Monty Hall II: Revenge of Monty Hall

[@NovaTheReaper]

It's always amazed me how much easier to understand math seems to become after introducing a few charts and some checkmarks and x's

[@singingbanana]

Good question. The show was called "Let's Make A Deal", hosted by Monty Hall. In the gameshow itself Monty would reveal 'a goat', but then wouldn't necessarily offer a switch. And he didn't offer what was behind the other door, but a lesser cash alternative. Monty himself knew of the 'Monty Hall Problem' and understood it. I've put a link in the description of an interview from 1991 with the man.

[@NoriMori1992]

2:15 "And it pretty much sucks to be you." I laughed so hard at that.

[@skycanth1969]

If my car was the prize you'd want to win the goat every time

[@singingbanana]

Adrian says it was shot with a Sony HVR-Z1E camera, edited with final cut pro but have a look at the new Panasonic P2 camera if you're interested in buying.

[@tajshoosh1196]

This is one of the better explanations of the original Monty Hall. The issue missed by those puzzled by the statistical logic is that the game host has “information”: she knows the location of the goats & the car. So the host is not eliminating a “random” option among the 3, but a “known” outcome (goat) from 2 identical outcomes (goat1 & goat2). This makes switching a better option with 2/3 probability, instead of the intuitive 1/3. Some are also confused by the gut feeling that 2/3 probability means 100% certainty. The random process of tossing a coin erases the impact of that information. Therefore, the probability is reduced from 2/3 to 1/3.

[@ericsurf6]

I've always wondered about this one. Thanks so much for the probability education. You're a genius! Can I subscribe twice?

[@singingbanana]

Glad you tried it. Some people need to be told to pause if they want to try it. You got the answer and it was wrong?

[@singingbanana]

Ah thanks, you have just made our AV guy very happy. The camera was a proper one, rather than my usual home camera, I will ask. I believe he edited it in final cut pro, I can probably get more tech details. At home I used cheaper editors.

[@seancunnion1438]

I want to see "Monty Hall II: Revenge of Monty Hall: Attack of the Goats"

[@janinipizzicato]

Would be the best movie title ever XD

[@phjorland]

Followed by "Monty Hall III: Return of the Car"

[@aaronfowler4921]

If you won a car in the past you must return it for a goat in equivalent condition?

[@sebastianespejoloyaga7603]

Peter Hjørland More like Monty Halll, am I right?

[@fadyserhan9537]

U should see the movie "the return of attacking tomatoes"

[@singingbanana]

D'oh! But really glad you had a play with it.

[@floyo]

This question is equivalent to Monty opening a random door and asking you to switch or not. If he happens to open the car then you always get a goat. This is similar to you picking a goat and him trowing tails. In this case, by symmetry, it is obvious that it is 50/50

[@MissMisstreater]

love it - this was my argument all along, even in the Monty Hall situation, you can never know if you're host is out to get you! Though saying that, it still probably makes sense to switch then as you either have 50-50 or 76-33, switching never hurts your odds

[@marcochimio]

I think you meant 67-33.

[@AInewsa3577]

Ok....but in the first case in which we have selected the car. Is there a need to consider getting a head or a take. It is sure that he will offer a switch. Why should we then take two sub cases? Plz clear my doubt.

[@s.ananthkarthikeyan4560]

The way you say "Hello everybody" cheers me up.

[@wisemantellsyousomething1134]

0:36 That is absolutely NOT a goat. It's a freaking Mogwai. And it's far more valuable than a car!!! I want one, period.

[@bentoth9555]

+Wise Man (tells you something) I dunno, the rules aren't very clear on those things. I mean don't feed it after midnight, sure. But is that midnight local time? And if you get the Mogwai in, say, California, then travel to London with it, does it still go off local time? Or is jet lag a factor and it goes off midnight Pacific time?

[@wisemantellsyousomething1134]

+Ben Toth What's your point? Just give it to me! :D ...Oh, wait, you're right. When to feed it then...? Well, I guess, its "magic" is triggered from the highest stand of the moon (midnight), until sunrise. Yeah, that makes sense... I'll go with that. :)

[@bentoth9555]

I'm just thinking it should be a bit clearer given the risks involved. :P

[@XJoschYZ]

Why would I loose when I pick door 1 and the car is behind door 1? Even if he throws tails he will just show me the car and I win. In this Version the only difference is the limited possibility to switch. You should still switch whenever possible.

[@singingbanana]

That's a genuine blackboard here at Cambridge maths department :)

[@peter8aus8berlin]

The reader's letter which the Monty Hall Problem is based upon does not specify wheather Monty offers the switching invariably (and if so whether the contestant knows this in advance). Looking for the average of the highest and the lowest chance a contestant can have I came of cause to the very same result: Monty offers only to switch if you pick the car. sticking=1/3 swapping=ZERO coin = 1/6 Monty offers you to switch invariably. sticking = 1/3 swapping=2/3 coin=1/2 Average is always: 1/3

[@TyYann]

1:40 "You lose and you go home with a goat". I don't call that losing, I call that winning a goat... ;~)

[@Radditz770]

Pretty simple when you think about it. There are 12 different ways it can play out (so 12 outcomes). 4 of those is you winning. 4/12 = 1/3 so you got one third's chance to win.

[@rossmccluskie]

Hey singingbanana... the quality of this video is excellent. I'd love to know what device you used to film this and what you use to edit and export your videos??

[@Nulono]

3:30 What's that in the middle?

[@KarstenJohansson]

One way I prove the monty hall problem is to use all 52 cards. I secretly write one down and then ask the spec what the chances are that they'll pick that card randomly from the 52. They correctly identify that it is 1/52. Then I get them to choose one and put their finger on it. I remove ALL the cards except 1. My card or theirs is the right card. What are the odds that they chose the right one now? 50/50? Not a chance in hell. They always recognize the odds are highly in their favour if they change to the card I singled out since the odds are still 1/52 that they chose the right card - 51/52 that they were wrong. Funny how using only 3 cards makes this so hard to recognize.

[@lets_see_777]

thats minutephysics or some other video

[@ghuegel]

Neat idea for a modified Monty Hall problem. A side-puzzle I tried was... in this coin-flip scenario, if I could see the result of the flip, I could boost my odds of winning to well above 1/3. I figure that by switching for heads and not switching for tails, my overall odds of winning would be 50/50.

[@Bratzbeerkuchen]

Would it be same with more doors?

[@EliosMoonElios]

The percent can't be added. If a factor change, all the ecuation change too. If the guest show the car you still have the 33%? And 30/70(any side) is the normal result of the 50/50 odds.

[@TheKartKiller]

It is now a genuinely random game because there is a coin toss (which is a random experiment) . In the previous one, because there was a human factor involved, the odds favored someone who switched. Am i right to reason it out this way?

[@hainsay]

It's not far off, but also the probabilities were chosen to specifically make it a 50% chance. If Monty had rolled a die and offered a door change if he rolled a 5 or 6 then you are reducing the probability that it is a good idea to switch and it is better to stay.

[@armacham]

0:42 If the original game was genuinely random, you would still have a 2/3 chance to win by switching. Genuinely random meaning the car is equally likely to be behind doors 1, 2, and 3, and the host is obligated to reveal a goat, and when the player chose the correct door to begin with, the host randomly chooses between the other two doors to decide which one to open.

[@aaronfowler4921]

I don't think he needs to go back stage or even know what doors have what prizes at this point though. You just add the rule if he opens the door with the car you lose. So the game would be you pick a door, he randomly chooses a different door if there is a car you lose if there is a goat you get the option to change your answer. This is the same probability right?

[@RonaldABG]

If the host does it randomly, once a goat is revealed the chances are not 2/3 for switching, and the reason is we are in a subset of the cases, and in a subset the proportions are not necessarily the same. Supposing you play 99 times, on average you choose the car in 33 and a goat in 66 of them. Then the host opens a door, but note he would reveal the car in half of the cases you chose a goat (in 33). So once a goat is revealed you know you are not in those 33; you could be in the 33 you chose a car or in the other 33 you chose a goat. Therefore you win half of these cases staying and half switching. As an analogy, since the host does not know, it is the same if who decided which door is going to be revealed was the contestant, because he is doing it without information. So suppose the contestant chooses the two doors that are going to remain closed and then the third one is revealed and it results to have a goat. Which of the two doors have more probability and why?

[@avinandanbanerjee9568]

Can we alternatively use Baye's theorem?

[@singingbanana]

Yes, but it's a little messier than typical Bayes Theorem en.wikipedia.org/wiki/Monty_Hall_problem#Direct_calculation

[@mustacheglasses5765]

The illustration was helpful. I heard of the Monty Hall Problem and the solution. Before I wrote out all the possibilities I couldn't believe that switching would net me a better chance of winning the car.

[@dushas9871]

you do understand that he's talking about different problem here, right? In the original problem you do have better chance of winning if you switch.

[@mustacheglasses5765]

...yes. The problem has changed. I was simply saying that when I heard others talk about The Monty Hall Problem they didn't explain it in a way that made sense to me. Whereas he has explained this problem (which is slightly different) in a way that was very clear and beneficial.

[@vikasmishra1190]

Mustache Glasses gggghgbbbbvgbbnnnbbhhbhhhh

[@mikejonsson7235]

even if the chance double you never know witch door has a car or a goat behind, so it dosent matter witch number you start or end with

[@arrow1archer]

what pc do you have and how did you get the background

[@yurenchu]

This variant is essentially the same as the Deal Or No Deal scenario (or, in other words, the scenario in which Monty would, with a 50% likelihood, reveal the car if your first pick was a door with a goat).

[@stormsand9]

And then... The third problem. Backstage, monty plays rock paper scissors with an employee. If he wins rock paper scissors, THEN he will get to flip the coin. Possibilities are endless

[@jvcmarc]

So, I've done the math: in average the probability of winning by sticking is of: 1/3. and the average probability of winning by switching is of: 1/2. with 1/6 being the instant lose case. that is considering that both the rock paper scissors and the coin flip are completly random (that is, the last outcome doesn't affect the next one. (as it could happen in a normal game of rock paper scissors.))

[@Zororo23]

explain it to me pls

[@Felixr2]

If the last outcome doesn't affect the next one, the chance of winning rock paper scissors in a 1v1 match is still exactly 0.5, the same as throwing heads with a coin flip. So the odds remain exactly as shown in this video.

[@C5nonZ]

Okay I've been on this hunt for hours now. I just want to understand this realistically. I've gone to a random number generator online and had it choose from 1-3 and hit the button 100 times and recorded the data in power point and used that number as the door the car would be behind. I came up with 53 1's, 19 2's, and 28 3's. Just to make things simple I always chose number 1 as my choice and kept swapping like said and lost 47 times. Is this because of a set rule? Can it be that my sample size wasn't big enough? Or is it simply because even though it's a 2/3 chance doesn't mean I can't land in the 1/3 more often ? This question maybe stupid but I'm just trying the figure this out. Thanks

[@derfunkhaus]

Small sample size that had a disproportionate number of 1s, which also happened to be the number you chose. Your random number generator gave you a 1 roughly 50% of the time when we would expect it to turn up 1 33% of the time. Can't make any logical conclusions about the MHP that way. The best way to understand the MHP is to map out the entire sample space, which can be done in 3 lines. Assume that you strategy is to always switch: 1. You choose the car, Monty shows one of the two goats, you switch and get the other goat. 2. You choose goat #1, Monty shows goat #2, you switch and get the car. 3. You choose goat #2, Monty shows goat #1, you switch and get the car. For each of the three possible things that you could choose with your initial guess, there is an inevitable outcome, two of which result in winning the car and one of which results in losing. The end.

[@zacharymueller2139]

I'm too tired to think right now, but is it possible to somehow get an advantage from the information if he offers a switch?

[@jaakkohintsala2597]

no its as good as picking randomly

[@alistairmuir5521]

+Zachary Mueller In the original Monty Hall problem, yes. But the advantage gained by that extra information is lost here by the two coin-toss outcomes which guarantee you to lose. (James deliberately chose for 2 coin toss outcomes to doom you to failure, since the chances of those two outcomes is 1/6, which is the same as the 1/6 chance you gain with the extra door information, so they cancel out).

[@zacharymueller2139]

Titus Pullo Yes, I see the flaw in my logic now. Probability is my weakest area in mathematics

[@alistairmuir5521]

+Zachary Mueller Like anything else, it just takes time. The problem with statistics is you don't want to take that much time on it because it's dry as hell. Alas, persevere and you'll find it to be extremely useful (and often quite fun, but that could just be me and my regrettable nerdyness).

[@zacharymueller2139]

Titus Pullo I'm fine with the rest of mathematics, it's just that statistics is particularly weird. I'm pretty sure that's where I lost most of my points on the SAT :/

[@stephenm6165]

What if I want a goat?

[@junkjunk2493]

ok funny guy , show your self out

[@SpectatorAlius]

In game theory, what Monty does in this version is called a "mixed strategy". The previous version of the Monty Hall problem, the one Numberphile did in their other videos, has both Monty and the player following a "pure strategy". What is a little surprising is that once Monty chooses the mixed strategy, it no longer seems to matter which strategy the contestant chooses.

[@singingbanana]

I know the one. I'll think about it and try and say something intelligent about it.

[@xXNuclearWarXx]

@peter8aus8berlin You're supposed to assume that the host always reveals a goat and always offers you the opportunity to switch, not just if you picked the car. Under those circumstances you *will* win 2/3 of the time by switching.

[@lucas_teo]

You made mathematics look so much FUN!!

[@AnandKumar-xi8qu]

+singingbanana I agree with you on saying we have a probability of 1/2 of either winning or loosing with either strategies... But considering the fact that at the time of choosing the doors, we have 2/3rds probability of choosing a goat, let's have a look at the following scenarios: 1.) 1/3rds probability of choosing a car --> 1/3rds probability of loosing is I swap 2.) 1/3rds probability of choosing a goat and getting a head --> 1/3rds probability of winning if I swap 3.) 1/3rds probability of choosing a goat and getting a tail --> 1/3rds probability of loosing for sure So, he offers us a swap only for 2/3rds of the times... Whenever we are offered a swap we have 2/3rds chances of having chosen a goat and 1/3rds chances of having chosen a car... I reckon, because of the above stated reason, we would have a higher chance of winning the car if we swap when we are offered one

[@Cream147player]

+Anand Kumar You're not thinking about it right. The coin toss is key. Sure, there is a 2/3 chance you pick a goat. However, the chance you pick a goat and then get offered a swap is 2/3 * 1/2 (the coin toss) which is 1/3. That is the same chance as picking a car and then getting offered the swap (also 1/3). Therefore if you get offered the swap, you have an equal chance of having selected the car or a goat.

[@2Truth4Liberty]

+Anand Kumar "he offers us a swap only for 2/3rds of the times..." You discovered the KEY. And that leads to the remaining options being a swap will win 8 out of 12 times (75%) CUT AND PASTE THIS INTO NOTEPAD to view it properly -- START 1st PICK CAR GOAT1 GOAT2 REVEAL x x CAR GOAT2 GOAT1 CAR SW OR STAY x x x x SW STAY SW STAY SW STAY SW STAY WIN WIN WIN WIN WIN LOSE WIN LOSE WIN LOSE LOSE WIN ALLOW/NOT ALLOW ALLOW ALLOW ALLOW NOT NOT ALLOW ALLOW ALLOW ALLOW NOT NOT END

[@yousefabdelsamie5983]

Which Uni did u graduate from

[@joeytje50]

I've actually thought of another way to make this 50/50 myself: in the original scenario, you pick a door, and a goat-door gets opened. Someone else then comes on the stage without knowing your first pick. You both say which door you'd like to open then. He will have 50% chance of winning, and you will still have 66% or 33% chance of winning. You're picking from the exact same 2 doors, but you can still have a better chance of winning.

[@rocykel]

Not so! Blackboards are still common at our universities. Whiteboards can't be wiped clean properly (at least not without dedicated solvents) and it's all too easy to use permanent markers instead of whiteboard markers on them. Whiteboard markers are expensive and don't last long, and they dry out quickly if left without a cap. Chalk is pretty dry to begin with! And chalk is cheap, and pure water will get rid of it every time.

[@singingbanana]

This was made on Adrian's machine. I got the background by spending eight years studying at university then getting a job at Cambridge. Then finally standing in front of the blackboard in the coffee room.

[@kyle7447]

I like this new version! As soon as you said flip a coin i got it haha.

[@scotland7yard]

thanks mate. btw, I'd be very interested in hearing your views on the unexpected hanging paradox, one of the few problems in which the more you think the more you wanna go "aaaahhhhggg!!"

[@Strijdparel]

That is so cool, I never got the Monty Hall thing, but after wachtching your vid i got it instantly

[@PowerStar004]

This is similar to the Monty *HELL* problem but with an extra coin flip to make it more mathematically interesting. The Monty Hell problem is a senerio in which Monty really wants you to lose. If you pick a goat, Monty will ALWAYS open your door, and you lose, period. If you pick the car... Monty will then offer you the chance to switch. Obviously, since he only offers then if you chose right to begin with, you should Never, ever switch. Unsurprisingly, your overall odds of winning Monty Hell are rather low, even if you know not to switch. The reverse, where Monty really wants you to win, "Monty Heaven", involves Monty only offering you the choice to switch if you choose a GOAT originally - if you choose the car, he will reveal the car and declare you the winner on the spot. Yes, your overall odds of winning Money heaven are high, in fact, if you KNOW you're playing Money Heaven you can't lose!

[@snakerattleroll6678]

Wouldn't the Monty Hell Scenario be all three doors having goats, but one is cleverly disguised as a brand new Car?

[@hainsay]

Monty Hell would just be a 1/3 chance of winning if you know what he's doing.

[@MauveTendingToBeige]

It is simpler to keep Monty in the dark. As in your proposition 1/3 of the time the game will end instantly when the uninformed Monty opens the door to the car. And 2/3 of the time a switch can be offered (or not, doesn't matter) with a 50/50 chance. No coin tossing, same result.

[@fishooi]

But I thought that if you picked the right door and Monty flips a tail and you win directly. If that's the case then you don't get to switch so chance of losing in this scenario is zero and you'll be back to same case as before.

[@KasabianFan44]

Hey singingbanana, I tried to make up a variation of this problem which looked like this: As before, you have to pick one of the doors. Behind one of the doors is a car and the other two contain goats. Once you've picked a door, the host opens one of the other doors (which has a goat) and lets you choose between sticking with your door or switching to the other one. But now, after you've made your choice, the host is going to toss a coin; if he throws heads, he will do what you wanted him to do (i.e. stick if you wanted to stick, switch if you wanted to switch). While if he throws tails, he will do the opposite (i.e. switch if you wanted to stick, stick if you wanted to switch). The question is: what are the probabilities of winning a car now? Is it still 2/3 if you switch and 1/3 if you stick? Is it now a 50/50 chance? Or is it maybe a 1/3 chance if you switch and 2/3 if you stick? (intuition tells me it's 50/50 but I'm not sure)

[@Whoeverheis11]

Alright, let's make the chart, assume you pick #1, and X marks the car: 1 2 3 Coin Stick Switch X H Lose Win X T Win Lose X H Win Lose X T Lose Win X H Win Lose X T Lose Win Your modified Monty Hall brings your odds to 50/50.

[@KasabianFan44]

Whoeverheis11 Okay, I understand now, thanks for the explanation :)

[@singingbanana]

Aargh! Gremlins!

[@JakobVirgil]

I have been saying this for years. I think it was partially the root of the misunderstanding of Marilyn's presentation of it.

[@donfrankmagic6442]

Ok After doing a sample space, I believe it is correct.

[@Markinpuff]

All that really matters is the first pick, the odds are against you. Switching is obviously the play.

[@rossmccluskie]

Many thanks man. This information is very helpful. (Say thanks to Adrain too!) Really love the singingbanana channel.... look forward to every new video! Take it easy!

[@edgunther8136]

I don't see how the coin alters the math. You should ALWAYS switch. Flipping the coin tails just prevents you from switching.

[@aj19bcx]

imagine if picking the first goat caused you to lose and picking the second goat or the car gave you the chance to switch, then having been given the chance to switch would narrow it down to one of two possibilities. now imagine that to make it as random as possible the host used a coin toss before the game to decide which goat would allow you to switch. now imagine a contestant picks a door, and the host says he forgot to flip the coin before and needs to do it real quickly. now the odds are 1 in 2 of a switch working and you have the situation he proposes in this game.

[@DaysDX]

It alters the probability of the contestant winning the game only from the get go. In the standard Monty Hall problem, Monty ALWAYS revealed a goat that didn't belong to you. Therefore you could justify saying switch every game, every game, because the opportunity in which swapping was the best choice (with 2/3 odds) came about every game. In this variable Monty Hall problem, that scenario is not always presented. From the start of the game your probability is drastically different than the standard Monty Hall. However, anytime a goat is revealed behind a door that wasn't yours, you then know for that specific scenario, from that point on the odds are 2/3 in favor of switching. We can weight that probability by how often that scenario arises based on the rules Monty uses for revealings (ie in the standard problem it is weighted 100%, in another scenario such as this we would weight it less), and know how much it contributes as a whole to the odds of winning any given game. But any given game that ends up looking like a standard Monty Hall problem will end up having 2/3 odds from then on. The probability of the scenario changes as it goes based on how any given scenario progresses and what it "looks like." Let me make an analogy. Imagine being asked to flip a coin 10 times and get at least 5 heads. You know from the very start you have a 50/50 chance of making that happen... no problemo. However by some freak chance (1/16) you flip 4 tails in a row... well shit. From then on, you only have 6 tosses and need 5 of them to be heads (That's only 7/64 odds btw). In that instance it was random bad luck that gave you 4 tails to start, but does it really matter if that happened by luck. If somebody started you off with 4 tails deliberately, you would still only have 6 chances to get at least 5 heads. At each junction one can evaluate the probability as a whole (Ie what are the odds it would come to this), or more narrowly excluding the past (ie what is my best choice, what are my chances now). So it changes the probability of the contestant playing any given game (ie you can't just say you'll switch every time... sometimes the option to switch isn't even presented!), but if the reveal turns out to be a goat not behind your door, you can feel confident the odds are the same as the standard problem in this specific scenario. So it changes it overall, but it stays the same in the specific case.

[@lamonthall82]

This is the comedian Monty Hall who can appreciate wisdom😀

[@jakepollen6839]

So this Makes the Monty Python game legitimate since it is truly random?

[@bentoth9555]

That wasn't a goat behind number 3. That's a mogwai, you're not gonna fool me. :P

[@XaxtonRevolution2]

I don't have to go home with a goat, I can just refuse to accept the prize, and it doesn't necessarily suck to be me, because I might want a goat.

[@hornylink]

+Xaxton Revolution don't "go home" with a goat, you'll ruin your bed

[@NoriMori1992]

The underlying assumption of the problem is that all players want a car, and any other outcome is undesirable. Obviously, not everyone in the world is going to feel that way. But we have to assume they do for the purposes of this problem.

[@badmanjones179]

xkcd.com/1282/

[@watchdogs1328]

Always account for variable change.

[@rymdmyggan]

why is it always goats and a car? why cant it be piles of shit and a bar of gold or something? does the monty hall problem only work with goats and cars and if its on a gameshow?

[@Awesomeness65828]

I don't get it

[@theaxehandle1]

how would this work as a real game? Because if Monty Hall only flips a coin when the contestant picked a goat, the contestant would know to stick with his choice when Monty doesn't flip a coin, or to switch whenever he does flip a Heads. And if he always opens your door whenever he flips Tails no matter what you choose, you now have a 4/9 chance to win instead of 1/3 like before.

[@1TieDye1]

He walks off stage, out of sight.

[@Nintendo101maniac]

Stick with your answer. If it's the car you're going to get the switch 100%, if it's the goat, you'll get the switch 50%. So you get 2 cars and 2 goats. Both cars you get a switch, one goat you get a switch the other you get. So staying, you have a 33% goat and 66%car. Basically, the coin trick made it a little more obvious by unbalancing the probabilities.

[@RonaldABG]

You got it wrong, 1) If it's car (1/3), you get the switch 100% of it, which is 1/3 of the total. 2) If it's a goat (2/3), you get the switch 50% of it, which is (1/2)*(2/3) = 2/6 = 1/3 of the total. So, you get the switch 1/3 + 1/3 = 2/3 of the total cases, that are constituted by 1/3 in which you have the car and 1/3 in which you have a goat. You win the car 50% of those cases staying.

[@TyYann]

Thanks!

[@Martymer81]

They still use blackboards at Cambridge!? I thought all universities used whiteboards nowadays... Hell, in Sweden even elementary schools have made the switch. I'll assume it's not a matter of money, so what's the reason? Tradition? Just curious :)

[@themeeman]

Wait what an HD video uploaded in 2009? This makes no sense.

[@AbdulrahmanMajash]

I like how he keeps tossing that "Monty Hall Problem" paper xDD

[@OMGFishoo]

If you could look at the camera whilst talking, that be great.

[@NoriMori1992]

He does.

[@OMGFishoo]

NoriMori Once? wow.

[@NoriMori1992]

OMGFishoo Uh, no. A lot more than once. Are we even watching the same video?

[@OMGFishoo]

NoriMori Oh rly?

[@NoriMori1992]

OMGFishoo Yes, really.

[@ManolitoMystiq]

If I'm not happy, what then?

[@oO_ox_O]

you might say it's misnamed but apart from that, so what?

[@GetMeThere1]

There's a problem with the entire logic of this that people are not giving consideration to: The value of a fine and sturdy goat is being severely underestimated. Just think, there are places where a car would be worthless, but in which you could trade that goat for a fine and sturdy wife.

[@onlynamelefthere]

you should defintivly clean your blackboard more often

[@WiseGuy508]

Hey James recommended your channel in a Numberphile video and then I subscribed to you.

[@donfrankmagic6442]

I believe this explanation is incorrect.

[@richardbuxton3546]

Why should the host go backstage and flip a coin? What's in it for the TV company - for Monty - for the advertisers ? For me this is an unnecessary and somewhat silly variation but I found it difficult to pay attention after the 1:14 point when he went backstage out of sight.

[@ReedoTV]

What's in it for them? They give away less cars

[@singingbanana]

We like blackboards.

[@mander40101]

There is a glaring problem with any explanation of the Monty Hall problem. On the big deal of the day there are two players, both of whom chose a door. If neither pick the big deal, Monty opens one of the contestants door, usually the one that's a zonk. I watched some videos, and after opening the first door, the contestant is not given a chance to switch. If they picked the big deal, then the door no one chose is opened first, and the big deal is revealed last. If the contestant didn't pick the big deal, the contestants door is revealed, followed by the big deal no one chose. The Monty Hall problem is based on one player, not two, and the contestants on the show aren't given the option to switch doors. I have yet to find a video of Let's Make a deal that's a true reflection of the actual problem as stated.

[@Hank254]

That's correct, the game was never played this way on Let's Make a Deal, it was only named after the game show host. You could call it the Bob Barker problem if you prefer... still a very good puzzle.

[@livedandletdie]

hmm I still think there should be yet another Monty Hall which is even more difficult to win. Let's say 4 doors. 1 Car 3 Goats. Door 1: Car, Stick. If Switch 2 lose. Door 1: Car, Stick. If Switch 3 lose. Door 1: Car, Stick, If Switch 4 lose. Door 1: Goat, Stick Lose. If Switch 2 Win. Door 1: Goat, Stick Lose. If Switch 3 Lose. Door 1: Goat, Stick Lose. I Switch 4 Lose. Door 1: Goat, Stick Lose. If Switch 2 Lose. Door 1: Goat, Stick Lose. If Switch 3 Win. Door 1: Goat, Stick Lose. I Switch 4 Lose. Door 1: Goat, Stick Lose. If Switch 2 Lose. Door 1: Goat, Stick Lose. If Switch 3 Lose. Door 1: Goat, Stick Lose. I Switch 4 Win. And now I consider the game fair. 3/12 times stick. 3/12 times Switch. Which means you are Ever so likely to lose. I mean 3/4 chance of losing doesn't sound too bad.

[@johnnyvaughan5841]

I don't think I'd call going home with a goat losing

[@Error081688]

Well done!

[@LukeNukem8]

Rather than go all through this trouble to get a 33% chance of winning, why not just have one of the contestants choose a door then open it?

[@psilocyberspaceman]

Because, just like sex, people like to make it complicated to get it.

[@1TieDye1]

The excitement of having to make a second choice. Good game shows are about anticipation and the illusion of the contestants influence on the results, not random chance.

[@allahfuckbar8466]

When thinking in terms of switching think of it as NOT NOT staying. When thinking in terms of staying think of it as not making a choice to stick. This will make things hell of a lot easier later.

[@gulgaffel]

Allah Fuckbar easier in what way?

[@multicolor0]

or you could just pick a door and have him open THAT door

[@whytauisrightandpiiswrong3296]

You're wrong, now just admit it and stop posting that same spam comment on every monty hall video.

[@scotland7yard]

yes, for some reason I applied the original MH problem scenario when the car was not behind door 1 and MH got heads with the coin... and I applied the new scenario for the rest of the cases... what made me think that (no car, heads) was equivalent to the original problem is beyond me, rats :)

[@enrique6587]

For the most obsocated with 50% (like me at the beginning): You never play alone with two doors, nor can a door ever have 50%, always play with three doors, which opens not the withdrawal of the game, the reuse to put anything. Therefore, it is always 1/3 present at each door. If they offer me two doors to change, the one not chosen and the one that they open, 2/3 always of probability of success. In which I choose, always only 1/3, As much as one of the other two is open. In all the two pairs of doors that remain, one to choose from, and another to open, there will be a total at the end of the simulations, 2/3. If I always change, I always win, use a group of two doors, of the three. People get 50% obese and are never playing alone with two doors. The possibilities are doubled not by changing, but by switching to two doors.

[@ArsonBeanTanks]

I'd feel stupid if I switched doors and it was the wrong door that time.

[@SomeRandomFellow]

Or, a better variation: Three doors. Pick a door. Thats it

[@charlieholloway627]

I still want the goat

[@romanr9883]

attack of the goats lol

[@denmartorlanda]

have you ever seen a singing banana? no, not the channel...

[@hatujemeletsplayeryheskyce6460]

or just pick a door and have 33℅ chance instead

[@bobevans-fk1fc]

It all comes down to the fact that if you picked the car door to begin with you don't win the car you have to reselect between two choices.

[@max5250]

Nope.... Host acts always the same, so, no matter what you picked, you will always get opportunity to switch. Reselecting would be the dumbest thing to do.