.3% Switching doesnt guarantee you the car but it increases the chances of you earning it. Like he said. It is based on the statistics. Which sounds more appealing to go with 33% of earning the car by staying with your choice or 66.7% by switching?
JesseAesthetic21 2 months ago
@MistyGothis then you don't truly understand the problem. also he's only offering him a chance to switch not suggesting him to switch this is not slum dog millionaire.
dimonai01 2 months ago
@dimonai01
@JesseAesthetic21
I do understand the problem. The conclusion given in the film is correct, provided one makes the assumption that the host will open a door with a goat behind it regardless of whether the contestant chooses a door with a car or not. I'm just pointing out that assumption is necessary to the traditional analysis of the problem. It's not stated in the film, and the teacher's suggestion the host might be trying to "trick" the contestant is actually contrary to it.
MistyGothis 2 months ago
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dimonai01 2 months ago
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@MistyGothis then you still don't understand the problem it doesn't matter if the host is trying to trick him or not because the odds will always be better for him to switch. the fact that the host knows exactly where the car is the reason why switching increases your odds. the point is no matter what. you're trying to add reverse psychology and intuition to a mathematical problem.
dimonai01 2 months ago
@dimonai01
What you're saying is correct provided we assume the game show host reveals a goat regardless of whether the contestant chooses a door with a car or a goat. If, for example, the game show host only reveals anohther door when the contestant picks a car, and if when the contestant chooses a door with a goat he awards them the goat without giving them a chance to switch, then switching doors when offered the chance is a losing strategy. That's the only point I'm making.
MistyGothis 2 months ago
@MistyGothis he doesn't imply that he has chosen the car.
tylerperson090 2 months ago
@MistyGothis Doesn't get it wrong at all. The motives of the host are not part of the equation, that is a human element not considered in the equation as it can't be measured by numbers. The scene is correct, the argument you throw up belongs in a psychology class, not a maths class.
rhinos111 2 months ago
@rhinos111 I think MistyGothis disliked the fact that the professor himself introduced the "possibility" that the game show host, who knows everything, is revealing the other door and offering the contestant the chance to change in order to "trick" him into changing his answer and choosing the wrong door. But the professor should know better because--as you said--he's teaching math, not psychology.
flavoredwallpaper 1 month ago
@MistyGothis Notice how he says, "I DON'T CARE", implying that he doesn't care about any psychological bullshit the host is trying to use. The post before me has it right, the mathematical model is correct, and the main character agree's with me.
Dillingerman1 2 months ago
@Dillingerman1 He SHOULD care about the "psychological bullshit." Numbers are predictable. But humans? Not so much. That's probably why you consider psychology to be bullshit to begin with. It's a difficult science. For the mathematical answer to have practical significance (and answer the "what would you personally do to win" question), we need to assume that the host has no ill motives, but is acting methodically. That is, he will ALWAYS reveal the door with "the other goat" in every scenario.
flavoredwallpaper 1 month ago
@MistyGothis Yup. Exactly right.
There is no longer so clear of an advantage once we start questioning the motives of the game show host. That's when we leave the ordered realm of mathematics to play around in the monkey house that is psychology.
flavoredwallpaper 1 month ago
@flavoredwallpaper
There's still some mathematics that can be done, even if the game show host is being tricky. Suppose the probability the game show host offers the contestant the opportunity to pick another is p if the contestant selects the correct door originally, and q if he picks a goat originally. Then the probability the contestant wins by switching doors is 2q/(p+2q). This reduces to 2/3 when p=q=1 (ie when Monty always offers a choice, as is usually assumed)
MistyGothis 1 month ago
I think the advantage is largely irrelevant if we consider the host's motives. For instance, if a host is aware of the "advantage" of switching and believes contestants will likely switch for this reason, then perhaps he will offer "the choice" if and only if the correct door was first selected. Thus, the person switching may be more likely to lose, depending on the psychology of the average host. My point: once we consider the "human factor," mathematics is largely lost in the fog.
flavoredwallpaper 1 month ago
@flavoredwallpaper
That would correspond to p=1 and q=0. Switching when given the opportunity would then have a probability of 2*0(1+2*0)=0 chance of winning. If the host always offered the contestant the opportunity to switch when he chose a goat, but only half the time when he chose the car, then switching would win half the time. As you say, we can't predict what the host will do, but we can still compute probabilities of winning under various assumptions of what the host will do.
MistyGothis 1 month ago
This only works if the game show host asks every time. Otherwise, you wouldn't know if he's just trying to trick you. If his strategy was only to show you a door if you picked the car (and you didn't know that), then staying would be a better strategy.
So really, there's an element of chance in this strategy itself, because you don't know that the host isn't just trying to trick you, but since in the game show Monty Hall they did this every time, it is correct.
evilerjolly 5 months ago
Isnt is a 50/50 thing when he opens door 3?
mathijs91 5 months ago
@mathijs91 With all 3 doors closed: 33% chance of picking the car & 66% chance of picking a goat (any of two goats)
This means that on your first pick, you have a 66% chance of picking a goat. When he opens the door to reveal a goat, the only options left are one goat and one car. Then you switch doors because you had 66% chance of picking a goat the first time, and after the reveal there's 66% chance that Door #2 is the car. If you don't switch your car chances stay 33%
Sean112Y 5 months ago
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the way they explained it is totally wrong. It most definitely matters if the host knows what's behind the doors and he's doing his best to cheat you. With these dumbasses playing the game the host will never have to give away one car. If they pick a goat with their first pick then the host says 'game over you lose', if they pick the car the host reveals a goat and asks if they want to switch to the remaining goat and they switch everytime. What a bunch of suckers.
JohnnyGeologist 5 months ago 2
BTW, this is no explanation at all.
Beriaal 5 months ago
A math student would already know the problem.
Beriaal 5 months ago
I just visited stayorswitch (dot) com and you can play the actual game that they're talking about and see animated explanations. Kevin Spacey would be proud
mumiscrunk 5 months ago
Soft math porn, no homo.
romale2307 6 months ago
he just spoke a load of crap for one minute, which seemed to overly impress kevin spacey for some reason. disappointed.
sbojify 6 months ago
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Choose Goat 1, Goat 2 is revealed, switch and you get a car.
Choose Goat 2, Goat 1 is revealed, switch and you get a car.
Choose Car, Goat 1 or 2 is revealed, switch and you get other goat.
You win 2 out of 3 times if you switch.
777Skeptic 6 months ago
a grasshopper wouldn't know all of his students names
martyxcallsiefer 6 months ago
Too bad you can't switch in the show, and there no zonks anyway on the big deal.
hungryhungrymetroid 6 months ago
it looks like he nailed this question!!
actualy my brother explained it to me afterwards and told me it's too easy that he doesn't get why kevin spacey was so impressed lol -.-
prinilen 6 months ago
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The problem is not explained correctly. He doesnt tell you that the host has to give you the opportunity to switch no matter what you initially pick. It makes it seem like the host could trick you and only give you the chance to switch if your first pick was the car.
TheFightens 7 months ago
Comment removed
TheFightens 7 months ago
the dude knew that he had an equal chance of getting the door no matter what door he chose. so he picked door number one. the host knew it was wrong so he pulled down door number three to reveal nothing. now, why did he decide to pull down door number three? becasue he knew the car was behind door umber 2 so he couldnt pull that one down, and he couldnt pull door number one down since the dude already picked it and it would have revealed the answer(that being he was wrong) so he chose door 2
killerxela 7 months ago
switching loses only if you originally pick the car, which has a probability of 1/3, so switching must have a probability of 2/3
Pinkerooprincess 7 months ago
a good way of thinking of this is by imagining the same scenario but with 100 doors. You pick one and the game show host opens 98 of them leaving behind your door and one other door behind. Now what is the chance you picked the right door to start off with, it is 1/100. The chance the other door is the right one is much more likely
lebagelboy 7 months ago 3
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If you're still stuck on 50/50 as an answer, take a look at my video response "Proving The Monty Hall Problem." It'll give you a way to prove to *yourself* just what the correct answer is.
kirkbocek 7 months ago
For everyone that doesn't get it, the easiest way to understand is if I rephrased the same problem like this:
"Would you like to choose 1 door or 2 doors? If you choose 2 doors, I’ll give you the better prize of those 2 doors."
133j0hn 7 months ago
This is as tricky as flipping 5 times a coin and getting all 5 times heads, and for a 6th toss you think the universe is "due to itself" to get a tails. But as every coin toss is an independent event, the probability always will be 1/2 for each side. The probability tends to its expected value (1/2) after 'n' coin flips when n tends to infinity (Law of large numbers)
The difference between the host problem and the coin toss is that one is dependent of previous events while the other isn't.
furby4bono 8 months ago
First, you have 1/3 of picking right, but the other 2 doors contain the other 2/3. The important part of the problem is seeing that the problem stays but the variable is the only thing that changes. When the host opens the 3rd door, the 2nd and 3rd door STATISTICALLY still have 2/3 of being correct, but you already know that 3rd one is a goat; from there you conclude the 2nd one gets all the odds from both.
Statistics are not always intuitive as they seem. BTW, the real Ben was asian: Racism!
furby4bono 8 months ago
You get one door, the "house" gets two. At least one of the house's doors is always going to contain a goat, that is a given. So revealing one of the goat doors makes no difference either way; it's just smoke and mirrors. The advantage in switching is simply this: you are trading your ONE chance for the house's TWO chances. Just like buying two lottery tickets instead of only one doubles your chances of winning. It's no more complicated than that so you all get some sleep!! ~Mommy~ :)
MommyofMany19 8 months ago
lets try this way. you walk in and theres 3 doors but one is already open with a goat and monty says pick one of the 3. now what are the odds?
exactly.
jubileeshine 8 months ago
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so you are sayin if it was 5 doors and monty open 3 then you got a 80% chance to be right if you switch to the other door.
so if its 100 doors monty open 98 then you are 99 times in 100 winning the car if you change your choice from between 2 options.
shoot tell monty i want a zillion doors i cant lose.
damn inert mathematicians cant relate to new conditions.
2 choices = 50/50. what came before has no bearing on now mathwise.
jubileeshine 8 months ago
actually. The problem would be much harder if the host did not open the door every time but only did so a certain percentage of the time. that would mean that you would have to change your chose a certain percentage of the time.
timonix2 8 months ago
i dont get it , how does it become 66.7% for the choice , theres only two doors left so does that not mean its 50 % for each ?????
crazimichael123 8 months ago
@crazimichael123
That's exactly correct. I wonder why this "problem" is still considered a problem. Once one of the door is open, your chance will stay at 50/50. Switching or not switching will give you equal chance of winning the car.
ahlpa06 7 months ago
@ahlpa06 I thought so too for the longest time, and it seemed simple enough. But it apparently isn't the case. If you chose a goat (66% chance), and the host reveals another goat and you switch, you get a car. If you chose a car (33% chance), and host reveals a goat and you switch, you get a goat. 66% car, 33% goat. If you do not switch in both scenarios, then you have 66% chance of choosing a goat and 33% chance of choosing the car. The misconception is that the events don't affect each other!
pixelsage 7 months ago
@pixelsage Doors 1 2 and3. You have 33.33% chance of getting the car. why? because there are 3 options. simple enough right? right, we both agree on that. Now, the host opens a door that reveals a goat, how many doors are left? TWO! At this moment, the opened door doesn't matter anymore BECAUSE IT IS ALREADY OUT OF THE GAME. There are 2 doors left. Only 2. Not 3. And with 2 options, i'll say the chances are 50/50.
ahlpa06 7 months ago
@ahlpa06 but it does not matter at that point, what matters is the initial choice. imagine if you had to choose betweeen 1000 doors. you have 1/1000 chances to get it right. now they open the wrong ones all the way down until you only have 2 remaining. whould you change or not? well you had only 1/1000 to hit it right at the first place, so you were wrong most likely. so you better change it. the probability is "of having it right at the first place".
mrmostarr 7 months ago
@mrmostarr ok..i give up. now that makes sense... you and @lebagelboy enlightened me. :)
ahlpa06 7 months ago
@ahlpa06 it took me a while as well... the caviat though is that stats do not take in consideration emotions. most likelly the suicide rate is higher in people who decided to switch and lost because they did hit it in the first place... so the choice is do you want to increase your chances to win or your chances to commit suicide if you lose :p because the impact is much higher if you know you had it right and lost it by switching.
mrmostarr 7 months ago
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@ahlpa06 i think i understand it now :P
A) You pick the car, host shows a goat, you switch to a goat and lose
(B) You pick goat #1, host shows goat #2, you switch to the car and win
(C) You pick goat #2, host shows goat #1, you switch to the car and win
that gives you a 2/3 or 66.6666.. % chance of winning
crazimichael123 7 months ago
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frostydisco 7 months ago
@ahlpa06 Mm nope.
NonisMonster 7 months ago
@NonisMonster Lol, ok. I understand your point. :)
ahlpa06 7 months ago
If you don't get it from this explanation, don't worry. It's not because it's incorrect, (it's not,) it's because it's simplified. Just look it up somewhere else for a more in-depth explanation.
MrEdLoop 8 months ago
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KRiiPTiiK 8 months ago
@KRiiPTiiK
Bradiils.
N1rOx 8 months ago
HAHAHAHAHAHAHA
The annotation top left hand corner.
2th.. Fucken stupid Americans.
KRiiPTiiK 8 months ago
@KRiiPTiiK
But he's Canadian?
N1rOx 8 months ago
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KRiiPTiiK 8 months ago
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KRiiPTiiK 8 months ago
@N1rOx Eat shit.
KRiiPTiiK 8 months ago
lol i learned this in 8th grade
krillzhd 8 months ago
wat
saladwithsalad 9 months ago
@samodoidi: Exactly why this leaves me confused.
BlockJuice 9 months ago
I like how he says it's based on statistics, but it's based on probability theory and doesn't touch statistics. The two are very different!
jimmayl1 9 months ago
@jimmayl1 Not exactly; probability and statistics are intertwined and studied together. We can't bash him for saying that it's statistics, where probability distributions are very often used.
MSJjoshua 9 months ago
@MSJjoshua Although the two are very closely related, I'd still maintain that the reasoning given is from a probabilistic standpoint. If it were from a statistical standpoint, the answer would've been along the lines of "In experiment, it seems to happen about 2 thirds of the time." I see your point though :)
jimmayl1 9 months ago
The 'professor' neglects to mention that the host has to reveal a goat for the Monty Hall problem to make sense.
RobinScott93 9 months ago
@RobinScott93 You fail. He clearly mentions it
bakedbean1232 9 months ago
@bakedbean1232 He doesn't, actually.
RobinScott93 9 months ago
@RobinScott93 "The game show host decides to open another door, behind which sits a goat." :)
bakedbean1232 9 months ago
@bakedbean1232 Depending on how you read it, that does not necessarily imply that the game show host *ensures* that he opens a door behind which sits a goat. :)
RobinScott93 9 months ago
@RobinScott93 Whatever, on an unrelated note, nice piano playing ;)
bakedbean1232 9 months ago
@bakedbean1232 Thanks, although it's really not great. Mostly simple music I played years ago. Great guitar playing, though.
RobinScott93 9 months ago
@samodoidi exactly. it's not really "explaining". and the "it changes everything" line is kind of weird, too.
and I wonder why he rounded the probability up to 66.7 :)
but it's suprisingly accurate representation of math anyway, compared to all that bullshit that's out there
runn1ng 9 months ago
Considering the fact that door number three was a goat and is now out of the equation that increases the chances of both doors 1&2 being the car. wouldnt switching doors not make a difference because its now 50/50.
my mind is full of fuck
TheMegaman321 9 months ago 4
@TheMegaman321 no, because you know that in 2 out of three cases you had initially picked the wrong one, now when one wrong one is taken out of the game, your chances that you are on the right one is still 1/3, by switching you increase your chances to 2/3. Incidentally, it doesnt even matter if the host knows what is behind what door as long as you find yourself in the situation that one door is opened and there is a goat behind it.
oringent 9 months ago
@TheMegaman321 if you chose a car first and then switched, you would get a goat. If you chose a goat first and then switched, you would get the car. Since there's a higher probability of you having chosen a goat at first, it is in your interest to switch.
rawoofi 8 months ago 35
@rawoofi YESSSSSS I FINALLY GET IT! THX BRO NICE EXPLANATION! :D
Bl4cKJack1 7 months ago
Guys guys, that a wrong math
ffiirraatt93 9 months ago
if u wana see one of the guys from the blackjack team in real life go to
YOUTUBE(DOT)COM/watch?v=solKiw10GUM&feature=related hes the one closer to the camera teaching the world champion of poker how to play poker (LOL)
brwnc 9 months ago
FAKE
like a professor would actually know his students name.
pru005 9 months ago 114
@pru005 moreover the professor would know the monty hall problem in great detail and wont look surprised.
Dathinkingman 9 months ago
@Dathinkingman The professor is playing Devil's Advocate. Of course he knows how the Monty Hall problem works in great detail. By acting surprised, he forces the student to defend his stance with conviction, whereas a less confident student would simply assume he or she was wrong. It wasn't just a test to see if he could guess right, it was a test to see if he knew he was right beyond a doubt, and could explain why.
jakebeanes 9 months ago