To be fair I knew the Monty hall problem but not how deal or no deal was played

@@elijahbuck6499 “To be fair, I am a nerd”

lol true!!

I had no Idea what the Monty Hall problem was, but what Deal or no Deal was. It could be because I'm from germany..?

@@ddudeace To be fair, I am too

From another comment I saw on here, this is in fact the apparent system which Monty used. He just kinda guessed at which door to open, and if he guessed the prize door then the contestant didn't get another shot. Which is hilarious.

A cute name for this variation is given on The Monty Hall Problem page on Wikipedia. They call it "The Monty Fall Problem" - the story being, that Monty is walking across the stage, accidentally falls against one of the doors and opens it. There is no purposeful revelation of a known result behind the opened door, the act of opening it was an accident - random.

I wonder why the drunken Hall lowers my chance of winning. Like yes in all cases my possibility drops, but in a scenario where Monty picks the goat, it doesn't matter if he picked it willingly or drunk, i had 1/3 of chance to pick the car before the opening of the other door, so when he picks the goat I should swap, shouldn't I?

@@mrsaxophone4765 It's because there's a difference in frequency of the scenarios. In the normal Monty Hall problem, there's only two scenarios: either you [first] picked a door that had the money, or you didn't. These two scenarios occur with 1/3 probability and 2/3 probability, respectively. If you're in the latter scenario, swapping always wins, and in the former scenario, swapping always loses. So because swapping is the better choice 2/3rds of the time, swapping is better. In the drunk Monty Hall problem, there's now three unique scenarios. The two scenarios from the normal problem still exist (with odds 1/3rd and 2/3rds respectively), but in the "you didn't pick the money" scenario, there's now a 50% chance Monty accidentally opens the door with the money, splitting into "you didn't pick money/Monty didn't open the money door" and "you didn't pick money/Monty did open the money door", each with 1/3 probability. So in the Drunk Monty Hall problem, "you did pick the money/Monty opened a door without money" and "you didn't pick the money/Monty opened a door without money", the two scenarios in the normal version of the problem, now occur with 1/3 probability each, instead of the former being 1/3 probability and the latter being 2/3 probability. The remaining 1/3 probability goes to the "Monty opened the money door and you automatically lose" scenario. So now because the two scenarios each have a 1/3 chance of happening, swapping doesn't change your odds, for better or worse.

As expected.

Precisely well said

That’s a familiar name… and profile pic…

To make it simpler: Picking a case then removing cases one by one until you leave one left is equivalent to just picking two cases at the start.

@@sploofmcsterra4786 that's actually a really interesting way to look at it. Assuming you start the game by picking one case to hold and mentally choosing one case to not open, would that make a difference? Or in the case of the Monty Hall problem (expanded to 100 doors), if you mentally pick one door in addition to the actual door without Monty knowing, he'll most likely open your second choice to reveal a goat. In either of these scenarios it doesn't change any of the outcomes as far as I can tell, but it does help to illustrate the way these situations are different problems, or rather that your random choices will always be random, but the Monty choice will have perfect information that skews the probability.

I couldn't find a good place to fit it in the actual video, but incidentally, this is why the actual game played on Let's Make a Deal also isn't the Monty Hall problem. Monty has stated in interviews that he doesn't follow a system of always opening a losing door--it's up to his whim whether or not he wants to give the contestant a second chance.

Monty: Opens a door with a car Monty: Do you want to stick or switch?

@@iamstickfigure Easy ... I like to switch with the oppend door °J° Nobody said I couldn't =P

good luck with that lol you aren't going to be able to

Well, there's a bit of a statistical challenge, but that's more for actually determining when you *are* ahead or not. There's also just deciding for yourself what amount of money you could gain from it that you would be satisfied with, which can vary a good deal between contestants. It's certainly fascinating, and there's a reason people put so much thought into it on a theory level, because there's a lot to consider for both the gameplay itself as well as more meta considerations.

@@IRNoahBody That's wrong, though. The expected value of 1mil and 1 cent is 500k, so picking that would put you behind.

@@nick012000 That depends on what your goal is. You're right that taking the offer has a lower expected value than going for the case, but expected values are only relevant if you're playing a game many times. You're not playing Deal or No Deal 100 times, you're playing it once. If you took the time to fly to LA, compete in a game show and you won 400k, you'd consider that a victory. 400k is a lot of money. If you did all that and won 1 million, you'd also consider that a victory. If you won 1 cent, you'd consider that a loss. So at this point you have two options, take the 400k and be guaranteed a victory and a large amount of money that represents multiple years of income for most people, or take a 50% chance of getting nothing. I know what I'd do.

@@every116 it reminds me of a Prisoner’s Dilemma situation, where how you choose between Trust and Betrayal dramatically changes depending on whether you are playing the game once or continually. If you are only playing once, each player is motivated to choose Betrayal, as it has an outcome spectrum with the most potential gain and least cumulative loss. In continued play, however, Trust is the better option as repeat games are played with the added knowledge of what choices the other player has made prior, and Trust is required to ever get a net positive result for either party. In this single instance of Deal or No Deal, taking the Deal is the outcome with the least possible risk for you, the player, and no way end up in the no-money scenario. If you can repeatedly play the game, of course, the law of averages catches up and the strategy of taking the deal will end up less profitable than always keeping your case.

It really just comes down to this: would you rather take the 1/3 odds that you happened to guess correctly the first time, or would you rather take the 100% chance it’s correct if you didn’t? 100% or the remaining 2/3 probability means switching gives you a 2/3 chance of being correct. People say to extrapolate this to a deck or cards or 100 doors or things like that, and I think that makes sense…would you rather have the 1/52 odds that you grab the Ace of Spades without looking, or would you like to give up the card you chose to be able to sift through the entire deck and find the Ace of Spades? 1/52 odds of you being right beforehand, but 51/52 odds of you being correct after.

That is an important part of the Monty Hall problem. The host knows the door that has a car. He will never show the contestant the door with a car ever. That changes things. In deal or no deal, (generally) the host shouldn't know, they're just offering you a chance to swap the cases.

"Monty from Purgatory": Monty offers to switch exactly enough of the time so that the outcomes are equally probable

As pimanrules pointed out in a reply to another comment, in the actual Let's Make a Deal, Monty Hall has stated in interviews that he doesn't actually follow a rigorous system and it's completely up to his whims if he wants to offer the contestant a second chance. So, assuming he doesn't choose to reveal the car (if you don't have the car behind your door), then you might have the Monty Hall problem, or you might just have a 1/2 chance of opening the right door either way. I suppose in that case on average you should still swap in the event that it is the Monty problem, but that certainly skews your chances and probability.

Nice. Be sure to check out the rehydrated lp for the ultimate nostalgia hit.

so dont choose 6 and 17 now?

That's the birthday paradox in action. In a huge series of random events, some events will happen twice before every event happens once.

@@Joker22593 in other words, if the gameshow aired many more times, and then we ran the data again, it is likely that the million dollars will indeed have appeared most frequently in a couple of the briefcases--but by then it may not necessarily be cases 6 and 17?

@@nefigah in other words, it showing up in 6 and 17 most frequently is a statistically-probable fluke

@@ferociousfeind8538 yup it could be a fluke since not every game was the same, most only had the top million bucks but there were generally a few each season where they changed the amounts such as the million dollar missions and multi-million dollar madness missions where they could have anywhere between 2 one million and 13 one million replacing the higher amounts.

No.

@@YaNoAwantoMas did you check for me. I appreciate that.

@@JeremyJudThe probability of a random KZfaq commenter having actually checked for you is close to zero. Most likely just a troll. With an unusual username like Pinmanrules, I would guess it is the same guy. Unfortunately, your comment probably got drowned in the other comments and he never saw it. There aren't that many, but not every KZfaqr read every single one of their comments, especially past the first week after a video has been posted.

@@PsychOsmosis Well, now this is the third comment from the top, so there might be an offhanded chance he checks the comments on some of his old videos

@@PsychOsmosis r/whoooosh

My question is why their dress straps hover above their skin.

@@pimanrules fuck i can't unsee it

While it's true that knowing that Monty is required to reveal a goat is a huge part of the problem, I think it's still pretty unintuitive until you apply it to a greater number of doors.

Wow, remarkably I never even thought of that

I mean, dealing with the Monty hall problem in Python is kind of perfect.

I mean, there could be a doodle on that last box, why does the probability raise? Is there a correlation between high numbers and the last box? or is there a correlation between human choice and low numbers? like i seriously do not understand and it feels weird like there is so hippy dippy stuff over here with Monty lmao but really why is that?

I think I explained it the best I can in the video, but the key difference (in the normal Monty Hall problem) is that the case you choose is essentially random--you have no idea what could be in it. But Monty does know, and *he's* the one who choses the case to leave closed. You need to think a bit further to get the actual probabilities (1/n and (n-1)/n), but that's the basic intuition.

Ah wow you did this exact example!

9:50 this answers your question i guess. the dude has 1$ and 1.000.000$ left his average win would be 500k not taking the offer (which is only 416k)

@@mrnoneofurbusiness7942 Probably still worth the offer though. Once you start getting into the hundreds of thousands like that, the money becomes more meaningless, 400k is still enough to live a comfortable life for a really long time. Id rather the guarantee of 400k than the coin flip of 1 mill.

At the scale he's doing it, it's possible that it's displaying an abusable trend in the rng on the machine itself.

im back bby

Same

Woot! RIGHT!

for what it's worth, money also has diminishing returns so going for pure expected value isn't always the smart choice. The banker offer is a guaranteed amount of money so it can be worth going for much lower than the expected value. Say the expected value of the game is 500 000 (it's not but for the exemple), This is a life changing amount of money, but getting a million won't change your life much more, you'll mostly be able to afford most of what you need either way and be safe for a very long time. Now if the banker offers 100 000, it's much lower than the expected 500 000, but it's still absolutely a life changing amount of money, maybe not "all my basic needs fullfilled for most of my life even if I quit my job" level but still a major improvement. So the downgrade from that to say 25 or 100$ is much steeper than the upgrade from that to 500 000$ or a million. So in this case accepting the deal might be smart. Of course there is a point at which you'll probably reject, like say you get offered 10 000$, yeah it's still very nice but it won't change your life to the same degree so at this point trying to go for the expected value is probably smarter but that also depends on the situation. That's true if you're financially stable, but if you are 8000$ in debt that you have to renew forever because you can't afford the interest rate, wiping that clean and starting from 2000$ and being able to start seeing your money increase might actually completely change your life, at which point it might be reasonable to accept such a shitty deal. But if you are say, 500 000$ in debt then odds are you won't be able to repay it even if you get a good deal like 400 000$ so since all the bad outcome are equivalent might aswell go for the risky plan and just try for the chance to repay your debt and start with extra money. I guess the point is, looking at money as if it was a good approximation of a score in game theory is often totally ignoring the material reality behiind it and so fails at explaining rational human behaviour as rational. Not sure why I decided to rant about that under this comment but you're welcome =p

Because I think you would be interested.

The chances of you having picked the million initially and the chances of you having not picked the million and happening to avoid it are exactly the same. You could make exactly the same argument you are making for the other value you have with you at the swap stage.

@@morbideddie Yep.

i can’t click your screenshot bro

Lol, good point. There's a clickable card, but those are easy to miss (and I think they don't show up everywhere?) so I'll add it to the description, too.

I'm pretty sure that still isn't the case. The only way Deal or No Deal can be a Monty Hall problem (I'm pretty sure) is if Howie was the one opening doors AND if he knew where the top prize was ahead of time. Simply having him opening one bad prize would only be beneficial if he opens the second-lowest case, because then you would know that you have the lowest case. This is still a benefit, to be sure, but since most cases are being opened randomly, you still have a high likelihood of opening the top prize, leaving you with nothing but goats.

There's no reason to select or swap a case in deal or no deal. It's just bias of ownership that one case got moved across the room.

You're playing against the host, but playing again someone isn't necessary for Monty hall problem

I'll go through the math here, since I happen to have gone through it in a different reply. So, when you pick a door, you have a 1/3 chance of picking the winning door, right? This leaves a 2/3 chance of picking the wrong door. 2/3 times Monty will have a 1/2 chance to open the winning door, so Monty has a |2/3x1/2 = 4/6x3/6 = 12/36 = 1/3| chance of opening the winning door. Therefore, 1/3 times the player will pick the winning door, 1/3 times Monty will open the winning door, and 1/3 times neither will pick the winning door. As a result, when given the opportunity to switch both doors have an equal probability of being the winning door, so it becomes a 50/50. I also did the math for a higher number of doors, in this case 100. The door the contestant chooses has a 1/100 chance of being the winning door and a 99/100 chance of not being the winning door. 99/100 times he has a 98/99 chance of opening the winning door. Therefore, Monty has a |99/100x98/99 = 9801/9900 x 9800/9900 = 96,049,800/98,010,000 = 98/100| chance of opening the right door. Therefore, 1% of the time the contestant will pick the winning door, 98% of the time Monty will open the winning door, and 1% of the time neither will pick the winning door, so no matter the number of doors the probability will stay the same.

The crux of your winning odds in the Monty Hall problem come down to 2:59 at "he has no choice."

The crux of this not working for Deal Or No Deal is 5:31, specifically "at random", meaning there was no host foreknowledge at play

A confused way of looking at it would be that the case you chose in the beginning had a 1 in 26 chance of having a million dollars. Therefore, there is a 25/26 chance that the million is in one of the other cases. At the end you have opened all the other cases except one, and if your case had a 1/26 chance, and all the other doors had a 25/26 chance, the last remaining case must have a 25/26 chance of having the million in it, and be a 25 times better choice than keeping the original case. That's the way it would be if it were a Monty hall problem and the host opened all the cases and knew which case was the million dollar case. What is wrong with this thought process is that as you open the non-million dollar cases, you are also gaining confidence about the case you chose, because they are being opened at random. Having said all that, I think the vast majority of people should take the banker's offer a long time before they typically did. Once the cash gets over $150K or so, especially if there are a lot of low cases left, it's probably advisable to take the money and run.

I don't know about the American show, but in the UK version the banker was one of the show's producers so would have likely known the content of the boxes. Though a lot of it is them is knowledge of the contestant. He was a friend of Richard Osman, who was a fellow TV producer who has since become a celebrity and author.

In the thumbnail of that "$1 fail video" that came up here, the banker offered a bit over $400k. This checks out with the banker not knowing where the money is, and offering just below the expected value of 500k. This might be a trick by the banker to pretend like they don't know the cases' contents, but seems unlikely.

@@pocarski correct, the Banker that we 'see' doesn't have a clue about anything, it's generally the Executive Producer who calls and relays the info, he may have an idea how the 3rd party randomizes stuff but doesn't know exactly where the values are at.

Don't hope for it to happen for all gens. Writing code just for gen 1 is a lot of work.

Indeed, in actual Let's Make a Deal, Monty is not forced to reveal a goat (and in fact has said himself that the Monty Hall Problem doesn't apply).

Ok, but if Monty only reveals one door the. How does the switch or stick round work? The scaling used makes sense, you start with n doors and 1 prize, the contestant picks and Monty removes n-2 non prize doors from the remaining doors then offers the contestant the chance to switch.

@@morbideddie If you're going to compare the Monty problem to that of Deal or No Deal, it doesn't make any practical sense to have the target probability be 90+. In Deal or No Deal the contestants decides the 25 or so boxes they want to eliminate. The host has no input in which of these are eliminated, hence my argument that the code for the Monty program should be altered for the comparison.

@@Pivot-Shorts Exactly, the difference between the problems is the targeted elimination of goats in MHP vs the random removal in DOND.

I think the video intentionally left the banker's mechanic out to keep it simple. It'd indeed get much more complex if you do try to factor in the banker's offers

Except what the Banker offers has nothing to do with the question of if this game can use Monty Hall logic or not. The Banker offers it based on an average and there is a point where you need to hop out or accept their offer based on the odds you have... but this never plays into you getting the winning case in any way at all.

Is having muscular and super ripped dudes in action games/fighting games objectifying? Didn't think so.. go cry somewhere else

@@foxysobek8109 To be fair, it makes sense that action/fighting game characters would be muscular. Scrawny people just wouldn't make sense there unless they had psychic powers to make up for their lack of physical strength. You can kinda just slap a hot chick on anything and it usually won't change much other than being eye candy. Zangief on the other hand, he makes me a bit uncomfortable in that speedo.

@@foxysobek8109 good point, they would normally put skinny guys in fighting games but the primarily female demographic pressured them to put in muscly guys. It’s political correctness gone mad.

It just requires the right explanation... 2/3rds appearing seemingly out of nowhere doesn't intuitively make sense, until you acknowledge that 2 times out of three (picking a losing door) forces the host's hand. The host has insider knowledge, and knows where the winning door is. By always choosing to swap to the host's choice in unopened door, you're taking advantage of that insider knowledge 2 times out of 3

it was also because the person who *did* explain it happened to be a woman, and back then, mathematicians did not think highly of a woman's intelligence.

"What if there are 2 contestants" MHP cannot work with 2 contestants because host might come to situation of being unable to follow the rules. "What if the sponsors gave away too many prizes and will only offer to swap if you picked the door with the car?" That's against the rules of MHP, hence can't happen.

The results would be exactly the same whether we turned the MHP into random prize values or deal or no deal into a single prize game. What you choose to put behind the doors doesn’t change the odds, it’s how the game is played. In the MHP Monty always reveals a non-prize and that is what makes the problem work, if he reveals randomly like in DOND the odds are 50/50 at the switch stage,

nah, it's still not Monty Hall because the "host" is never revealing any "privileged" information to you. He goes over exactly this in the video at 6:25

So, your thought experiment is the Monty Hall problem, and like I explained in the video, the problem is that Monty is an important actor in the scenario. The Bayesian intuition is that, as Monty opens cases, the game *always* looks the same from the contestant's perspective. It doesn't matter if you picked a winner or a loser--Monty is only going to open losers, giving you nothing to update the probability of your own case winning, so it stays 1/26. Meanwhile, you *are* learning things about the other cases. With Deal or No Deal, since the contestant is the one opening cases, every case they open teaches something about their own case. If they open a loser, it increases the chance of their case winning, and if they open a winner it sends it to zero. It's simply not valid to consider just the scenario in which the contestant is left with two cases, one winning, without also considering all paths they could have gone down to get here. The probability distribution simply isn't the same. With Monty Hall, it's useful to think about more a scenario with more cases. For Deal or No Deal, you can try it with just three cases--few enough to list out all the possibilities on a sheet of paper, perhaps.

@@pimanrules I see what you are saying. But, in the simple Monty scenario where you are just dealing with the three doors, and the contestant opens one door, and Monty opens the 2nd door for you"knowing" that it is one with a goat, I understand that Monty is an important actor there... But, what if Monty didn't actually "know" which door had a goat in it? Even If he randomly guessed and reveals a goat for the contestant, the contestant should still switch because his door would still only have a 1/3 chance of winning while switching to the other door would provide a 2/3rds chance of winning. All that matters is the contestant made it to the end of the game with the opportunity to choose.

@@caseygordon3323 "You should ALWAYS switch cases if you make it to the end with one case left in the stands. In the beginning, the contestant has a 1/26 chance of holding the million dollar case. That is a low chance." There is equally low chance of not picking winning case among all other cases, since number of cases decreases. At the end, there are two cases, and each has equal chance of being the case with 1 million dollars. In other words, it is equally likely (or unlikely) to pick a winning case initially, as there is a chance that you are going to miss that winning case all the way to the end of the game. MHP is different for the fact that host knows which door to open, and always open a door with a goat. If host was opening the doors without that knowledge, there would be no use to switch, since odds would be the same for both doors.

@@caseygordon3323 the error in your reasoning is that you assume the probabilities are not updated when cases are randomly taken away. yes for 26 cases, you have a 1/26 chance of getting the right one (and 25/26 of not getting the right one). BUT, when you randomly take away a case that is revealed to be a non-winner, the probabilities change - with 25 cases left, you now have a 1/25 chance of having chosen the right case (and an 24/25 chance of choosing the wrong one). why? that's what RANDOM means : each case has an EQUAL chance of being the right one. You don't have to take our word for it, you can demonstrate this empirically. just do it with 3 cases to save time (initially a 1/3 chance of getting it right and 2/3 chance of getting the case wrong). take one away randomly and if it's not a winning case, note down how many times you get the winning case by switching. it's not 66%. it's 50%. i'm happy to bet ANY amount of real world money that this is correct. are you willing to bet real money on it?

This isn't how concentrating probability works. I'll use a variation of the Monty Hall problem where Monty picks a random door to make my point clear. When you pick your door, there is a 1/3 chance of picking the right door and a 2/3 chance of picking the wrong door. When Monty plays, he has a 1/2 chance of opening the winning door if you pick the wrong door, which gives him |4/6x3/6 = 12/36 = 1/3| odds of opening the winning door, identical to the odds you had of choosing the winning door. Since the door you pick has a 1/3 chance of being the winner and the door Monty opens has a 1/3 chance of being the winner, there is no bias towards the last door being the winner or not. The odds are now 1/2. If we extend this to 100 doors, we see a similar pattern. The door the contestant chooses has a 1/100 chance of being the winner. When Monty randomly opens all of the doors except one, he has a 98/99 chance of opening the winning door. Therefore, Monty has a |99/100x98/99 = 9801/9900 x 9800/9900 = 96,049,800/98,010,000 = 98/100| chance of opening the right door. Therefore, the contestant has a 1% chance of picking the winning door and Monty has a 98% chance of opening the winning door. This leaves a 1% chance for the unopened door to be the winning door, which once again leaves no bias towards or against a switch. DOND works the same way, except the contestant is the one opening cases. When you reach the final two cases, the probability of the case you chose or the final remaining case containing the million will be the same, because the low odds of choosing the right case in the first place will be offset by the high odds of opening the million dollar case. The reason the Monty Hall problem concentrates probability is because Monty will never choose the winning door. You have a 1/3 chance of picking the winning door, same as before. However, Monty now has a 0/3 chance of picking the winning door. This is why the probability is concentrated into the remaining door, because you have a 1/3 chance of picking the winning door and Monty has a 0/3 chance of picking the winning door, there is a 2/3 chance for the remaining door to be the winning door.

You have a 1/3 chance of picking the car. When he opens the door, it's always going to be a bad door he opens, so you learn nothing new. The chance you picked the right door is still 1/3, and the chance the other door is a car is 2/3. The python simulation demonstrated this.

@@gamerdio2503 Yeah the python did, but I'm saying the python is coded on a false assumption that you're choosing 3 doors. You're not. You have an irrelevant choice and then are presented with the real problem after he opens one - chose door A or door B, which is 50/50. There was never a choice of 3 doors. The first step is a misdirect.

If it was only ever a choice between two doors which two were they? You pick a door at a 1/3 chance, monty then always reveals a goat and offers the switch. The only time sticking is beneficial is if you picked the car on your first pick which happens 1/3 of the time. The other 2/3 of the time Monty was forced to leave the door behind because it was the car.

@@barretprivateer8768 He coded it exactly as its played. There are three doors, you pick one, Monty opens one, then you switch or stay. If it was truly only a choice between two doors, like you claim, the simulation would have taken account of that. However, that did not happen, so there is clearly an error in your reasoning. There are three doors to pick in the game. Monty opening one of the doors after the decision doesnt change that. You'd be right if Monty opened one of the doors *before* you picked a door, but that isnt the case.

@@barretprivateer8768 We can analyze each possibility to make it clearer. Let's say door 1 has a car, and the other two have goats. You start by picking one out of the three doors, so we will analyze all 3 cases, all of which have equal probability. Case 1 --- you initially pick the car. Monty opens either of the other doors showing a goat. In this case, it is better to stay. Case 2 --- you pick door 2. Because door 1 has a car, Monty will open door 3 revealing a goat. In this case, it is better to switch. Case 3 --- you pick door 3. In this case, Monty opens door 2 revealing the goat. In this case, it is better to switch. As you can see, in 2/3 situations, it is better to switch

Very much so, you play against the host. If you have already reasoned the mechanics of the game out beforehand, you can guarantee a win for yourself 2 times out of 3

Thanks me too

Monty always selecting a goat is crucial to the MHP and the reason it’s not a 50/50. The standard MHP is laid out like this 1/3 you picked the prize, monty opens an empty door. Sticking wins 2/3 you picked an empty door, monty reveals the remaining empty door, Swapping wins. Without Montys knowledge the problem looks like this. 1/3 you picked the prize, monty opens an empty door. Sticking wins 1/3 you picked an empty door, monty reveals the remaining empty door, Swapping wins. 1/3 you picked an empty door, monty reveals the prize, game over. As you can see without Montys knowledge and him always selecting the goat then 1/3 of the time you don’t reach the switch and the 2/3 times you do the probabilities are evenly weighted. Deal or no Deal works like this. Because the boxes are selected randomly without knowledge of the contents if you end up with a high and low value box (or really any two values) the chances are 50/50.

@@morbideddie Perfectly explained.

"The fact that the host opened door two, revealing a goat, is not changed by the fact that he knew that that door had a goat behind it." This statement implies that two hosts, one who knows what's behind the doors, and an other, who doesn't know what's behind the door, would have equal chance of always revealing the door with a goat, which is plane wrong. Second host (who don't know what's behind the doors) would need some superpowers to be able to "randomly" always open the door with a goat. "Why would his knowledge that a goat is behind that door change the situation? " Because, only due to that fact (knowing what's behind each door) he is able to always open the door with a goat, and never a door with a car (which would happen in 1/3 of situations) and ruin the game completely. "If the host were to simply open door two, not knowing whether there was a goat behind it, or not, he nonetheless did, in fact, open door two, revealing a goat behind it. If we take this as a pre-condition, it is irrelevant as to whether or not he knew beforehand whether there was a goat behind that door, or not." Yes, of course, if you take specific possibility as precondition, then we are not talking about probability any more, since you excluded all other possibilities, but we can still see that it is very relevant for the host to actually know what's behind the doors, snice only then he well be able to always open the door with the goat. "Furthermore, this reveals information to the contestant in the same way that additional information is provided by the opening of additional cases in Deal or No Deal." On the contrary, random opening cases in Deal or No Deal will always lead you o 50%/50% chance of winning, while selecting other door in Monty Hall Problem (where the host always know what's behind the door) will always lead you to 66.6%/33.3% chance of winning. "So, if we take it as a given that a contestant selects a case which is not the million dollar case and that twenty-four additional cases are opened, leaving only the original case and the million dollar case, the fact that you opened the other cases "randomly" does not change the character of the situation." Cherry-picking specific possibility again tells us you are not exactly familiar with probability calculation, if you was, you would easily understand that, probability of this specific situation is equal as picking the winning case in your first try, so there would be no use switching the cases (their probability would be the same 50%/50%). And, again, if someone who knew what's inside each case opened these cases, your chance of winning by switching would be 25/26 (96%), as opposed to 1/26 (4%) if you stay with the first case. "As long as it is taken as a pre-condition that this occurs, it matters not how you got there." It surely matter how you got there, because, if you choose randomly, you will come into that specific situation very, very seldom. "When controlling for this factor, computer simulations will show that 25/26 times the million dollars will be behind the other case." Actually, computer simulation will show you that 1/2 of times the million dollars will be behind the other case, since your cherry-picking possibility of 24 empty cases opened in a row, leaves only two valid situations, million dollar is either in the case you picked initially, or in the last unopened case, so probability for both cases will be 50%/50%.

I did a simulation with python, and if Monty just picks a random door, switching and staying is 50/50. So yes, whether or not Monty knows which door the car is behind matters

If the host knows where the goats are, it makes "Monty selects the car" an impossible outcome, even before he has had to pick a door.