Was searching for this kind of explanation using bayes theorem. Wonderfully explained

I'd like to mention that we can go beyond the Monty Hall problem and touche on a fundamental issue in probability and statistics: the comparability of events with different sample spaces or magnitudes. 1. A first event with a magnitude of 3 (three doors) 2. A second event with a magnitude of 2 (two doors) While a great introduction into probability, the Monty Hall Problem only works if one accepts the comparison of two events of different magnitudes as logical. To dismiss people who cannot agree with this comparison as they do not get it is a problem if you ask me. I see the importance of highlighting both ways of thinking. In many contexts, comparing probabilities from events with different magnitudes or sample spaces is indeed problematic or even meaningless. For example, to provide another example of fundamentally different events: Comparing the probability of rolling a 6 on a die (1/6) with the probability of flipping heads on a coin (1/2) doesn't make much sense in isolation. In more complex scenarios, like comparing stock performance across different markets or time frames, not accounting for differences in magnitude can lead to serious misunderstandings. Additionally, people think if your initial choice was the car (which happens 1/3 of the time), then switching would be the wrong move. In this case, the host's reveal of a goat door doesn't help you at all. You've already won, and switching would make you lose. If one accepts the comparison of two events of different magnitudes, the Monty Hall strategy isn't about "always switch" but rather "switch because it's more likely you initially picked a goat." The host's reveal doesn't create new winning chances. The host doesn't change the fact that you probably (2/3 chance) guessed wrong at first.

@EffectiveMuscle

20 күн бұрын

Great explanation. Thanks for this 👏👏👏

@fordland08

18 күн бұрын

Exactly, you go from picking 1/3 of the doors to 2/3 of the doors. It’s more about choices….

Well I think I can say that I'm now confident in my understanding of Baye's theorem and Bayesian inference.

Well explained!! Your videos have been helping me with my Stat! Great thanks!! :D

I am confused about the part when you say p(B) = 1/2.

@laxisharma8545

3 жыл бұрын

The game show host can either open 2nd or 3rd door for showing us a goat. That means having a goat in 2nd door is half a chance.

What if the contestant WANTED a goat?

@RishabhPatil

5 жыл бұрын

👏👏👏

@rantlord8373

5 жыл бұрын

Then he is middle Eastern.

@user-mz7cn9hq8v

4 жыл бұрын

R I am very smart

@maximus9053

3 жыл бұрын

If he wanted a goat he/she should have stuck with the door they chose at first

@BlackZephyrReal

3 жыл бұрын

if the contestant wanted a goat, he should get a car; then sell it off to buy the 2 other goats

this is my 2nd rodeo in this playlist

I am confused by p(a,b)= 1/3 x 1/2. Does this assume A and B are independent? If they are independent, p(a|b)=p(a); no need to go through all the rest of the derivation steps.

When this problem was first posted in the 70s, scores of people refused to accept the solution as it is so counterintuitive, even famous mathematician Paul Erdos. However, this solution is correct.

Here's another way to look at it: the contestant only has to choose either to switch a curtain or not. If he chose not to switch, than it's obvious his chances of winning (i.e. getting the car) is 1/3, which was his chance in the beginning (since no matter what curtain you choose, monty could always show you a goat). Now - what's the possibility of winning if he does switch? Well - if he chose the car in the beginning (1/3) then there's 0 chance of winning in case of switching. But if he didn't choose the car (2/3) than there's 1 (=100%) chance of winning because Monty will open the other goat, and then the other curtain has to be the car. So the chance of winning if you don't switch is 1/3, while the chance of winning if you do switch is 2/3 - i.e. 2x more => SWITCH!!!

@timopheim5479

7 жыл бұрын

D. Refaeli just explained it this way to one of my classmates today but he couldn't accept it lol. The proof is solid though

@chirpbirds924

Жыл бұрын

Isn't it 50% chance now? Because you're either right or wrong? It was a 1/3rd chance, but then the host offers you an opportunity out of a possible two doors, one with it, one without it.

Best explanation!

I am sorry, but I have to point out that your method is totally wrong. I don’t want people to get more confused by seeing this video. First of all, event B can not be described as “a goat shows up behind door 2”. That is also very tricky and critical insight of the Monty Hall Problem. The most correct description of event B is “the host had to choose a door between door 2 and door 3 to reveal a goat, and the door 2 was chosen”. The difference is that event B is INDEPENDENT of event A! I want to mention there are other mistakes you made in the later part of this video. I would say, basically you used a wrong method to get a correct answer.

@dihan6130

6 жыл бұрын

Okay I got you point now. Basically you only failed to make a very good description of the event B. Other parts are good if you correct the definition of the event description. Apparently, based on your description, you cannot conclude to P(B) = 1/2 and P(A,B) = 1/6 (which are correct answers). But basically if you correct the description of event B, you can figure out that A and B are independent and don’t have to use the Bayes formula I think.

What is marginal probability?

@faaiquealimehmood

2 жыл бұрын

Marginal probability is the probability of an event irrespective of the outcome of another variable. Conditional probability is the probability of one event occurring in the presence of a second event

i don't really understand where he got the 1/2 at 3:31.Can someone please explain to me in more detail.

@jaymefosa2003

6 жыл бұрын

Yeah I think here he got a little fuzzy. Event A, the car IS behind door 1, means that it can't be behind door 2 or 3. He starts to talk about maybe the car being behind door 2 or door 3 but that confuses the point. The 1/2 comes from: If the car is behind door 1 then the host has 2 choices. He can show door 2, or he can show door 3. What is the probability that he chooses door 2? It's half.

@klaus7443

6 жыл бұрын

"What is the probability that he chooses door 2? It's half." Does it matter if he shows door 2 1/2 of the time? Since the probability of choosing the car from three doors is 1/3, then the chances of winning by switching is 2/3. If the contestant has a 1/3 chance of picking the car, then there's a 1/3 chance two goat doors are left over for the host, which means there's a 1/3 chance he leaves a goat to switch to.

@Hamzaben9

6 жыл бұрын

The car is behind door 1 therefore theres only two doors left to choose from, so the chance that goat is behind one of the two doors left is 1/2 .

@klaus7443

6 жыл бұрын

Hamza that is not what the original comment was about. The host can show one of the goats 1/2 of the time and the other goat the other 1/2 of the time when both doors left over from the contestant's guess have goats.

@yuuisland

6 жыл бұрын

I agree, he could've made this explanation more clear. Although he only wrote P(A,B), what he actually did was expand that probability using the Chain Rule of Probability: P(A, B) = P(B | A) * P(A) We know P(A) = 1/3, and P(B | A) = P(host opens door 2 | car behind door 1) = 1/2, since he's equally likely to open door 2 or 3 if the car is behind door 1. Hope that helps.

Very well done m9, finally a proper explanation. Get rekt Monty Hall.

@sourabhjadhav7929

4 жыл бұрын

I'm 4 years late but lmao

@finnstewart8033

Жыл бұрын

I’m 6 years late but lmao

The probability of p(B) is 1/2 there can only be 2 choices to make and it's independent from your choice youre using that information in your solution to P(A,B) hence p(a)p(b) so why do you over complicate it by then going into great detail about the marginal value for p(B) ? p(a|b) = p(a) because of independence .

@jackgarand7284

5 жыл бұрын

Idiotic explanation. The answer to this riddle is EXTREMELY SIMPLE. "If you switch, you get two doors for your one". WTF?

@omid_tau

5 жыл бұрын

learn some manners!

@michaelgarrow3239

2 жыл бұрын

These people are idiots - but fun to argue with. You have a 50% chance of winning a delicious goat no matter what you do….

@matthewrigby6089

Жыл бұрын

Even easier, if we make the event B to be that the game show host shows a goat in one of the remaining doors (which he always does), then P(B) is 1, and P(A,B) is 1/3, and you're done.

Wow His explanation is very straightforward. But I am kinda afraid of his approach cuz it cannot go further. Actually, you should define 3 events. A_i: Your choice. (i = 1, 2, 3) B_j: Host's choice (j = 1, 2, 3) C_k: Where is Car. (k = 1, 2, 3) 2/3 can be derived by P(C_1^c/B_2,A_1) , (^c is complement of a evernt) Now, let our strategy be more complicated. If door 2 opens you do not switch. If door 3 is opened, you switch. (Host randomly chooses the remaining door) The answer is 1/2 so this strategy is worse than the previous one.

Over complicated. Here's a very short proof: 1. Whenever you switch, you get the "opposite" of you original choice. (ie car to goat/ goat to car) 2. 2/3 of the time your initial choice will be a goat. (2 goats/ 1 car) 3.2/3 of the time you will get the car by switching.

@sarpcakc56

6 жыл бұрын

Roger Bodey Nice explanation.

@iwtwb8

6 жыл бұрын

That's a compact explanation but the logic isn't very generalizable.

@rogerbodey9475

6 жыл бұрын

Thanks SC.

@rogerbodey9475

6 жыл бұрын

What does that mean?

@iwtwb8

6 жыл бұрын

This game is taught to illustrate properties of conditional probability. Your explanation is good but doesn't really relate the game to the larger pedagogical objective.

I've always hated this question because there are a lot of assumptions behind it that are never addressed, namely that 1) the host will always open a door, 2) he has knowledge of where the car is and is acting adversarially to the contestant (i.e., he will never open door with the car behind it), and 3) he's either not aware of conditional probability or he is and believes the contestant is not. For example, if the host has the ability to decide whether to open the door but is aware of conditional probability and believes the contestant is also aware of conditional probability, well good luck in that case.

@matthewrigby6089

Жыл бұрын

Agreed, though the issues can be addressed just by saying the game show host always opens one of the doors you didn't choose.

Noooow i get it

Maybe thinking this way....... Try to understand it from the perspective of the host. The player provides his prediction to the host. The host opens a pair of doors with sheep, leaving only two pairs of doors, one sheep and one car, for the player to choose. If the player predicts the door of the car, what will the host do? of couse Want the player to change their choice. The host asks you to calculate the odds with misleading you already have once choice before. you swapping your choice, 2/3 will win. On the contrary, the player predicted the door with the sheep. What will the host do to make the player still choose the door with the sheep? you can tell. ........ Dawyer's door problem, calculate the chance of the host winning.

does this apply in the real world or just theoretically

@RonaldABG

7 жыл бұрын

How could it apply just theoretically? The contestant chooses a goat door 2/3 of the time. Then, since the host knows the positions and must leave the car hidden, in those 2/3 the other door he leaves closed is the car door. In other words, the host chooses the correct more times than the contestant.

P(B), that a goat is shown regardless of whether or not there is a car behind door A is 1/2. I believe it, but so counterinteruitive...

Each sheep is also an independent and unique individual. This game cannot mislead players by treating every sheep as the same thing. There are three options, two of which also stand on their own. If the player can tell them apart, there is no chance of winning like 2/3

This solution omits an important piece of information. Assume three doors: 1, 2, and 3. You pick 1, Monty opens 2 revealing a goat. That leaves doors 1 and 3 unknown. Doors 2 and 3 have a .67 probability of having money. But, importantly, there is a second combination, doors 1 and 2, which also have a .67 probability of having money. At this point you have two combinations, each with a .67 probability. It becomes a choice between two equally probable combinations. Effectively this is the same as a probability of .5 There is no advantage to either staying or switching.

@georgcantor3603

3 жыл бұрын

Did you look at the video? Did you read the description? It's a conditional probability problem so the probabilities are calculated before the reveal, not after.

Idiotic explanation. The answer to this riddle is EXTREMELY SIMPLE. "If you switch, you get two doors for your one". WTF?

@couldntcareless146

4 жыл бұрын

It's a statistics tutorial for Conditional probability... it's made for university students to understand and visualize what they study. Not for your drunk ass that scrolls in youtube videos at 2AM. Also your point can raise debate! Example: Yes, but one of the 2 doors is busted already. Guess what mathematics can't raise...

Incorrect. you make the mistake of adding, magically, the 1/3 of the door opened to the door not opened and not chosen. By the same token, I could add that same probability, magically equal, to the door we chose. The reality is that it does not add to one or the other, but it is recalculated, giving 50% to each door that remains unopened.

@RonaldABG

5 жыл бұрын

You are not taking into account that the host could not reveal the door that the contestant chose despite of its content, so seeing it in the second round is not a surprise. Instead, the other that remains closed survived a possible elimination, because if it had a goat it could have been removed. The host must always reveal a goat from the two doors the contestant did not choose, and he can because he knows the positions. That means that he will never reveal the car, it must remain hidden. But only two doors are going to meain closed: 1) the one the contestant selects and 2) the other the host decides to not reveal. In 2 out of 3 games on average the contestant would choose a goat door, but since the car must remain hidden anyway, the host is who purposely avoids revealing it from the other two doors in those same 2 out of 3 games, in order that it can remain hidden. So, you will always end with two doors, but one is which you selected (randomly) and the other is which the host left on purpose, with knowledge. Yours will have the prize in 1 out of 3 games; the other will have it in 2 out of 3.

@knightthanatos

5 жыл бұрын

@@RonaldABG Thanks for the reply :). But, yes, I have taken into account that the presenter will open an unselected door and where it does not contain the prize, but that does not change the percentages of 50%, it only reduces the samples. The error originates in the language: when the option is described where the chosen door = the door with reward, it is said that, no matter what door the presenter opens, if you change you lose. And with that you write a single line of reasoning. Error. There are 2. I describe it to you: I always choose the door 1 - Prize at the door 3. Presenter opens the door 2. I do not change, I lose - Prize at the door 3. Presenter opens the door 2. Change, I win - Prize at door 2. Presenter opens door 3. I do not change, I lose - Prize at door 2. Presenter opens door 3. Change, I win And, here people say that if the prize is in the door 1 and the presenter opens "a door without a prize", if I change I lose and if I do not change I win. Giving that fictitious 1/3 of winning if I do not change and 2/3 if I change. But the following lines are these: - Prize at door 1. Presenter opens door 3. Change, I lose - Prize at door 1. Presenter opens door 3. I do not change, I win - Prize at door 1. Presenter opens door 2. Change, I lose - Prize at door 1. Presenter opens door 2. I do not change, I win Count them. 50% win if I change, 50% win if I do not change.

@RonaldABG

5 жыл бұрын

@@knightthanatos Ok you are talking about the two cases depending if the host chooses to open goat 1 or 2, but the problem is that they don't have the same probability as the others. In the beginning you choose the car 1/3 of the time, so the two cases depending of which door he reveals must fall into this 1/3, and they end having 1/6 probability. For example, let's say you played 900 times. Since you are 1/3 likely to select each content, you should pick each one in about 300 games. Let's put the cases: 1) In 300 games you select he door with the car. The host can reveal any of the two goats, and since we don't know if he has preferences, let's say he opens each with 1/2 probability (half of this case): 1.1) In 150 of them he reveals goat 1. 1.2) In 150 of them he reveals goat 2. 2) In 300 games you select goat 1. The host is forced to reveal goat 2. 3) In 300 games you select goat 2. The host is forced to reveal goat 1. So, if goat 1 is revealed, you know you can only be in case 1.1) or 3), a total of 450 games. You win by staying in 150 (case 1.1) and by switching in 300 (case 3). You cannot compare one to one cases that do not occur with the same frequency. Cases 1.1) and 1.2) are half as likely as cases 2) and 3).

@knightthanatos

5 жыл бұрын

@@RonaldABG And what does it matter 900 times? My explanation is independent of repetition, it is valid for a case like for infinity. You have just written 4 lines of possibilities, but, humanly, you decide to convert those 4 lines to 3, distorting the final result. I could modify the result in the same way, if I write as I decide to write: 1) I select the door with the car, presenter opens door with goat 1 2) I select the door with the car, presenter opens door with goat 2 3) I select one of the goats and the presenter discovers the other goat And, suddenly, as I have decided to write the problem (wrongly) I could say that if I change I have 1/3 of probabilities of winning and if I do not change 2/3 chances of winning. And you'll say "but that's not how the paradox describes it", but I've used exactly the same way of defining it, I've just changed the way I say it a bit, but I've actually mentioned all the options. Because the way I just said it is wrong, but the way the paradox says it is the right one? Neither of the two would be correct, these 2 ways are poorly written.

@klaus7443

5 жыл бұрын

Dani, I already told you that this is not an unconditional probability problem. It's 50/50 if the host opens a door that has a 1/3 chance of having the car and IF a goat was revealed. In that case the door he is leaving has the same chances as the one you first picked and the probabilities are recalculated from 1/3 vs.1/3 to 50/50, or 1/2. This is a conditional probability problem because the host must reveal a goat therefore the door he WILL open has 0 chance of having the car and the other door he WILL leave has a 2/3 chance.

You have a 50% chance of winning a delicious goat- no matter what you do. Final answer….

@max5250

2 жыл бұрын

You have 66.6% of winning a goat. Will it be delicious or not, is not random function, but it depends on the knowledge of the person who will prepare goat for eating.

@michaelgarrow3239

2 жыл бұрын

max5250 - why don’t you give up. You are obviously out of your element. Let me know when you figure out causality… Moron!

Is the reality of quantum mechanics more logical than what the math? The player never knows which goat is the one revealed.。.. If life is an endless choice, then if we face it with this way of dealing with things, then what aspect of human nature is eternal? Some people always want to control the world and act as the so-called "God"?

I do not believe a word of this - it's over-complicated and difficult to follow - difficult for me to follow. If there are three and you reject one of them - that leaves two. And two doors are better than one. And between two doors there will always be at least one goat. And the opened door carries the same weight in the equation as it did when closed. The probability of the car associated with any door is determined by there being three doors and only one car. The numbers set the probability. The probability is not determined by what lies behind the door. Else we would need to say - of a closed door - that is has a 1/3 chance of hiding the car unless it doesn't hide the car - which is preposterous. The 1/3 versus 2/3 probability of car or goat allows for either eventuality in proportion - one does not cancel the other . They exist together. Even the door with the car has a 1/3 chance of the car because - wait for it - *there are three doors and only one car!* For a single door to have a 2/3 chance of something (like a goat for instance) there needs to be two of them. There is only one car - a door can not have a 2/3 chance of a car. In the MHP the best a door can do is to have a 2/3 chance of a goat and only a 1/3 chance of the car.

@HumptyDumptyOakland

8 жыл бұрын

+Richard Buxton Why do you bother coming to a video on conditional probability, acknowledge that you don't actually understand its contents and then go on to say _I do not believe a word of this"_ ? Practically every sentence in your comment above regarding what probability is, and how it is calculated is WRONG. How many times do you have to be told: *you don't know what you're talking about* ?

@imranhirani5017

8 жыл бұрын

+Richard Buxton Think of it this way. You choose Door 1. There is a 33% chance that your door has the car and a 66% chance that another door has the car. But this 66% is between two doors. Once a goat is revealed, your door still has a 33% chance of having a car, and there is still a 66% chance that another door will have the car. But now since that 66% chance is on one door, it makes sense to switch.

@HumptyDumptyOakland

8 жыл бұрын

+Imran Hirani _" But now since that 66% chance is on one door,"_ That is very true, however *+Richard Buxton* believes that the opened goat door STILL has a 33% chance of being the car door - which is just about the most stupid comment you're ever likely to read about the MHP

@richardbuxton3546

8 жыл бұрын

Imran Hirani You are sort of right... *This is correct*.. _There is a 33% chance that your door has the car_ *But this is incorrect*... _and a 66% chance that another door has the car_ Incorrect because there's a 66% chance that the prize is between *two doors* - each with its own 33% chance i.e. 66% = 33%+33%... The 33% exists because of the numbers of things in the mixture - the 33% is not conditioned (arrived at) by what lies behind a door. A 33% chance-of-the-prize door (all of them) is by no means certain to have the car - indeed it is more likely *not* to hide the car. That's what it means a 1/3 chance - more likely to be a dud - and when its seen to be a dud it's still a 33% versus 66% door. The overall 66% chance is derived from two 33% chance doors combined. A single door can never have a 66% chance of the car because there is only one car - for a single door to have a 66% chance of a car there needs to be two cars. Many people find this difficult to accept. For them common sense and logic evaporate - they fist consider the opened door to have zero chance of the car because it shows a goat - which is wrong - and they are then forced to imagine the combined 66% chance to arrive at a single door - which is ludicrous. All the doors begin with a 66% chance of a goat and a 33% chance of the car - and all the doors end as 66% and 33% doors. Their chance of car or goat is not determined by what they hide - that would be an impossibility despite what others might say. .

@HumptyDumptyOakland

8 жыл бұрын

Richard Buxton Well look who it is ........Professor _"Bozo"_ Buxton from Clown College, here to lecture us all on a subject he admits he knows nothing about: Probability. _"Many people find this difficult to accept"_. That's because it isn't true Richard - you don't know what you're talking about.

*I can prove the true nature of this 'problem'!* .. they trick you into thinking you have 'chosen' the first door. Think of it this way: the first round of the game is the fake 'choice' where you don't actually choose, you 'reserve', to protect one door and let one goat be eliminated from the game, 1st round is now done! Now they start the second game where the REAL choice is allowed, there are no longer any 'reserved' doors and the game parameters have now changed to allow 50/50 odds because you get to play this game from scratch with two doors and for the FIRST time get to actually 'choose' instead of 'reserve/protect', which is very different than the first game's round of choices..it's TWO different games NOT one, so if you do not account for this by creating two separate math problems then you are NOT accurately representing the true nature of the game, as you have to change the odds for all 'unknowns' every time you eliminate possibilities. This logic applies to different examples of this 'problem' as well. The key is understanding that the only real game is when the final choice is made, and that everything before that is just changing the parameters, you have to make the math adjustment for the new parameters as they change, it all comes down to the *state of the parameters* when you *actually choose* and NOT when you are simply 'negotiating' the parameter changes, ..thanks and you're welcome;)

@Dosino_

4 ай бұрын

Now test that theory in real life or in a simulation

@Araqius

Ай бұрын

Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Switching means the opposite. It's just basic math/logic kids understand. Sadly, it's far too hard for idiots.

@jaybefaulky4902

Ай бұрын

@@Araqius exactly, the illusion comes in when you just look at what is actually in front of you at the time and disregard the past process. In either case you are looking at, what appears to be a binary equation with the statement 'it HAS to be inside one or the other' being 100% true so some people assume: "how could two identical scenarios have different properties? it has to be 50/50 no matter what!" I like to say:"think of splitting a shuffled Deck of cards into two piles, but then you remove half the Spades from one pile and balance the amount out using just Diamonds from the other half. One pile has a lot more red cards in it, so it is more likely one pile will give you a red card and the other will give you a black card, even though there are only two piles" eliminating the door in the game is what 'takes the spades away', so lol, yeah, everyone is trying their best to describe being right so i took the other-side and did my best at describing the logic 'in being wrong'..did i win? (P.S. people who actually don't get this should never gamble)

Yeah, except this is the probability of which door Monty opens, not the probability of you getting the car. The difference is that you counted only once on the either/or ⅓ side, but twice on the ⅔ side. The ⅔ side is also either/or, because the probability to get the car remains the same; Monty always opens the goat door. The chance for you to get the car remains the same with him opening either doors. So which is it, counting either/or door once, or twice? Either way it is ⅓ vs. ⅓ or ⅔ vs. ⅔ or 50/50. Guys, you’re wrong. You’ve made a mistake. Quit with your Reductio ad absurdum.You’ve allowed semantics to completely change what probability actually means. You’ve ignored the truth and continue to spread false propaganda. It’s not about debunking anything. Just get it right. 🤙

@Araqius

Ай бұрын

Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Switching means the opposite. It's just basic math/logic kids understand. Sadly, it's far too hard for idiots.

@TristanSimondsen

Ай бұрын

@@Araqius And here's cyber bully #1 spreading the wrong info. How much are they paying you to do this?! 1/3 of the time the car is behind Door A. 2/3 of THAT time the car is behind Doors B and C. Way to use the probability of where the car IS with the probability of where the car ISN'T to complete the equation. 👌 Also when Monty opens the goat door, you ignore, fail and REFUSE to re-attribute half of the probability of you getting the car back to you since you haven't done anything further than making your initial pick. Good one, man. Keep up the wrong work, dude. You and your drone are quite pathetic. 🤙

@Araqius

Ай бұрын

@@TristanSimondsen All you can do is bark? Not that i am surprised though, considering your parents. "Either way it is ⅓ vs. ⅓ or ⅔ vs. ⅔ or 50/50." Only a complete idiot like you or your parents would make a super stupid sentence like this. Imagine the host say "I am going to give you both door", what is your winning chance? Tristan, the idiot among idiots: The chance for my door VS the other door is 2/3 VS 2/3 so 2/3 + 2/3 = 2/3. Tristan, the idiot among idiots: This means my winning chance is 4/3.

@Araqius

Ай бұрын

@@TristanSimondsen Tristan, the idiot amonmg idiots: Either way it is ⅓ vs. ⅓ or ⅔ vs. ⅔ or 50/50. Tristan, the idiot amonmg idiots: The chance that my door is the car is 2/3 and the chance that the other door is also 2/3. Host: I am a good guy so I will give you both doors. Tristan, the idiot amonmg idiots: Now, the chance I get the car is 4/3. Tristan, the idiot amonmg idiots: I am a genius. Hoooraaay!!!