"21" explains the Monty Hall problem

The character, Ben Campbell, from the movie "21" explains the Monty Hall problem.

[@nitishchauhan8655]

For all the people saying that it is 50% not 66%: Imagine if instead of 3 doors there was a billion. You choose the 1 door out of a billion. What are the chances that you chose the correct door? 1 out of a billion. Then imagine that the host opened every other door except for one (now only two doors closed). Now you're given the choice to re-choose. Would you bet on the fact that you got it correctly the first time or would you swap? You would definitely not think that you got it correct with a 0.0000001% success rate. This is where you might now say "Well now it's just a 50/50." Well is it really? The probability of your first choice hasn't changed. It's still 1 out of a billion. But the dealer has helped you out by eliminating all the other choices, leaving you with just two doors. So now you have a choice between 1 out of a billion and 999,999,999 out of a billion. Which do you choose?

[@somatotrophin1535]

This is correct. The host consiously removes the baddy and rveals that it is a baddy. Imagine if the host decided to randomaly remove a door but did not reveal what was behind it. Then, the remaining door would be 1/3 chance of having the car, not 2/3, and switching will not make a statistical difference.

[@StuUngar]

I always looked at it this way....assuming you initially picked the wrong, switching will always give you the correct answer...and in your scenario, choosing the wrong door has a 999,999 chance out of 1,000,000.

[@daniellann4902]

Thanks, this was the first explanation that made it so clear to me.

[@GrammeStudio]

this is probably the most unique explanation i've found in the comment section and really challenges our math intuition. and i love it. this is a reasoning tactic that has a name that I've forgotten. it's basically reframes ideas in its extremes. if someone could tell me the name, that'd be great

[@shsaa2338]

⁠​⁠​⁠@@StuUngarit is one out of billion (not million) that your first chose would be correct. Nevertheless, imagine that in fact the host would open other doors and would offer you to switch only when your first chose is correct. Should you switch in such a case?

[@arnoldrotundo7979]

It comes down to whether you think the second decision is dependent on the first, or whether the second decision is independent on the first. The second decision is, in fact, dependent on the first. If your first decision chose a goat (66% chance) always swapping will get you a car, because Monty opened the door with the other goat, and always keeping will get you the goat you originally chose. If your first decision chose a car (33% chance) always swapping will get you a goat, cause you already chose the car, and always keeping will get you the car you originally chose. So, would you rather get a car 33% of the time when you always keep your original decision, or would you rather get a car 66% of the time when you always swap your original decision. It's easier if you start with the probabilities of your second decision first, and then the probabilities of your first decision second.

[@larrybud]

This is very well explained.

[@VegasRush]

Arnold Rotundo You need rules to the game, If you use statistics to answer the questions every time you will not win the car 100% of the time. If you choose door 1 and the host opens this door right away because the car isn't behind door 1 you lose, if you choose door 1 and the car is there the host simply opens another door and offer's you to change, using statistics you again swap to the wrong door and therefore have 100% chance of never getting the car.

[@MshariAlBarazy]

I've been searching for hours, and this is the best answer of all

[@vishwanathkoijam7912]

@@VegasRush well yeah..

[@JonnysGameChannel]

I saw an explanation in an other vid that was quite easy to understand: The door you initially pick has a 1/3 chance to have the price, the remaining 2 doors have a combined chance of 2/3. The host eliminates one of them, so the 2/3 chance concentrates on one door.

[@vemundbrandtzg4575]

Shouldnt it just be a 50/50 chance with the remaining choices? Why do we count with the % from the reviled door?

[@shemmy95]

@@vemundbrandtzg4575 Think of it this way. let's expand. you pick one door out of 20, host opens 18 that are not the one. 2 left. Host asks you to keep yours or go with the other what do you do? your door has 1/20 chance. other one had pretty much 19/20. In the door you pick you are 33% (of the total). while you are right it is 50/50 of the 2 left it is still 33% of the total and cannot improve. The other door is 50% of the 2 left but overall 67% of the total 3. So do you chose 50% and 33% or 50% and 67%?

[@HeyEveryoneHi]

That's not simple explanation

[@gryphonburdman7]

think about it this way: after you guess, the game show host opens up all doors except for the correct one and your guess. if you have guessed the correct door, he will leave an incorrect door closed. since it is more likely that you initially guessed the incorrect answer, switch more a greater chance of success

[@vagrant87]

I got it. There's 1/3 chances the car is in the door you pick, and 2/3 chances it's in the other two. But since the show host removes one of those, now there's 2/3 chances it's in the remaining door. The host changes the odds because he is choosing with the knowledge of where the car is

[@kigulamugambe1380]

Nice explanation

[@shsaa2338]

Nice explanation. Nevertheless, that works only if host of such show is always opening one of two remaining doors. Imagine that you know that in all previous shows host gave players opportunity to switch ONLY when their first chose was correct. Would you still switch in such case? 😜

[@vagrant87]

@@shsaa2338 right! also true. To me in that situation, the host offering another door is a somewhat strange turn of events that makes you think you chose the correct door and the host is trying to trick us. But if you don't have information about the host behavior in the past, and this situation happens for the first time without further data, then you should go with the one with better odds

[@shsaa2338]

@@vagrant87in that movie Ben definitely gave wrong answer. Professor clarified that it is possible that host may be playing against Ben [what is contrary to classical Monty-Hall problem]. Nevertheless, Ben replied that he doesn’t care about that and that he based his chose on: 1) - statistic [which is absent in his case, and could be very different from classical Monty-Hall] and 2) - “variable change” 🤓🤦‍♂️🤦‍♂️🤦‍♂️😜

[@shsaa2338]

@@vagrant87 “Unnatural generosity” of the host makes original Monty-Hall problem so counter-intuitive. But here the professor warned Ben that host may be playing against him. It would actually make no difference whether or not Ben would change his first chose (the probability to win would remain at 33%), if we reasonably suggest that host plays fairly: a) in 50% of the games host plays agains the guests - host would offer to change original chose ONLY if the first chose was correct (in 1/3 cases of those 50%); b) in other 50% of the games host doesn’t play against the guests and randomly (NOT depending on their first chose) offers to make change in 1/3 of those 50% of the cases (to make change offers equally in 1/3 of all cases a and b). That would be the fair approach by the host. No extra chances. If you may have a reason to believe that host may be giving you extra probability - use that. You are not losing or wining anything in such case if host is only “fair” and not giving extra favour. But if host is playing against you, then you would lose what you had. Common sense prevails. Real life.

[@Dragoncro0wn]

For those still wondering You can do this yourselves. Get a sheet of paper then write 15 scenarios of 3 doors. then go to a randomizer in google and use it to place the car randomly in each of the 15 scenarios. Then you pick randomly don't think about it just quickly say 1 through 3. After that open all the others doors with a goat and switch you will promptly win 7-8 times. Now do this all over again but do not switch. You will win 2-3 times depending

[@imnotfrommilkyway]

This explained the problem better than wikipedia ever did.

[@jangtheconqueror]

That is a really good classroom.

[@willoughbykrenzteinburg]

Very true. The Monty Hall problem is based on the premise that the host must ALWAYS reveal a goat door. If the host acts randomly (meaning he could potentially reveal the car door - voiding the game), then in the cases where the host reveals a goat door, the probability of you having a car in your door is 1/2. When the host is required to reveal a goat door, the probability of you having a car is 1/3 - - 2/3 that the car is in the other door.

[@SuperScrubGaming]

the easiest way i can think to explain this problem is this: every time that you choose a goat, and then switch, you will win the car, and the odds that you will choose a goat primarily are 66.66%, whereas when you stick with your original choice, and there is a 66% chance your original choice is wrong, you will win a goat.

[@monhi64]

His answer is such a classic example of someone being too smart/different to explain their correct answer. But the professor is the opposite, he’s explaining the wrong answer so well. They went as far as making the kids most intuitive argument wrong as well (variable change sounds kind of correct but he’s using it incorrectly). Really makes me wonder what they what they wanted the overall message of this scene to be because it seems like a scene intentionally written poorly to drive their message better than something more accurate. Which is pretty interesting and super smart in a way. I really really hope it wasn’t just lazy writing but I doubt it.

[@klaus7443]

As you said the scene was poorly scripted. I would have preferred that the script remained the same for the professor but have the student point out the error of his question. That would have been more effective.

[@thomthum2000]

One of my professors went over this concept in class the other day. He showed us how we can set up a program that proves the Monty Hall problem. Neat stuff!

[@nogoodbastid]

you have to take into account that the host is going to open and show you another door with a goat behind it no matter which door you choose. he is showing you the incorrect door of the remaining 2, which means by switching to the 2nd door you double your odds of getting the car because you are getting the odds of guessing 2 doors at once rather than your original odds of only guessing 1 door out of the 3.

[@abuchanan821]

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.

[@timrussell5110]

Let's clear this mess up right now. 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!

[@freddieorrell]

+Tim Russell "3) Monty always picks a goat..." If this means 'whenever we describe the MHP game' it is tautological. If it means 'in any series of games in which the MHP takes place' it is incorrect. Monty need not ALWAYS reveal a goat as long as he acts non-randomly. Monty reveals his deliberate object either (a) because it is the only car, (b) because it is the only goat, or (c) because it is one of two goats. Whether it is (a) or otherwise is known to the contestant - they can see a car or a goat - so the odds of that are resolved. So there are no additional odds on which the remainder is conditional. It's still (b) in 2/3 of cases or (c) in 1/3.

[@AtaOfTheShard]

There are 6 possible outcomes (assuming the host is forced to reveal a goat). 1: Pick door 1(Goat 1), keep door, get goat 1. 2: Pick door 1 (Goat 1), switch door, get car. 3: Pick door 2 (Goat 2), keep door, get goat 2 4: Pick door 2 (Goat 2), switch door, get car. 5: Pick door 3 (Car), keep door, get car. 6: Pick door 3 (Car), switch door, get goat 1 or goat 2. Switching wins 2 out of 3, Keeping wins 1 out of 3.

[@LylWren]

52 cards in a deck. The goal is to find the Ace of Spades. You randomly pick a card and place it face down on the table without looking at it. The Dealer goes through the rest of the deck and removes all card but one and places the remaining card on the table facedown next to your card. One of those cards is definitely the ace of spades. Which is it more likely to be? The random card you pulled from the deck or the one the dealer picked out? If you have a deck of cards and a friend, try this out. You'll see it is better to switch every time.

[@vinnieof1]

This is what made it properly click for me. Thank you!

[@willoughbykrenzteinburg]

In every game, the car is either in your door, or the only other remaining door. Now, the important thing to understand is that you are not actually making a second choice. You are just asking yourself if you think you picked the car or one of the goats. The car is either in your door...or it is NOT in your door. When the host reveals a goat, he is just reducing the number of doors you did NOT pick. He's not changing the odds. There is still a 33% chance the car is in your door; 66% NOT.

[@oHInsane]

An easier way to understand that is : You have 100 doors, behind only one of them you have gold and behin the others you have trash. ->You pick one door, you have 1% chance to win. ->98 other doors are revealed to you with trash behind. ->You have now two doors. That means you still have 1% chance to win, but the other door have 99% to win so you should change :) In fact the chance you found the right one right from the start is still thin so it's pretty intuitive with a high number of doors.

[@BJDGames]

Consider this statement: If you pick a goat door on your first choice AND you then decide to switch, you will end up with the car. Now on your first choice, there is a 66% chance that you will pick a goat, right? So because there is a higher chance that you will pick a goat on your first choice, it makes since to always switch. Unless you actually want a goat not a car :)

[@Only1DennisBergkamp]

Interesting concept if the host did have the choice whether to open or not. It would be a more of a physchological battle with the host on 'why isn't he opening the door?' or if he does, 'does this mean he is tempting me to stay with my option and not switch' etc. etc.

[@RhoadsLivesOn]

It's hard to explain this well in 500 characters. However, this is a very well known problem and is all over the internet, just search "Monty Hall". And next time, don't assume you're right and call anyone who thinks otherwise dumb, because in this case, you are wrong.

[@MrRoverno]

1/2 is the probability of winning the car if sometimes you change and sometimes you don't. So you could say it is the overall chance of winning the car when given an OPPORTUNITY for change. However, if you change every time then the chance of winning the car is 66.67%

[@dubiumguy]

No. Its the fact that the host always reveals a goat that changes everything. Door 1 = goat. Door 2 = goat. Door 3 = car. Contestant chooses door 1. Host has to reveal other goat. Contestant switches and the only other door has the car. Contestant chooses door 2. Host has to reveal other goat. Contestant switches and the only other door has the car. Contestant chooses door 3. Host reveals one goat. Contestant switches and the only other door has the remaining goat. 66% win chance.

[@aftermath4096]

No, he cannot open the door with a car, but he could choose to not open a door if he had a choice, and that changes everything...

[@fokke1234]

Ben picks door nr. 1, so at that point he has a 33.3% of being correct a 66.6% chance of being wrong. The host reveals one of the doors that was definately wrong Remember, before door 3 was revealed there was a 66.6% that the car was in door 2 or 3. The host HAD to pick a wrong door out of 2 and 3, and since it is MOST LIKELY in either door 2 or 3 by eliminating door number 3 the chances are greater that is in door nr 2 cuz ORIGINALLY the chances were better (66.6%) that it was in door 2 or 3.

[@nicholaslance7947]

This is wrong, I knew it was when I saw the movie, but it bounced around in my head for a while and I figured out why it was wrong, then went looking for why it was presented as a fact. Basically, they leave part of the problem out. The way the problem is described in the movie means there is exactly the same chance of the car being behind either of the remaining doors. For it to work, the game show host always has to pick a door other than the one you pick, and it always has to have a goat.

[@drglove614]

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!

[@phizzy56]

i researched this for hours, and i finally understood it by watching 'the monty hall problem' video there in the links on the right. basically, it's like this: if you do swap, then the only way to lose is if you had initially picked the car. so, rather than having a 33% of winning the car, you instead have only a 33% of LOSING by selecting the car door first.

[@monhi64]

I find it mostly simply explained that we know the host has to avoid the winning door so any door he avoids opening is probably the winning one. The movie explains the correct answer the worst way I’ve ever seen for this problem lmao. Sorry for old reply

[@hokiefan104]

Think of this way. Say there are 1 million doors (1 with the car, the rest with goats.) You pick a door and the host reveals 999,998 doors with goats behind them. Do you still think it's equal odds that the door you picked the first time has the car behind it? It's much more likely it's the other door that the host skipped.

[@flavoredwallpaper]

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.

[@hopelansing7229]

Your odds of picking a car are 33%. The odds you did not pick a car are 66.6%. This never changes. It is important to realize this fact. At the begininng of the game, that 66.6% chance was divided by the 2 doors you did not pick giving THOSE doors also...a 33% chance each of having a car. The only thing the host does by revealing a goat is reduce the number of doors remaining. It is still 66.6% that you did not pick a car, but now it is only divided by 1 remaining door giving it 66.6%.

[@phomarphoto]

Ok, this is true, however its not getting you a car in reality... Imagine I present you 3 cups, and there is a $100 dollars under one cup. Its my game. Choose a cup. You choose no.1. ...If it is not a $100 I'll give you that cup and you loose. Get it? If it is a $100 cup, ill ask you to switch because i don't want to give you $100... so odds are in my favor

[@ashurtz]

it never changes.... you could still switch to the door he opened if you really wanted. ie that door is still there and is never taken away, but you know it's not there so the odds still are 33% you picked that car. You just combinded that 66% of the two door you didn't pick into one because you got a sneak peak at one knowing it is not it.

[@Pinkerooprincess]

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

[@megamule064]

you are twice as likely to pick a goat first than the car, thus if the host opens up a door with a goat, you are more likely to have the goat, thus by switching you have doubled your chances.

[@lebagelboy]

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

[@cptwhite]

No he IS NOT transferring the probability evenly across the two remaining doors. If you don't believe me conduct a physical test, one where you always stick with your answer and once where you switch. Conduct each test 15 times, you'll see you get a much better win ratio on the tests where you switch.

[@aftermath4096]

That is because the guy playing the game has prior information that the other one showing in the middle of the situation doesn't have, namely that he made a first choice and that the host subsequently chose a door with a goat. If the game starts with 3 doors and 1 is opened with a goat, you obviously have 1/2 chance of choosing (or equivalently switching for) the door with the car behind, but that is precisely because you don't have the prior information of the problem.

[@kappakappa123]

If you work out all the possible outcomes by choosing from the initial choices 1. You picked the car 2. You picked the first goat 3. You picked the second goat of which the host shows you the other goat, you will see that if you had chosen one of the goat to begin with - 2/3, the host will "weed out" the other goat every single time, so switching will always give you the car in that case. Otherwise, if you picked the car first - 1/3, then switching will always give you a goat.

[@seanmft]

of course that this only works if the game show host knows what's behind the doors and always opens a door with a goat.

[@RedJoker9000]

really there is another part to this. When choosing a door like number 1, the host needs to open door 3, which would make the person switch. if he opened door 2 it is more ideal to stay with your choice. so it is all about what door the host opens for the final decision.

[@bladehea]

it only works if we choose one door earlier than the TV presenter

[@DozaArchives13]

If you pick a goat (66% likely) and you switch you will always win the car. Chances of winning the car... 66%.

[@jonathanturner198]

Your list of possible outcomes is bang on. If you were to keep playing the game, randomly choosing between switching and not switching, then you would get all these outcomes and you would win 50% of the time and lose 50% of the time. But the question is 'how likely are you to win given that you switch?' If you look at all three outcomes where you switch, you will see that 2 are wins and 1 is a loss. IE 2/3 'switches' is a win. And likewise, 1/3 'not switches' is a win.

[@croxleylad]

@wcoolside - Always best not to shout, especially when you get it wrong. At the start all chances are equal (1/3) so it is twice as likely we have chosen a goat than a car. So we pick door one but realise we are most likely picked a goat. When the host reveals a goat in box 2 he has altered the odds - door one is still 1/3 chance of containing the car - nothing here has changed -- door 2 is now 0 chance of containing a car (we have seen the goat) so the remaining box has 2/3 chance

[@johndeerishere]

Another way of thinking about it is you have a 1/3 chance on your side and 2/3 a chance on the other side. So, if one of the doors on the other side is revealed all of the 2/3 goes to the other side that still has an unrevealed door. So, the door you did'nt originally pick, but is still unopened has a 2/3 chance of being a car.

[@l1meyman]

Because your chances are higher from the beginning that you would have picked a goat. So switching actually increases your chance of getting a car.

[@babyybluebear]

@pru005 In the movie, he actually answered a smart question first. The professor then asked for his name and asked him his choice for this problem.

[@AimRobot]

Switching door makes no difference, as soon as he gives you the choice to chose again you dont have any of the doors so picking door 1 again would be same as switcing

[@sagajagan]

@priveprive It did change from 33.3 to 66.6 At the beginning each door had a 33.3% chance of having the car. Now he picked 1. Door 1 has a 33.3% probability. Doors 2 and 3 combined have a 66.6% chance. We now know that door three can't be the car, so door 2 alone now has a 66.6% chance. Get it?

[@mariolakis]

there is a 2/3 chance of picking a goat at first. so it's more likely the first time you chose a door it was a goat. so when the offer to change the door comes you have a statistically greater chance of winning the car:)

[@pthombeaux]

Don't look at it as a 33% chance of picking the car door first. See it as a 66% chance of picking a goat door first. If you did, and the host opens the other goat door, the only door left is the car. Switching will then win. Of course there is the 33% chance that you will have picked the car door first and switching will lose. But that only happens 33% of the time. If you played 100 times, you'll see you'll win about 2/3 of the time if you always switch.

[@georgeapplegate3535]

The logic is sound if the host is scripted to reveal a goat regardless of what's behind the door chosen by the contestant. Switching is then like letting the contestant choose two doors instead of only one. In the real world, however, the host will only make such an offer if he knows the contestant has initially chosen the door with the car. In that case, the probabilities suggest that you have a 100% chance of winning the car if you don't switch. Most people intuit this.

[@MrRoverno]

Tis not a half, since every time you pick a goat initially, and then switch, you will get the car. The odds of picking a goat first time is 66.67% and thus the chance of winning the car when switching is 66.67%

[@wallacehiggins7748]

Quickest non-expert explanation: On average, 2 out of 3 times you will lose; 1 out of 3 times you will win, only a 33% chance of being right. The host removes the possibility of being wrong if you were originally wrong and you were to switch. So, if you switch every time, those 2 out of 3 times of being wrong will always be right, which is now a 66.667% chance of being right after switching.

[@Longhorn_Legend]

let me explain the math...when you play this game you need to think of the probability that you will LOSE, not win. There are 3 doors with 1 prize so statistics says by majority rule you will not win. When you pick a door and the host reveals another door, it changes the odds so that the other door has a higher chance of having the prize since originally it was 2/3rds in your favor. So in this example let's assume door 3 has the prize and door 1 and 2 have goats. Continued...

[@SKL1919]

Think of it this way. The chances of you getting the car in the first place is 33.3%. Therefore you are more likely to pick a goat 66.7%. If you choose the goat then switch, you will win every time. Fact.

[@snois2]

are u stoned?

[@SKL1919]

You just don't understand it. Most people think there's a 50/50 chance when the host asks if you want to switch doors, so they don't. There's actually a 66.7% of winning the car if you do switch doors on the final 2 doors. 33.3% if you stayed with your original choice. Fact.

[@reslllence]

snois2 are you? hes correct.

[@reslllence]

sam batida yeah thats what he said, you win every time you switch when youve already chosen the goat (which is 66.7% of the time)

[@sillydillydokieo]

It's not a fact you will win every time, your chances are just higher.

[@HumptyDumptyOakland]

No it isn't. DoND is completely random, if i comes down to 2 suitcases at the end (yours and 1 other) they're equally likely to contain the $1M

[@WhiteVanyo]

Yes and no. In this scene, it is because this was not the Monty Hall problem. The Monty Hall problem changes this by adding a forced reveal. If you open a door, the host MUST open a door with a goat and can only choose a random door if you chose the car. This alters your odds.

[@TheCheweeRevolutions]

it explains this in the curious incident of the dog in the night time. once with a fucking complicated equation and then with a diagram. THE MONTY HALL PROBLEM

[@Ramatganski]

@theseanze It works with anything more than two, Because under this game's rules, by first choosing a door, you are actually saying: "I have forfeited my option to get any information about what's behind THIS door. So now, what can you tell me with certainty about these other TWO?". The host knows what's behind BOTH of the doors you allow him to give you information about. You can trust that the door that's left untouched by the host is the one most likely to hide the prize. Try it out.

[@BoneBaron]

......... I'm in an advanced math class and almost all of my classmates dont understand it. What we always believe is that each door becomes 50/50 since there are only 2 doors left. What we tried to do. Say that D1 Goat D2 Car D3 Goat. There is a 66.66% that you pick goat, and 33.33% for car and you pick D1, the host opens D3 which is a goat and asks you if you want to change so the 33.33% will be split amongst the 2 remaining but why does the other door which you did not choose get more %?

[@Ganymede2048]

I'll make it simple for everyone: I choose Goat 1: I am shown Goat 2, I switch and get Car I choose Goat 2: I am shown Goat 1, I switch and get Car I choose Car: I am shown Goat 1 or 2, I switch and get Goat 2 or 1 66.7%

[@shsaa2338]

For all the people believing that it is always better to switch: ​​⁠​imagine that in fact the host will open one of two other doors and will offer you to switch only when your first chose was correct. Should you switch anyway?

[@sgnewman2000]

Go to wikipedia and look up "Monty Hall Problem" for a detailed discussion of this problem.

[@TheMercDeadpool1]

thx. i spent the day i wrote tht looking up the reasoning behind the answer to the monty hall prob. i now hav a full understanding of it

[@barrsftw]

Its easier to think on a larger scale. If there are 100 doors for example, and you pick door 47. The host says "well its not behind these other 98 doors (leaving door 91 and door 47). Of course you switch... what are the chances you picked it originally? 1%. But on the reoffer? 51%

[@HumptyDumptyOakland]

So you admit that there's a 66% CHANCE of winning the car if you switch, a 33% CHANCE of winning the car if you stay, and a 50% CHANCE of winning the car if you make a random choice of staying/switching. So WTF are you arguing about?

[@dayjun]

The problem is badly explained in this film, the host is FORCED to reveal a goat in the real problem

[@boliussa]

Well, they're not 2 equal chances. Here with logic, The probability that you chose a goat is 2/3. If you chose a goat then monty will leave the remaining door as a car, as monty never opens a door that is a car, he has to open the goat door. So the probability of a car is 2/3 if you swap. As it's the same event as choosing a goat on the first time. If you do that then you have a car in the remaining one.

[@km9OOO]

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.

[@chriscxd]

event A: His first choice was a goat. event B: Switching the door leads to a car. You can see event A = event B, therefore the probablility that event B happens would be the same as the probalility event A happens, which is 2/3

[@lazylink]

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".

[@rocketman584]

@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. :)

[@Carbosful]

Problem is, that's the exact same thing I said earlier, Yet the kid still doesn't get it and calls it a bad explanation.

[@maurocoruzzi5464]

I feel bad for people who do not understand this

[@TheRenekruse]

I feel bad for people, who think it's correct.

[@MarcKloos]

9 billion is a bit much. But if you Google on 'Monty Hall problem simulation' your first link is a simulator that does 10, 100 or 1000 goes. And it gives a 66% chance of winnnig if you swap. Then there is a Mythbusters ep. here on YT showing it. The chance of picking a goat is 66% because there are 2 goat doors. The host will always pick a goat door himself. The chance of you picking a goat still remains 66% instead of 50% (2 doors) so swapping gives you a 66% chance of getting the car instead.

[@Only1DennisBergkamp]

If the host always shows the goat, despite what you pick then unless you pick car (33.33% chance), then the door left (that the host offers you to swap to) will always be the car.. At no point are you making a 50/50 decision..

[@Stalky24]

Oh I see! So the chance of first pick stays and after one of choices is revealed, the other chance is 100-33.33. Thank you :P

[@Zralf]

well sticking with door number 1, still makes you have 66% in light of one goat being revelead. revealing the goat only cuts 33% off the full scale 100% options.

[@bloosea123]

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.

[@HumptyDumptyOakland]

Why do you need 3 or more tries? If you understood mathematical probability you'd realise you can derive the solution directly from the rules of the game. After all you don't actually need a pack of cards to work out that the chances of drawing 2 Aces are 1 in 221

[@PurpleNymph]

... you chose a goat because the doors are closed. BUT the probability of you choosing a goat on your first try is 66%, double than choosing a car. thus you should always switch because you probably have chosen a goat on your first try. this will work approximately 66.6 percent of the time and you will lose by switching 33% of the time.

[@badass447]

Well said. So if you decide on staying or switching 50% of the time you will win 33.3% plus half of 33.3%, which is 16.6%. 33.3% + 16.6% = 50%. In this case a logical person, with your term of logic, gets beat out by a person using math 16.6% more often.

[@mitchconn]

@rockstarrist It's only 50/50 if one losing door is shown BEFORE you pick your door and THEN you pick among 2 that's left. If you pick one door and never intend or plan to switch, your still is always 33%. The two doors you aren't picking collectively have a 66.7% chance. That's why the losing 33% is transferred to the other door you didn't pick to become 66.7%, and not yours.

[@RandomAssStufff]

the monty hall problem was always confusing to me. i get the whole 66.7% as well as the 33%. but by eliminating one option, wouldn't it just reset to a whole new 50/50? as then you started with 3 options but you really only had 2 after one was freely eliminated with no consequences

[@TedManney]

This is a piss-poor "explanation" of the problem and solution. And whether the host was trying to use "reverse psychology" or whether he was scripted to always reveal a goat and offer the switch absolutely make a difference in the outcome of the game. God damn Hollywood, always sloppy with the details.

[@Ben-pp5tt]

This is incorrect. If it’s within the rules of the game for the host to only ask the contestant to switch if the contestant guesses correctly at first (implied at 1:55) then the odds of winning by switching will always be 0%.

[@Stubbari]

You must be confusing MHP, which is a brain teaser, to some actual game contest.

[@Ben-pp5tt]

@@Stubbari What Spacey explains in this movie is supposed to be MHP, but isn’t explained correctly.

[@Stubbari]

@@Ben-pp5tt Yeah that's Hollywood for you.

[@morbideddie]

Yeah, I think part of what throws people off is that the problem is often poorly explained. Once you know monty always opens a door, it’s always the wrong door and he always offers you a switch then the problem is easy to grasp.

[@KufLMAO]

Spacey probably just ab libbed that in as a joke, or because he was too busy thinking about underage boys while shooting the scene. In the official Monty Hall problem the host ALWAYS opens another door and asks if you want to switch regardless of which door you pick the first time.

[@sundeenj]

I just finished teaching this material in my grade 12 math course and this is the best way I can explain it. View the two possible options STAY or SWITCH 1-Assuming you STAY we're all in agreement that it is 1/3 of winning. 2-Assuming you SWITCH You will always LOSE if you choose the car at the beginning (1/3 chance). It won't matter which door the host opens. You will always WIN if you choose the goat at the start (2/3 chance). The only remaining goat will be revealed, guaranteeing you the car.

[@MistyGothis]

@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.

[@aftermath4096]

If the host has a CHOICE of opening/not opening a door (with a goat), the monty hall solution isnt correct, assuming the host doesn't want you to win. Just beleive me man, i have a degree in mathematics, i know what i'm talking about. Though the monty hall solution is correct if it is explicitly mentionned that the host MUST open a door after you make a choice because then the host cannot have any strategy

[@lapisdust]

This is a good example. This is how I initially thought about it - by expanding the number of doors, you make the change in odds much easier to grasp.

[@flavoredwallpaper]

@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.

[@MrLEOLOVEME]

AT FIRST HE HAS 33.3 % OF WINNING .IF HE SWITCHES HE HAS 66.7% OF WINNING BUT WHY AT THE END DOES HE SAY THANK U FOR THAT EXTRA 33.3 %?

[@markrankin4214]

Don't see the issue here. When you choose one of the three doors you have a 33.3333.....% chance of being right. If the host opens a door and it's a goat you are left with 2 doors behind which there is an equal chance of there being a car. Your original choice now has odds of 50% of being right. Your odds have improved because of the elimination of one of the doors. There is no statistical reason to change your mind at all.

[@flamer101]

The obviously exaggerated way it was taught to me was that: Say you had to pick 1 out of 100 doors. You have 1/100 chance of being right and 99/100 chance of being wrong. Now say someone opened 98 of the doors and all had goats. Now there is one door with a car and one with a goat remaining. Would it be better to switch to the remaining door or stick with your original door that had a 1/100 chance of winning. Obviously you would switch.

[@JAYRAY00]

That is ridiculous. You think because you have a 1/100 chance of winning when picking out of 100, that when it's down to 2 doors, your original selection still has a 1/100 chance of winning? By that theory, when presented with the last 2 doors, if you stick with your original selection you're only going to win 1 out of 100 times?! Try it and see, i'm positive that staying with your original selection will equate to 50% success, and switching will equate to 50%.

[@HumptyDumptyOakland]

JayRay00 You do realise that the person opening the doors knows where the prize is and will never open the prize door?. Out of the 99 doors you didn't pick, 98 of them (at least) don't contain the prize, and if the person who opens those 98 doors knows for certain that they're all empty - why do you think your chances of winning have suddenly increased from 1/100 to 1/2? Get a deck of cards and see how often you can pick the Ace of Clubs, it should average out at 1 in 52 right? Turning over 50 of the 51 cards you didn't pick that you know for certain aren't the Ace of Clubs doesn't change your chances from 1/52. Try it and see.

[@Muzikman127]

JayRay00 correct. If you stick, you have only a 1 in 100 chance of winning. Think about it, you had a 1 in 100 chance as the beginning, and then instead of just telling you boo, you lose, he shows you all the other losing doors first. Why would showing you the other losers increase your chance of having a winning door?

[@chillbabe104]

JayRay00 Yup. You would still have 1 out of 100 chance if you stick with the original door. There's 99 out of 100 chance that the doors you didn't choose is the correct one. Now if you have the chance to choose 99 of the other doors at once versus 1 one your current door. Would you switch? Of course because your chances has just jumped from 1/100 to 99/100. That is basically what is being presented. When taking out 98 doors and giving you the option to switch, you have basically been allowed to choose 99 doors at once rather than 1 door. Hopefully that makes sense. This isn't theory. It's statistical fact.

[@flamer101]

HumptyDumptyOakland short version, given a 1 in 51 chance, in most cases you will pick wrong... therefor the other card remaining has to be right 51 out of 52 times after the person who already knows which card is right eliminates all 50 other options.

[@GonnaBeALongLongTime]

@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?

[@JerryCheng22]

..i forgot the name of the book but you can find it at any bookstore just say the movie "21" (hope they know that movie" and that's it.. BTW the movie was based on a book that was based on a true story..

[@HumptyDumptyOakland]

The 66% kicks in on the 3rd trial. LMAO. And yet each trial has the same set of possible outcomes with the same probabilities

[@mumiscrunk]

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