This is the most clear explanation of this problem that I have ever seen.

When I saw the columns SWITCH and STAY I understood everything. Even after seeing many other videos only this one gave me the answer, thanks!

His explanation is simple, direct, and multiple. That's awesome.

Best explanation. Clear and thorough. Thank you!

Clear and thorough explanations. Thanks a bunch!

Jimmy Li, well done. very very decent explanation my favourite from all u tube. i also like ur seriousness it reflects the explanation effectivness. Many thanks!

Really after that many videos this lecture totally cleared my concept.

YEP. This is the best video explaining it!!!! Well done Jimmy Li.

I didn't understand this until I watched your explanation. Very good job sir

Thank you Jim for clarifying the rules of this problem or in engineering talk, the boundary conditions. In the absence of applying proper boundary conditions to the problem ALL other solutions on the internet are WRONG!!!!! Although the name contextualises the problem so common folk can understand an application of Bayes theorem, Monty's meddling messes up the math. You must further apply the probability that Monty will extend the option and under what probabilistic context he provided such an option, as Monty controlled this mechanism and hence the posterior probability requirement of Bayes Theorem. In lay terms, IF Monty only extends the switch option when he knows the person is right, this makes the probability of winning given a switch, ZERO. I have had issues with the Monty Hall Problem since it is a terrible example and makes my job of teaching Bayesian probability so much harder, further compounded by the fact I was a mediocre stats prof. The option to switch doors was not always and unconditionally given by Monty, therefore Bayesian probability does not exist in this scenario. There is no need to go any further. If you qualify that Monty ALWAYS gave the option to switch, Bayes math works, BUT HE DID NOT. This is not a minor point and this makes ALL the assumptions of Bayes theorem invalid in this application. My office mate was a geophysicist and brilliant stats prof and his job before teaching was to check to see if the stats actually say what the author says they say. The Monty Hall problem is what us engineers call a divide by zero error, and results from having just enough knowledge to be dangerous. This is actually a great example for students, demonstrating that it is extremely important to define preconditions of the problem before you jump into the math. A+ Jim.

I finally understand now. It doesn't matter about which door is or isn't open. All that matters is, that your initial choice has a probability P(C) = 1/3 of being correct, and 1-P(C) = 2/3 of being incorrect. And you only ever have one other choice to switch to, which means that the other choice has a chance of 2/3 to be correct.

@klaus7443

7 жыл бұрын

"It doesn't matter about which door is or isn't open." If you removed that statement then the rest of your comment would be correct. As it stands it is not.

@dienekes4364

Жыл бұрын

_"All that matters is, that your initial choice has a probability P(C)"_ -- Yes, this is where the calculation goes off the rails, which is why the model fails.

@xxJing

Жыл бұрын

@@dienekes4364 I suppose it is easier to understand without mathematic language. You just have to understand that no matter how the problem is presented it's equal to being asked "Which has the greater chance of finding the winner, one door or two doors?" Since if you switch you are effectively given access to both of the doors you didn't originally choose.

@dienekes4364

Жыл бұрын

@@xxJing So, you are completely ignoring the 2 scenarios? Why do you think you are doing that? Is it because they prove the model wrong?

@dienekes4364

Жыл бұрын

@@xxJing Your logic fails at 5:00, when you talk about switching. The problem is that you assume that both of these 2 options are available, but they are not. Only 1 of the options is available. If the car is behind door 2, then the host opens door 3. If the car is behind door 3, the host opens door 2. These are mutually exclusive events. Both of them do not and can not apply to the same solution. This is where the logic fails.

best explanation seen so far, can follow the listing all possible outcomes case but find the tree strategy a bit confusing

@betobayona7812

8 жыл бұрын

+Anna Abdulla If it helps... The tree cross-and-checks are wrong, the correct order should be: X X V V, because you are giving V's if you win by switching and X´s if you dont win by switching... Therefore, if you multiply you're left with prob 1/3 of not winning by switching versus prob 2/3 of winning by switching

I like the idea you wrote on the board. I got it easier when I saw that. Thanks

I finally got it, thank you so much

good job. Now i got the intuition behind this problem

When you simulate this in python, do you get the Monty Python problem? :)

@MalgosO

2 жыл бұрын

I think you found the holy grail right there

@dienekes4364

Жыл бұрын

Only if you are a shitty programmer and design the program for this outcome.

Great explanation, thank you. The only thing that made me confused a bit is your initial table. The way two different outcomes (for every choose) were combined into one: for example - (1, 1, 2/3) is in fact two outcomes (1, 1, 2) and (1, 1, 3). After doing this there are 12 outcomes instead of 9 and there are 6 outcomes of winning without switching. This might seem as the probability of winning without switching becomes 6/12 = 1/2. But it's not and important thing here is that after we split that outcome into 2, the sample space is not uniform anymore so that we can't calculate probabilities by simple counting. So you combined those 2 outcomes into one and got uniform sample space. Great idea!

@hamidnikbakht1295

6 ай бұрын

Can you please elaborate on this part of your comment: "But it's not and important thing here is that after we split that outcome into 2, the sample space is not uniform anymore so that we can't calculate probabilities by simple counting. So you combined those 2 outcomes into one and got uniform sample space. Great idea!" I wonder why he did not make the same decision tree that he did for the scenario when the player has a strategy, for when he did not (the original game play), as you also mentioned, and ending with 12 options! 😕

@eugeneshcherbo9821

6 ай бұрын

​@@hamidnikbakht1295 He could, but calculating probabilities in that case would be more complicated, it would require to identify cases where the player wins and calculate the probability of the event using chain rule and total probability law (you can try this and you will get the same results). But he decided to go different way and combine outcomes that lead to the same situation for the player into one, so that all outcomes in the sample space are equally likely and you can use simple counting to calculate probability. Why combining makes the outcomes equally likely: in the beginning you have the same probabilities for choosing any of 3 doors, then the same probability for choosing any door out of 3 to place a prize and after this things go different. If we choose the first door and the prize is under the first door, then there are 2 doors to open for your friend (with probability p and (1 - p)), but if the prize is under the door 2 or 3, then your friend does't have a choice, therefore (1, 1, 2) is less likely then (1, 2, 3) or (1, 3, 2). But (1, 1, 2/3) is equally likely to (1, 2, 3) or (1, 3, 2). I hope this helps :)

Best explanation, thank you

First up, 1/3 the time you will pick the car and 2/3 the time you will pick a goat. The host always removes one goat that's also not your pick. Thus by switching you are trading off your 1/3 chance with the 2/3 chances and with one door removed, switching gives you the prize 2/3 of the time, as you only have one door remaining to switch to.

I remain mystified why so many people reject the simple truth that the only way to lose with the strategy of switching is if you happen to choose the prize door first, which is a 1 in 3 chance. It's a clear, direct, concise explanation and yet the MHP attracts comment from people who, on the one hand, are interested in probability, but on the other can't grasp the correctness of that simple proof.

@liulucy6051

3 жыл бұрын

If you don't lose, you must be win. so If you switch, what you lose? if you stay, what you lose?

@maksphoto78

2 жыл бұрын

The crux is not "win or lose" but believing that there is 50/50 chance when presented with the choice to "switch"

@derfunkhaus

2 жыл бұрын

@@maksphoto78 from my previous comment, it is clear that those who think the odds are 50-50 are in error.

@healthquest4823

Жыл бұрын

@@derfunkhaus What confuses people is that they assume that if one door is eliminated and the contestant is then given a new chance to choose, that constitutes a new game with just two doors giving naturally 50:50 odds. They don't realize that the two choices are not independent, and the only reason for that is because the host could not have revealed a goat from the first door chosen. Nearly 100% of people will get this wrong, so you shouldn't be so hard on them.

@dienekes4364

Жыл бұрын

_"I remain mystified why so many people reject the simple truth that the only way to lose with the strategy of switching is if you happen to choose the prize door first, which is a 1 in 3 chance."_ -- It's because the model doesn't hold up when applied to slightly different scenarios. It fails horribly. That's why intelligent people reject it.

I have an easier way to explain: imagine instead of 3 doors, it's 100 doors with 1 car and 99 goats. The game show host asks you to pick the door with the car. You randomly pick one knowing that you have a 1% chance that it's the car (basically, zilch). The host opens 98 doors that all have goats. So, the doors that are left are yours and one other. Obviously, that other door has the car (your door's odds of being the right door is still 1%). Switch!

@dAbaPEsTA123

9 ай бұрын

This relies on the assumption that the host wants you to switch to the door with the prize. AND combining two scenarios into one (first choice and second choice after other door(s) is revealed)

@thomasvontom

9 ай бұрын

@@dAbaPEsTA123 the host always asks you to play round two. Regardless of your choice in round one. So their is no you picked right and im going to trick you. Plus the math still stands. Odds are you picked wrong and the other door left behind is the door with the car that he is forced to give you a option to switch to.

@UNABRIDGED_SCIENCE

2 ай бұрын

THATS NOT TRUE. IN THE EVENT THAT THERE ARE TWO CHOICES, YOU HAVE A 50% CHANCE OF GETTING IT

@IdunDied

2 ай бұрын

@@UNABRIDGED_SCIENCE As explained in the video, because of the condition that for the 2 doors you didn't choose, 1 with rocks is always revealed, the probabilities of 2 choices at the START, which is choosing 1 door, or the other 2 doors, is the relevant choice. Because the 2 you didn't choose, get turned into 1 door by the revealing of the rocks. There is a 2/3 chance it is in one of those two doors you didn't pick, and the rock reveal means you know that in the 2/3 chance it is in one of those 2 doors, it has to be in the unrevealed one. So the door you didn't pick has a 2/3 chance of having the prize. TLDR, you're looking at the wrong 2 choices due to semantics of game rules, and these choices are weighted differently. Like an unbalanced coin flip.

@susanhanji1189

Ай бұрын

Deciding to switch or not switch is not the main issue of the monty hall problem, because obviously you should switch( as your comment explaination), the main issue is to find the probability of finding the car if you switch, which can be very confusing

Best explanation. Period.

Why do you continue to use the third door in the equation after it is eliminated as a "choice" The Math is changed when an option is eliminated.

@morbideddie

3 жыл бұрын

Think of it this way, the scenario is always the same, you pick a door, monty reveals a dud then offers you the chance to switch. If you pick the correct door on your first try (1/3 chance) you should stick, if you pick the incorrect door on your first try (2/3 chance) you should switch. 2/3 of the time switching gets you the car. Just because the door is eliminated that doesnt mean it cant have an effect on the match or logic of the situation.

Try clipping the mic higher on your shirt, it will drastically improve the quality of the audio.

Inoppugnabile e chiarissima dimostrazione che il cambio della scelta ha il doppio di probabilità di vittoria che rimanere sulla scelta fatta: enumerazione di tutte le possibilità con i relativi esiti.

It's a conditional probability puzzle and more a challenge of deductive reasoning than probability ( which is the easy part ). More specifically if your last selection is random, then your odds are in fact 50/50, however this does not change the odds of switching Vs not switching which is 2/3 and 1/3 respectively. Thus there are actually 3 scenarios, however if you count doing nothing at all, then your odds are 0% You always initially had a 1/3 chance of selecting the car and 2/3 chance of selecting goat. Thus by switching doors you're more likely to end up with the car, because the door you picked first up, was always more likely to be the goat and with one door and goat removed after your initial pick, the only remaining switch option is the car 2/3 of the time and the goat 1/3 of the time. Thus a random selection between the last two doors ( 50%), offers better odds than not switching (33.33%) at all, but not as good odds as switching (66.66%).

@JoeyBlogs007

6 ай бұрын

Put alternatively, every door has 1/3 chance of the prize. One door/goat is eliminated by the host, who is not going to reveal the car until the final selection. If you unknowingly picked the car ( which occurs 1/3 of the time ) and switch, you lose ( as you switch to a goat ). However if you unknowingly picked a goat (which occurs 2/3 of the time) and one door/goat is eliminated, then when you switch you win ( as you switch to the car ). Thus 2/3 of the time switching wins and 1/13 of the time switching loses.

Great. How about the case where there are 5 doors and your friend opens just 1 empty door after you select one of the doors. Should you change?

@ZphyraRyuu

9 жыл бұрын

ThePositiev3x Yes.

@rogerbodey9475

4 жыл бұрын

1/5 chance of picking the car first time. Picking any of the 3 remaining doors gives you a 4/15 chance of success. 4/15 is greater than 3/5.

@rogerbodey9475

4 жыл бұрын

sorry...... 4/15 is greater than 1/5.

@rogerbodey9475

4 жыл бұрын

So, yes, you should stilll switch.

Can someone please explain how to solve this using SAS programming

Everyone seems to forget the fourth dimension. Calculating the probability of the game state at the end based on the state at the beginning seems incorrect to me. Try moving to the end of the 'game' show. You now have two doors, two options. Let's up the ante. The choices are death or life. You are asked, 'which of the two doors you want? Stick with option one or switch?'. You call a friend, vos Savant says' switch, your odds or 2/3 of living if you switch!' You look back at two doors with two options and scratch your head.

@klaus7443

3 жыл бұрын

"You look back at two doors with two options and scratch your head." And it would be your fault if you scratch your head because you didn't realize that the host didn't have to open another door at all. That condition was included in the rules in order to fool those who do not understand probability. As long as the host knows where everything is he could have left that losing door closed and offer the other one for you to switch with as an alternative condition. Switching will either increase your chances of life from 33% to 100% or decrease your chances of life from 33% to 0%. And even if you do happen to switch to 'death' at least you will be remembered for being smart.

@morbideddie

3 жыл бұрын

It might seem incorrect but it isn’t. Whether sticking or switching wins or loses is entirely dependent on your first choice. You can only win by sticking if you have the car behind your initial pick, you can only win win by switching if you have a goat behind your initial pick. The latter is twice as likely therefore twice as often that will win you the prize. If you bring in a new contestant it’s a 50/50 problem since they lack the information we do, however that doesn’t really prove anything about the regular game.

wow now I see why good schools like MIT are important finally understand it

Very good point

takeaway message is awsome

This was the best explanation of this problem ever.

To explain it simply: Your first pick has a 33% chance of winning. Those odds remain the same. When door #3 is picked, since the odds of number 1 remain the same(33%) and the 3rd door is revealed, through logical deduction, the remaining door has the highest chance of winning.

@ronwloutzenhiser5953

2 жыл бұрын

Why? The prize is still behind one of two doors and the odds just went from 1 of 3 to 1 of 2.

@stefanohamam5805

2 жыл бұрын

@@ronwloutzenhiser5953 It's still 33% because of when you chose it. This is purely playing the odds. When it's 1 out of 3 and you select 1, the odds don't change. Your original selection had a 1 out of 3 chance. Now, when the goat is revealed you have a 2 out of 3 chance because you know where the goat is. So, you can use your original pick which is 1 out of 3, or select your second pick which now has a 2 out of 3 chance.

@thomasvontom

Жыл бұрын

@@ronwloutzenhiser5953 I have watched these break downs many many times. The first round doesn't matter. At the end of the day no matter what you choose he will show you one of the two doors with a goat. Leaving a door with a goat and a car. Switch don't switch. It doesn't matter. The second round is the only round that matters. Which is a choice between two doors. Sometimes a problem is so simple it does not matter if someone has a phd and you don't. The first round does nothing but add extra math to round two that has nothing to do with anything. Now. If the game worked the following way it would matter. If he revealed the car if you choose correct the first time. Then ya you would switch. The first round is nothing but a fun show. It's all about guessing between two doors no matter how you break it down. Unless the host declares you the winner and gives you the car if you choose correct the first time.

@Araqius

Жыл бұрын

@@thomasvontom 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.

@Araqius

Жыл бұрын

@@ronwloutzenhiser5953 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.

I like the listing method

9:24 if you're familiar with the problem and just want the "click" moment.

@mars8957

4 жыл бұрын

you're more helpful then my math teacher

thanks

Why 2/3 1/3 1/2 just one case.must be 2 cases ??????????????????

Your chance of winning does not change no matter what else is done. You can not add the the fact the one door is a loser to the other Monty door just because you want to. If I get to pick two doors then I have a 2/3 chance of winning even though I know that both of my doors aren't winners. Just as I know that both of Monty's doors aren't winners. So the fact that Monty's shows me a loser door does not change my odds of picking correctly in the first place. The confusion comes from people seeing it one equation 1/3 or 2/3 chance of winning. When in reality it is two equations with the first being 1/3 or 2/3. And the second equation being 50/50. If you are shown door 1, door 2 and a goat. you don't have a 1/3 chance of being correct in picking door 1.

@klaus7443

6 жыл бұрын

"The confusion comes from people seeing it one equation 1/3 or 2/3 chance of winning. When in reality it is two equations with the first being 1/3 or 2/3. And the second equation being 50/50." The confusion sure came from you all right. The host will always show a goat. Do you honestly think that half of the contestants have a car behind their door? The contestant wins by switching if both he and the host pick goats. Multiplying those probabilities you have 2/3x1=2/3. The staying chances are 1/3x1=1/3. If the host does not know where the car is and the door he opens happens to be a goat, then you would be correct in saying it's 50/50. That is because the probability of the host picking a goat is not always 1 as in the MHP, but rather 1 one-third of the time, and 1/2 two-thirds of the time. "So the fact that Monty's shows me a loser door does not change my odds of picking correctly in the first place." Of course not. That is why 67 contestants out of 100 who first picked a goat are stuck with them behind their doors forever no matter what the host does. "When in reality it is two equations with the first being 1/3 or 2/3. And the second equation being 50/50." How do you arrive at 50/50? If the contestant has a 1/3 chance to pick the car, then what do you think are the chances that two goat doors remain for the host? 1/3!! So the chances of the host revealing a goat and leaving one to switch with is 1/3. If the contestant has a 2/3 chance to pick a goat, then what do you think are the chances that a car door and a goat door remain for the host? 2/3!! So the chances the host reveals the goat and leaving the car to switch with is 2/3. The contestant's chances are equal to the host's chances of what is left over.

@MetaphoricalMountaineering

6 жыл бұрын

Look at it like this ---- Once you pick any door, you have a 50 50 chance that the host is showing you the only remaining goat, or one of two goats remaining.

I had a lot of trouble working at this problem in my head in terms of maximizing my chances of winning. At some point, i switched to thinking about it in terms of trying to minimize my chances of losing. Let me share with you my thought process: Say you choose Door 1 as your first pick. (You know that your first attempt at choosing the right door will fail 2/3rds of the time because 2 of the 3 doors are losers). Then next thing to happen is that Monty reveals one of the losing doors. In this example, say Monty opens Door 3 to reveal Door 3 is a loser. Remember, your first choice has 2/3rds chance of losing. That means that the one remaining door has only 1/3rd chance of losing. So switch to it to reduce your chances of losing. I've never heard anyone explain it this way, and right now I'm thinking I'm a genius!! :) So tell me where my thinking went wrong, because there's a real good chance I'm not seeing something that I'm supposed to (even though I don't think that is the case cuz my logic seems so sound in my mind right now).

@klaus7443

3 жыл бұрын

There is nothing wrong in your method of understanding the problem. Although it hasn't been explained in that way from what I have read it is somewhat similar to the explanation that goes like this....if the probability of the contestant in picking a door with a goat is 2/3, and for the host it's 1 then the probability of them both picking a goat is 2/3x1=2/3 and there's a 1/3 chance that a goat is left to switch with.

@ronwloutzenhiser5953

2 жыл бұрын

Why? The prize is still behind one of two doors and the odds just went from 1 of 3 to 1 of 2.

@xp_money7847

Жыл бұрын

@@ronwloutzenhiser5953 Think of it this way. I would rather bet on 2 doors(BUC) rather than on 1 door(A) even before the game begins. U stands for set union. But Monty only allows you to choose 1 in the beginning. He opens a door with a goat, you dont get any new info about the door you choose, but you get some info about B and C. So the B U C is now just B (if monty opened C). B U C has concentrated over B(as C=0). So wont it be better to choose B? Basically Monty is giving you an option to switch to B U C from your initial choice A.

What about when you originally choose the right one, there are two possibilities and if you switch what you get is wrong, I think you just covered one of those scenarios so for every door there are 4 possible scenarios not three. If I am wrong could you please explain to me why thx.

@klaus7443

2 жыл бұрын

The host can show either goat but there is still a 1/3 chance you picked the car. The probability of picking the car with the host showing Goat A is 1/6, and the probability of picking the car with the host showing Goat B is also 1/6.

@santiagovelasco966

2 жыл бұрын

@@klaus7443 I think probability has to do with the number of scenarios that can happen. Being this said, we have 4 possible outcomes for each time the car is behind a different door, 2 in which if you switch you win and 2 that if you switch you loose; this gives us 2/4 per door which gives us 6/12 when translated to probability is 50%.

@klaus7443

2 жыл бұрын

@@santiagovelasco966 Did you read my comment???? The host can show either goat but that doesn't mean you will pick the door with the car two times.

@santiagovelasco966

2 жыл бұрын

@@klaus7443 It actually does because if it doesn't why would picking the wrong one if you selected the wrong door count as two times you fail.

@klaus7443

2 жыл бұрын

@@santiagovelasco966 The probability of picking the door with the car is 1/3, the probability of picking the door with Goat A is 1/3, and the probability of picking the door with Goat B is 1/3. The TOTAL probability must equal one (1). In your explanation you have the probability of picking the car is 1/2, the probability of picking Goat A is 1/4, and the probability of picking Goat B is 1/4. That is wrong. The probability of picking the car is 1/3 and HALF of the times you do pick it the host can show Goat A, and the other HALF of those times the host can show Goat B. What you are doing is DOUBLE counting the picking of the car.

I will probably never understand how switching the door affects the probabilities : (

@Stubbari

Жыл бұрын

The probabilities are tied to the doors. You are switching probabilities.

@xp_money7847

Жыл бұрын

Think of it this way. I would rather bet on 2 doors(BUC) rather than on 1 door(A) even before the game begins. U stands for set union. But Monty only allows you to choose 1 in the beginning. Entire game you would be more concerned about the 2 unchosen doors than the one you chose( or you should be as a rational person). He opens a door with a goat, you dont get any new info about the door you choose, but you get some info about B and C. The B U C is now just B (if monty opened C). B U C has concentrated over B (as C=0). So wont it be better to choose B? Basically Monty is giving you an option to switch to B U C from your initial choice A. He has even narrowed your decision to B by revealing that C=0.

@willoughbykrenzteinburg

Жыл бұрын

It doesn't affect the probabilities. At all times, there is a 1 in 3 chance you picked the car, and a 2 in 3 chance you did not. This probability never changes; the DISTRIBUTION of the probability changes. When the host removes a "losing" door, he isn't changing the probabilities. There is still a 1 in 3 chance the car is behind your door, and a 2 in 3 chance it's in a door you did not pick. When the host removes a "losing" door, he merely reduces the number of doors that represent that 2 in 3 chance from 2 to 1. So now, you are left with a single door that you picked (still 1 in 3) and a single door you did not pick (2 in 3).

@ivantsanov3650

8 ай бұрын

​@@willoughbykrenzteinburg It does change probability, there's something called 'conditional probability'

@willoughbykrenzteinburg

8 ай бұрын

@@ivantsanov3650 The only thing I can surmise from your comment is that you don't actually know what "conditional probability" means.

Hey Jimmy. Have you figured out where the failure in your logic is yet? You don't get to include ALL possible outcomes, you have to only count EACH possible outcome individually.

@klaus7443

Жыл бұрын

Just look at what you picked. You already know if staying will win or not.

I get the math and I even tried an experiment but what I can't shake off is why is the second choice(to switch or not switch) is not an independent event? Why is the 1/3 probability from the revealed goat door transmitted to the other non-chosen door? The objective probability should not change unless we are somehow including the psychology of the game host Monty. Oh wait... isn't that the correct way to approach the problem? elimination of falses instead of search of correctness?

@hatim178

7 жыл бұрын

here is another way to think about it. if I showed you a gun that can hold 6 bullets and I put just one bullet inside. the chance I fire the bullet is 1/6. Then I fire the gun 5 times and all of them were a miss. what is the probability of the next time I fire actually shooting the bullet out? is it still 1/6 or now that I have removed all the the other possibilities, has the chance gone up to 6/6 or 100%? now say I bring this gun to a random stranger who has NOT seen me fire the gun 5 times and I ask him what are the chances that I will fire this gun next time I try? he will say 1/6 which is different to what you know which is 6/6 as you have already seen me fire the gun 5 times where as he has not. and that differrence of information is why your answer is different to the others guy. same with this monty hall problem. because you were there when there were three doors, the probability is 1/3 and so when one door is removed, the probabilities still remain from the start because of the information you have. if a new person came and did this puzzle at the two door step, he would say it's 50/50 but because you have more information then he does, you know it's not 50/50 but rather 1/3 and 2/3. I hope that explains it better.

@bmd4534

7 жыл бұрын

Stretch it out to the extreme. Imagine you had 1000 doors. 999 doors are nothing and 1 door is the prize. You pick one door and the host will open 998 doors and leave 1 door and the door you picked left. Unless you picked the 1 door out of 1000 with the prize, it makes mathematical sense to switch. You didn't pick 1 out of 2 doors... you picked 1 out of 1000 doors. The odds that you picked the right door is 1 out of 1000 regardless of what happens with the other doors.

@dnsmithnc

7 жыл бұрын

Great explanation.

@spiritualanarchist8162

7 жыл бұрын

I got actually got stuck, using the same train of thoughts and even words you did use...Then i changed 'goats, cars and doors into x and Y 's and i just had to except the outcome. This goes against one's feeling of' realty' right? It just show's how we humans perception is genuinely flawed (somehow why people voted for Trump made more sense to me ;-)

@titelmund2670

4 жыл бұрын

I was actually asking myself the very same question - even going through the problem and understanding the solution there is still a wierd awkward feeling about the solution because it is just so counterintuitive. What about the problem or what event makes switching better than staying and at what moment does that occur? Well think of it this way: there is actually not one choice being made, but three choices: your first choice, then the game show host choosing the door with the goat and then your choice whether to switch or not. On a "non-skewed", equal probability game, the game show host would, if you've picked a door with a goat, choose either the remaining door with the goat or remaining door with the car with an equal 50 % probability (the distinction doesn't happen when you chose the door with the car, because there is no difference, from your perspective, whether he chooses one goat or the other, either 50/50 or with a conscious bias). But he doesn't. He CONSCIOUSLY CHOOSES THE DOOR WITH THE GOAT RATHER THAN CHOOSING ONE OF THE REMAINING DOORS WITH EITHER THE GOAT OR THE CAR WITH EQUAL PROBABILITY and therefore REVEALS ADDITIONAL INFORMATION about the point you are in the decision tree before you make the final decision, whereby we know that the outcomes where he chose the door with the car ARE ELIMINATED and that happens with a skew of 2/3 towards being better switching. I'm trying now to make a triple decision tree showing this. Or think of it this way: what is different in the Monty Hall problem compared to the case where you just choose a door and then have the option of switching or not? Or put yet another way: the game show host is biased, but only 2/3 of the time from your point of view.

Thank you SO much! Other people made similar videos, but the solution never made sense to me. I love how you used multiple teaching techniques to explain why you'd have a 2/3 chance instead of 50/50.

@dawyer

4 ай бұрын

It is the same as The Monty Hall Problem, but there is one more player B. When player A chooses the door with the car, player B can open the door with the sheep and get the sheep. From Player B's perspective, no matter what others do, when Player A chooses one of the two pairs of doors, there is only a 1/2 chance of getting the sheep. Will Player B's chance of getting the sheep be the same as Player A's chance of getting the car?

@Araqius

4 ай бұрын

@@dawyer lmfao If player A get the car, player B cannot get the car. If player B get the car, player A cannot get the car.

@dawyer

4 ай бұрын

Is the reality of quantum mechanics more logical than mathematics? Players never know which goat is revealed...don't let the Jews fool you. If life is a never-ending choice, then what aspect of human nature is eternal if we approach it this way? 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

Lose first, then must win then.

Simply you are more likely to get it wrong in your first time than get it right

I could be wrong about it but there's something which I noticed which makes the probability 50% If I choose door 1 and it is the right choice or I choose door 2 and it is the right choice or I choose door 3 and it is the right choice, in all of these three cases there should be 2 cases whether the friend opens any of the two gates(the probability of him opening the other two gates should be taken separately instead of clubbing the probability that he chooses either door 2/3,1/3,1/2 respectively). This makes the total cases 12 instead of 9 with a 50% probability of winning(as the probabilities are taken separately). This is what I could think but maybe it doesn't matter what door the friend picks(when my chosen door is the correct one) or maybe it does.

@klaus7443

Жыл бұрын

It doesn't matter which door your friend opens because you still have a 1/3 chance of picking the door with the prize.

@darkness1130

Жыл бұрын

Yes..the list seemed to have combine the possible events of pick(1)prize(1)open(2) and pick(1)prize(1)open(3)..but open(2) and open (3) should be treated as different event (50:50) with same outcome, unless the game host always open a specific goat door when the prize door is already chosen. Can anyone clarify on this? 😵‍💫

I know I'm wrong, but the way I look at it is that Monty Hall always eliminates one door and then gives you a choice between two doors. So your choice to begin with is NOT THREE DOORS. It's two! He ALWAYS takes one away. Therefore the game starts after he takes the door away and asks you to choose between two doors. When you have that choice your probability is 50/50. There are always two doors he never takes away. Your pick and the one with the car. He then asks you to choose between a car and a zonk. But again, a lot of people with glasses say I'm wrong so I guess that's my lot in life.

@klaus7443

6 жыл бұрын

"Therefore the game starts after he takes the door away and asks you to choose between two doors." Switching already wins when the contestant picks a goat before the host does, a 2/3 chance.

@jfox11000

6 жыл бұрын

Hugh Reilly I'm a math professor and still don't get it I'm with you on it being 50 50

@klaus7443

6 жыл бұрын

Joshua, so how come you don't know how to calculate probability? Switching wins if the player and the host both pick goats. So the chances to win by switching is calculated by their respective probabilities of picking a goat....2/3x1=2/3. I would like to see how you calculate 50/50..... ....and how come you don't use the Bayes Theorem?

@klaus7443

6 жыл бұрын

"Hugh Reilly I'm a math professor" Judging by your uploads you must be teaching at Harvard.

@Luis-mh7qg

6 жыл бұрын

66% you will pick a goat bc there's 2 of 3. 33% will win you the car. Say you stay with you choice in 3 scenarios which are 1. Goat 2. Goat 3. Car Conclusion: you will get a goat 66% if the time Now let's say you switch 1. Goat. Reveals the other. If you switch you will get car 2 . Goat. Reveals the other. If you switch you will get a car 3. Car. If you switch you will get goat Conclusion: 66% of time you will win if you switch Compare both conclusions 66% lose if you stay 66% win if you switch What will you do 🤔?

Why do you count the first row as one count? Shouldn’t it be 2 counts?

@Araqius

Жыл бұрын

Assume the car is in door #1. If your pick door #1 twice, the host will open door #2 once and open door #3 once. If your pick door #2 twice, the host will open door #3 twice. If your pick door #3 twice, the host will open door #2 twice.

@Raven-bi3xn

Жыл бұрын

@@Araqius Right. So the first row in the table should be counted as 2, not as 1. This means 2 out of 4 wins, not 1 out of 3.

@klaus7443

Жыл бұрын

@@Raven-bi3xn Possibilities and probabilities are two different things. Just because the host can show either goat when the contestant picks the door with the car it doesn't mean that the contestant will pick that door two out four times. If it did then two of the three doors have a 3/4 chance of winning by switching.

@Araqius

Жыл бұрын

​@@Raven-bi3xn You need to put the probability of each possibilities into calculation. Let's say we roll a dice. If you get 1 or 2, you win. If you get 3 or 4, you lose. If you get 5 or 6, you roll again, any number = you win. Here are all the possible scenarios. 1 win 2 win 3 lose 4 lose 5 - 1 win 5 - 2 win 5 - 3 win 5 - 4 win 5 - 5 win 5 - 6 win 6 - 1 win 6 - 2 win 6 - 3 win 6 - 4 win 6 - 5 win 6 - 6 win You win in 14 out of 16 scenarios. Does this means your winning chance is 14/16 = 87.5%? OF COURSE NOT!!! If you take a look carefully, you will see that you will always win if your first roll is 1, 2, 5, or 6 and lose if it is 3 or 4. So this means your winning chance is 4/6 = 66%. In order to get to scenario "1", you need your frist roll to be "1" (chance = 1/6). In order to get to scenario "5 - 1", you need your frist roll to be "1" (chance = 1/6) and you need your second roll to be "1" (chance = 1/6) so the total chance for scenario "5 - 1" is 1/6 * 1/6 = 1/36. Scenario "1" and "5-1" doesn't have the same probability.

@Raven-bi3xn

Жыл бұрын

@@klaus7443 I hear you, but that's not why I'm asking. Possibilities are not probabilities, but in a discrete system like this, possibilities contribute to calculating probabilities. The possibilities that I asked about are, 1) you choose door 1 (the correct door), the host reveals door 2, you stay on door 1 and you win, 2) you choose door 1 (the correct door), the host reveals door 3, you stay on door 1 and you win. Why these two scenarios count as 1?

Once the third door was removed AND the option to switch was also included, the probability now changes to 50:50. Had the option to switch not been offered, then the odds would have remained 66.67% the prize was behind the other door.

@klaus7443

3 жыл бұрын

"Once the third door was removed AND the option to switch was also included, the probability now changes to 50:50. " You don't make any sense. If both doors have equal probabilities, which they don't anyways, then it wouldn't matter if the switching option was given.

@morbideddie

3 жыл бұрын

If the probabilities are 1/3 to 2/3 how does being offered the switch change that? It doesn’t change the location of the car and they aren’t shuffled. A choice of two doors is not 50/50.

the rules of the game is important. you failed to clearly state one important premiss. that you are always given the chance to switch

So Monty Hall only operates on the rule of the game-in a scenario when he opened a the door random, then would it 50/50 for the player that switching or staying would be equal probabilities of getting prize?

@klaus7443

10 ай бұрын

That is correct.

not really, host is offering to switch the door by the guest , only because the car is behind chosen door , if it was not, the host would open the chosen door to avoid giving the reward , dealer always wins and sometimes manipulates the mind of losers

Is this considering the contestant will play the game like 9x in a row?

I think the probability of winning a car (stay or switch) is 50%. because at the first choice, it doesn't matter if there is a car behind the door that i chose. After my friend open a door, there are 2 unopened doors in front of me. A car behind one of them. I can choose the left one or right one. It means stay if two choices are same. It means switch if two choices are different. Obviously, the answer is 50% Since my friend can only open a bad door, we can remove a bad door before the game started. And then i choose a door but don't open it, just tell people which one of them. After that, i choose a door and open it. It means stay if i don't change my mind. Am i wrong? This problem has troubled me for a long time.

@klaus7443

4 жыл бұрын

So if I am the contestant and you are the host, and I picked a door with a goat, how would my chances of winning the car by staying improve if you revealed the other goat?

@Araqius

4 жыл 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 win 1 out of 3 games. Assume you switch to the last door. If your first pick is Goat A, you get the car. If your first pick is Goat B, you get the car. If your first pick is the car you get a goat. You win 2 out of 3 games.

Hello Again, PLEASE READ PREVIOUS COMMENT FIRST I thought I would show the error in your own work. On the check and X chart to the left you prove that no matter what combination of doors picked, related to doors opened results would be 3 checks for staying and 3 x for swapping. Not only did you prove my point but you did it three times giving nine checks for staying and giving nine x for switching. Those totals clearly show a 50/50 result. As for your chart on the right, you failed to convert the probability to 50% after not having but two doors in question now. Your flow chart will prove that out if you hadn't ended it to abruptly. After showing which door was opened another bracket should have been added to show your last option , stay or change. These results will again add up to be the 50% probability as I stated.

@max5250

3 жыл бұрын

Maybe you should apply your work on probability for Nobel prize... but I wouldn't hold my breath if I was you... But before that, try to count number of check for staying and switching, to see if 3 equals to 6. Or, maybe you also have new maths for applying to Nobel prize?!

look closely at 1/2 and 2/3 and 1/3 those aren't fractions their whole and deserve separation adding equality ! disagree

@seankennedy7276

3 жыл бұрын

huh?

The strategy here was not stated correctly. It wasn't about which door was the better choice, it was about is there any point in switching. That's different. The former assumes one door is better, while the latter assumes the possibility the odds are just 50:50.

Dawyer's door No.2 problem..... t is the same as The Monty Hall Problem, but there is one more player B. He just standing there and waiting to open the last closed door. When player A chooses to opened a door with the car, player B can open the door with the sheep and get the sheep. From Player B's perspective, no matter what others do, when Player A chooses one of the two pairs of doors to open, there is only a 1/2 chance of getting the sheep. Will Player B's chance of getting the sheep be the same as Player A's chance of getting the car?

@Araqius

4 ай бұрын

lmfao If player A get the car, player B cannot get the car. If player B get the car, player A cannot get the car.

I began to understand this a lot better when I realized that if there were only two doors, your initial chances would be 50/50. But when the other door is uncovered, your chances of getting it right become 100%. A very simple explanation, but I think it also carries over to games with more doors like this one.

El problema o la paradoja de monty hall es un error: Se basa en un error de suma: Cada puerta tiene tiene 1/3 de probabilidad, si escojo una, las otra dos tienen 2/3 de probabilidad, eso es correcto, se ha sumado las probabilidades de las otras dos puertas cerradas, pero cuando una de las dos se abre, las suma no es válida, porque es como sumar peras con manzanas… es decir la probabilidad de la puerta abierta no se le suma a la puerta cerrada… Vámonos por otro camino, supongamos que las cabras son de oro, unos 50 kg de oro, y valen mas que el carro… asi quiero es la cabra y el abre una puerta con una cabra … La probabilidad de cada puerta es de 2/3, si escojo una, las otras dos valdrán 4/3 Una probabilidad de 4/3 es superior a la unidad, lo cual es infalible, pero no es real … la probabilidad real es 1/2, porque la probabilidad cambio a abrirse una puerta … se podría decir en este caso se redujo y en el anterior aumento, ambas quedaron en ½ se igualaron entre si …

why did you write the 2 paths 1-1-2/3 as 1 path and not 2/ that is not correct probability, you collapse 2 paths where you lose the car to 1. they are 2 distinct paths and should be counted as such. in the same way for paths 2-2-1/3 and 3-3-1/2. when you add these 3 lines the odds become 6 to 6 so 50%.

@klaus7443

3 жыл бұрын

"they are 2 distinct paths and should be counted as such" You are counting the paths without considering their probabilities, which are 1/6 and 1/6.

This is wrong. The game starts and ends not as a 1:3 chance (33%) but a 1:2 (50%) game. The rules of the game always reveals a non-winning door with the chance to chose either of the winning remaining doors. It is a 50:50 game from start to end.

@klaus7443

8 ай бұрын

The very reason switching doubles the chances of winning is because the host knows where everything is and always reveals a non-winning door.

@RonaldABG

8 ай бұрын

You may always end with two doors, but that does not mean that each will have the same chance to be correct. One was randomly chosen by you, who had no clue about where the car is, while the other was purposely left by the host, who already knew the location of the prize and was not allowed to reveal it (and neither the door that you picked). As you only start picking the car door 1 out of 3 times, then the host is who is forced to leave it hidden in the other that keeps closed the remaining 2 out of 3 times that you start failing. What you are saying is like pretending that if we put a random person from the street in a 100 meters race against a world champion in that discipline and we have to bet who will win, then it is a 50% chance because they will always be two options, one winner and one loser. But the point is that it is much more easier that the champion is who results to be the winner, not both equally likely.

@Araqius

7 ай бұрын

Let's say the game start with 1 door (1 car, 0 goat). After you make a pick, the host add a goat door and ask whether you want to switch door or not. What is your winning chance if you stay with your first pick? It's 100%, right? Chuck: The game starts and ends not as a 1:1 chance (100%) but a 1:2 (50%) game. It is a 50:50 game from start to end. Chuck: Even if the host open your door, for a few second, to show that's it's the car. After the host add a goat door, you door has 50% chance to be a goat.

@philip5940

4 ай бұрын

This teacher here who received a state award for his dedication just ruined his reputation and credibility The probability is simply 50/50 . A host of paradoxes can arise if the other answer is claimed . Paradox is not good , it indicates a wrong answer .

@Araqius

4 ай бұрын

@@philip5940 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.

Need help from a big brain: The problem itself seems like a problem, it should not be considered having three doors, because one wrong answer is already shown to us, so the main decision starts with two doors. When you start from having two closed doors in front of you (no matter which example you give this is always the case in Monty Hall problem), the chances seem to be 50% all the time.

@Hank254

Жыл бұрын

Mert!! You posted this to like 5 different videos and don't seem to care about any of your replies. No wonder you can't understand it :)

@merttanatmis

Жыл бұрын

@@Hank254 I read them, but still doesn't make sense to me. I'm also running around, don't have chance to reply every comment but thanks to everyone who tried!

@Araqius

Жыл бұрын

@@merttanatmis In MHP, you make a pick when there are 3 doors. Assume you stick 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 stick with your first pick. If you make a first pick when there are only 2 doors, it's not MHP.

@merttanatmis

Жыл бұрын

​@@Araqius chance of winning is 2/3 only when there are three closed doors and I can pick two of them. After I learn there's one goat behind one of the doors, we have two doors left which leaves us with 50-50.

@klaus7443

Жыл бұрын

@@merttanatmis "After I learn there's one goat behind one of the doors, we have two doors left which leaves us with 50-50." Irrelevant. The problem is already solved before the host opens a door.

Monty never offered to open a door or allow a swap all the time. What if Monty only offers a swap if you selected a winning door? Go to ny and you will lose your money.

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 Predicted 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.

@Araqius

4 ай бұрын

lmfao again Assume the player pick choose the same door as his prediction. If he predict door 1, he get door 1. If he predict door 2, he get door 2. If he predict door 3, he get door 3.

The chance is 50:50, as the actual game begins only after the host opens one of the door. prior to opening of the it was 33.33 % probability. but as you eliminate the one of the door, the 33.33% probability of the eliminated dor is shared between the other two doors , making each doors probability 50.(33.33+16.67). the way the problem is narrated makes you feel that probability never changed once one of the door opened. while it actually changes the probability. on the calculation provided it shows three scenario for each choice.while in reality the scenario is only two. Take the 1st scenario. if you choose door and stay. the host has option to open the door only once and not twice( either door 2 or door 3). so the scenario can only be two.

The only important takeaway is that two doors are better than one door - as Jimmy points out at 8:35 in the video - all the rest is just a gross over-complication to justify his salary.

@HumptyDumptyOakland

8 жыл бұрын

+Richard Buxton Two doors are only better than one door if you actually *want* to win a goat

this is wrong .The chance is 50:50, as th actual game begins only after the host opens one of the door. prior to opening of the it was 33.33 % probability. but as you eliminate the one of the door, the 33.33% probability of the eliminated dor is shared between the other two doors , making each doors probability 50.(33.33+16.67). the way the problem is narrated makes you feel that probability never changed once one of the door opened. while it actually changes the probability. on the calculation provided it shows three scenario for each choice.while in reality the scenario is only two. Take the 1st scenario. if you choose door and stay. the host has option to open the door onlu once and not twice( either door 2 or door 3). so the scenario can only be two.

@derfunkhaus

2 жыл бұрын

No, you're wrong.

@altruisticlemur

2 жыл бұрын

If you test all the possibilities, it makes sense. We can do this because the problem only has a definite amount of outcomes. Door 1: Goat | Door 2: Goat | Door 3: Prize Scenario 1: You pick door 1. Door 2 is revealed. You switch to door 3 and WIN. Scenario 2: You pick door 2. Door 1 is revealed. You switch to door 3 and WIN. Scenario 3: You pick door 3. Door 1 or 2 is revealed. You switch to door 1 or 2 and LOSE. It's a 2/3 probability of getting a prize when you always switch.

Not really! After opening the vacant door, the probability for the other doors is 1/2 each.

@max5250

2 жыл бұрын

Not really. Why are you trying to "explain" something you can't even grasp.

@Araqius

Жыл 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.

3:35 Bravoooooooo, if you guess GREAT!!! you win if not, you don't win!!!!! Applause for Mr. Teacher what a great answer.

You're trading one door for TWO! There, I just saved you 15 minutes of explanation.

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.

@Stubbari

4 жыл бұрын

The Monty Hall problem is a fictional math problem named after Monty Hall. Now you are confusing the show Let's Make a Deal with a simple math problem.

So: If I am offered two doors the choice is 50/50 but if offered three doors and one is removed from the selection choice then the odds are 2 of three? Even when one door is removed? I don't accept the premise that removing a choice still has odds of three rather than odds of two. ---- Wait!! I just heard you say something big. If you choose door number one and the prize is behind door number two, then "Monty" MUST reveal door number three. Hmmmmm.

@RonaldABG

2 жыл бұрын

Yes, because each of the two doors that remain closed were left by a different person, where one had advantage over the other on being who left the correct. It is not the same an informed choice than an uninformed one. Remember that one option was randomly chosen by the contestant while the other was deliberately left by the host, who already knew the locations and had to avoid revealing which he knows that contains the car. So, since the contestant only picks the car door 1 out of 3 times, the host is who leaves the car in the other closed door 2 out of 3 times. You can also notice that the switching door is equivalent to the selection that a second player that was cheating would make, a cheater that already knew which is the option with the prize and only had the restriction that cannot repeat the selection that the first one makes. In this way, it is obvious that the second player would win as long as the first one fails, which has probability 1 - 1/3 = 2/3. The revealed door is like which neither of them chose, that is always going to be incorrect of course. So, when you are deciding between the two doors at the end, you are basically betting who won: the first normal player or the cheater. You are more likely to win if you bet on the cheater's option. Very different to having to select one of two when you don't know anything about them.

@max5250

2 жыл бұрын

"but if offered three doors and one is removed from the selection choice then the odds are 2 of three?" Wrong, If you pick 1 of 3 doors without opening it, and, someone who knows what's behind doors, opens a door without prize, then the odds of remaining door is 2 out of 3, while odds of your initial door is 1 out of 3. "Even when one door is removed?" The door is not "removed" but opened, and it doesn't affect upper odds in any way. "I don't accept the premise that removing a choice still has odds of three rather than odds of two." If you chose to not to accept basic premises of probability, do not expect to calculate probability correct.

@derfunkhaus

2 жыл бұрын

If you are offered two doors the odds could be 50-50, but there may be more going on that makes the odds something other than 50-50. Such is the case in the MHP.

Wrong. 3 more written possibilities are missing on the board, which coincide with hits if you do not change the door.

@knightthanatos

5 жыл бұрын

6:46 There you mention 2 possibilities of winning written in a line. but there are 2 chances to win. that is the error, not writing all the possibilities, and that error is repeated 3 times.

@klaus7443

5 жыл бұрын

If you are considering possibilities instead of probabilities then the mistake is on your part, not on the part of the video. There is a 1/3 chance of winning by staying, a 2/3 chance of winning by switching.

@knightthanatos

5 жыл бұрын

@@klaus7443 Possibilities or probabilities, In that minute he is saying 2 possibilities written in a sentence, it is one of the errors that the paradox presents when explaining it, which is badly explained. I should say, in 2 sentences, that if you choose door 1, the presenter opens door 2 and does not change, you win, AND, in another line, you choose door 1, the presenter opens door 3 and you do not change, you win. If you count the options that way, you get 50%

@klaus7443

5 жыл бұрын

No one cares about how many possibilities there are. There is a 1/3 chance of you picking Goat 1, host shows Goat 2 and switching wins. There is a 1/3 chance of you picking Goat 2, host shows Goat 1 and switching wins. There is a 1/6 chance, NOT a 1/3 chance of you picking the car AND host showing Goat 1, and a 1/6 chance, NOT a 1/3 chance of you picking the car AND the host showing Goat 2. So switching DOUBLES your chances of winning.

@knightthanatos

5 жыл бұрын

@@klaus7443 How does it not matter how many possibilities there are? It is precisely what we are talking about. 1) Prize at door 1 1.1) I choose the door 1 1.1.1) Open the door 2. Do not change. Won 1.1.2) Open door 2. Change. I lose 1.1.3) Open the door 3. Do not change. Won 1.1.4) Open the door 3. Change. I lose 1.2) I choose door 2 1.2.1) Open the door 3. Do not change. I lose 1.2.2) Open the door 3. Change. Won There are no more samples in this type of options since the contestant, according to the premise, the presenter must open a door without a prize AND different from the one chosen by the contestant. 1.3) I choose the door 3 1.3.1) Open the door 2. Do not change. I lose 1.3.2) Open door 2. Change. Won 2) Prize on door 2 2.1) I choose the door 1 2.1.1) Open the door 3. Do not change. I lose 2.1.2) Open the door 3. Change. Won 2.2) I choose door 2 2.2.1) Open the door 1. Do not change. Won 2.2.2) Open the door 1. Change. I lose 2.2.3) Open the door 3. Do not change. Won 2.2.4) Open the door 3. Change. I lose 2.3) I choose door 3 2.3.1) Open the door 1. Do not change. I lose 2.3.2) Open door 1. Change. Won 3) Prize at door 3 3.1) I choose the door 1 3.1.1) Open the door 2. Do not change. I lose 3.1.2) Open door 2. Change. Won 3.2) I choose door 2 3.2.1) Open the door 1. Do not change. I lose 3.2.2) Open the door 1. Change. Won 3.3) I choose door 3 3.3.1) Open the door 1. Do not change. Won 3.3.2) Open door 1. Change. I lose 3.3.3) Open the door 2. Do not change. Won 3.3.4) Open the door 2. Change. I lose There you have ALL the possibilities. Tell them If you want, I'll tell you. Out of a total of 24 possibilities, 12 win and 12 lose. fifty%.

Monty need not even know where the car is. The way that the puzzle is set up we ignore the cases where Monty exposes a car, which could happen if he did not know. Your original choice separates this into two populations. Later information does not change those populations. It only tells us more information about the population where the door was opened. Your original chance of winning was 1 out of 3. When one of the other two doors is opened our odds remain the same that you were right or wrong. The odds that the prize was in population two is still the same. The switch of course improves our odds because now we know which one of the remaining doors has a 2 out of 3 chance of being right.

@klaus7443

2 жыл бұрын

Of course he needs to know where the car is. If he opened another door that has a goat without knowing what was behind it then it would be 50/50.

@subductionzone

2 жыл бұрын

@@klaus7443 Incorrect. The initial odds are made when the person chooses. At that point his odds are one out of three. How does Monty no knowing change that? Remember the set up. We are only counting the cases where Monty does not mind a car when he opens a door. That was a given. If he does not know and reveals the car it would not count according to our set up. It would result in a "do over". The two doors are one population, they represent a 2 out of 3 chance regardless of what Monty does.

@klaus7443

2 жыл бұрын

@@subductionzone Incorrect. If you played 99 games with a host not knowing where the car is and he revealed it 33 times then we only need to access the 66 games that you can continue. In 33 of them you have picked the car and in 33 of them the car is behind the door neither you or the host picked. 50/50.

@subductionzone

2 жыл бұрын

@@klaus7443 Sorry but you are not approaching this properly. The strategy of changing only applies in cases where the host does not reveal a car. As a result you are applying the odds incorrectly. If Monty reveals a car that cannot count for this problem. The strategy only applies when he does not reveal a car. And for that the odds are the same as if he knew or did not know. You are trying to claim that if Monty did not know and did not reveal a car that that somehow affects the odds of the first choice. You would need to explain how that makes a difference. Maybe we will have to break this down more.

@klaus7443

2 жыл бұрын

@@subductionzone "The strategy only applies when he does not reveal a car." That is EXACTLY what I did. Read!!!!!!!

Quick and easy: the assumption is that the odds are based on the contestant choosing a door. They do not. When the game starts, you have 3 doors and each door has a 1:3 chance of being right. How do those odds get calculated? 3 unopened doors, each having an equal chance. Once a door is eliminated, why would you change the way you calculate the odds? What does the contestant's choice have to do with anything? Once a door is eliminated, the odds are calculated exactly the same as when the game starts: by how many doors are available. In this purposed explanation of the failed model, the number of rounds and available doors in each round are completely ignored. In round 1: each door has a 1/3 chance of winning. But these odds are meaningless because, according to the rules, the host will ALWAYS open a goat door and will ALWAYS push the game to round 2. The round will NEVER be resolved, the contestant CAN NEVER WIN. You can't take the chance of them winning in round 1 into account, because that is AGAINST THE RULES.

@klaus7443

Жыл бұрын

"You can't take the chance of them winning in round 1 into account, because that is AGAINST THE RULES." Lol...so ignore what you saw as a host and continue like the dum contestant you are.

@klaus7443

Жыл бұрын

"Once a door is eliminated, why would you change the way you calculate the odds?" The probabilities aren't recalculated after the reveal dum-dum!

@dienekes4364

Жыл бұрын

@@klaus7443 You are an idiot, so I'm going back to ignoring you.

@klaus7443

Жыл бұрын

@@dienekes4364 So you know I must be right.

@dienekes4364

Жыл бұрын

@@klaus7443 I know you are too stupid to be right and too stubborn to admit that you could ever be wrong about anything. You are delusional. You are a fucking moron. Trying to debate with you is like trying to play chess with a pigeon.

Look at it this way: once you pick a door, you have a 50 50 chance of him showing you the only door with a goat or one of two doors with goats. switching isn't better.

@Luis-mh7qg

6 жыл бұрын

66% you will pick a goat bc there's 2 of 3. 33% will win you the car. Say you stay with you choice in 3 scenarios which are 1. Goat 2. Goat 3. Car Conclusion: you will get a goat 66% if the time Now let's say you switch 1. Goat. Reveals the other. If you switch you will get car 2 . Goat. Reveals the other. If you switch you will get a car 3. Car. If you switch you will get goat Conclusion: 66% of time you will win if you switch Compare both conclusions 66% lose if you stay 66% win if you switch What will you do 🤔?

Stupid answer. The answer is EFFIN simple. "If you switch, you get to see behind two doors instead of one". Jesus Christ, what is more simple than that?

@klaus7443

5 жыл бұрын

If you stay you get to see what's behind two doors as well.

@jackgarand7284

5 жыл бұрын

You are so totally wrong again. If you stay you get to see behind THREE DOORS. What are you smoking over there?

@klaus7443

5 жыл бұрын

"you get to see behind two doors instead of one" "you get to see behind THREE DOORS" Lol!!

@jackgarand7284

5 жыл бұрын

At the end you find out what is behind all THREE doors. HELLO.

@jackgarand7284

5 жыл бұрын

Well if you find out what is behind all three doors, then in theory you could win three cars. You could then keep one, sell one, and give one to your mama.

Confusing explanation, you make combination calculations using 3 doors (9 choices ) which does show merit if you switch. But the choice is between 2 doors so the combination calculations are (4 choices)50% . 1 door has been taken out of the calculations, so you are left with 2 doors, why are the 3 doors calculated still applied?

@Araqius

5 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.

@Jm-bp5zw

5 ай бұрын

@@Araqius goat A you get goat A=30% Goat B you get goat B=30% Car. You get car. = 30% You have a 30% chance of winning when you start the puzzle. You can see that this is indisputable. Now the host reveals a goat so 1x 30% of the odds has been removed, is this correct? So now we have 1x goat and 1 x car, you have to decide your original choice (30%) and the other choice (30%) So why is it your choice stays at the 30% and the other choice odds change from 30% to 60% even though it has been revealed as a goat leaving a choice os 2, the goat or the car? You start with a 30%,30%,30% odds of being right. Which I agree. ☝️ Then you go to 30%,60% when one of the options is removed ? A: you only have 1 out of 3 games if you stay with your first pick Yes this is correct 30% odds, but didn’t they remove one of the choices? Doesn’t that leave me a choice between 2 50% And why does the other choice gain an extra 30% odds. When the odds for the third choice was removed? Your explanation sounds like this is what they tell me and I must believe it! It doesn’t look like you fully understand it. I am not saying that their probability calculations are wrong, just the explanation could be a bit clearer? If you followed this discussion you would have seen a lot of highly educated and intelligent people smarter than you and me disagree with the initial odds, but they had the resources of a lot of smart people and discussions to come to a conclusion. They were not followers they were investigators which need proof! Switching means the opposite is not very scientific,

@Araqius

5 ай бұрын

@@Jm-bp5zw lmfao How are you going to stay with your first pick and win the car if your first pick is a goat?

@Jm-bp5zw

5 ай бұрын

@@Araqius probability is not an absolute, how do you Know he doesn’t have 2 goats? Although I accept the theory behind it now after watching many different videos. It is still has to do with luck🙏

The situation is that "decisions without consequences" and "choices with practical benefits" are confused and regarded as having the same meaning. It has been clearly decided. If there are really two choices, shouldn't we get both things? In fact, the rule changes from one of three choices to one of two, misleading you into thinking that you have already chosen once, but in fact you cannot get any of the so-called choices at all. When you finally make the real choice, in order to prove how smart and knowledgeable you are, of course it should be counted as 2/3 when there are only two choices. This is a testament to the Goebbels effect. If scholars and the educational community do not correct the extension of Monty Hall's problem and still believe that 2/3 is correct, then what is the point of educational scholarship? ? It was ruined.

@Araqius

4 ай бұрын

lmfao Let's say the game start with 1 door (1 car, 0 goat). After you make a pick, the host add a goat door and ask whether you want to switch door or not. What is your winning chance if you stay with your first pick? The answer is obviously 100%. Super easy, right? D: No. In fact, the rule changes from one of one choice to one of two, misleading you into thinking that you have already chosen once, but in fact you cannot get any of the so-called choice at all. D: Even if the host open the first door for a few second, tho show you that it's the car. After the host add a goat door, the first door has 50% chance to be a goat.

Marilyn Mach Vos Savant solved the problem - smart lady

To me, the 3 choice tree is only good for the first choice, before the door is opened. When first door is opened, you now have a new choice, and you will need a new, 2 choice tree. There are 2 choices, not one and the odds are different in each choice. Does not matter if you are right or wrong, the first door opened will be a loser, so you have a new choice between 2 other doors, not 3.

@klaus7443

2 жыл бұрын

You are completely fooled by the wording of the problem. There is no reason for the host to even open a door.

@Araqius

Жыл 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.

There is only one way to answer this and it's way, way simpler then anyone can imagine. First, it's not a 33.3% question, nor is it a 67% right or wrong. It's ALWAYS, and I mean ALWAYS 50/50, because in order to understand the 50/50 chance, you need to understand the rules of the game show. Regardless of which door you choose out of the 3, the game show host will ALWAYS open one of the two remaining doors. This is where people get confused - if the show host NEVER offered to open one of the remaining doors, then pretty much the 33.3% and 67% idea would be correct - what everyone forgets is the rules of the game states the host will always open one of the goat doors REGARDLESS of which door you originally picked. Meaning - stay with me here - even you stayed with your choice or decided to switch, makes NO DIFFERENCE IN THE EQUATION, it's 50/50 whether you stay or change your mind to the other door. It's ALWAYS a 50% question regardless of which door you choose, knowing you will see what's behind one of the 3 doors anyway. IT"S THAT SIMPLE.

@max5250

3 жыл бұрын

"Regardless of which door you choose out of the 3, the game show host will ALWAYS open one of the two remaining doors. This is where people get confused" Actually this is where you got confused, since, depending on the choice of a player, game host will be able to choose between 2 available goats (in 1/3 of cases), or he will be limited to a single goat, leaving you the door with a car to switch to (in 2/3 of cases). " if the show host NEVER offered to open one of the remaining doors, then pretty much the 33.3% and 67% idea would be correct" In that case, player would still have only 33.3% chances of winning by staying, and the same probability if he switches to any of two unopened doors. "what everyone forgets is the rules of the game states the host will always open one of the goat doors REGARDLESS of which door you originally picked. Meaning - stay with me here - even you stayed with your choice or decided to switch, makes NO DIFFERENCE IN THE EQUATION, it's 50/50 whether you stay or change your mind to the other door." It surely makes all the difference, since staying with your initial choice, you stick to 33.3% chances you had at the beginning, and if you switch the door, you get the probability (100-33.3=66.6%) of the two doors yo couldn't select initially. "IT"S THAT SIMPLE." Yep, it's quite simple, and you still got it all wrong, that's quite an achievement...

@morbideddie

3 жыл бұрын

So to be clear you have a 50% chance of picking the car on your first guess? And this chance of picking the car depends on whether you know that after the pick Monty will reveal the door? I’ve seen some wacky explanations from people refusing to accept the truth but this is one of the most weird and convoluted ones that’s come up. It’s also not particularly simple. Let’s try a simpler explanation for why your odds do improve. The game is always the same, you pick a door, monty reveals a goat and then offers the the switch. You only win by sticking if you picked the car which happens 1/3 of the time, hence the other 2/3 of the time switching wins.

@MrLou345

3 жыл бұрын

@@morbideddie Monty ALWAYS reveals one of the other doors that you didn't pick. Knowing that, gives you a 50/50 chance - HOW? - simple - it doesn't matter which door you chose, you take one door away knowing that Monty will show you 1 out of the 3 doors ALWAYS, you are then left with the door you picked - and this is important - you will ALWAYS be given a choice to either keep the door you first picked OR switch to the only other door left. Since you are left with two doors ALWAYS, AND YOU ARE ALWAYS ASKED TO KEEP YOUR FIRST PICK OR SWAP IT TO THE ONLY OTHER DOOR LEFT - THAT IS A 50/50 CHANCE.

@morbideddie

3 жыл бұрын

@@MrLou345 knowing Monty will remove a door gives you no new information that allows you to improve your odds with your first pick, there are three doors and you have an even chance of the car being behind any, there is no reasonable way you can derive 1/2 chances from that. It’s only after Montys reveal that you gain new information that informed your odds. Ignoring the fact that millions of simulations show your wrong In your scenario what are the odds for each door? 1/2, 1/2, 1/2 for doors a, b, and c? Think about what your saying here. Let’s say we have two games, one where you pick a door and get whatever’s behind it and one where you pick a door, Monty reveals a goat and you always stick with your choice. Both games are functionally identical, you pick one doors in three with no way of knowing where the prize might be but your claiming that in one game I have a greater chance of winning?

@klaus7443

3 жыл бұрын

@@MrLou345 Your usage of mathematics is terrible. "Monty ALWAYS reveals one of the other doors that you didn't pick." If the host always picks a door with a goat then there is a 2/3x1=2/3 chance, not a 1/2 chance that they both did.

There are 12 possible combinations obtainable from the Monty Hall Problem. The probability remains 50/50 if switching or staying. 1. Combination 1 You chose door 1 The Prize is in door 1 Host opens door 2 If you STAY, you WIN If you switch, you lose Combination 2 You chose door 1 The Prize is in door 1 Host opens door 3 If you STAY, you WIN If you switch, you lose Combination 3 You chose door 1 The Prize is in door 2 Host opens door 3 If you stay, you lose If you SWITCH, you WIN Combination 4 You chose door 1 The Prize is in door 3 Host opens door 2 If you stay, you lose If you SWITCH, you WIN Combination 5 You chose door 2 The Prize is in door 1 Host opens door 3 If you stay, you lose If you SWITCH, you WIN Combination 6 You chose door 2 The Prize is in door 2 Host opens door 1 If you STAY, you WIN If you switch, you lose Combination 7 You chose door 2 The Prize is in door 2 Host opens door 3 If you STAY, you WIN If you switch, you lose Combination 8 You chose door 2 The Prize is in door 3 Host opens door 1 If you stay, you lose If you SWITCH, you WIN Combination 9 You chose door 3 The Prize is in door 1 Host opens door 2 If you stay, you lose If you SWITCH, you WIN Combination 10 You chose door 3 The Prize is in door 2 Host opens door 1 If you stay, you lose If you SWITCH, you WIN Combination 11 You chose door 3 The Prize is in door 3 Host opens door 1 If you STAY, you WIN If you switch, you lose Combination 12 You chose door 3 The Prize is in door 3 Host opens door 2 If you STAY, you WIN If you switch, you lose TOTAL WINS IF YOU STAY = 6 TOTAL WINS IF YOU SWITCH = 6 PROBABILITY = 50% / 50%

@Stubbari

Жыл бұрын

Maybe you should have checked your math first before copypasting that nonsense to all MHP videos?

@Araqius

Жыл 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. Basic math/logic kids understand, idiot among idiots doesn't.

@Araqius

Жыл бұрын

......................................... Olajide Adelusi: 1. You chose door 1 - The Prize is in door 1 2. You chose door 1 - The Prize is in door 1 3. You chose door 1 - The Prize is in door 2 4. You chose door 1 - The Prize is in door 3 5. You chose door 2 - The Prize is in door 1 6. You chose door 2 - The Prize is in door 2 7. You chose door 2 - The Prize is in door 2 8. You chose door 2 - The Prize is in door 3 9. You chose door 3 - The Prize is in door 1 10. You chose door 3 - The Prize is in door 2 11. You chose door 3 - The Prize is in door 3 12. You chose door 3 - The Prize is in door 3 Olajide Adelusi: Let's rearrange it. 1. The Prize is in door 1 - You chose door 1 2. The Prize is in door 1 - You chose door 1 3. The Prize is in door 1 - You chose door 2 9. The Prize is in door 1 - You chose door 3 Olajide Adelusi: I can choose the door with the car 50% of the times despite of the fact that there are 3 doors. Olajide Adelusi: This is because I have magic power. Hoooraaay!!! Olajide Adelusi's parents: Hoooraaay!!!

@olajideadelusi1151

Жыл бұрын

@@Araqius I guess you are the idiot as you just exposed your poor upbringing. Learn to debate without insults. Cheers.

@olajideadelusi1151

Жыл бұрын

@@Araqius Your parents didn't teach you that there will be no Monty Hall paradox without the action of the host? I guess not. You can't give what you don't have. Generational curse.

Each game has different factors that affect each person's values ​​and calculation of his chances. If the car behind the door is a Bugatti "La Voiture Noire", how many opportunities can the host provide to play with it? What if you had to make a choice with your child's life?

It always reduces to a 50/50 choice after he opens his door. There was never a one in three option for the contestant. The first round only suffices to either eliminate one of the two non-winning doors that he will open or for it not to matter which one he opens.. It makes no difference to the contestant, who will always have a 50/50 choice. Simple as that. Good example of how to confuse with probability theory.

@RonaldABG

Жыл бұрын

You are only focusing in the fact that you will end with two doors, but the question is: will the correct be the staying one or the switching one? That depends on the first selection, and more importantly, most of the time the correct will be the switching one. To make an analogy, would you accept this bet about who will manage to say the number that will win the next lottery contest? You must first guess a number and tell me which it is. The day when they give the results I will see them but you won't yet. If I see your number was not the winner (which is more likely) I will say the winner number, but if I see you were lucky and your number was the winner, I must say any other incorrect I can think of, because I cannot repeat your choice. In this way, there will always be two possible numbers remaining: which you said first and which I said later, and one of them will necessarily be the winner of the lottery. Do you think each of them has 50% chance to be correct, which implies that if we repeat this many times you would have guessed the winner about half of them? Or do you think almost always I will be who says the correct, because I already saw the results? This is basically what occurs in Monty Hall game. Since the host knows which is the winner door and takes care of never discarding it, then the other door he leaves closed is like the number I would say: correct as long as you start failing. The only difference is that in Monty Hall you can actually switch to it. And the revealed door is like the rest of numbers that neither of us said.

@klaus7443

Жыл бұрын

Think about it, if you tested this with three cards why would reveal a 'Goat' at all? You could simply write down what you first picked.

@Araqius

Жыл 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. Basic math/logic kids understand, idiot among idiots doesn't.

@willoughbykrenzteinburg

Жыл бұрын

No it doesn't. Think about this carefully - - - the ONLY scenario where you LOSE by switching is if you pick the car to begin with. You pick the car, the host reveals one of the goats, you switch to the other goat, you lose. If you pick a goat first, the host will reveal the OTHER goat, and you will switch to the car. You will pick the car initially 1 out of 3 times, and a goat 2 out of 3 times - ergo, you will be switching to a goat 1 out of 3 times, and to the car 2 out of 3 times. Quick example : You choose CAR : Host reveals a goat - doesn't matter which one - you will switch to the other goat. You choose GOAT A : Host reveals GOAT B - you switch to CAR You choose GOAT B : Host reveals GOAT A - you switch to CAR You win the car 2 out of 3 times.

Its an illusion it always 50/50

@klaus7443

2 жыл бұрын

You would have to be delusional to think the actions of the host changes anything when it takes only one card flip to determine as to whether it would win or lose by staying.

@superyahoobrothers

2 жыл бұрын

@@klaus7443 It does but you are too stupid to realize it

@klaus7443

2 жыл бұрын

@@superyahoobrothers It takes you more than one card flip to realize it will win or lose by staying???

@klaus7443

2 жыл бұрын

@@superyahoobrothers "It does but you are too stupid to realize it" Lol...I wasn't the one who spent hours trying to figure out which door the car was in after knowing which two doors didn't have it.

@superyahoobrothers

2 жыл бұрын

@@klaus7443 No flipping one card out of the 3 cards change the odds from 1/3 per card to 1/2 per card

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.

@klaus7443

3 жыл бұрын

Why are you even calculating probability after the reveal when this isn't an unconditional probability problem to begin with?

@max5250

3 жыл бұрын

"There is no advantage to either staying or switching. " Both, probability calculation and empirical results disagree with you. Can you guess now, who is wrong?!

BUT.......playing a random game with the prize placed randomly behind different doors......it is possible to randomly pick the right door 100% of the time by sheer luck. It is also possible to pick the wrong door every time by either switching or by staying. However......the odds are in your favor if you switch doors. Overall you are more likely to win by switching doors.

@max5250

3 ай бұрын

"it is possible to randomly pick the right door 100% of the time by sheer luck. " Nope, that's not possible. If you play many games in a row, and you consistently win with considerably higher probability that the calculated one, you are cheating. That's how casinos get a clue if someone is cheating or not.

@rael5469

3 ай бұрын

@@max5250 Well....in reality anything is possible......but not "likely."

@max5250

3 ай бұрын

@@rael5469 Well, in reality, probability tells us how likely certain possibility is, while you claim we could randomly pick the right door with all the time by sheer luck, which is a complete nonsense.

@rael5469

3 ай бұрын

@@max5250 No it is NOT complete nonsense because anything is possible even if it is not probable. After all....somebody DOES eventually win the lottery even though the odds are millions to one. So even though the odds are against it, it is POSSIBLE to pick the car every time even though it is unlikely.

Oh my god. This doesn't require 16 minutes of explanation. Whatever door I pick, there is a 33% chance the car is behind my door, and a 66% chance of the car being in one of the other 2 doors. The host eliminates a door and a goat from that group of "other" doors, but the 66% chance still belongs to that group (which now contains only one viable door). Do I want a a door with 33% chance of a car behind it (my door), or a door with 66% chance of a car behind it (the group of the other 2 doors now whittled to one)? Force the dissenters to do their own grid, and save your breath.

@DanielMaidment

6 жыл бұрын

You realise that telling people save their own breath and force everyone else to do their own work from first principles is a complete cop-out? Tutorials are meant to show you how to do the problem if you don't know how. By your logic we should all be forced to reinvent every idea in all of science every time we don't understand anything, instead of just learning from someone who knows better. You in all of your wisdom and might didn't have to sit through the whole video if you didn't want to. I did my own grid before watching the video, and solved the problem correctly, but it didn't occur to me that expanding the problem to 1000 doors would make the problem more intuitive, even if I could correctly do the math. So even though I can solve it from basics by "doing my own grid", I still learned something. Also, this isn't a question of dissent, this lecturer probably gets paid to "waste his breath".

@thomasvontom

9 ай бұрын

The problem with explaining this is two things. One people hyper focus on round two. The fact that 50% of the doors win and 50% lose. Two they forget that the host asks you reguardless of if you choose right or wrong. It really helps to see it with a big ass number first. It so quickly can feel like a witty man is selling you snake oil. I for about a month was in the camp of coin flip. Till I backed up and looked at it from a larger number of first doors. Then it made sense.

I can prove, beyond any doubt (to anyone who is intellectually honest) how this video is complete and utter bullshit. What is presented here is called, in mathematical and scientific nomenclature, a "model". There are 2 ways to prove a model: direct testing and observation and applying it to similar yet slightly different scenarios and having it be consistent. Since testing isn't possible in this venue, I will prove this model is wrong by the second method. Core Scenario: A contestant is presented with 3 doors. They get to choose 1. For simplicity, we'll say they pick door 1. Each round, the contestant get to switch to a different door or stay with the door they have. There is only 1 door that has a car, but 1 door _always_ has a car. So, at this point, each door has a 1:3 chance of winning. If the round is resolved, the contestant has 1:3 chance of winning because they only got to choose 1 out of the 3 doors. If anyone disagrees with this, please comment below. The host opens one of the doors the contestant didn't choose (door 3). If the car is behind the door the host opened, the contestant loses. If the host opens a door with a goat, then we move to round 2. The contestant is allowed to switch to door 2 or stay with door 1. I think we are all in agreement up to this point. According to the model presented here, the odds of the eliminated door gets shifted over to door 2. This is based solely on the fact that the contestant picked a door in round 1 (an action that played absolutely no role in establishing the odds in the first place). According to the standard model, the odds of the eliminated door get evenly distributed between the existing doors. This is based on the number of doors remaining. In order for a given model to be correct, the outcome must equal 100%. There is a 100% chance that the contestant will win or lose. All odds needs must equal 100%. So, in round 2, according to the presented model, there is a 1:3 chance you win if you stay with door 1 and a 2:3 chance you win if you move to door 2. This equals a 3:3 chance, which is equivalent to 100%. We're good so far. But what about the standard model? In round 2, with an eliminated door, since there are 2 doors, the odds of either door having the car is 1:2 (also described as 50%). Since each door has a 1:2 chance, that adds up to 2:2 or 100%. So, both models are seemingly viable at this point. So, let's apply these models to 2 slightly different scenarios: Alternate Scenario 1: In round 2, the round isn't resolved. The host opens door #2, which has a goat. What are the odds of the door the contestant has picked having the car? Let's go back to our models: In the presented model, remember that the ONLY criteria of the odds for being correct are the doors that the contestant DIDN'T pick. So, according to this model, if the contestant stayed with door #1, there is a 1:3 chance of winning. But the 2 other doors, having been opened and showing goats, is clearly 0:3. What does that add up to? 1:3 + 0:3 = 1:3. So, if the car isn't behind door 2 or 3 and is only behind door 1 1/3 of the time, where is the car the other 2/3 of the time? The total for winning and losing here is 33%. SO this model fails. What about the standard model? Once door 2 is eliminated, the odds of each remaining door get recalculated and distributed evenly. So, if there is only 1 door, that means that the chance of that door being correct is 100%. So the standard model is superior and proven correct so far. But let's move to alternate scenario 2 and see what happens: Let's play the exact same game except with 2 contestants: In the 1st round, contestant A chooses door 1. Contestant B chooses door 2. The host opens door 3 with a goat). Where do those odds get shifted? In the presented model, the odds can't be shifted because there is no door that hasn't been picked, so the odds are: Door 1 -- 1:3 Door 2 -- 1:3 Door 3 -- 0:3 (shown to be a goat) This adds up to 2:3 or 66.6%. So this model fails because it doesn't add up to 100%. 1/3 of the time contestant 1 wins, 1/3 of the time contestant 2 wins, and 1/3 of the time they both lose. If they both lose, where is the car? *_OR_* The contestants both switch to the other door. According to the model, "the other door" gets the additional odds. Okay, so the odds end up like this: Door 1 -- 2:3 (because contestant 2 didn't pick it, so it gets the odds from door 3) Door 2 -- 2:3 (because contestant 1 didn't pick it, so it gets the odds from door 3) Door 3 -- 0:3 (exposed and eliminated as a possibility) So, the odds (2:3 + 2:3 + 0:3) equals 4:3, or 133%. Once again, a failure. 1/3 of the time contestant 1 wins, 1/3 of the time contestant 2 wins, and 1/3 of the time they both win. We've established that there is only 1 car behind 1 door, so it is impossible for both contestants to win. So, what does the standard model say? The odds get evenly distributed to the remaining doors. How do those odds look? Door 1 -- 1:2 (one out of 2 remaining doors) Door 2 -- 1:2 (one out of 2 remaining doors) Door 3 -- (not in the calculation because it was eliminated by being exposed as a losing door) 1:2 + 1:2 = 2:2 or 100%. Once again, the standard model is victorious. So, in both of these scenarios, the presented model fails miserably and the standard model stands.

@Araqius

Жыл 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. Basic math/logic kids understand, idiot among idiots doesn't.

@dienekes4364

Жыл бұрын

@@Araqius _"Assume you stay with your first pick."_ -- I get it. You're embarrassed because I keep proving you wrong. Being stupid isn't something you should be ashamed of.

@dienekes4364

Жыл бұрын

@@Araqius _"Assume you stay with your first pick."_ -- I think it's HILARIOUS that you have to keep coming back to this. I keep proving it wrong, but you are just incapable of learning anything. _"Switching means the opposite."_ -- No, it's not. The opposite is ONLY true if you get to pick TWO doors that are *unopened.* When you "switch", you aren't picking 2 doors, you are only picking ONE door. _"Basic math/logic kids understand"_ -- Yes, but some of us grow up and learn how to think rationally. If you want to think like a toddler, there is no help for you.

@Araqius

Жыл бұрын

@@dienekes4364 All you can do is bark?

@klaus7443

Жыл бұрын

@@dienekes4364 Dinkass, why are you opening a door in the first place??? If the END RESULT is that you will always be able to switch to the prize or away from it then there is absolutely no reason why you need a door open nor do you need to know where a goat even is.

Why do the probabilities not take into account a door that is opened midway? That door is no longer part of probabilities from the point of opening it. because it is a KNOWN. When the door is revealed (with the goat or whatever) it sets up a NEW probability between two unknowns (50/50) Being shown one of the two non-winning doors simply narrows the probability of the last two left to..... 50/50. Why is there even talk now of 1/3 or 2/3rds once one door is a known thing and no longer part of the probability equations? WHAT other third are you taking into account after that? Chuck the third door out of the probabilities once it is no longer part of the equation. It is 100% known. Imagine that when one door is revealed, leaving only two doors to choose from, another person comes onto the scene and has to choose between the two doors. They would now (with this new situation) have a 50/50 chance of choosing the right door, right? OK. Think about that. The first participant has to choose between the two doors also. Has he chosen the right door, or the wrong door? He/she does not know which door offers the prize either. The third probable door is DONE WITH at that point and cannot magically affect the probability of two unknown door outcomes.

@mrpad0

2 жыл бұрын

One has to approach this in different ways What happens if the presenter of doors opens a door with a goat behind it right from the get go. How would that alter things? Are you going to count that door and goat in the equation from that point onward too? HOW is that any different? Whatever door the contestant chooses, one goat door is GOING to be revealed. The presenter might have two goat doors to choose from or one goat door (depending upon the contestant's door choice). It is a GIVEN that a 3-way choice is going to become a 2-way choice. There are TWO SEPARATE probabilities here. One before and one after one of the probabilities is known (therefore no longer part of any probability equation). The contestant is still faced with two doors. He might choose one or the other (50/50 still) Faulty math will still be considering the third, KNOWN door into the equation results? Math is logical, but that is not logical. "Here is $90, but I am going to take away $30 somewhere along the way. This will happen at the start of my calculations, or somewhere mid-way through them" So, one is really dealing with $60. "Here are three choices, but I am going to remove one of those choices along the way" How is that different?

@RonaldABG

2 жыл бұрын

@@mrpad0 You must take into account that since the host must reveal a door that is not which the contestant chose and neither which has the prize, that means that when the player's has a goat the host is forced to reveal specifically which has the only other goat. Instead, when the player's door has the car, the host is free to reveal any of the other two, we don't know which in advance, they are equally likely for us, because both would have goats in that case. In this way, the 1/3 case in which the player's door is the correct actually splits in two sub-cases (two halves of 1/6 each) according to which door the host reveals next, and therefore when a door is opened we must discard the half in which he would have revealed the other. I mean, not only the 1/3 in which the car is in the opened door must be discarded; also 1/6 of the car being in the contestant's door must be discarded. This is better seen in the long run. If you played 900 times, the car should tend to appear about 300 games in each door. Now, if you always start picking door 1, the possible cases would be: 1) In 300 games the car is behind door 1 (yours). If the host reveals each of the other two with 1/2 probability, then: 1.1) In 150 of these games he reveals door 2. 1.2) In 150 of these games he reveals door 3. 2) In 300 games the car is behind door 2 and in all of them the host reveals door 3. 3) In 300 games the car is behind door 3 and in all of them the host reveals door 2. So, if door 2 is revealed, you could only be in case 1.1) or in case 3), which is a subset of 450 games. You win by staying in 150 of them (case 1.1), that are 1/3 of 450, but by switching in 300 (case 3), that are 2/3 of 450. The error is to think that after door 2 is revealed all the 300 games of case 1) should still be possibilities. But for that to be true you should be sure that if the car was in yours the host would have definitely revealed door 2 and not door 3. With that condition, it would be 50% for each door at that point, but note that it would imply that in the other case when door 3 is opened, the car is definitely not in your door, because that would only occur when door 2 is the correct and therefore you would win by switching with 100% probability. You cannot make one case 50% without sacrificing the other. And the average of both cases will always be 1/3 for staying, because in total the car is only 300 games (1/3 of 900) behind your option, which does not change by revealing a door. The contents were not shuffled after the revelation.

@mrpad0

2 жыл бұрын

@@RonaldABG No, I'm sorry but that is a logical fallacy. Because the host WILL open a door with a goat, the contestant knows from that start that he will ALWAYS have a 50/50 chance of getting the car. There is no probability the that host will choose the car. Ever. Build that into the equation and one realizes that one third of the choices WILL be removed. It is not a probability but a given. 100%. As I said. Bring on a new contestant and have them choose from the remaining two doors and their chance of getting the car is 50/50 The host opening a door with the knowledge that he will open a goat door blows away and 'probability of three' immediately. There never was going to be a choice of three. ONE of the two dud doors was ALWAYS not going to be part of the equation. How can someone have a 2/3 chance of something when the probability only ever is between two viable doors. From the get go, one of the three doors will be opened. Doesn't much matter when that happens, it does not alter the probability equation one jot. It is a logical fallacy to think of one in three chances of winning (or 2 in 3 later) because the door is NEVER immediately opened. Before that one in three chance can play out, the whole thing is reduced demonstrably to one in two - as it always was.

@RonaldABG

2 жыл бұрын

@@mrpad0 Your reasoning is like saying that since we all know that we are going to die someday, then it is the same as if we didn't exist. So, how is this written? Who did it? The point is we do this before we die. First, to answer your question of what would happen with a new contestant, that person has 1/2 chance to select the winner but not because each switching or staying wins 1/2 of the time, but because that person cannot force the selection to be specifically the switching or the staying option as he/she does not know which is which, so has 1/2 chance to pick each and the probability averages to 1/2: 1/2 * 1/3 + 1/2 * 2/3 = 1/2 * (1/3 + 2/3) = 1/2 If that person played multiple times, about half of them would end picking which was the staying door for the first player, and the other half which was the switching one, so the extra wins that switching provides are compensated with the extra loses that staying provides. It is very different to always deciding to switch. If you find this strange, think what occurs in exam questions. In a classroom normally the exam is the same for everyone, but why do some people get better grades than others? Because some have study more, and therefore they can know which answer is more likely to be correct instead of leaving the decision to the random. In a true/false question, for example, it is 1/2 likely to get it right for someone who makes a random selection, like flipping a coin, but not for someone that has idea of the possible answer. Imagine if those who know more would base their decisions on what those who know less would do. Now, the problem with the Monty Hall game is that you are only focusing in the fact that you will end with two doors, but the thing is if the correct will be the "staying" or the "switching" one, a distintion that would not exist without the first selection. The first selection determines if the car will be in the staying or in the switching door, and the most important thing is that most of the time it makes that it will be in the switching option. To make an analogy, consider another game in which there are three tables and each one has two face-down cards; one says "car" and the other says "goat". It is a rule that in two of the tables, the card on the left says "goat", and only in one table the card on the left says "car". For example, one possible configuration could be: Table 1 Table 2 Table 3 goat-car car-goat goat-car (unique case) Since the cards are face-down, you don't know which table is which has the different configuration. You must choose a table, and then select one of the two cards on it in order to hit whicht says "car". As you see, regardless of what table you pick you will later be deciding between two cards, one on the left and one on the right. But the important thing is that it will be 2/3 likely that you would have selected a table in which the card that has written "car" appears on the left. You cannot assign 1/2 to the two positions left/right because the amount of configurations is not the same for them. And if the game started with more tables but still only one had the different configuration, then the probability for the "car" card appearing on the right would increase. In the Monty Hall game the first selection is similar to when selecting a table because it determines in which kind of configuration you will be, only that the distinction is not left/right, but staying/switching. Selecting each of the two doors that hide a goat will make the configuration to be {staying: goat, switching: car}, and only selecting the door that hides the car will make the configuration to be {staying: car, switching: goat}.

@mrpad0

2 жыл бұрын

@@RonaldABG I can see how the probability is not 50/50 now (having sat and gone through it in my head so many times!) so thank you. On the other hand I am STILL baulking at the 2/3rds thing. As one of the doors is a known, it can no longer figure in the calculations because there is now only a choice of two doors - so 2/3rds is going to be slightly less (the way I am seeing it), and yet the previous probabilities are FULLY carrying over? I can now see the probability is not going to be 50/50 and that switching is going to make sense - but not quite to the degree that is suggested. Not twice as likely. It's the carrying over of a whole third to what then becomes a two door option. I actually have done this problem with cards and a random number generator (set 1-3) and of the many, many times I have tried this I am falling shy of 66% wins. It comes out a little closer to 50% but I can see the results are never quite hacking 66 wins out of 100 tries. I shall STILL need to consider this more because the math on this part is foxing me! Thank you for your very full response/explanation.

You have a 50% chance of winning a delicious goat- no matter what you do. The best strategy would be random.

@Hank254

2 жыл бұрын

No matter how many videos you post that to it is still wrong.

@michaelgarrow3239

2 жыл бұрын

Hank - that is a gratuitous assertion. No! Your wrong!!! What are you 5

@Hank254

2 жыл бұрын

@@michaelgarrow3239 I will be 5 next month but I still wouldn’t go posting the same wrong answer to multiple videos. Do you think your initial pick from the three doors is a 50/50 shot at getting the car?

@michaelgarrow3239

2 жыл бұрын

Hank - I have explained it to you many times. You write pretty well for a four YO…

@max5250

2 жыл бұрын

@@michaelgarrow3239 "You have a 50% chance of winning a delicious goat" If you stay with your initial door, you will have 66.6% chances of winning a goat, and if you switch to the door offered by the host, your chances of picking a goat will drop to 33.3%. Where do you see 50% chances?! In your imagination.

Rubbish. What everyone is ASSUMING is that the SAME person is going to be on the show EVERY DAY but they are NOT. The contestant is only playing this game ONCE. So these repetitive odds everyone is imagining DO NOT EXIST. The prize is in a fixed location. It doesn't matter if there are 3 doors or 3 million doors. No matter how many doors you open or close the location of the prize remains constant. The contestant only has one shot at it. If you don't pick the winning door you loose. If you only play this game ONE TIME there is absolutely no reason to switch doors. Simple as that.

@Stubbari

Жыл бұрын

Let's say you had a severe medical condition. There were 2 medicines for you. Medicine A would cure you with a 10% chance and medicine B would cure you 90% chance. You only have 1 shot at it. Which one would you pick? Surely it wouldn't matter?

@morbideddie

Жыл бұрын

Just because something only happens once that doesn’t mean all outcomes must be equally likely. Are you suggesting that 1/2 of initial games the contestant picks the car initially but that changes to 1/3 in subsequent games?

@Araqius

Жыл 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.

Sorry My Friend but you are incorrect. The fact stated that the "probability" is maintained and used to explain your answer is incorrect. Every time a door is opened the "probability" is changed. No matter how many doors you start with in the problem of which door hides the car comes down to two doors, period, which is then a 50/50 chance. Stay or change it is still 50/50. Expand the problem to five doors and use your method of analysis and switch doors every time allowed. The odds of getting the car will never be greater than even. In fact your odds increase as multiple doors are opened. With five doors your odds increase as fallows: 20%,25%,33%.50%. Change or stay, it ends up being 50%. Probability is a term used to explain the randomness of an action. The hosts action was NOT random, but in fact, already increasing your odds of winning by approx. 17% on the three door problem. 33+17=50 Final guess two equal options. You could even pick two doors at onset and the final decision will be between two doors, a 50 50 shot on which one to keep.

@klaus7443

3 жыл бұрын

You would have been right if the host didn't know where the car was.

@derfunkhaus

2 жыл бұрын

The first door is chosen at a disadvantage. It is selected randomly from a set of 3. We know that the prize is somewhere in that set of 3, i.e. the odds are 1 that the prize is in that set. So the odds that the first door is the prize are 1/3 * 1 = 1/3. (1/3 for selecting 1 door in 3, and 1 for the fact that we know the prize is in fact among those 3 somewhere.) The door that Monty selects to keep closed is chosen under the conditions created by the player's selection. Monty chooses his door at an advantage: he knows where the prize is. The only hinderance to Monty is that he does't get to play first, and by the time Monty is allowed to select his door, the odds that the prize is in what he has to choose from have been reduced from 1 to 2/3. Thus Monty is presented with a set which has 2/3 chance of having the prize, but if it _does_ have the prize then Monty will get it with 100% certainty. It's a two step process and so to calculate the odds we multiply the the odds of the individual steps: 2/3 x 1 = 2/3. Thus Monty has 2/3 chance of having selected the winning door. The player is then compensated for his disadvantage by being offered the chance to swap. If he has the brains to see that this doubles his odds, good for him. If he thinks it makes no difference then tough luck for the player. The fact that we are down to two doors at the end does not make the odds 50-50 at all. An analogous situation would be if you were to select a card randomly from a deck of cards and place it face down without looking at it. Then Monty is allowed to LOOK at the cards and select the highest card left. Then all the other 50 cards are discarded. Who has the higher card? The odds are certainly not 50-50. The odds would be 51/52 in Monty's favor since the only thing that would prevent him from wining would be if you got very lucky and drew the Ace of Spades, the odds of which would be 1/52.

The probability of choosing at the start is 1/3, there are 3 objects, and 3 doors. You are given a choice again between 2 objects and 2 doors, and that is 1/2. Your 1st choice is irrelevant to the 2nd choice. The problem is that you are combining both goat reveals when you pick the car into a single scenario. Monte actually could reveal either goat, given him 2 scenarios. You should be using 1/4 for each goat reveal...not 1/3. When you pick the car, he can reveal 2 different goats. It is 2/4, 1/4, 1/4. When you pick the car...you can switch to both incorrect goats....not just one. "1/3 of the time staying would win, 2/3 of the time switching would win" - Wrong 2/4 (there are 2 scenarios when goats are revealed...not 1) when of the time staying would win and 2/4 (1 for each of the other goat reveals) of the time switching would win.

@max5250

3 ай бұрын

"You should be using 1/4 for each goat reveal..." No he shouldn't, since player can't pick the door with a car twice as likely as the door with a goat.

@Hank254

3 ай бұрын

You can't have a denominator of 4 when there are only 3 doors, therefore it is impossible that it is 50/50.

@Araqius

3 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.

This is just a game show on TV. Don’t over-beautify and package it as great. What ideology ​​does it export through entertainment? It is normal and inevitable for people to pursue maximizing interests!! If you are more calculating than others, you are smarter. !! Utilitarianism! Opportunism! But most people have been manipulated by others. The person in charge of the game is always the winner. What's the difference between predicted and already selected? If you know, that's clearly explained. On the contrary, if you think there is no difference, I can only say that your views and perspectives are very different from mine. The player tells the host his prediction, then removes a door with a sheep, and leaves the rest to the player to choose from, that's it! ! It seems that the host can use many methods and means to manipulate your true choices.

The teacher is 100% wrong, if the contestant died after seeing the goat and a new contestant was picked. And the contestant had no knowledge of what happened before he stepped on stage. Then was was told he could have door #1, or 100$, if he picked door #1 would have a 50 percent chance of winning the car. Yes, the flaw in the teachers logic, is that a new game has started with new odds after the goat was shown. The same logic holds in roulette, the wheel has no memory of previous games. You need to apply the new odds to the Monty Hall problem once the goat is shown.

@Hank254

2 жыл бұрын

What you are describing is not the Monty Hall problem. If you change contestants it is a new problem... the host revealing a goat is part of the same problem.

@RonaldABG

2 жыл бұрын

Ok, what about if I put you this question: "what is my KZread username?", and you have two options: A) Ronald B) Mario Since you can see my username, you know that option A) is correct with probability 100%, so you cannot fail. But if I put the same question to someone random from the street that does not know me, the probabilities that that person picks the correct are 1/2. That's because that person is only 1/2 likely to pick option A), as he/she could also think that B) is the correct, but that does not mean that for you, who already knows the right answer, the probabilities must also be 1/2 for each option. The point is that the chances depend on the information you have, and that's the reason why people who have study more get more correct answers in exams than those who have not. In Monty Hall game, remember that the host must always reveal a door with a goat from the two that you did not pick at first. That means that if you select a goat door he is forced to reveal specifically which has the other goat, but if you pick the car door he can reveal any of the other two, we don't know which in advance, they are equally likely for us, because both would have goats in that case. So, the 1/3 case in which your door has the car actually splits in two sub-cases (two halves of 1/6 each) according to which door he reveals next. For example, if you start choosing door 1, the possible scenarios are: 1) The car is in door 1 (yours) -> 1/3. But it is divided in: 1.1) The host then reveals door 2 -> 1/6 1.2) The host then reveals door 3 -> 1/6 2) The car is in door 2 -> 1/3. Here it is sure that the host will reveal door 3. 3) The car is in door 3 -> 1/3. Here it is sure that the host will reveal door 2. If he then reveals door 2, you could only be in case 1.1) or in case 3), where case 3) is twice as likely as case 1.1). Since the probabilities must always sum 1, applying rule of three you get that case 3) is 2/3 likely at this point, and case 1.1) is 1/3 likely. So you have 2/3 probability to win if you switch. You would only start picking the car door 1 out of 3 times, and since the host can never reveal it, then he is who leaves it in the other closed door 2 out of 3 times. And the contents are not shuflled after the revelation. If you already had a goat in your door, that goat will remain in there. If a new contestant who does not know what was the original door that the first one selected was about to choose, that person would be 1/2 likely to win, but not because each the staying door and the switching one are correct 1/2 of the time, but because that person would select each with 1/2 probability as he/she does not know which is which, so the chances to win average to 1/2: 1/2 * 2/3 + 1/2 * 1/3 = 1/2 * (2/3 + 1/3) = 1/2

@max5250

2 жыл бұрын

"The teacher is 100% wrong" Nope. You are wrong 1000%! " if he picked door #1 would have a 50 percent chance of winning the car." Wrong. Door initially chosen by player always has 1/3 probability, while door offered by the host always has 2/3 probability, even for a new contestant. The only problem is, new contestant doesn't know which door is which, so he can only guess it and randomly pick both of the doors 50% of time, which makes his odds to be (0.5 * 2/3) + (0.5 *1/3) = 1/2 although odds of the doors never changed.

@healthquest4823

Жыл бұрын

Steve you have no idea what you are talking about. The two choices are not independent. There is no new game. The first choice gets locked in at 1/3 because the host can't touch the contestant's door. The host can't reveal a goat from behind it even if he wanted to. The second choice is part of the same game.

@healthquest4823

Жыл бұрын

@@max5250 max5250 is correct. The odds of the revealed door don't split 50:50 to the two other doors because it was not revealed randomly and the contestant's door could not be opened. The choices are dependent not independent.