Math that will make you think. Twice....maybe 

Can you make a video on Construction Engineering and management please..!

I can see now why 90% of the people get that one wrong.

hey you made a mistake, with regards to the cards and flipping them over your question was "which cards must be flipped over to determine the truth" this question wanted to know if the statement "if there is A on one side of the card there is definitely 5 on the other side" it didnt say anything about if there is 5 on one side there must be an A .... it didnt say anything about B or 6 either ..... all that info is irrelevant ... to determine if behind A is a 5 ... not behind 5 is an A flipping card A determines the truth .... if 5 is on the other side statement is correct

@Jack Black Logic isn't your thing.

@@jackblack5082 lets say you flip the 6 and behind that there is an A. therefore the earlier statement is false. Do you see why you have to flip the 6 now?

Yeah but looking at it from the alcohol example the letters stands for a "person" And the numbers stands for a "drink" So it MUST be on one side of each card a letter and on the other side is a number. To be fair as you said he didn't mention this but when you compare it to the alcohol example then you have to presume this, right? But he didn't give us both of them at the same time so you are right.

Yeah you’re right, totally forgot to explicitly mention that (just pinned a comment about the mistake).

ur just bad

@@anno96 I know this comment is 9 months old, but he's right, in the a b 5 6 example it's important to note that the wording is different to the beer example. In the a b 5 6 example, they say that an A will have a 5 behind it, whereas in the alcohol example it essentially says A (the underage person) can't have 6 (the beer) behind it meaning that A can technically have either B or 5 behind it.

@@anno96 It really isn't the same thing as the alcohol problem. He never specified that each card had a letter on one side and a number on the other.

Aaand you don't stop running and you don't stop running and you don't stop running .... (all stars version)

What's running?

Mathematicians aren't that fit

True, but if you have a heavy hand luggage (which I usually have) then you might want to stop and rest somewhere every once in a while, in order to have the stamina to do sprint dashes in between each rest. Then the optimum place to rest is on a walkway.

@@jojoyear e

Thanks pal. Logical proof without the need of a mathematical one

F54DF4 A044 o

I guess.... that makes sense. In essence, the relative distance you are sprinting on the walkway is 'less' than the distance you are sprinting when you aren't on the walkway, at least when the distance on the walkway and the distance you are traveling off of the walkway are the same. Makes sense to me.

I like this logical explanation. I had been using a less rigorous intuitive explanation: You get more benefit from speeding up slow activities than from speeding up fast ones.

This explanation is so much easier to understand

Well maybe, but they just decreased the chance of the show having to give away a car so 🤷

@@aidan8858 Not really, because most contestants are stupid thinking that when the host opens a door with a goat their probability of getting a car increases to 1/2 so they won't think it's worth switching. But the drunken member of audience has actually increased that stupid contestant's chance...

@@heimdall1973 But logically if they all think they had a half chance already, then half will randomly choose to stay and half will randomly choose to switch, since they have an equal chance of being the correct strategy by their logic? The half that chose to stay will then see their odds improve by this intervention, by your own arguement, but the half that stick will see a proportional decrease in their odds, and overall the show will see no increase in those in the 'stupid' category winning.

@@MrDannyDetail y'all seem to be forgetting about the 6 people who just had the friend open the door with the car behind it... rip those people, rip the friend, and rip the guards' jobs.

The guard's job is probably fine. If the didn't get fired after 17 times, it seems pretty safe to assume they don't care.

NPC Grian ! Looking for this

Or was it?

Or was it?

Yeah, that's what I was thinking, too. (By the way, if I had pancakes instead of eggs, I would have had 10 or so... But as I was typing the last sentence, I just realised I don't know what sort of pancakes the author had in mind. If you say "pancakes" to me, I think of the sort that are also called crepes. But in the USA pancakes are thick.) But then you *can* make pancakes without eggs, they're just not as nice... Maybe the problem should be worded as "4 eggs or 2 rashers of bacon", then it would be clearer.

There are eggs in pasta

Thank you for that knowledge cat kook

No shit Sherlock

@@catkook543 not in my pasta!!

Agreed

Why y’all correcting him? That’s highkey funny😂

If those pancakes are made from no-egg, I'm not having any.

Sean Bean: what if I told you they might not?

Flippy Studios who doesnt love goats?

Wait so a goat in the car? Is that the joke

Underrated comment.

@@michaeladdis3323 goats are good bois and everyone wants goats

@@cube4923 “…are goats, and one of which has a car…” one of the goats has a car.

Intaanetto he got the math wrong. It shows them running the same distance even though the person on the left is on the walkway. So technically the person on the left should moved farther using the same energy thus making it there faster than the person on the right

@@ThatOneDude80085 you are wrong. The math checks out.

@@ThatOneDude80085 the math is correct, the animation might not be.

If you switch you will have 50% chance of winning (depending on which door you switch to).

No, but the rules dictate that Monty can't reveal the car

Nope. The entire logic relies on the host knowing which door is which, and not revealing the car door. If it was random which door was eliminated, then it's just randomness upon randomness. You would have a 33% chance and switching wouldn't change that. If we take your question literally however, switching wouldn't do anything, as your friend has already revealed the door with a car and you have a 0% chance to get it

@@Mr_Doon well in here he expected that friend opening door will in 50% result in goat which is wrong(when he cut down group of 6 in half)...its 66% which will leave 8 for win, 6 for lose and 4 for repeat so 57%?....no, all of this is just stupid ..it would be as if question of winning probability for switching door from the past was asked once the final door with lets say goat was openned..it should not change 66% to 0% just cuz u now know that u didnt win...it doesnt matter..switching was still right way to go...the probability of winning with switching is decided the moment door is picked not the moment friend opens random one....so switch

a worrying number of people are missing the joke here

good catch, the information that all cards contain one letter and one number on either side is supposed to be given at the start of the problem.

No thats not an intuitive explanation

@@onethegogd5783 You have a 2/3 chance of getting a Goat when choosing one of the doors. therefore the chance of your door having a car is 1/3, so the chance of the second door having a car is 2/3

@@tjgdddfcn that's not intuitive is it

@@onethegogd5783 The intuitive part would be that since you have a 1/3 chance that the door you picked is not a car, the chances are that the car is in one of the two doors you did not pick. It's intuitive to realize you have likely picked the wrong door, much like when playing the lottery it is intuitive to realize you've likely picked the wrong numbers.

@@jaypee9575i would not come to this conclusion during a game show because i would look at the choice in isolation.

Thanks! That finally helped me understand this!

Honestly my logic for understanding it was you CAN choose the goat if you wanted so you know one option and you can choose the other since you dont want a goat but they arent preventing you from choosing the goat they just cut to the chase because they know you would prefer the sports car instead

math

Tie your shoe? Wait till your on the plane.

@@bulldozer8950 trying to tie your shoe? flick your feet in a certain way like some harry potter shit while running and dont stop running

same

What did it to me the first time I saw it was giving an example of 100 doors and a single car. If the host always opens the door with the goat (except conestant's door), obviously surely I wasn't that lucky to pick a car first try, it must be more beneficial to me to open some other door the host didn't open. Obviously, the probability difference in the example with 100 doors is marginal, but it helped me see the concept for the 3 door example where the difference is huge.

how is the friend pulling the door 50/50? he said 6 and 6 in the video but isn’t he forgetting the 6 that also lost immediately after the car was revealed so it would still be 2/3?

@@robertbruce5335 no because all times the drunk friend reveals the car don't count. The question is the drunk friend at random revealing a hidden door given that they don't reveal the car, not "the car being revealed is an automatic lose".

@@robertbruce5335 Im not very sure but honestly theortically I agree that its a 50/50 chance that you win so changing wont affect your outcome BUT if your friend reveals a goat should it not be treated the same as the host revealing a goat as one goat is removed your initial probablity of picking the other goat was still 2/3 and removing a goat would essentially be making the odds the same but it would only apply when the friend reveals a goat I dont understand why we would factor in the people who would have chosen the correct door in that scenario as we already know one of the doors have a goat so the probabilties should change to become the same(clarification below) as in that case it essentially doesnt really matter about intent anymore as the outcome of the goat being revealed has happened. (Clarification: I mean that the probability becomes same as if the host had revealed the goat)

@@Diamonse Late response here, but there is a difference between the host revealing the goat and your friend doing it. The host knows the locations and is forced by the rules to reveal a goat from the two doors you did not pick. That means that when your option has a goat, he is restricted to reveal one specific door: which has the other goat. Instead, when your option has the car, he 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. For example, if you start picking door 1 and he open door 2, only two possibilities remain: 1) Door 1 happened to contain the car, and the host decided this time to show door 2 instead of door 3. Two conditions must have been fulfilled: 1/3 x 1/2 = 1/6 2) Door 3 happened to contain the car, which forced the host to show specifically door 2: 1/3 x 1 = 1/3 So, he was more likely to open door 2 in case the correct were #3 than if the correct were #1 (your choice), and that's why it is better to switch. As 1) and 2) are the only possibilities now, their probabilities must sum 1, so applying rule of three you get that case 1) is 1/3 likely at this point, and case 2) is 2/3 likely. Instead, your drunk friend would be equally likely to have chosen door 2 if the car was in door 1 as if it was in door 3, as his choice didn't depend on the car location, so there is no reason to think that seeing a goat in #2 makes door 3 more likely than door 1.

I would like to know too since I feel it wasn't explained very well here, if anyone's big brain let me know.

I would, too, but I doubt he'd interact or engage with a comment that isn't just praising him about how right he is all the time, lol.

Yes I agree, his explanation didn't match his original question at all.

He asked one question, and then answered a completely different question.

@@christianlapp8306 "What fraction of the population is expected to be girls?" For me this is a different question than the question you answered: "What is the expected fraction of the population to be girls?" The difference is, do we look at the expected fraction, or do we look at the expected population?

It's not about intent, it's about the cases when the door opener reveals a car, and then these cases being discarded.

It’s not about intent; it’s that if he opens doors randomly, he’ll sometimes reveal the car by accident, and those situations have to be discarded, screwing up the ratios.

+1 for the first part intuition -switching guarantees swapping the item in the original problem-. But, if your drunken friend opened a random door and it was a goat, you won't benefit from the change, because, although you are still guaranteed to swap your item, the probability of getting a car is still %50. That's because the probability of ( you picking a goat initialy (2/3) AND your friend opening a goat door (1/2) ) is 2/3 * 1/2 = 1/3. And the probability of you picking a car initialy (1/3) AND your friend opening a goat door (1/1) ) is 1/3 too. So whether you changed the door or not, the probabilty of getting a car is the same as getting a goat.

The difference from the original problem is that the host %100 will open a goat door, where as your friend will do that in %50 of the cases if you have already picked a goat door initialy

@@nosignal5804 But surely, given he did reveal a goat, we can simplify it down? 2/3 of the time I chose a goat, and in that scenario my friend having by chance revealed a goat, would guarantee I had to switch to the only unopened door, just as I would have done anyway under the original problem if Monty had revealed one of the two goats. The other 1/3 of the time I had picked the car and it was already guaranteed that I had to not switch, so my friend revealing a goat has not changed this. 2/3 of the time I need to switch either way, and 1/3 of the time I need to not switch either way. Either way my friend revealing a goat doesn't change my strategy, so combining the two with with each other, and especially with an alternative group of scenarios (my friend revealing the car) which we already 100% know didn't happen, doesn't seem to make sense.

I've just figured out a better explanation for why your odds don't change, yet why the video argues that they do. Say Monty instead removes an option by simply switching the spotlight off from that door, so that only the two doors Monty leaves lit are still in the game. Our stage indvader also uses the same method of switching out a light, not opening the door. Monty switches the door off with the knowledge it doesn't contain a car, and your original probability stays where it is at 1/3 chance you picked the car already, so the other door now has a 2/3 chance of having the car behind it. However say the stage invading friend switches out a spotlight without knowing what was behind it, well now all three doors have a 1/3 chance still, so the two doors remaining in the game do have an equal chance to each other, but only because the remaining 1/3 still exists, but outside of the two doors that are still in the game. But if you have the knowledge that your friend eliminated a goat (say the door was actually opened rather than unlit, or else Monty confirms that the friend happened to eliminate the door Monty himself was going to elminate anyway), then you have the same knowledge and position as if Monty had done it, and thus the same odds.

@@MrDannyDetail I am one year late, but I want to defend No Signal’s argument: if a person that does not know the positions randomly reveals a door from the two that you did not pick and just by chance it results to have a goat, then you don’t have advantage by switching. Firstly, suppose you played 900 times. Since at first you are 1/3 likely to hit the door that hides each content, in about 300 you would pick which hides goat 1, in 300 which hides goat 2, and in 300 which hides the car. In total, about 600 times a goat and 300 times the car. Your door The other two doors ================================= 1) In 300 games ---> car two goats 2) In 600 games ---> goat a car and a goat If Monty knows the positions and always reveals a goat from the two doors you did not pick, in the 600 games that they had a goat and the car, after the revelation the remaining closed one from them must be which has the car. So, if you always switch, you win in those 600 games, which are 2/3 of the total 900. (And of course you lose in the other 300). Instead, if your friend is who randomly makes the revelation, in the 600 cases that the other two doors have a car and a goat, your friend has 1/2 probability to reveal each, so in about 300 of those 600 he would reveal the car, and in about 300 would reveal the goat. In total, this is what would occur after the random revelation: Your door Revealed door The other remaining closed door ====================================================== 1) In 300 games ---> car goat goat 2) In 300 games ---> goat goat car 3) In 300 games ---> goat car goat So, if a goat results to be revealed, you know that you could only be in case 1) or in case 2), which is a subset of 600 games. You win by staying in 300 of them (case 1), and by switching also in 300 (case 2), so each strategy wins in 1/2 of this subset. In fact, they are each 1/3 of the total trials, but represent 1/2 of the subset in which a goat is revealed. Moreover, if Monty later says that he would have revealed the same door that your friend revealed, as you similarly suggested in your example, the probabilities are not 1/3 vs 2/3 again; they are still 1/2 vs 1/2, because we know that Monty is not able to say this in the 2/3 times that you fail first, but only in half of them. That is, from the 600 cases in which you pick a goat (2/3 of 900), it won't occur in all of them that your friend reveals a goat and then Monty says that his revelation would have been the same, but only in 300. So, the information is not equal as in original Monty Hall problem.

Yup. You spend less time overall moving at a speed of 2m/s and 3m/s by 4.23s. Although you cover 5m less. By average you move 0.433m/s faster for the 100m while sprinting before the walkway. I'm trying to find exactly where that number comes from. If that number can be solved then we find exactly where it is more inefficient/efficient. But the logic I'm content with is that it's more efficient to speed up the slower activity than the faster one which was stated above somewhere.

He asked a question about the percentage of girls in a single village. He then gave the answer to a completely different question about the average percentage of girls in different villages.

Thats not the point. The one on the walkway still win anyways

@@prumchhangsreng979 well, they still didn't mention that they stop walking when they get on the walk away

@@catkook543 the video said stop so u 1digit iq brain understand that one is ahead of other. Nothing to do with race anymore. Its just mean pause for u viewer to inspect. Its not part of the race. Its hard to explain when its so fking simple omg

@@prumchhangsreng979 no need to insult me personally, that'll get us no where I'm saying that they didn't mention the person who stopped before getting on the walk way stops again on the walk way

@@catkook543 hey, kinda late but i got your point. i guess you based you argument on animation, thats why you're confused. let's assume your speed as 2m/s, and walkways as 1m/s and tying shoes takes you 5s. when you stop to tie them before walkway and your twin immediately on walkway, in these 5 s you moved 0 m, but your twin moved 5 m. after these 5 s both of you start walking again but he has 5 m advantage.

Ya, I also freaked out

No, who reveals the car and with which intention determines how often it will occur, and in consequence how often of those times you can win by switching or by staying. Take for example a variation that some people call Monty "Hell" problem. In there, the host knows the positions and only reveals a goat and gives the opportunity to switch if the contestant selected the car first, because his intention is that he switches and so does not get the car. Instead, if the player selected a goat, the host purposely reveals the car forcing the game to end. It implies that despite you start choosing a wrong door 2/3 of the time, under these rules it is impossible to win by switching. Once a goat is revealed you know you can only be inside the 1/3 in which you start picking the car door, so 100% chance to win by staying. Note that it is completely different to the normal Monty Hall problem, in which you can win by switching everytime that you start choosing a wrong door -> 2/3. In the same way, if the friend randomly reveals a door, we know that not always that you start choosing a goat (2/3) he will manage to reveal the other goat, only in half of those cases (1/3) because in the other half he will reveal the car. So, once a goat is revealed you must restrict yourself to the subset in which that occurs, not to assume that the revelation of the goat would occur in all those 2/3. If you like a more evident example, suppose you are in front of 30 persons, where 20 of them are engineers and 10 of them are doctors, but you don't know the respective profession of none of those people. You have a list of their names, so you call one by his/her name with a speaker and that person approaches you. We both agree that it is 2/3 likely that the person is an engineer, right? But what about if you know that all of the doctors are using blue jacket, but only half of the engineers are using blue jacket (the other 10 are using white jacket)? In total we have: 10 doctors use blue jacket --> 1/3 of the total people. 10 engineers use blue jacket --> 1/3 of the total people. 10 engineers use white jacket --> 1/3 of the total people. So, if you see that the person that approaches you is using blue jacket, would you still say that he/she is 2/3 likely to be an engineer? Note that in total there are only 20 people using blue jacket, from which 10 are doctors and 10 are engineers (50% chance for each). As you see, once you see the blue jacket, what you have to do is to restrict your sample space to the only 20 people that use blue jacket, not count as if all the 30 people used that same colour. This is the same thing that differentiates normal Monty Hall problem to the variation with the friend. In normal Monty Hall problem you will always see a goat revealed, so it is like if all the 30 persons used blue jacket. With the variation of the friend only in half of the cases you started picking a goat door you will see a goat revealed, in the same way that only half of the engineers use blue jacket.

Well, it wouldn’t make for a very good game show if the host instantly revealed which door leads to the prize, would it?

It would remain the same since you would want to avoid going too close to the speed of light while at the same time trying to maintain the highest average speed

The only knowledge he has that you gain, is that there is a goat behind the door he opens. You gain that knowledge no matter if its your friend who opens a goat door og the host who does it.

Marcus Mikkelsen it is not YOUR knowledge that changes the problem. It is the knowledge that the host has. The host KNOWS where the car is and so will always open a door with a goat. Your friend doesn’t know where the car is so only opened the goat by chance.

Oliver Foggin yes, that is true!

_"But 3.3k of the games you just lost before you even got to make a decision."_ No you don't.

@@Stubbari yes you do as 1/3 of the times your friend will open the car door, leaving you with no chance of winning

@@orbmac No he doesn't.

@@Stubbari read up on it. Your friend opens a random door which is not your picked door, and it can be the car door they open. It's called Monty fall.

@@orbmac Ah, my bad. I thought you were talking about the Monty Hall problem.

Yeah. I thought of that insantly

If you decide to change the door in the monty hall problem, the only possible way you can end up worse is if you've chosen the car in the first place, so it's always 1/3. Accounting for the times when the drunk person opens the car door or opens your door just makes no sense because the game wouldn't continue from there so that is an impossible situation to be in. The only way the game can continue is if the drunk person reveals a goat AND it's not the door you've chosen, so it's exactly the same as the original version.

@@venenifer3569 It is not the same as in the original version. At first you are 2/3 likely to select a goat door, that is true. But if your friend reveals a door by random, does he reveal a goat in all those 2/3 cases in order that you can win by switching in all of them? The answer is no; since he is doing it by random, he will only reveal a goat in half of those 2/3, which means that you only win by switching in half of the total amount of times in which you picked a goat -> 1/3. On the other hand, when you have picked the car (1/3) he will always reveal a goat, so you can still win by staying in those 1/3. This is better seen supposing you played 900 times. In the first selection you would pick about 600 times a goat and 300 times the car. Your door The other two doors ===================================== 1) In 300 games ---> car two goats 2) In 600 games ---> goat a car and a goat If your friend is who randomly makes the revelation, in the 600 cases that the other two doors have a car and a goat, your friend has 1/2 probability to reveal each, so in about 300 of those 600 he would reveal the car, and in about 300 would reveal the goat. In total, this is what would occur after the random revelation: Your door Revealed door The other remaining closed door ======================================================== 1) In 300 games ---> car goat goat 2) In 300 games ---> goat goat car 3) In 300 games ---> goat car goat So, if you are in the case in which a goat results to be revealed, you know that you could only be in case 1) or in case 2), which is a subset of 600 games. You win by staying in 300 of them (case 1), and by switching also in 300 (case 2), so each strategy wins in 1/2 of this subset. In fact, they are each 1/3 of the total trials, but represent 1/2 of the subset of 600 games in which a goat is revealed. Moreover, this is the reason why some people wrongly get 50% chance results when trying to simulate the Monty Hall problem. They think that they can emulate Monty's knowledge by randomly revealing one of the other options and discarding and repeating the game when the prize option is revealed. This increases the probabilities of staying because if they got the correct at first, the game necessarily continues as a goat will be revealed. Instead, if they selected a bad option, there is still the chance that the car is revealed and so the game is repeated, giving them a new opportunity to get the car in the first selection.

The travelator solution is wrong! I am assuming you continue to walk after stepping onto the travelator. Consider distance lost per second taken to tie your lace. Before travelator OR on travelator, you will lose distance = walking speed for each second taken to tie your laces. So if you are walking at 2m/sec and it takes you 1 sec to tie your laces, then you will lose 2m in scenario A. And 2m in scenario B also. Assume travelator speed is 5m/sec. Then, if you don't tie your laces, you will travel 5m + 2m = 7m in 1 sec. If you DO tie your laces, you will travel 5m in 1 sec. The distance lost is still 7-5=2m. This is the same as in scenario A where you tie your laces before tge travelator. Hence, no difference. I am assuming you are continuing to walk even on the travelator.

Yeah, that was explained in the pinned comment.

@@aj_style1745 thanks

He alrady apologised for that

@@clemenss.8641 Then add a caption in the video? Instead letting everyone baffled?

I like this method, thanks

You should switch if p > 0. If p=0 it makes no difference which door you choose.

Yeah, I’m thinking that if there’s any chance at all that your friend might have known, no matter how slight, then you should always switch, because your chances of winning if you switched would still be 50%, at worst, even if he didn’t know, and that’s the same as they would’ve been if you hadn’t switched, but if he did know, then they’d go up to around 66.666666666666666666666666666666666666666666666666667% or so, approximately, give or take (I rounded up, so it’s actually a little bit less than this). In other words, there would be no reason not to switch, because your odds of winning by switching couldn’t get any worse, regardless, and they could possibly get better. Any possibility at all of your drunk friend knowing where the car is makes switching worthwhile, because there’s no downside to it, either way. Therefore, you should switch, and if you win, you can reward your intoxicated pal, by letting him drive the car y’all rode up there in back home, and if you lose, since you’ll be feeling depressed, your wasted buddy can drive the car home anyway, except with you in it, of course. That way, at least, you might not have to deal with your depression much longer 🤷🏿‍♂️ Just make sure that he keeps it floored, and he doesn’t stop for any reason, especially the popo. If for some reason he does stop for the popo anyway, just jump out of the car, and start running towards them, while pointing your phone at them, as if it were a water gun, and you were trying to get them wet 💦🔫 That should help the pain go away, for sure. You should trust me on this. I’m a guy on the Internet, and your well-being is very important to me. Good luck 🍀 👍🏿

He did explain it well but it’s not a math problem it’s a psychological problem.

@@nathansahy983 He did explain it well (better than all the other problems anyway), and whilst there is potentially an element of pyschology in a real world version of this, in the form it is stated here it is defnitely a maths problem, or more specificially a probability problem. In the first version the probabilities show that switching, after the first goat is revealed, is more likely to lead to you winning, which is a purely mathematical strategy that is, over many games, guaranteed to improve your chances to win (note that I didn't say that it would guarantee you won any specific game).

He did say that... but only AFTER finishing the problem statement (as he got into the example scenarios). As narrated, it sounded to me like an unfounded logical conclusion rather than an assumption or part of the problem statement. Better would be to overtly state that the host knows where the car is and will never reveal the car.

This is definitely math, not psychology. Psychology does not tell you there is a 2/3 chance you will win if you switch. Math also works on variations of the puzzle. If there are 20 doors and the host opens exactly 17 wrong doors (leaving 3 unopened) after you have picked one, I can tell you that there is a 5% chance your door has a car, and a 47.5% chance for each of the other two remaining doors. Again, that is math.

Truee lol,I knew there was more chance if you switch,but this man simplified that to atoms bruh

Exactly!

How is the other guy going to catch up? You are going to walk at a constant speed, not accelerating.

The easiest way to make this intuitive for people when they don't understand is to increase the number of doors. Let's say there are 1,000,000 doors. You pick one, then the host reveals 999,998 goats leaving only 2 doors. Now do you switch? ;)

@@DJ-Bean I never claimed as such?

Why 1/999 and what do you mean none of the doors had goats? Then what did they have?

John Dinner I mistyped, I meant none of the doors had the car behind them. The 1/999 is probability that out of the 999 doors, the door with the car didn‘t get picked.

@@Supremebubble Do you know why there's different results from the following? - The chances for the car to be in any door is 1/999. You go through 998 doors with that and you get 998/999, which gives you the 1/999 you mention. - The chances for the goat to be in any door is 998/999. You go through the AND of probability and therefore multiply the chances of each door to show a goat by 998/999 998 times. I know I'm doing something wrong but afaik both give different results and my brain can't work it out right now.

John Dinner What do you mean by „you go through all of them“? There are 1000 doors but actually at every moment, only 1 of them is getting picked. First I pick a door, then the host picks a door that doesn‘t get opened and opens the rest and that was it essentially.

No. I may be wrong, but what I understand is that he is specifically testing the statement. B can have anything under it and it wont test whether A ONLY has the number 5, while 5 isn't only on the other side of A.

Yeah I forgot to mention that. The problem as originally stated said that you know there’s a letter on one side and a number on the other. Pinned a comment regarding the mistake.

@@_godsl4yer_ Yes. "B" can have anything under it. INCLUDING "A". So it must be checked as well.

Honestly I stopped watching at that point. Considering he also completely ignored the B card, these are 2 huge logical mistakes in a video about logic, not worth losing your time.

I have to disagree. It was a great video. No one is perfect and everyone makes mistakes, plus he doesn't have some large team behind him to back him up.

The question asks about the expected proportion, not the expected number of boys and girls. While the expected numbers of boys and the girls are the same, the expected proportion is surprisingly, not 50%. This is a tricky question and is unfortunately poorly explained in the video, but what he said is true.

@@haidonggong9745 that's what he said verbatum, yes. That however, makes no fucking sense. If there are 50 girls and 50 boys, the proportion is 50%. He never spoke about other villages

@@haidonggong9745 But the proportion is a direct consequence of the number, surely? The way he poses the problem at the start means that the answer tends towards 50%, with the only reason it doesn't get there perfectly in at any finite point being that there will always be some families who haven't quite had enough time to manage to get a boy, as the tree diagram shows. Honestly I don't know what the presenter was trying to say, but one sentence was even a direct contradiction of itself. Also he started with our population being within one village (which I personally took to be contextualisation rather than a necessary part of the problem), then suddenly later on he is trying to justify a illogical answer because we are somehow now working with multiple villages, and this somehow changes the answer whilst simultaneously keeping it the same?

You do walk on the treadmill though, just not while tieing your shoes obviously (I think he just made a mistake while animating). But it doesn't matter, the example still works, after your both done with tieing your shoes, your twin already has a head start and you're not gonna catch up, since you both walk at the same speed

I think using the term "head start" doesn't help at all, because you can still think that when they tie their shoes you might have a chance to catch up. Instead you can think of it like that while tying your shoes you increase the time you spend on the treadmil relative to how long you spend off the treadmil, you want to spend more time on the treadmil because during that time you move faster than during the time you spend off it. This exact same reasoning helps explain why the second example is optimised if you sprint off the treadmil in the second setup and cuts to the core of the problem in general instead of relying on the specific version of the question asked.

@@MagicGonads But they tie their shoes at the same time you do, so there really is no way to catch up

@@Mmmm1ch43l Nowhere does it say they tie their shoes at the same time, that implies they'd reach the end at the same time because their playing field is the same.

@@MagicGonads Well, you start right before getting on and they start right after, so you'd be a split second apart, so you would also finish at pretty much the same time

You have increased resistance while moving at higher speeds due to fluid dynamics. The point wasn't to look at it from a physics perspective, but from a mathematical void.

That'd be funny and cool at the same time.

@Dhyan Jay tf presh doesn't do real math

FinetalPies I think that was the point of the video a little fun thing to see if you can spot the mistakes.

only if a letter can have another letter on the other side

@@johnnycochicken Of course there can be. There can also be a troll.

@@Rekko82 It dosen't fit with the alcohol analogy.

@@coolguy284_2 I don't use alcohol.

@@Rekko82 That's got nothing to do with anything, only the analogy mentioned in the video matters.

Yep. Totally forgot to mention that. The problem as originally stated said you know there’s a card on one side and a letter on the other. My bad for forgetting that.