Monty Hall, yeah yeah, but with me. It's a mathematical curiosity and, in general, maths is intuitive - an important statement which we cut out of the video.
There have been studies that tell you exactly how many people stick or switch, I don't know the figure off-hand but the vast majority stick. And it's not their fault, this problem is notoriously tricky.
True. The most elegant explanation I have encountered is that Monty is actually offering you a choice between the door you selected and both of the other doors, and opening one of those other doors ahead of time.
This is by far the best explanation of this puzze ive ever seen. The visual is what convinced me. I was very much in the "it is still 50-50" boat until i saw that part. Great job man!!! I see how it is smarter to switch now.
Before the contestant chooses a door the probability that he picks the door with the car is 1/3. This means there is a 2/3 chance that one of the other two doors has the car. Since Monty Hall knows where the car is, the odds are still 1/3 that your door has the car after he opens one door, but the odds STILL remain 2/3 that the other door has the car even though there is only one door remaining. Amazing How Many Contestants Didn't SWITCH DOORS, isn't it?
The way you approach a mathematical problem is quite amazing. Earlier I used to not explore mathematics but I got interested in it by watching the various problems.
I'm glad that you call it the monty hall problem and not the monty hall paradox because it is NOT a paradox. It actually makes sense if you think about it and understand it.
This is the best explanation found in the universe for the Monty Hall Problem. I was not able to accept anything other than 50 50 odds.. But now I am clear. Thank you very much!
/_FR0STB1T3_/ Basically it is about probability of 1 in 3 for the first chaice and 2 in 3 for the second because the host has helped you by eleiminating the chance of getting one of the original goats that were available in the original 1 in 3 choices. (remember the host cannot reveal the car when he opens the door).
I think I understand this now. I first heard this from the movie '21' and I did not understand why it was not 50/50. I can see now (possibly) how what we are doing here is saying ...we have 3 cards. You pick one. I get the other two. My ah ha moment came when you explained the difference between the knowing and the randomness. That is very key. So, you only have a 33.33% of the card being right, while I have a 66.66% chance of my cards being right. Now, while I reveal one of my cards to you, since it is not random but my choice, I will always show you the wrong card. Therefore I am always hiding my 66.66% from you. I still have a 66% chance of being right versus you. So, if you pick the unseen card I am still holding, you will take my 66%. I am looking at my cards and so I will not turn over the CAR if I have it, because I want to win. This allows me to have an edge over you, and is not really fair, because you can only win if you picked the CAR to begin with, at a 33%. However, if I don't know what card I am showing you than I have a 33.33% chance of showing you the right card or hiding the right card. At that point if the CAR is shown, the game is over. We both lost. Randomness allows us both to loose. Allows the game to be fair. In the other game when I can see and turn over what I want, there will always be a winner and therefore the 33% loss which would have gone into us both loosing, is now gained by me. So, in randomness I've given away my 33.33% gain into us both loosing. In knowing I've kept my 33% and therefore it is statistically smarter for you to switch to my side. However, I could still not have a Car;)
What I find most amazing in this is the certainty of those who say it is 50:50. I would think a normal person has some doubt of his position when the see several state it is wrong but no. They have the arrogance to think they are right. In fact there is something scary of people who think so. Some people do not even get at the hundred door example.
"Some people do not even get at the hundred door example." Touche, and they aren't even suspicious as to how can everyone pick out the one correct door from a hundred 50% of the time yet no one can pick it at 22%, 73%, or 2% of the time, always 50%!!
If you follow the chart, along with the assumption of the game show host "knowing" what is behind the doors, it's a simple decision to to switch, therefore, increasing your odds to 66.7%. Only when the game show host "doesn't know" what's behind the doors do you flip a coin; that is if he doesn't reveal the car by accident. Great job on the video. Easy to follow and understand.
ingenuous move to rank it up to 100 doors. That meant that I totally got it. You are in that case so very unlikely to have picked the correct door and the game-show host have to remove all the remaining 98 incorrect doors. Then adding the chart was a nice bonus.
One of my favourite statistical problems, made famous by Marilyn Vos Savant in her column, when she calmly stated the correct answer to the initial puzzle submitted by a reader, and was subsequently subjected to an inundation of outraged mathematics professors claiming she was full of s***. All her critics were later forced to swallow their words when they eventually worked out that Marilyn was, in fact, totally correct. When I first came across this problem, I was skeptical about Marilyn’s answer, so before even delving into its philosophy, I derived an experiment to determine whether she was full of s*** or not. I took three playing cards: one was an ace (“car”) and the other two ordinary number cards (“goats”). I shuffled them and placed them face down on the table. I then chose one at random, then turned over another one (just like Monty Hall), but if it turned out to be the ace I turned it back over and turned over the remaining card (which is effectively what Monty Hall was doing by always choosing a goat). Then I wrote down whether it would have been better for me to switch from my original choice or not. Guess what the results were after trying this 100 times? In 67 times out of 100, it turned out that it was to my advantage to switch, as opposed to 33 times when it would have been better to stay with my first choice... Well, obviously, I was now convinced that Marilyn was right, so I thought about the problem for about 5 minutes and realised why...
+Idling dove So... instead of thinking about the problem for 5 minutes *first*, you spent probably an hour doing 100 iterations? I'd give you shit for that if i didn't do something similar years back when i was also suspicious of the switch strategy. Was it really 67/33? Being that close to 2/3 / 1/3 must be a statistical improbability. Or... mustn't it?
For programmers, it's easier to explain like this: Let's say that I will wrote an algorithm which always plays the not switching scheme: 1) pick any door. If the goat is there, return true, else return false. It will return true only 33% of the times is called, obviously :)
For a long long time I though you were the numberphile. I thought you and brady were like a team. of course, I didn't know brady so I thought "numberphile's friend". so glad I found this channel. This is a time of my life when the maths I'm supposed to do is significantly more difficult but I love maths more than ever because I understand it thanks to your videos.
That's the essence of it in a small number of words. It reduces the potential for confusion and I think there may have been some deliberate confusion put into this problem over the years to make more of it all.
What if there were TWO contestants and each of these two contestants were ardent mathematicians, both with an innate understanding of the principle behind the Hall paradox. After each choose a door, and the third unchosen door was opened to reveal a zonk, would each of the two players, if given the chance, leap at the opportunity to accept the other fellow's door? (And in this scenario the 2 scholars are rivals who would not agree to sell the car and split the proceeds) Hmm? ~Johnny Radionic ™
@RationalConclusion This may help? In a probability distribution, the sum of the probabilities of each possible outcome must add up to 1, so the P(staying w/ initial choice wins the car) + P(switching doors wins the car) + P(the door the host opens wins the car) = 1. The P(staying) is 1/3 because you have 1 chance of choosing right out of 3 doors. The P(host wins) is 0/3 because he will always choose the door with the goat. The P(switching) must be 2/3 because 1/3 +2/3 + 0/3=3/3 or 1.
People are maybe reluctant to change doors because they feel as if they are being tricked. But I think this problem work so well because most people get 'drawn in' to the blind alley that is the 50/50 choice. The host opens a door and you are left with TWO doors (you are now thinking 50/50), then he offers you to stick with your original choice (50/50) or change to the other remaining door (by now you are sure it's 50/50), so when he offers you the possibility of it being exactly the same odds - you are so relieved and convinced you now have the obvious correct answer that you jump for that. At this point most people are locked in and can not see anything other than the 50 /50 odds. It is very hard to change someone's mind even with a detailed explanation.
I absolutely LOVE this problem--and suffered with it for DAYS (by trying to force my intuition into line) with no success. Finally, I saw it in the way you suggested--I imagined the case where there were a million doors, realized that he could ALWAYS bring it down to 2 doors, yet, there's no way my choice could ever be better than one in a million. Note: Of course I mean that I had seen the problem long before ever seeing this video.
I remember reading that article in Parade. I was among the people who disagreed with the genius columnist because I just didn't get it; I thought the odds change to 50-50 after one door is opened. But after your explanation I get it! The chart really helped.
@ion010101 yes, there is one winning door, one goat door that will stay for the final check, and one goat door that will be opened before the final. Whichever you choose, one goat door will always be opened. It's sure. The scenarios shown as 2 and 3 in this video are actually the same.
It was interesting to find that according to the Wikipedia article for the Monty Hall problem, it isn't known for a fact that this problem was used on Monty Hall's program, though similar ones were. The difficulties with this proof make it interesting to consider the nature of proofs and there are probability axioms (wiki), from Kolmogorov. This problem seems to fit them. With the roll of dice, introduced below here, the Monty Hall problem becomes like a Monte Carlo treatment used in physics.
I want to post to admit I was wrong. I did a test with cards, two jokers and an ace, and switching does improve your chances to 67%. I think the key, and where I was wrong, is that your chances of picking right out of three is 33%. So the chances of the ace being one of the others is 67% and knocking out one of the jokers does not chance those odds. Live and learn.
Yeah you gain from switching if you do the same thing every time. But this is very unrealistic in a real life scenario and BTW isn't even mentioned in this video, nor the letter to Parade nor in other videos on this topic. The solution to this artificial problem will never help you in real life because the situation will never occur.
What everybody omits to note in the problem description is that the host never(!) opens the door with the car behind, since in that case the show would be over. In other words the host knows where the car is. Because of his knowledge you get 2/3 chance to win by switching.
In bridge, this is called the Law of Restricted Choice (or principle of RC). It works the same way. If you need to play a suit that contains AT32 opposite your K9876 without losing a trick, you play the Ace and say the Jack appears as the last card, its 2/1 that the player who played last didn't have a choice. It's probably the biggest application of the Monty Hall problem I can think of.
For instance, start watching at 2:55 he says "he will open door 2 or door 3" and lumps them into one scenario, however these are 2 different scenarios, one where he opens door 2 and one where he opens door 3 (because he has a choice). Meaning you had one win from not switching to 2 (when he pulled door 3) and one win from not choosing door 3 (when he pulled door 2), so staying should deem 2 checks and switching should deem 2 checks or 50/50 odds. There should be 4 scenarios total on his chart.
@Anon909 That you just said, as also our "Host" said, is easily bypassed as it is only a detail, but a crucial one for this problem. The reason so many mathematicians had trouble realizing this is becuase they ignored the fact that it's not a pure mathematical problem, but actually has the influence of a (tricky) human mind. Btw @singingbanana I loved your 100 door example, it made me understand this issue alot better as I've tried to understand this problem before :)
You explained case #1 very quickly. I split this case into two: Case 0) You select door #1. Monty opens door #2. Case 1) You select door #1. Monty opens door #3. The result is the same. If you stick to door #1 you win. If you switch you lose. So they are 4 cases.
I decided to make a second counting (sorry if your tired of reading it all) 1. I choose Door 1, he opens Door 2: a) the cars are in Order #1 b) the cars are in Order #3 You can make two cases for you choosing Door 1, him opening Door 3; you choosing Door 2, him opening Door 1 etc. etc. I'm still interested in this theory, so if anyone has something to back it up, you can write me a letter.
The difference between the Monty Hall problem and games such as Deal or No Deal is that the game show host in the Monty Hall problem knows which door the prize is behind and thus his choice to reveal a door with a booby prize is not random. In Deal or No Deal the choice is completely random as nobody knows where the 250,000 is. For these reasons it is correct that you should always swap in a Monty Hall scenario as you stand a 2/3 chance, however in DoND, the choice of swap or no swap is 50:50
I come across this problem so often and I can never remember what made me understand the probabilities last time I saw it, but this time the way I'm thinking of it, is your initial choice still has 1/3 because it was chosen out of 3 choices. The contents never change so regardless of whether the other options increase or decrease, you know the car must've been under one of the initial 3 and thus is just as likely to be behind your initial choice - 1/3. Since they take away a choice, you now have the combined chance of both the removed choice and the remaining, but unpicked choice, as 2/3rd of the time, it is behind one of the two, and since the one removed is ALWAYS a bad door, all 2/3rds worth of that chance goes to the unpicked remaining choice. Alternatively if you can understand that your current choice remains at 1/3 because it was not impacted by the removal of one door (since they would only ever remove a bad door regardless of your pick), you can just think that it must add to 1 probability total so the other door must have 2/3 chance.
The fact that the host knows where the car is doesn`t affect the odds because you are the one that have to choose between the only 2 doors left.So that makes it 50/50.
I remember *_hugely_* heated pub arguments over this (amongst the planetarium employees at the pub)... I always held the 2/3 chance against everyone else insisting the 50/50 chance. The 100 door scenario really drives the point home better than I ever did (tho it does feel good to be vindicated) !!
It's actually 33/66, as he says. One way to think about it is as follows: suppose you choose your door having openly declared that you wouldn't swap beforehand. Then you could simply tell the host to skip all the rest of the procedure and straight out open the door of your choice, since your chances of winning wouldn't change. That's how I figured it out. I hope I made it clear for you, or at least didn't confuse you anymore. Cheers.
I can see that what you have said might be simple for you but because I came at this with a different set of words from what you have here, I would have to displace my set of words and integrate your set to understand it in your way. Well, I was thinking there might be a somewhat ideal set of words in English that would be best for most people and presented them below here.
I think that your not counting the cases right: 1. I choose Door 1, the cars are in Order #1 he opens Door 2 (don't change) 2. I choose Door 1, the cars are in Order #1 he opens Door 3 (don't change) 3. I choose Door 1, the cars are in Order #2 he opens Door 3 (change) 4. I choose Door 1, the cars are in Order #3 he opens Door 2 (change) Because he could open Door 2 or 3 if I choose Door 1, and that should be counted as a different case.
Aha! Finally this gives me the simple answer I was looking for. The second guess is different because it's really a guess about the *first* guess. It says "was my first 1:3 guess wrong?". So naturally saying yes is an answer with flipped odds. Stick-or-switch means "keep your odds or flip them?"
I'm not sure if you were replying to my comment but, anyway, I made it because although the explanation here is clear, it raises questions about how people reach incorrect conclusions and how different people can decide differently from the same facts. Freud's psychological ideas, fractional electric charges, relativity and the big bang theory, with 'dark energy'! How could we agree on such things if we can't even agree on this three-box problem?
Mr Banana,you say that, in general, maths is intuitive, but it is well known that most people are very poor at intuiting questions of probability. One example is the "Gambler's Fallacy" which leads many to think that a coin that has come up heads four times in a row is more likely to come up tails on a fifth toss. Another is the answer that many give to the question of how many people are needed for there to be an evens chance of two sharing a birthday. What really surprises me about the MHP is that so many are unable to accept that there is (as in all elementary logic problems) one, and only one, correct answer.
I had a hard time wrapping my head around this, but I think I finally figured out why it's such a difficult concept to grasp. It's designed to take advantage of a person's intuition and use it against them (like the birthday problem and other logical conundrums).
Well done. You've caught me. I didn't invent mathematics. Does that mean you win? I did reference the Parade article in the video. Yet the problem was old when Marilyn was asked about it in 1990. At least 15 years old, if not older. It's a modern classic, I said that in the video too didn't I? The proof is my own, but would be similar to any mathematician's proof. If you prefer, next time I will make a video about my own maths research in combinatorial representation theory.
I actually had a really long discussion on this topic yesterday/ I'm of the opinion that there's a little more to it, and the statistics are only true if the host HAS to give you a choice of changing your mind. If he has to give you that choice then sure, but say you're playing 3 card monty and you're the dealer, if the player gets it wrong to start with you're not going to risk him winning with a choice, therefore if I'm offered a choice I'll assume it's a tactic :P
If we can accept that a roll of dice can represent the problem, where 1 or 2 and 3 or 4 represent losing initial choices and 5 or 6 represent an initial winning choice, we can easily list results on paper, say with L and W. Two thirds would be Ls - losers. Representing a decision to switch every time, we add to each L a W and to each W an L. Here, the "overall" situation for each test is represented by two letters side by side on the paper. In the final result, two thirds would be LW - winners.
The web site of the German newspaper, Die Zeit, has an article about this problem, called, 'the goat problem' -- 'das Ziegenproblem'. Their treatment considers the fact that Paul Erdős, a famous mathematician whose name appears on many published papers, seemed to have difficulty with the problem. The article's title is: Sind Sie schlauer als ein Genie? (It's easy to translate with Google Translate)
From the Wikipedia article for G.H. Hardy: "In an interview by Paul Erdős when Hardy was asked what his greatest contribution to mathematics was, Hardy unhesitatingly replied that it was the discovery of Ramanujan." Ramanujan said that he had visions from his family's goddess. (See Wikipedia article for Namagiri Thayar.) Paul Erdős was very attached to his mother and concentrated on mathematics to the exclusion of almost all else after she died. In connection with the problem in this video, maybe some people have difficulty with it because of how parts of the brain work somewhat separately. Goats, boxes and cars are features of the universe; substantives. Probability doesn't exist in the same way. Maybe there is a connection between Erdős' closeness to his mother, Ramanujan's family goddess, probability, Lady Luck, religious belief and emotions of love and madness. Maybe they are more to do with one part of the brain and can help to produce great insight, such as with Ramanujan and probably Erdos, but they don't necessarily link easily to substantive elements in the universe. In connection with this I would like to say that mathematics could tend to lead someone to believe that if you have no elephants and take 2 elephants away and do that minus 3 times, you would have plus 6 elephants. But really, you would have no elephants. When people apply complex mathematics to the real world, they won't easily know what it is telling them.
I can see some of what you're saying here but it seems to rely on what you said before and I would have to integrate the various points. For this problem there is just enough space to give a complete, self-consistent explanation with the 500 characters allowed but I don't think I could do it with what you've said. I have developed another explanation which also draws the problem together visually, as my second one did and better than my first one - search this page for "overall".
Yes it is basic stuff: probability prize is behind door you picked to begin with is 1/3, probability it's behind the door the host didn't open is 2/3. If you realise this you'll switch, if you don't you'll flip a coin for a 1/2 chance
ahh now i get it, thanks for explaining this, it's been haunting me for years. funnily enough, in order to optomise your chanses of winning you must decide to switch before hans and then 'try' to get the wrong door. (which is a chance of 2 in 3) instead of deciding to stick before hand, in which case you must 'try' to guess the right door (which only is a chance of 1 in 3) I'm not sure if i dare to watch 'the revenge of monty hall' now.
A very easy way to see the solution is to to think each door has a likely percentage of having the car behind it. (Let's say the actual order is Goat, Car, Goat). So Door 1=33%, Door 2=33%, Door 3=33%. We can re-write this as Door 1= 33% and Doors 2+3=66%. If door 3 is revealed to be a goat, its percentage of being a car is now 0%. So Door 1=33%, and Doors 2+3 now = (66% + 0%). Hopefully this way makes it easy to see!
A factor in these disagreements may be that people have been bred to stick with a decision so they will be loyal to a religion or a leader. A brain can be split by injury so that one half acts based on facts that the other half doesn't know. If that other half controls speech and is asked why it acted as it (thought it) did, it rationalizes - invents some reason that sounds plausible. Different reasoning processes go on in a normal brain and we don't always know why we make the choices we make.
@swcomer I believe he does mean Deal or No Deal, as he's talking about a situation where you can accidentally reveal the prize, the million dollars. Unlike in this problem, when you have two cases left in DOND, the chances are genuinely 50% each, because there was no rigging by the host.
I saw this problem in my maths book and I immediately came up with the answer. Maybe it was easier since in that example they had four doors instead of three. The right answer is the intuitive one in my opinion.
By the way, that was 321 characters. I was thinking that a lot of words, different explanations and different ways of expressing things can get in the way. It was a lot of work for 321 characters but it is often a kind of effort that is well spent.
Thanks. I wondered if you were using a dialectal technique where you would take part of one person's argument and part of someone else's and run them together, as a way of keeping the discussion rolling. I'm sure it's been done. About the problem, if anyone liked my idea of doing a Monte Carlo treatment but didn't have dice, they could get random-like numbers from the last 2 digits of square roots on a calculator. Then, e.g., 1-33 gives 1, 34-66 gives 2, 67-99 gives 3 -- 00 doesn't count.
Let's use the notation P(door) = x, which means the chance that the car is behind door 'door' is x. If you choose door 1, the chance you picked the car is 1/3: P(1) = 1/3. Therefore the chance that the car is behind door 2 or 3, is 2/3: P(2 or 3) = 2/3 .... which is P(2) + P(3) (Using an 'or' means that we must add the changes together). Now Monty opens door 3, which has a goat. Because P(2) + P(3) remains 2/3 and P(3) appears to be 0, P(2) must be 2/3.
Yes, exactly, however he said we were down to box 1 and 2 when asked if it was a 50/50 or 33/66 probability. The scenario that door 3 has a car behind it cannot exist, he already opened that door. Or inversely if we are allowed to go back to having door 3 where is the scenario in which we choose box 1 and he eliminates box 2 so we can stay or switch to box 3 (again causing a loss)?
This was a much better explanation than the numberphile video, thank you.
numberphile? What's that?
singingbanana did something happen between you and brady?
singingbanana The other channel you're on?
Walter Kingstone
SecretMan TwentyFive I think what he means here is that in 2009 (when this video was uploaded) there was no Numberphile.
mursie100 Oh. I see. :D
Even if it's late, thanks for sharing your thoughts. :D
Simple. If you pick a goat and switch you'll win a car. If you pick a car and switch you'll win a goat, and you're twice as likely to pick a goat.
Finally an intuitive explanation!
The chart gave me a much better understanding. Thanks a lot.
There have been studies that tell you exactly how many people stick or switch, I don't know the figure off-hand but the vast majority stick. And it's not their fault, this problem is notoriously tricky.
Basically, the question is: "Do you think the car is behind your door, or any of the other doors?", which makes it more intuitive.
True. The most elegant explanation I have encountered is that Monty is actually offering you a choice between the door you selected and both of the other doors, and opening one of those other doors ahead of time.
I've seen this problem explained many many times, and only now do I finally understand it. Thank you so much!
This is by far the best explanation of this puzze ive ever seen. The visual is what convinced me. I was very much in the "it is still 50-50" boat until i saw that part. Great job man!!! I see how it is smarter to switch now.
Deal or No Deal does give you a 50-50 chance at the end. Deal or No Deal is random, Monty Hall is skewed by the host.
Before the contestant chooses a door the probability that he picks the door with the car is 1/3. This means there is a 2/3 chance that one of the other two doors has the car. Since Monty Hall knows where the car is, the odds are still 1/3 that your door has the car after he opens one door, but the odds STILL remain 2/3 that the other door has the car even though there is only one door remaining. Amazing How Many Contestants Didn't SWITCH DOORS, isn't it?
THE BEST EXPLANATION YET!!! IVE NEVER UNDERSTOOD THIS TILL RIGHT NOW! THANK YOU SOO MUCH
I really want to take a wet sponge to that blackboard
The way you approach a mathematical problem is quite amazing. Earlier I used to not explore mathematics but I got interested in it by watching the various problems.
I'm glad that you call it the monty hall problem and not the monty hall paradox because it is NOT a paradox. It actually makes sense if you think about it and understand it.
Exactly!
You can easily tell where the doll is in this video by looking at the shadow it casts to the viewers left (the same direction as Dr. James' shadow)
I love this stuff. Great video!
If he shows you the car, you just switch to the door that he showed you had the car underneath. You then have a 100% chance of picking correctly.
Came here to look for jokes about the ' booby prizes'. Was highly disappointed.
Heh, boobie.
Came here to look at young james
This is the best explanation found in the universe for the Monty Hall Problem. I was not able to accept anything other than 50 50 odds.. But now I am clear. Thank you very much!
Brill.
Best explanation I've heard.
I haven't seen the film myself, but this problem has become quite well know, it's a very good problem.
Are you the same person from numberphile?
Yes I am.
Thanks!
singingbanana So basically its mostly about chance?
/_FR0STB1T3_/ Basically it is about probability of 1 in 3 for the first chaice and 2 in 3 for the second because the host has helped you by eleiminating the chance of getting one of the original goats that were available in the original 1 in 3 choices. (remember the host cannot reveal the car when he opens the door).
No they are two different people
This the best and simplest explanation I found on youtube
I think I understand this now. I first heard this from the movie '21' and I did not understand why it was not 50/50.
I can see now (possibly) how what we are doing here is saying ...we have 3 cards. You pick one. I get the other two.
My ah ha moment came when you explained the difference between the knowing and the randomness. That is very key.
So, you only have a 33.33% of the card being right, while I have a 66.66% chance of my cards being right.
Now, while I reveal one of my cards to you, since it is not random but my choice, I will always show you the wrong card.
Therefore I am always hiding my 66.66% from you.
I still have a 66% chance of being right versus you.
So, if you pick the unseen card I am still holding, you will take my 66%.
I am looking at my cards and so I will not turn over the CAR if I have it, because I want to win. This allows me to have an edge over you, and is not really fair, because you can only win if you picked the CAR to begin with, at a 33%.
However, if I don't know what card I am showing you than I have a 33.33% chance of showing you the right card or hiding the right card. At that point if the CAR is shown, the game is over.
We both lost.
Randomness allows us both to loose. Allows the game to be fair.
In the other game when I can see and turn over what I want, there will always be a winner and therefore the 33% loss which would have gone into us both loosing, is now gained by me.
So, in randomness I've given away my 33.33% gain into us both loosing. In knowing I've kept my 33% and therefore it is statistically smarter for you to switch to my side. However, I could still not have a Car;)
Very good, my British friend. The 100 door example is absolutely the best way to reason anyone out of the 50/50 argument.
What I find most amazing in this is the certainty of those who say it is 50:50. I would think a normal person has some doubt of his position when the see several state it is wrong but no. They have the arrogance to think they are right. In fact there is something scary of people who think so.
Some people do not even get at the hundred door example.
"Some people do not even get at the hundred door example."
Touche, and they aren't even suspicious as to how can everyone pick out the one correct door from a hundred 50% of the time yet no one can pick it at 22%, 73%, or 2% of the time, always 50%!!
If you follow the chart, along with the assumption of the game show host "knowing" what is behind the doors, it's a simple decision to to switch, therefore, increasing your odds to 66.7%. Only when the game show host "doesn't know" what's behind the doors do you flip a coin; that is if he doesn't reveal the car by accident. Great job on the video. Easy to follow and understand.
ingenuous move to rank it up to 100 doors. That meant that I totally got it. You are in that case so very unlikely to have picked the correct door and the game-show host have to remove all the remaining 98 incorrect doors. Then adding the chart was a nice bonus.
What the devil?! I thought this upload had failed.
Thank you!
One of my favourite statistical problems, made famous by Marilyn Vos Savant in her column, when she calmly stated the correct answer to the initial puzzle submitted by a reader, and was subsequently subjected to an inundation of outraged mathematics professors claiming she was full of s***. All her critics were later forced to swallow their words when they eventually worked out that Marilyn was, in fact, totally correct.
When I first came across this problem, I was skeptical about Marilyn’s answer, so before even delving into its philosophy, I derived an experiment to determine whether she was full of s*** or not. I took three playing cards: one was an ace (“car”) and the other two ordinary number cards (“goats”). I shuffled them and placed them face down on the table. I then chose one at random, then turned over another one (just like Monty Hall), but if it turned out to be the ace I turned it back over and turned over the remaining card (which is effectively what Monty Hall was doing by always choosing a goat). Then I wrote down whether it would have been better for me to switch from my original choice or not.
Guess what the results were after trying this 100 times? In 67 times out of 100, it turned out that it was to my advantage to switch, as opposed to 33 times when it would have been better to stay with my first choice...
Well, obviously, I was now convinced that Marilyn was right, so I thought about the problem for about 5 minutes and realised why...
idlingdove Very nice approach!
+Idling dove
So... instead of thinking about the problem for 5 minutes *first*, you spent probably an hour doing 100 iterations?
I'd give you shit for that if i didn't do something similar years back when i was also suspicious of the switch strategy.
Was it really 67/33? Being that close to 2/3 / 1/3 must be a statistical improbability. Or... mustn't it?
@Tolstoievsky No paradox, just counter-intuitive.
For programmers, it's easier to explain like this:
Let's say that I will wrote an algorithm which always plays the not switching scheme:
1) pick any door. If the goat is there, return true, else return false.
It will return true only 33% of the times is called, obviously :)
For a long long time I though you were the numberphile. I thought you and brady were like a team. of course, I didn't know brady so I thought "numberphile's friend". so glad I found this channel.
This is a time of my life when the maths I'm supposed to do is significantly more difficult but I love maths more than ever because I understand it thanks to your videos.
That's good to hear!
Finally! This puzzle has stumped me since 3rd grade. Thank you for finally providing an answer!
That's pretty cool. That column became infamous in the world of mathematics.
Simplest explanation and most intuitive i've ever heard so far thanQ =]
From all the explanations in youtube, your chart made me understand, thank you!
I was viewing it wrong the whole time! The chart really helped. I thought I was going mad with thought. Thanks for clearing it up for me.
That's the essence of it in a small number of words. It reduces the potential for confusion and I think there may have been some deliberate confusion put into this problem over the years to make more of it all.
The 1 in a hundred example really illustrates the point well.
What if there were TWO contestants and each of these two contestants were ardent mathematicians, both with an innate understanding of the principle behind the Hall paradox. After each choose a door, and the third unchosen door was opened to reveal a zonk, would each of the two players, if given the chance, leap at the opportunity to accept the other fellow's door? (And in this scenario the 2 scholars are rivals who would not agree to sell the car and split the proceeds) Hmm?
~Johnny Radionic ™
Which is why I said this doesn't work for Deal or No Deal because the boxes are taken away at random.
That grid does the trick. best explanation I've seen.
I had problem accepting any proof, but the table really did it! I think that the table is the best way to explain it.
Even though I already understood this, your explanation was by far the best i've heard.
Thx for explaining ..saw it in 21 but was not convinced till I came here
The chart was VERY good!
Great way to explain it !
@RationalConclusion
This may help?
In a probability distribution, the sum of the probabilities of each possible outcome must add up to 1, so the P(staying w/ initial choice wins the car) + P(switching doors wins the car) + P(the door the host opens wins the car) = 1.
The P(staying) is 1/3 because you have 1 chance of choosing right out of 3 doors. The P(host wins) is 0/3 because he will always choose the door with the goat. The P(switching) must be 2/3 because 1/3 +2/3 + 0/3=3/3 or 1.
the Monty hall problem NEVER WORKS I swear every time I switch I get a goat and vice verca
+Mei Yifu nothing
People are maybe reluctant to change doors because they feel as if they are being tricked. But I think this problem work so well because most people get 'drawn in' to the blind alley that is the 50/50 choice. The host opens a door and you are left with TWO doors (you are now thinking 50/50), then he offers you to stick with your original choice (50/50) or change to the other remaining door (by now you are sure it's 50/50), so when he offers you the possibility of it being exactly the same odds - you are so relieved and convinced you now have the obvious correct answer that you jump for that. At this point most people are locked in and can not see anything other than the 50 /50 odds. It is very hard to change someone's mind even with a detailed explanation.
I absolutely LOVE this problem--and suffered with it for DAYS (by trying to force my intuition into line) with no success. Finally, I saw it in the way you suggested--I imagined the case where there were a million doors, realized that he could ALWAYS bring it down to 2 doors, yet, there's no way my choice could ever be better than one in a million. Note: Of course I mean that I had seen the problem long before ever seeing this video.
I remember reading that article in Parade. I was among the people who disagreed with the genius columnist because I just didn't get it; I thought the odds change to 50-50 after one door is opened. But after your explanation I get it! The chart really helped.
That's good. A larger number of doors with one 'winning' door makes it clearer.
@ion010101 yes, there is one winning door, one goat door that will stay for the final check, and one goat door that will be opened before the final. Whichever you choose, one goat door will always be opened. It's sure. The scenarios shown as 2 and 3 in this video are actually the same.
It was interesting to find that according to the Wikipedia article for the Monty Hall problem, it isn't known for a fact that this problem was used on Monty Hall's program, though similar ones were.
The difficulties with this proof make it interesting to consider the nature of proofs and there are probability axioms (wiki), from Kolmogorov. This problem seems to fit them.
With the roll of dice, introduced below here, the Monty Hall problem becomes like a Monte Carlo treatment used in physics.
I want to post to admit I was wrong. I did a test with cards, two jokers and an ace, and switching does improve your chances to 67%. I think the key, and where I was wrong, is that your chances of picking right out of three is 33%. So the chances of the ace being one of the others is 67% and knocking out one of the jokers does not chance those odds. Live and learn.
Correct.
Yeah you gain from switching if you do the same thing every time. But this is very unrealistic in a real life scenario and BTW isn't even mentioned in this video, nor the letter to Parade nor in other videos on this topic.
The solution to this artificial problem will never help you in real life because the situation will never occur.
What everybody omits to note in the problem description is that the host never(!) opens the door with the car behind, since in that case the show would be over. In other words the host knows where the car is. Because of his knowledge you get 2/3 chance to win by switching.
Finally I understand the why-part. Watched some other videos which just told how it is. Very good!
In bridge, this is called the Law of Restricted Choice (or principle of RC). It works the same way. If you need to play a suit that contains AT32 opposite your K9876 without losing a trick, you play the Ace and say the Jack appears as the last card, its 2/1 that the player who played last didn't have a choice.
It's probably the biggest application of the Monty Hall problem I can think of.
Dude that chalkboard is massive I’m jealous
"Dis Mogwai"
"We'll put it down, he doesn't like the bright light"
In the UK it takes three or four years, but mostly four years which is how long it took me.
For instance, start watching at 2:55 he says "he will open door 2 or door 3" and lumps them into one scenario, however these are 2 different scenarios, one where he opens door 2 and one where he opens door 3 (because he has a choice). Meaning you had one win from not switching to 2 (when he pulled door 3) and one win from not choosing door 3 (when he pulled door 2), so staying should deem 2 checks and switching should deem 2 checks or 50/50 odds. There should be 4 scenarios total on his chart.
The chart was a great way of explaining it. thanks.
@Anon909 That you just said, as also our "Host" said, is easily bypassed as it is only a detail, but a crucial one for this problem. The reason so many mathematicians had trouble realizing this is becuase they ignored the fact that it's not a pure mathematical problem, but actually has the influence of a (tricky) human mind.
Btw @singingbanana I loved your 100 door example, it made me understand this issue alot better as I've tried to understand this problem before :)
You explained case #1 very quickly.
I split this case into two:
Case 0) You select door #1. Monty opens door #2.
Case 1) You select door #1. Monty opens door #3.
The result is the same.
If you stick to door #1 you win.
If you switch you lose.
So they are 4 cases.
4 cases but Case0 and Case1 each have a probability of 1/6 whilst Case 2 and Case 3 each have a probability of 1/3
I decided to make a second counting (sorry if your tired of reading it all)
1. I choose Door 1, he opens Door 2:
a) the cars are in Order #1
b) the cars are in Order #3
You can make two cases for you choosing Door 1, him opening Door 3; you choosing Door 2, him opening Door 1 etc. etc. I'm still interested in this theory, so if anyone has something to back it up, you can write me a letter.
i actually saw that in the movie "21" but didn't fully understand it , but now i do thanks to you :)
The difference between the Monty Hall problem and games such as Deal or No Deal is that the game show host in the Monty Hall problem knows which door the prize is behind and thus his choice to reveal a door with a booby prize is not random. In Deal or No Deal the choice is completely random as nobody knows where the 250,000 is. For these reasons it is correct that you should always swap in a Monty Hall scenario as you stand a 2/3 chance, however in DoND, the choice of swap or no swap is 50:50
I come across this problem so often and I can never remember what made me understand the probabilities last time I saw it, but this time the way I'm thinking of it, is your initial choice still has 1/3 because it was chosen out of 3 choices. The contents never change so regardless of whether the other options increase or decrease, you know the car must've been under one of the initial 3 and thus is just as likely to be behind your initial choice - 1/3. Since they take away a choice, you now have the combined chance of both the removed choice and the remaining, but unpicked choice, as 2/3rd of the time, it is behind one of the two, and since the one removed is ALWAYS a bad door, all 2/3rds worth of that chance goes to the unpicked remaining choice. Alternatively if you can understand that your current choice remains at 1/3 because it was not impacted by the removal of one door (since they would only ever remove a bad door regardless of your pick), you can just think that it must add to 1 probability total so the other door must have 2/3 chance.
Best explanation that I've seen, thank you for posting this.
The fact that the host knows where the car is doesn`t affect the odds because you are the one that have to choose between the only 2 doors left.So that makes it 50/50.
I remember *_hugely_* heated pub arguments over this (amongst the planetarium employees at the pub)... I always held the 2/3 chance against everyone else insisting the 50/50 chance. The 100 door scenario really drives the point home better than I ever did (tho it does feel good to be vindicated) !!
It's actually 33/66, as he says. One way to think about it is as follows: suppose you choose your door having openly declared that you wouldn't swap beforehand. Then you could simply tell the host to skip all the rest of the procedure and straight out open the door of your choice, since your chances of winning wouldn't change.
That's how I figured it out. I hope I made it clear for you, or at least didn't confuse you anymore. Cheers.
I can see that what you have said might be simple for you but because I came at this with a different set of words from what you have here, I would have to displace my set of words and integrate your set to understand it in your way. Well, I was thinking there might be a somewhat ideal set of words in English that would be best for most people and presented them below here.
I think that your not counting the cases right:
1. I choose Door 1, the cars are in Order #1 he opens Door 2 (don't change)
2. I choose Door 1, the cars are in Order #1 he opens Door 3 (don't change)
3. I choose Door 1, the cars are in Order #2 he opens Door 3 (change)
4. I choose Door 1, the cars are in Order #3 he opens Door 2 (change)
Because he could open Door 2 or 3 if I choose Door 1, and that should be counted as a different case.
Aha! Finally this gives me the simple answer I was looking for. The second guess is different because it's really a guess about the *first* guess. It says "was my first 1:3 guess wrong?". So naturally saying yes is an answer with flipped odds. Stick-or-switch means "keep your odds or flip them?"
I'm not sure if you were replying to my comment but, anyway, I made it because although the explanation here is clear, it raises questions about how people reach incorrect conclusions and how different people can decide differently from the same facts.
Freud's psychological ideas, fractional electric charges, relativity and the big bang theory, with 'dark energy'! How could we agree on such things if we can't even agree on this three-box problem?
Mr Banana,you say that, in general, maths is intuitive, but it is well known that most people are very poor at intuiting questions of probability. One example is the "Gambler's Fallacy" which leads many to think that a coin that has come up heads four times in a row is more likely to come up tails on a fifth toss. Another is the answer that many give to the question of how many people are needed for there to be an evens chance of two sharing a birthday. What really surprises me about the MHP is that so many are unable to accept that there is (as in all elementary logic problems) one, and only one, correct answer.
I had a hard time wrapping my head around this, but I think I finally figured out why it's such a difficult concept to grasp. It's designed to take advantage of a person's intuition and use it against them (like the birthday problem and other logical conundrums).
Well done. You've caught me. I didn't invent mathematics. Does that mean you win?
I did reference the Parade article in the video. Yet the problem was old when Marilyn was asked about it in 1990. At least 15 years old, if not older.
It's a modern classic, I said that in the video too didn't I? The proof is my own, but would be similar to any mathematician's proof.
If you prefer, next time I will make a video about my own maths research in combinatorial representation theory.
I actually had a really long discussion on this topic yesterday/ I'm of the opinion that there's a little more to it, and the statistics are only true if the host HAS to give you a choice of changing your mind. If he has to give you that choice then sure, but say you're playing 3 card monty and you're the dealer, if the player gets it wrong to start with you're not going to risk him winning with a choice, therefore if I'm offered a choice I'll assume it's a tactic :P
If we can accept that a roll of dice can represent the problem, where 1 or 2 and 3 or 4 represent losing initial choices and 5 or 6 represent an initial winning choice, we can easily list results on paper, say with L and W. Two thirds would be Ls - losers.
Representing a decision to switch every time, we add to each L a W and to each W an L.
Here, the "overall" situation for each test is represented by two letters side by side on the paper.
In the final result, two thirds would be LW - winners.
The web site of the German newspaper, Die Zeit, has an article about this problem, called, 'the goat problem' -- 'das Ziegenproblem'.
Their treatment considers the fact that Paul Erdős, a famous mathematician whose name appears on many published papers, seemed to have difficulty with the problem. The article's title is:
Sind Sie schlauer als ein Genie?
(It's easy to translate with Google Translate)
From the Wikipedia article for G.H. Hardy:
"In an interview by Paul Erdős when Hardy was asked what his greatest contribution to mathematics was, Hardy unhesitatingly replied that it was the discovery of Ramanujan."
Ramanujan said that he had visions from his family's goddess. (See Wikipedia article for Namagiri Thayar.) Paul Erdős was very attached to his mother and concentrated on mathematics to the exclusion of almost all else after she died.
In connection with the problem in this video, maybe some people have difficulty with it because of how parts of the brain work somewhat separately. Goats, boxes and cars are features of the universe; substantives. Probability doesn't exist in the same way.
Maybe there is a connection between Erdős' closeness to his mother, Ramanujan's family goddess, probability, Lady Luck, religious belief and emotions of love and madness. Maybe they are more to do with one part of the brain and can help to produce great insight, such as with Ramanujan and probably Erdos, but they don't necessarily link easily to substantive elements in the universe.
In connection with this I would like to say that mathematics could tend to lead someone to believe that if you have no elephants and take 2 elephants away and do that minus 3 times, you would have plus 6 elephants. But really, you would have no elephants. When people apply complex mathematics to the real world, they won't easily know what it is telling them.
I can see some of what you're saying here but it seems to rely on what you said before and I would have to integrate the various points. For this problem there is just enough space to give a complete, self-consistent explanation with the 500 characters allowed but I don't think I could do it with what you've said.
I have developed another explanation which also draws the problem together visually, as my second one did and better than my first one - search this page for "overall".
OMFG! I love this problem! This problem was on Numb3rs and Charlie explained this to his students, it was so awesome!
thank you for explaining this better then others.
Yes it is basic stuff: probability prize is behind door you picked to begin with is 1/3, probability it's behind the door the host didn't open is 2/3. If you realise this you'll switch, if you don't you'll flip a coin for a 1/2 chance
ahh now i get it, thanks for explaining this, it's been haunting me for years.
funnily enough, in order to optomise your chanses of winning you must decide to switch before hans and then 'try' to get the wrong door. (which is a chance of 2 in 3)
instead of deciding to stick before hand, in which case you must 'try' to guess the right door (which only is a chance of 1 in 3)
I'm not sure if i dare to watch 'the revenge of monty hall' now.
A very easy way to see the solution is to to think each door has a likely percentage of having the car behind it. (Let's say the actual order is Goat, Car, Goat).
So Door 1=33%, Door 2=33%, Door 3=33%. We can re-write this as Door 1= 33% and Doors 2+3=66%. If door 3 is revealed to be a goat, its percentage of being a car is now 0%. So Door 1=33%, and Doors 2+3 now = (66% + 0%).
Hopefully this way makes it easy to see!
I like that a lot.
A factor in these disagreements may be that people have been bred to stick with a decision so they will be loyal to a religion or a leader.
A brain can be split by injury so that one half acts based on facts that the other half doesn't know. If that other half controls speech and is asked why it acted as it (thought it) did, it rationalizes - invents some reason that sounds plausible.
Different reasoning processes go on in a normal brain and we don't always know why we make the choices we make.
0:16 You just threw that paper? Litter bug! I am going to keep this video on pause until you pick that paper up Mr!
@swcomer I believe he does mean Deal or No Deal, as he's talking about a situation where you can accidentally reveal the prize, the million dollars. Unlike in this problem, when you have two cases left in DOND, the chances are genuinely 50% each, because there was no rigging by the host.
I saw this problem in my maths book and I immediately came up with the answer. Maybe it was easier since in that example they had four doors instead of three. The right answer is the intuitive one in my opinion.
Fascinating. Many thanks! Perhaps this has an application in a magician's arsenal? Must put my thinking cap on.
By the way, that was 321 characters. I was thinking that a lot of words, different explanations and different ways of expressing things can get in the way. It was a lot of work for 321 characters but it is often a kind of effort that is well spent.
Thanks. I wondered if you were using a dialectal technique where you would take part of one person's argument and part of someone else's and run them together, as a way of keeping the discussion rolling. I'm sure it's been done.
About the problem, if anyone liked my idea of doing a Monte Carlo treatment but didn't have dice, they could get random-like numbers from the last 2 digits of square roots on a calculator. Then, e.g., 1-33 gives 1, 34-66 gives 2, 67-99 gives 3 -- 00 doesn't count.
Let's use the notation P(door) = x, which means the chance that the car is behind door 'door' is x.
If you choose door 1, the chance you picked the car is 1/3:
P(1) = 1/3.
Therefore the chance that the car is behind door 2 or 3, is 2/3:
P(2 or 3) = 2/3 .... which is P(2) + P(3)
(Using an 'or' means that we must add the changes together).
Now Monty opens door 3, which has a goat.
Because P(2) + P(3) remains 2/3 and P(3) appears to be 0, P(2) must be 2/3.
Yes, exactly, however he said we were down to box 1 and 2 when asked if it was a 50/50 or 33/66 probability. The scenario that door 3 has a car behind it cannot exist, he already opened that door. Or inversely if we are allowed to go back to having door 3 where is the scenario in which we choose box 1 and he eliminates box 2 so we can stay or switch to box 3 (again causing a loss)?
That's interesting. I don't play bridge, but that sounds right.