This Simple Puzzle Tricks Mathematicians -- Monty Hall Problem in 5 Levels 

Yeah, this is what tripped me up when I was initially wrapping my head around it. It can be easy to overlook that the door with the car will never be opened by the host, and so they're forced to help you by always removing the incorrect door in the event that the car is in one of the other 2 doors.

From the first choice the probability you win is obviously 1 out of 3. Once a door is open the odds do change. The game starts over. Which one of these 2 doors have the prize? 2 possibilities only 1 right door. 50/50.

​@@TravisMcGee151 I know it's tough, a lot of people don't get the right answer on this one. I think it's not intuitive because there are so few doors. So let's reframe this scenario. Like he does in the video, but we'll go even bigger. Let's say, no matter how many doors the host has (whether it's 3 like the original problem, or however many more), he's required to open all the remaining doors except your original choice, and one other door. And there must be a goat behind every door he opens, never a car. He knows where the car is, so he has no issue doing this. Now let's picture that instead of three total doors, this place has a million doors in a gargantuan warehouse, and you pick one. Say, Door #1. You'd agree with me that you have a one in a million chance at this point, right? And now, by the rules of the game, he opens 999,998 doors with goats behind them. Only your originally chosen Door #1 and the one other door the host didn't open remain closed. "Stay or switch?" he asks you. So there are two possibilities: either you picked the car door correctly originally, and it didn't matter which door he left unopened because the others were ALL goats, so you should STAY (remember, there was a one in a million chance of this), OR... you originally picked one of the 999,999 goats, so he opened the remaining 999,998 goat doors and left the car door closed in that massive sea of doors, because the rules forced him to and he knows where the car is... and you should SWITCH. When you see the problem this way, it becomes obvious that when he is forced to open everything except your original door and one other and he's only allowed to reveal goats, the part of the probability that wasn't on the door you originally picked (whether that's 2/3 in the original problem, or 999,999/1,000,000 in the latter example) gets congealed behind that remaining door the host leaves unopened. It's definitely not a 50/50. Hope this helped.

@@TravisMcGee151 That is incorrect because the host can reveal either Goat when the Car is picked. If Goat A is revealed then either... 1) Goat B was picked, a 1/3 chance or 2) Car was picked and the host chose to reveal that one, a 1/3x1/2=1/6 chance If Goat B is revealed then either... 1) Goat A was picked, a 1/3 chance, or 2) Car was picked and the host chose to reveal that one, a 1/3x1/2=1/6 chance You are still twice as likely of having the other goat behind your door.

I submit this possibly flawed logic for analysis: What if we work this backwards: You have a valuable object and put it under a cup. I put down a matching empty cup and ask you to close your eyes. I move the cups around and have you pick one… what are the odds you will get the valuable object? 50/50. Now put the object back down under the cup and I mix up the cups again and then take out a 3rd cup and show you it is empty and show you where I place it on the left. I do not move that cup. If you choose that cup, you will ALWAYS see an empty cup. Now… which cup is the valuable object under? What are your odds of choosing your prize correctly? remember since this is backward you have knowledge in a different order … but I submit that your odds are the same 50/50. You did not lose any information when shown the 3rd cup. So therefore when done forward, you will always know that the first pick reveals andempty cup, so the first pick is a false choice…. Only the second choice matters. 50/50.

It took me a while, but it seems that way.

Works much better if you expand the doors to 100 or a million, and you can see that switching is correct.

This is an awesome way to think about it that also makes sense intuitively without resorting to stochastics!

The host must know where everything is, otherwise there is no advantage in switching if he reveals a goat. In your explanation the contestant couldn't care less if the host knows where anything is or not.

Most people miss the host always opens the dud door.

Thanks so much, I'm glad you like them!

@@DrSeanGroathouse I however think that you are a lowlife academic populist who repeated mathematical nonsense. You are a purveyor of mathematical garbage. You can't count to 2 correctly maybe because your media content is more valuable to repeat a confirmation bias rather then to tell the truth. Either that of you really can't count to 2. The contestant could only pick from 2 doors, not 3. Are you dense or are you malicious. That choice is yours.

I'm glad you liked it!

@@DrSeanGroathouse I however think that you are a lowlife academic populist who repeated mathematical nonsense. You are a purveyor of mathematical garbage. You can't count to 2 correctly maybe because your media content is more valuable to repeat a confirmation bias rather then to tell the truth. Either that of you really can't count to 2. The contestant could only pick from 2 doors, not 3. Are you dense or are you malicious. That choice is yours.

Thanks for the idea, I think that would be a great video! I added it to my list

@@DrSeanGroathouse I however think that you are a lowlife academic populist who repeated mathematical nonsense. You are a purveyor of mathematical garbage. You can't count to 2 correctly maybe because your media content is more valuable to repeat a confirmation bias rather then to tell the truth. Either that of you really can't count to 2. The contestant could only pick from 2 doors, not 3. Are you dense or are you malicious. That choice is yours.

Yeah, this guy is awesome and straight to the point. Been here since day 1, under 50 subs

That's exactly why this problem got so popular; The host is forced to open the door with the goat in.

You are wrong. If there was 50% chance to win by switching, there should also be 50% chance to win by staying, as the car is 100% likely to be behind any of those two doors, so the sum of their chances must add up 100%. In fact, if staying is only correct 1/3 of the time, then switching will be correct 2/3 of the time (66.66%). You are confused because you see two doors and you want to include the 1/2 somewhere, but the point is that the important thing is not how many options there are, but how frequently each will result being correct.

There are clearly 2 camps in this debate. The populist conception that believe the contestant had 3 doors to choose from and the minority opinion that believe the contestant only had 2 doors to choose from. I'm commenting to your post because you seemed very timid to clarify who you believed was eligible to be shamed. I'm in the minority who believes that the contestant only had 2 doors to choose from and therefore vos Savant was wrong. I'm a 50/50 advocate. I feel no shame. But let those who ate that toxic vos Savant's cake get hammered for being so stupid. Dr Sean has a higher responsibility and should be more subject to being a scientist, not a populist.

​@@williams.benjaminiii9043it depends on whether it's clear to the contestant that the host must and will always act the same way. If So, then vos Savant is indeed correct. But her assumption that this should be the case is wrong in my opinion. If we drop this assumption we're back at 50:50.

The best counter argument is to reduce it to 2 doors. 1 goat, one car. The contestant is told to choose a door and no matter which of the 2 doors they pick, rather than open a door, they are asked to choose again. Did the odds of winning change by asking twice?

@@MurderMostFowlThat case has no interaction of someone that 'knows' the answer and removing answers.

@@MurderMostFowl You're completely lost. There is a 1/3 chance the host chose which goat to reveal and a 2/3 chance he revealed the only one he had.

@@MurderMostFowl 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? M: The best counter argument is to increase it to 2 doors. 1 goat, one car. Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Switching means the opposite. It's just basic math/logic kids understand. Sadly, it's far too hard for idiots.

Glad you like it! I added your idea to my list for future videos

@@DrSeanGroathouse I however think that you are a lowlife academic populist who repeated mathematical nonsense. You are a purveyor of mathematical garbage. You can't count to 2 correctly maybe because your media content is more valuable to repeat a confirmation bias rather then to tell the truth. Either that of you really can't count to 2. The contestant could only pick from 2 doors, not 3. Are you dense or are you malicious. That choice is yours.

It does. If one of the two remaining doors (not originally picked by a contestant) is _randomly_ revealed (both doors equally likely to be revealed), and it _just so happens_ to be a goat, then the probability of winning by switching is 1/2.

I think the problem with the sleeping beauty problem is the question being ambiguous of what it's even asking you to find the probability of.

Just looked up the sleeping beauty paradox because I was curious upon reading your comment. It looks like the accepted solution is that, even though the result of the coin toss is 50/50, you are twice as likely to be in the state of being woken up when a tails is thrown; thus upon being woken up, it is 2/3 likely that it was a tails. Weird problem though, it is curious to think about ways of adapting the problem and see how it lies with your intuition. As an example, suppose it was determined by 3 coins tosses, and only if it came up TTT would you be woken 100 times. Where as were it to land any other combination, you would be woken only once. By the logic presented, it would be far more likely upon being woken up, that it had landed TTT, but that's certainly not wholly intuitive.

@@nickronca1562From what I read, the question seemed fairly clear. What are the odds that the coin landed heads or tails, upon being woken up?

@@CarlosSamuel-ms9ee But is it "what are the odds it landed heads or tails?" or is it "what are the odds we are waking you up on a heads or waking you up on a tails?"

@@nickronca1562If you are woken up on a head or a tails, then it must have landed as such no? The coin is only tossed once per sleep.

If you pick a goat, you WILL switch to a car. Because the host specifically removes a goat. Hence, P(pick a goat) = P(switch and win) = 2/3

Maybe it helps you if you flip the logic. You pick one of the doors, the chance that you picked incorrectly is 2/3. What happens in this scenario? The host still has to open a door and will never open the one with the prize so they have to open the other wrong door, only leaving the winning door unopened. So 2/3rd of the time you were initially wrong which means the remaining closed door at the end has to be the prize. 1/3rd of the time you were already correct and the remaining door will be the goat which means in two out of three scenarios, switching at the end gets you to the correct door

It works because the host already knows what's behind each door - the cup being removed is not something that's happening randomly. There is no possibility of either your cup or the cup that has water being "blown away", and a cup being blown away is absolutely guaranteed to happen. The host does not really have any choice of what to do (the only thing the host can actually choose is.. which door to reveal if both of the other doors are empty, which isn't really a meaningful choice). The kind of example you're giving is treating it as though it's something that's happening randomly - and if it did happen randomly, then it's true that it wouldn't make one of them any better than the other.. but in the monty hall problem, it is *not* something that's happening randomly. If you want to take something similar to your example, let's say there are 3 cups, and it's a really windy day and instead *every* cup that doesn't have the water (other than the first one you picked) would get blown away. In that case, it's very obvious that if a cup isn't being blown away then it must have water in it, and if you hadn't picked the correct cup initially then you'll always be left with 1 cup that doesn't get blown away. and you can obviously tell which cup has the water in it guaranteed based on that, and you're guaranteed to pick the correct cup by switching if a cup still exists. This is obviously a bit different because in this case there's a possibility of 2 cups being blown away if you had picked the correct cup initially which would never happen in the monty hall problem.. but remember - there's only a 1/3 chance of that possibility happening if you had guessed the correct cup initially, the other 2/3 of the time it's doing the exact same thing as the monty hall problem.. which means that if you treated the monty hall problem as though it were the same as that, you'd be correct 2/3 of the time. Given that you're guaranteed to guess correctly if all the empty cups get blown away, and there's a 2/3 chance of the monty hall problem doing the same thing as blowing all the empty cups away, then you should have at least a 2/3 chance of winning the monty hall problem by using the same strategy you would as though all of the empty cups get blown away (well, exactly 2/3 when you do more math on it).

That is a different situation, because the wind could also blow away the cup you picked, which would make it clear you picked the wrong cup. Or it could blow away both empty cups. A strong wind could even blow away the cup with the water. In the Monty Hall problem, the host always opens one door with a goat, and never opens the door that you have chosen.

What you're failing or seem to be failing to understand is that the host will ALWAYS pick a door with a goat. Think about it like this if you really really have a hard time understanding. You pick a door with the intention of sticking with that door. The ONLY thing that changes of whether it's revealed you won right away or whether a goat behind one of the other doors is revealed first, then it's revealed that you won is the order at which the doors are opened in. There is a ⅓ chance of picking the correct door. So there is a ⅓ chance of winning. Being purposely shown a goat behind one of the other doors first and then being revealed whether or not you won doesn't suddenly make you go from winning ⅓ of the time to winning ½ the time. Because it doesn't matter what order the doors are revealed in, and by saying the probability is ½, you're saying the equivalent of "if the doors are revealed in a slightly different order, I go from winning ⅓ of the time to winning ½ the time!", even though that's clearly stupid.

But by that same logic, if the revealed door is always the same content, then was there ever really a 3rd door? ( I’m not being a smart alec here ) The host could have just as easily had 2 doors and told the contestant to “guess again”. How is that not 50/50?

@@MurderMostFowl There were and are 3 doors. 3 possibilities.

@@MurderMostFowl 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? M: The host could have just as easily had 2 doors and told the contestant to “guess again”. M: How is that not 50/50?

You are right

There is no advantage in switching if the host revealed a goat without knowing what is behind the doors.

@@klaus7443 I thought so too but running the simulation - there is. Yeah these things really are strange. The reason is quite simple: if the host chooses at random and reveals a goat it is the same as him choosing deliberately - so switching helps. And if he would hit the car - then switching or not has no impact anymore. draw the probability-tree your self and check it.

@@ABaumstumpf You're wrong. These are the four possibilities if he reveals a goat without knowing where anything is and they all have the same probability. Pick Car, host shows Goat A Pick Car, host shows Goat B Pick Goat A, host shows Goat B Pick Goat B, host shows Goat A Probability of winning by staying and switching is equal when a goat is revealed randomly.

@@ABaumstumpf "draw the probability-tree your self and check it." You should have.

@@klaus7443 i have done the maths - you didn't. "These are the four possibilities if he reveals a goat " So if you reduce it to just a subset of all possibilities... so you are still wrong. You really should map out or draw down the probability graph.

The reason that it happens over many cases, is because those are the odds in an individual case. If the odds of each case are independent of each other, then they mean the same thing.

It does matter though, we're trying to find the best possible strategy such that if you were in this situation you maximise your chances of winning. Ok you only get 1 go so the long term effect averaging to 2/3 doesn't happen, but that doesn't make it 50/50 now. Unless you're gonna extend this logic and claim that if you play the lottery 1 time ever it's a 50/50 chance of winning

@@wilflava Yes, but we are talking about a strategy going into the game for players, and in that case I think switching is smart. But the question is after you are already in the game. You already picked the door, the host already opened a door. All those possibilities you thought about before the game simply don't exist now. It is either 100% you picked the car or 0% you picked the car. The only possibilities now are do you stay or switch. The problem is about understanding how new information changes probability. It removes certain possibilities. But in the game after the doors are set and opened all the possibilities but one are already removed. You have limited knowledge and don't know this, but the reality is from the host point of view who knows where the car is. The solution given is from an observer standpoint but the question is for the player. Basically the strategy isn't for you to win, but for the group of players to win. As you are one of the group it might make sense for you to play that way but they are telling you this not because of you the individual but you a player. In life you should always consider this distinction.

_"if you switch and are wrong you will hate yourself for more than if you stayed and are wrong"_ - definitely not. If you switch and are wrong, you have the comfort that you at least maximised your chances. If you stayed and are wrong, you have the additional burden of the knowledge that you made a dumb choice.

@@christopherkopperman8108 right I see what you mean now. Yeah in hindsight everything happened with 100% probability, the only reason we say a dice roll is 1/6 is due to our own ignorance and inability to solve a differential equation before the dice has shown it’s face etc. But this is why Probability is defined such that you are unaware of what will happen, and that a probability is defined by its long-term effect. It’s a fun philosophical exercise to think about, but this is a Maths question at the end of the day

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

@@Araqius How can the contestant win only 1 out of 3 games if the contestant is only able to play 2 games? How many games can the contestant play? By my count it's 2 games, not 3. The wrong method of this simple math problem is not idiocy. It's confirmation bias. It's regular human psychology to go with the flow even if that means believing that the contestant had 3 games to play when there were only 2 available. This widespread theory of how math works is simply one of the finest examples of confirmation bias I've witnessed. It snookered a RU-vidr Doctor of science as well as many other's who repeat it with a passion seemingly empowered because it's wrong and will never make any sense. But it's all just for fun. Nobody's going to get hurt for this kind of debate. But is it a harbinger to how people vote or manage their finances, or are fine with innocent people being convicted? Probably. Pier pressure can be a force to line up to aqueous like Dr Sean did even though it will never make any sense.

@@williams.benjaminiii9043 "How can the contestant win only 1 out of 3 games if the contestant is only able to play 2 games?" lmfao No matter how many game the player play, his winning chance is 33% if he stay with his first pick.

@@williams.benjaminiii9043 Let's say you buy a lottery ticket (Assume there are 1,000,000 numbers with 1 jackpot.). After the result is out (you don't know the result yet.), your friend (who know the winning number) say your number and another number, and then tell you that one of them is the jackpot (He basically remove 999,998 loser numbers.). What is your winning chance? William: My initial odds is 0.0001%. William: When my friend remove 999,998 loser numbers, my odds becomes 1/2. William: This is proven by the fact that there is only 2 numbers left, not 1,000,000. William: My lottery winning chance is 50%. I am a genius. Hoooraaay!!!

@@Araqius This is where I disagree with the whole 1/3 premise. Since the player is only able to pick from 2 doors and one of those 2 doors is guaranteed to win a car, the actual probability of winning a car is 50%, not 33%. The idea of the player being able to pick from 3 doors is false and therefore so is the idea that the player ever had a 1/3 chance. It's 50/50 so long as there is an understanding that the player was only capable of choosing from 2 doors which is true. This idea that the player was capable of choosing between 3 doors is purely a prejudice...

I mean, if you have more doors there is an increased chance to win by switching. The more doors, the more chance you are wrong initially. In the scenario you are wrong initially the host will eliminate EVERY OTHER incorrect option leaving only your initial wrong choice and the correct choice. Increasing the number of doors increase the strength of switching. Also honesty if you refuse to believe the math, just simulate it. Whether that be via programming it or just getting three (or more if you want to see the effect grow stronger) cards and running it yourself recording the results. One thing you can also do is since your initial choice has no information and thus completely random, you can just always choose say the left door initially. This can help visualize things in your head. Now play out the game with the car in the left door, middle door, and right door. Which scenarios win by switching and which win by swapping?

"lets reexamine the odds using different numbers of doors with the rule that the contestant will only have 2 doors to choose from." lmfao If the player pick a goat door, what will the host do? Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Switching means the opposite. It's just basic math/logic kids understand. Sadly, it's far too hard for idiots.

@@Araqius As the predicate the host will always expose a goat door regardless of the contestant's initial pick. If the player picked a goat door then Monty Hall will expose the other goat door. The crux of this widespread perversion of basic math is trying to make the initial odds + the alternative choice = 1. The initial odds of winning were 1/3, but the host's action of exposing one of the goat doors improved the contestants initial choice from 33% to 50%. 50% + 50% = 1. Mystery solved.

@@williams.benjaminiii9043 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? William: The initial odds of winning were 1/1, but the host's action of adding one goat door decreased the contestants initial choice from 100% to 50%. 50% + 50% = 1.

@@Araqius That's actually a really neat way of showing things, by adding a door instead of removing one. Not something that would click for everyone, but I feel it would help some get the idea that 2 doors isn't always 50/50.

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

Everything in your comment is incorrect. The host reveals the other goat when one is picked, and reveals either goat when the car is picked.

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