Math that will make you think. Twice....maybe
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Alright well everyone’s commented this already but might as well add 1 more. Regarding the logic problem with the cards it’s supposed to be assumed that every card has a letter on one side and a number on the other. I totally forgot to mention that so my bad! Without that being stated then yeah the B card would need to be flipped over. Had i included that like the original problem though then the A and 6 are still the right answer. Thanks to those that caught that.
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?
For the A-B-5-6 one you didn't specify that there is a letter on one side and a number on the other for all cards because, for all we know, there is an A on the other side of the B which would break the rule.
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.
The reason you want to stop on the walkway and sprint off it instead of while on it is because you want to maximize the time spent on the moving walkway. Sprinting on it just decreases the additive property the walkway’s boost would have given to the man if he had sprinted earlier and was walking.
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
The Vsauce "Or are they?" was so perfect
NPC Grian ! Looking for this
Or was it?
Or was it?
7:55 - 0% Eggs? What if I told you eggs are an ingredient of pancakes?
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?
I know who the real loser from the Monty Hall problem is... It's the security team who lets various drunken audience members rush the stage 18 TIMES... and is about to lose their jobs.
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.
"... behind two of which are goats, and one of which has a car..."
seems like you can't lose in this instance
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
@@cube492 “…are goats, and one of which has a car…”
one of the goats has a car.
Airport: *exists
Mathematic puzzles: "It's a free real estate."
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.
Suck it up and sprint the whole way, geez.
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
2:57
How are we supposed to know that there can't be an "A" on the other side of the card with a "B"
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.
14:25 How about my friend opens a door and reveal the car behind, does switching increase the probability to win a car?
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
The simplest explanation of the Monty Hall solution is: when you make your choice, it is most likely wrong. When the host opens one of the other doors, your choice is still most likely wrong, but now there's only one other choice, which consequently is most likely correct.
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.
Your explanation of the classic Monty Hall problem (from 13:24 to 13:50) is the only one that has allowed me to fully understand, seeing the outcome of each scenario. Thank you. 🙏
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.
the two walkway problems are really actually the one same problem phrased in two different ways: it is always more efficient to move as fast as you can off the walkway (whether that means walking instead of stopping to tie your laces, or sprinting instead of walking) and save the slower bits for the walkway because the walkway is moving you forward faster than the ground (again, whether the slower bits means stopping entirely or walk vs run) regardless, which means that bit is always shorter: it isnt effectively 50 meters because the ground basically moves, making itself shorter.
Yeah I don't get it. Unless you're slowing down your walking speed when you get on the walkway, it shouldn't matter whether you tie your shoes before the walkway or on the walkway.
If i’m late in the airport i don’t stop running.
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
The best way to intuitively understand the Monty Hall problem is to realize that because Monty will never open the door with the car, if you switch you are guaranteed to change what is behind your door. If you start with the car you will switch to the goat, and if you start with the goat you will switch to the car. You cannot switch from the goat to the goat because Monty just showed you where the other goat is. Since you had a 2/3's chance of picking a goat at first, that means that switching gives you the car 2/3's of the time. This also means that if your drunken friend opens the door to reveal a goat you should still switch, you are still guaranteed to swap which item you picked. But your friend might have opened the door that showed the car, in which case you are guaranteed to lose because you can't switch to the open door.
+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.
He will ask a question with a given set of rules and then changes those rules as he goes.
I remember somewhere someone had shown me some intuition into the Monty Hall problem. The reason for the 2/3 probability is because the host KNOWS the information behind the door and PURPOSELY chooses the one with goat. 2/3 and 1/3 probably may not sound too "distinctive" from our perspective so let's switch to another example
Suppose you have 100 doors with 99 goats and 1 car. Initially you chose 1 door which only give you 1% chance of winning. Now the hosts open 98 out of 99 remaining doors and of course they are all goats. Now the question is should you switch? The answer is obvious that you should. You have to wonder why, out of 99 doors, the host chose those 98 doors SPECIFICALLY? You wonder why he didn't open the remaining door. After all, initially you have 1% chance of winning. Now 98 doors are open, meaning the probability of winning with remaining door is 99%. Therefore you should switch
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
Thanks for the video Zach! For the populating question, I think that I fall into the controversy. Presumably the controversy is around interpreting the question. Your question asks ‘What fraction of the population is expected to be girls’ and you then state correctly that the fraction of boys to girls is 50/50 in the limit of infinite couples (there will always be slightly more girls due to that last couple that just keeps at it). I believe this is the answer to the question you pose, as quoted above. The explanation that you provide, which does not yield 50%, seems instead to answer the question: ‘what is the average percentage of girls within a household’. I do not know which question was posed within the book, but struggle to link your answer to your question. I also do not see the relevance of invoking multiple villages; they are not part of the original question. Similarly I do not see the relevance of the coin flip picture. You do not ask the ratio of male ears to female noses (equivalently to pancakes and eggs this will yield 66.6•%). Would love to know if I am simply missing something
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?
The "B" card should be included since the question never states that the cards have a letter on one side and a number on the other. With the information given, the B card could have an A on the other side.
Edit... and then I saw the comments.
For the A, B, 5, and 6 question, wouldn't the B card need to be flipped over as there is no requirement that there are a number and letter on each card?
Couple years ago I came up with how the odds in monty hall problem somehow depend on the door opener's intent, and it's been evading my intuition since, but I think you've just pushed me a lot closer, thanks for this video!
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.
Something important to understand about the Monty Fall problem, is that 33% of the times you will just lose as your friend opens the car door. A bit hard to wrap your head around, but if you play 10000 games, 3.3 k of the times you will win if you stay, 3.3 k of the time you will win if you swap. But 3.3k of the games you just lost before you even got to make a decision.
_"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.
For the boy/girl village question, the proportion is averaged over villages (which disproportionately weights the village with only 1 boy, relative to if we averaged by child).
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.
The very first problem assumes you are NOT walking on the walkway itself? Because if you were it really shouldn't matter when you do it.
No you do walk on the walkway (unless you’re tying your shoe)
A very little proof(I think, I have not checked or thought about it much) of the first problem:
Sprinting will cover more distance than walking in the same amount of time.
If you do this while going slower(i.e walking) then you will spend less time at that speed and more time on the walkway where you are going slower.
If you do this while going faster(i.e on the walkway) then you will spend less time at this faster speed and more time at the slower speed.
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.
*****The Monty Hall problem is incorrect!!! If your drunken friend randomly opens a door and it is a goat, you have a much higher chance of winning by keeping the door you have
if you have chosen correctly, your drunken friend has a 100% chance of showing you a goat. If you have chosen incorrectly, your drunken friend has a 50% chance of showing you a goat. Since you now see a goat randomly, it is much more likely that you have already chosen correctly.
Keep the door you have. Yes. It DOES matter.
About the village: Suppose you have a village with infinite couples. They have their babies, and 50% are girls. Then some bureaucrat shops up and tells them that actually, their infinite village had been split up in an infinite number of 1 couple villages many years ago. Even though the sexes of the babies didn't change by receiving that message, all of a sudden only 30.69% is female?
That's not what they are trying to say, but rather that despite the proportion of girls is 50% in all the village, that does not imply the the average proportion of girls must be 50% per family. To see why, let's put an easier example in which there are two boxes; in one of them there are 300 blue balls and 200 red balls, while in the other there are 0 blue balls and 100 red ones:
N° Blue balls N° Red balls % Blue balls %Red balls
=====================================================
Box1 300 200 60% 40%
Box2 0 100 0% 100%
In total there are 300 blue balls and 300 red ones, so there are 50% of each color. But if we take the average percentage of blue balls in each box, we get: (60%+0%)/2 = 30%, while the average percentage of red balls in each box is: (40%+100%)/2 = 70%.
As you see, the average of percentages for each color per box is not 50% despite there are in total the same amount of balls of each color. This occurs because the boxes don't have the same amount of balls, so in each case the percentages represent different quatities. For example, in Box1 the blue balls win 20% of percentage over the red ones, and that 20% represents 100 units in this case. In Box2, the red balls win 100% of percentage over the blue ones, which is bigger than the 20% that the blue ones won previously, but that 100% still represents 100 units, no more. In other words, in one case the 20% is respect of a big amount, and in the other case the 100% is respect of a smaller amount.
The same occurs with the families.
On the puzzle with A B 5 6 you didn't say a card has a letter and a number so shouldn't you have to turn over B as well?
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.
This video states the Monte Hall paradox better than most people have done. It works only if Monte is required to open one of the doors. It does not work if he has the option of not opening one of the doors. For example, suppose that Monte believes that you understand the Monte Hall paradox. You pick the door in front of the car, so Monte shows you one of the goats knowing you will switch. If you pick one of the goats, he does not open one of the doors but asks you to choose blindly.
For the cards, you never stated that one side is always a letter and the other is a number, thus we would still need to check to make sure there wasn’t an A on the other side of the B
The method i used to explain the Monty Hall, goat/car problem to those who didn't get it, is to take a pack of 52 cards, turn them face down, then ask them to pick out, say e.g. the ace of spades, ok? Now, after they choose a card, cover it so neither of you can see it, ok? Now, his chances of having actually picked the ace of spades is 1 in 52 right? Ok, now you take the other 51 cards and look through them face up, (only you can see them) you select one and place it face down, then you turn over the other 50 to show no ace of spades, ok? Now, ask them if they think they are still a 1 in 52 chance or have they suddenly become 50/50? (after all, either they have it or you have it) Of course they are still 1 in 52, this is the same as the Monty Hall problem, as the host knows which door has the car or goat, he is never going to open the door with the car is he? the game would be over then.
I like this method, thanks
The second part of the Monty Hall problem is wrong. The only reason it comes out to 50% is because you are ignoring the data that makes it the true 66%
Each individual still has a 2/3 chance of winning if they switch. The only difference is a small portion dont get to play anymore because their friend revealed the car.
Essentially you are asking a different question. You are playing a different game. The door being opened at random instead of revealing a goat is the new rule of the game. This is no longer the Monty Hall game. The people who dont get to switch are now losers. This means you now have only a 1/3 chance of winning no matter what you do.
this man is right, when they open the first door is a whole different story
Actually, the "true data" makes it to be a 33% win rate. If you include the other 6 who got the car revealed, then they can only pick a goat (1 behind door they picked, and 1 begins closed door they didn't). So, that makes total of 6 people winning, and 12 people losing. Basically... the original 1/3 chance...
The door question contradicts itself, the question said “your friend randomly opens a door with the goat on the other side” than in the explanation it took half away because the door with car got opened(so no it doesn’t matter who opens the door, it just matter what door was opened)
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.
The interesting thing about the walkway, (and I wonder if Tao had considered this or even was aware of it at the time), is that many cyclists and runners have known about this for quite a while. If you consider your baseline power output and are in a situation where you face a headwind/uphill you can greatly increase your overall speed by pushing harder when the going is toughest, (into a headwind or uphill and in the walkway example while on the unmoving ground), than you would maintaining normal output during those harder times and instead adding your increased output during times like going downhill or with a tailwind. This guy named Millar wrote a piece about a race he was in from point A to B then back to A where his main competitor was nearly equal to him physically only Millar won the race by almost a minute. When he talked to the other cyclist he learned that while he himself held normal power output during the easier portions and increased power output during the slower portions the guy he solidly beat did the opposite and tried to push the hardest going downhill and while he had a tailwind.
i was just about to fall asleep when i heard michael‘s voice, opened my eyes, saw him disappear behind that screen again and nearly had a stroke oOF
The people who's friend revealed the car would DEFINITELY switch. Meaning a total of 12/18 would win. Or 2/3
The simple answer to the drunk monty hall problem is this:
Your friend chose the door randomly, with a 2/3 chance of picking a goat.
The host knows the answer, and has a 100% chance of picking a goat.
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.
I'm sure this comment on a two year old video will not be seen by anyone. But here's how I convinced myself that if your drunken friend opens the door then it doesn't matter if you switch.
Consider an expanded example. Monty has 1,000 doors. You pick one, he reveals 998 others, all guaranteed goats, then asks if you want to switch. Yes, of course you do. He just removed 998 goats.
--
Enter the drunken friend. they had the sense to skip opening your door, but opened 998 other doors at random. Now there are only two unopened doors left, as before. Monty asks if you want to switch.
Lets compare the chance that your initial pick was correct with the chance that your friend randomly opened 998 goat doors, avoiding the car by drunken luck.
Okay, your initial chance of picking the car is 1/1000. Easy.
Your friend's task is to pick 998 consecutive goats, at random. The first door has a 999/1000 chance of being a goat. The second door has a 998/999 chance of being a goat. The third has a 997/998 chance, etc. The final, 998th, door your friend opens has a 1/2 chance of being a goat. There were two doors left and one lacked a goat.
Multiplying all these contingent probabilities together, we get the product of (999-x)/(1000-x) as x goes from 0 to 998. Which Wolfram Alpha conveniently tells me is exactly 1/1000.
There were two possible routes you could have taken to reach the specified end state, both of which have a 1/1,000 chance. You don't know which occured, so the best you can say is that your odds of going home with a goat, or a car, is 50/50. Either you picked the door with the car, or by process of elimination, your friend did.
You are correct, and I will put another way of thinking about it, just in case you want:
Since the drunk friend is revealing randomly, then it would be the same if you (the contestant) were the one who also revealed the other 98 doors. By the end both are doing it without knowledge of the locations, so in the long run the outcomes shouldn't tend to be different.
For example, you could pick door number 1 and then reveal all the others except door number 100. But in this way what you are doing is basically selecting which two doors will remain closed for the second part. I mean, picking door 1 and then revealing all but 1 and 100 is the same as saying: "I want to keep doors 1 and 100 closed", and then you proceed to reveal all the rest. Now, supposing that all the others result to have goats, those who say that the probabilities are the same as in the original game would think that door 1 has 1/100 chance, because it is like the staying door, which you declared first, and door 100 has 99/100 chance, because it is like the switching door.
But what about if you had said it in the opposite order: "I want to keep doors 100 and 1 closed". Should the chances be reversed now? Of course not, the doors "don't know" which one you declared first, and neither which should be the staying door and which the switching door.
Something you kinda touched on but didn't state, like in every monty hall problem article or video I've seen, is that the reason the drunk friend doesn't change the odds but the host does is because the host has restricted options. He *can't* choose the winning door, but your drunk friend can. This restriction is what provides the extra information to your guess.
I always have to explain to people that the Monty Hall problem isn't the same if the host "decides" that he will open another door and let you switch, it has to be a rule of the gameshow. Because then he could rig the game to where he only lets you switch after you chose the car. Hence why in 21, the presentation of the problem isn't quite right, and the answer from the kid should be to ask for clarification on the rules.
The monty hall problem: excellent example with the friend opening the door! But it is about knowledge. The talk show host KNoWS which door has the car, your friend doesn’t know: he was just “lucky/unlucky” opening the goat door. You didn’t get the benefit of the talk show hosts knowledge.
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!
The math behind the walkway basically boils down to the fact the function 1/x is decreasing but at a slower and slower rate
4:30 the b card could have an a on it, no where is the rule stated the a letter always has a number on the back, therefor the bar anology doesnt hold up because it was stated everyone at the bar is drinking
I think the first one is quite intuitive for strategy game lovers. you maximize the the efficiency of your buff (the runway) by maximizing its duration (i.e. slowing down on it)
Great video! One thing bothered me though: In the card problem, you either have to state that each card has a number on one side and letter on the other or flip the B card too cause if the B card has an A on the back then the rule is broken.
The problem of the monty hall problem is that they usually don't say if the game show host knew. He/she could also open randomly.
Well, it wouldn’t make for a very good game show if the host instantly revealed which door leads to the prize, would it?
11:13 was terrifying. 😱
I didn't expect that!
Ya, I also freaked out
The coin flip thing is also why people believe that you can just bet 1 dollar at roulette, then 2 dollar if you lose, then 4 if you lose, etc.
Yes, you might win one dollar in theory. In practice you will lose all your money.
Well, you will always win your initial bet that way (1 dollar). The problem is that Casino's have a max bet to prevent that. For example 100 dollars. So you can only try it 1/2/4/8/16/32/64->7 times.
got it at around two minutes ; sprinting on the ground minimizes your time on the ground and maximizes time on the walkway. ground sprinting multiplies your speed by 7/2, walkway only multiplies your sprinting speed by 8/7, walkway multiplies your walking speed by 3/2. for the highest average speed you want to maximize your average speed multiplier.
It's A, 6 AND B that you have to flip over... man this is going to bug me all night. The 22 year old could have been a Vampire.
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.
Q1. It would be more efficient to tie your shoe on the walkway.
Example:
There are 10 meters of walkway, 10 meters of normal floor, and it takes 10 seconds to tie your shoe.
You walk at 1m/s
The walkway increases your speed by 1m/s
Tying your shoe on the walkway: 10s to travel on floor, 10s to travel on walkway: total time 20s
Tying your shoe on the floor: 20s to travel on floor, 5s to travel on walkway: total time 25s
Q2. It would be more efficient to sprint off the walkway.
Example:
There are 10 meters of walkway, 10 meters of normal floor, and you can sprint for 10 seconds.
You walk at 1m/s
The walkway increases your speed by 1m/s
You can sprint at 2m/s
Sprinting on the walkway: 10s to travel on floor, 1.33s to travel on walkway: total time 13.33s
Sprinting on the floor: 5s to travel on floor, 5s to travel on walkway: total time 10s
Michael's head popping up from behind the screen was so creepy lol
The 3 doors problem starts with the phrase: "There're 3 scenarios", when there's only 2: the door you picked is right or wrong. It doesn't matter will you know anything about other door, or other 2 doors, or other mln doors. Until you're allowed to change the door you picked and it turned out to be wrong.
What would be really interesting is restating the airport puzzle but with all the speeds as a significant percentage of the speed of light.
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
To the First Problem, think of it like this, the more time you spend on the walkway the longer you get the additional distance travelled due to the wall way. So you want to be faster of the walkway, and slower on the walkway, so you get the benefit of +1m/s for longer
Concerning the card "puzzle": How do you want people to solve it if you are forgetting the essential informations like 1 card = 1 side with a letter, 1 side with a number? :-/
He alrady apologised for that
@@clemenss.8641 Then add a caption in the video? Instead letting everyone baffled?
so the idea behind the card "trick" is that if condition A necessarily determines that B is true then if we negate the logic here we get:
if B is false then condition A must not be met.
14:28 Wow in my case it was the opposite. This moment gave me the realization I needed to fully understand that paradox. Thank you
You would still have to flip over the b card, cause if there is an A behind it, then the rule is broken.
Monty Hall question : What if there's a nonzero probability p that your friend has been told by the production where the car was, so that they don't accidentally open the door to the car ? Should I switch now ? Or does it depend on p ?
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 🍀 👍🏿
The ab56 excercise says that if you have a on one side there is definitely 5 on the other, but it doesn’t specify that 5 also has to have an A on the other side so you would have to check : A - to see if there is 5 on the other side, B - to check if there isn’t an A on the other side, and 6 to check if there isn’t an A on the other side. You don’t have to check the 5 because if there is an A the rule is true, and if there isn’t the rule is also true.
3:34 WRONG. Your answer leaves the other side of B ambiguous. If an A is on the other side of the B card, then your answer is insufficient. It was never established that every card has a number and a letter.
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.
For the last one, falling on 50/50 is still better than the original 1/3 from when you picked, right? Even though it’s not 2/3
Missed one crucial detail while explaining the Monty hall problem.. the host doesn't just reveal a door. He reveals a losing door every time
the shoelace and running problems is really just the same problem: should you go slower on or off the walk way. Stopping to tie your shoe lace is just an extreme case of going slow. And you should always go slow on the walkway, since it moves you forward anyway
Here's my interpretation of 4:20
BOTTOM:
A = under 21
B = NOT C
C = Alcohol
A leads to B
"leads to" is only false when A == true and B == false, it is true whenever A == false OR B == true
This means you don't need to check anyone who is >= 21 (A == false) OR not having alcohol (B == true) (Second and third person)
TOP:
A = A on one side
B = 5 on the other
A leads to B
this is true when A is false OR B is true (check my table), so you don't need to check these cards: "B" (A is false), "5" (B is true), but you do need to check these: "A" (A is true, B is unknown), "6" (B is false, A is unknown)
(This is assuming each card contains one "A" or "B" and one "5" or" 6", otherwise you would need to check "B" as well (B is false, A is unknown))
'leads to' logic gate for clarity bc i dont know what its called in english (sry)
0 leads to 0 = 1
0 leads to 1 = 1
1 leads to 0 = 0
1 leads to 1 = 1
In my calculations, the solution to the village problem is wrong. It shouldn't vary based on the number of families. It should be 50% even for a single family.
The calculation goes as number of girls = sum (1/2)^n.(n - 1), n = 1 to infinity. The percentage of girls is number of girls/(number of girls +1), cause there's always 1 boy per family.
To put it into english, every family has a 50% chance of having a boy, 25% chance of having a girl then a boy, 12.5% of having 2 girls then a boy, 6.75% of having 3 girls then a boy, and so it goes until infinity. It's nice to visualize it as a square being cut in half infinitely. Interestingly enough, that sum = 1. Which means the families will have 1 girl on average. So 1/(1 + 1) = 50% are girls. No matter the number of families.
Maybe I'm wrong though, let me know if you spot some error.
On the first one I thought you would walk on the treadmill thing so you would be slowed down by exactly your walking speed no matter where you tied them, so it wouldn't matter.
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
Wearing slip-on shoes has apparently saved me time AND brainpower.
I really liked that you put in the thing about the Monty Hall problem that I also didn't realize until earlier this year. I have some intuition to offer: Imagine that you don't have 3 but 1000 doors. You pick one and your drunk friend opens 998 doors at random. And "surprisingly" none of the doors that got opened had the car behind them. I say surprisingly because clearly if one of the 999 remaining doors had a car behind them, then it would be pretty odd if out of the 998 doors that got opened randomly, none have the car behind it. It would not be really surprising of course if the door you picked already has the car behind it. Turns out the chance of picking the correct door out of the initial 1000 is 1/1000 and the probability of not opening a door with a car IF there is one out 999 remaining doors is 999/1000 times 1/999 = 1/1000. Both are the same so we are back to 50:50.
@@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.
When you decide to tie your shoe doesn't matter as the conveyor belt adds the same amount of bonus distance regardless if you're stopped or walking during the conveyor belt.
You'd need to flip over the B to check if there is an A on the other side. If there is, the statement is false.
for the first one, you want your speed at any time to be as close to your average speed all the way through.
similar to how a square has more area than a rectangle with the same perimeter. because the sides are equal in length.
so you want to accelerate when you are slow, and slow down when you are fast
monty hall twist is easy as the friend could have easily picked the car.
I want to see a puzzle death battle between MajorPrep & Presh Talwalker.
That'd be funny and cool at the same time.
@Dhyan Jay tf presh doesn't do real math
For the analogy to work...
A would equal Alcohol,
B = under age, unknown beverage
5 = 21+, unknown beverage
6 = under age, unknown beverage
Or are they 😂 …Vsauce is such a legend
He randomly picks a door but can’t pick your door or the car door, so it isn’t random
With the airport problem, it would actually be fastest to start just before the end of the walkway, to maximize your acceleration before you hit the solid ground.
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.
The question for AB56 did not specify that each card has a letter on one side and a number on the other. He just said that there were 4 cards and they showed A-B-5-6. So, without the knowledge that every card has a letter on one side and a number on the other, the correct answer is actually to flip over A, B, and 6
I have so many minor grievances here but I'm too tired to explain them. I don't think you're ever explicitly wrong or anything, and in fact it was a pretty informative video
FinetalPies I think that was the point of the video a little fun thing to see if you can spot the mistakes.
You need to flip A,B and 6. How do you know there's not an A on the other side of the B?
7:39 My body doesn’t want to get heads
The airport puzzle makes bad assumptions.
1) I never stand still on the walkway.
2) People don't sprint a fixed distance, they should be able to sprint further on the walkway.
I think you need to re-watch that bit.
1. No clue what you're talking about what assumption did he make about standing still?
2. He says that the person can sprint for 5 seconds, regardless of distance, therefore, they ARE sprinting further on the walkway.
Basically, you should maximize the time spent on the walkway, because it grants an additive bonus rather than a multiplicative bonus.
@@fangere thanks you were right about the second point.
The first point was about tying the shoes. When person A (left in video) is tying shoes, person B (right) is walking, but not vice versa. If both people keep walking, it would not matter where you tie shoes
@@kaskilelr3 The person on the right isn't walking, they're tying their shoes but on the treadmill, so they still move forward. After they're both done tying their shoes they can start walking, but the person on the right still wins, because they have a head start
14:29 I don't understand how you can not understand the difference in the two scenarios... in the first scenario, the host knows where the car is, and there will always be a goat to show you, which wasn't behind your door. In the second scenario, your dumb friend just happened to pick a goat... no extra information is revealed.
An elegant way of thinking about the first problem:
Walking is the same as staying still on an imaginary conveyor belt that has your walking speed. The question is then, is it more efficient to sprint on this imaginary conveyor belt, or the real conveyor belt? Well it has to be the slower conveyor belt, because the faster a conveyor belt is, the less efficient adding speed becomes.
You tie your shoelace before getting on the walkway because they might get caught in the walkway mechanism and you would DIE HORRIBLY
4:48 but wouldent it be incorrect if A was on the other side of B. I think you forgot to mention that there has to be a letter on the other side of a number and vice versa
I don’t really buy the Monty Hall explanation... I mean, you first had a 66% chance of picking the wrong door. If somebody opens a wrong door, be it because of pure luck or knowing it is a wrong one, you still have chosen a door with a 33% chance of being the good one vs the other door (wich now can only have a 66% chance of being the correct one).
You are mostly right as the problem was stated in the video. But there's a minor detail that he omitted from the explanation that changes the story: the host knows where the car is. That is why he said "The host will always open a door with a goat."
To make this more intuitive, imagine that there are 1,000 doors and there is a car behind only one. You choose a door, then the host (who knows where the car is) opens 998 other doors with goats behind them, leaving just two doors. One has a car, and the other has a goat. It isn't coincidence that the host didn't reveal the car, since he knew where it was. Now... do you still think there is a 50/50 chance you guessed right the first time, or is it just possible that the host has done you a massive favor by throwing out nearly all of the wrong answers?
Under this scenario, you have only a 1/1000 chance that the car is behind the door you picked, meaning there is a 99.9% chance you will win if you switch doors.
The new scenario is now that 1/3 of the players automatically lose if the random door is the car.
In his explanation he is ignoring one group of losers. If you are in the remaining group that didnt lose when the second door was open you now have a 1/2 chance of winning regardless of what you do, but you have a 1/3 chance of winning overall. He frames this as 'well the original question asks what happens when you switch so we can ignore the 1/3 of people who never get the chance to switch.
mike ard Hmmmm that’s what my statement said. If you picked a random door and another person opened 998 doors, just happening that none of them had the car (it is now less likely than the first case, but applying the same logic), there is still a 99.9% chance of the car being at the door you didn’t pick first time.
Skadragon Well, that makes sense. However, in the video he says that your case is that your friend without knowing anything picked a wrong door, which now is information that you know, so you obviously have to take into account that fact (the door he picked has a goat) into your analysis.
In any case, I guess that solving the problem in the first stage (before your friend opens any door, therefore not having any extra information and leaving the scenario where he opens the correct one still open) does leave you with the 33% chance and a very entangled way of formulating a trivial probability exercise.
Anyone with a Brilliant subscription, How good is the annual plan? Worth it?
strdst. Very, yes
Well it’s not something I’m going to consider after seeing this video as they got the Monty hall variation wrong...
Like any thing, it depends how much you use it. I’m currently on a binge and enjoying it.
In the 2. question i think we have to also flip the B card. Because if there is an A behind it then the rule is broken.
The rule says if there is an A on the one side of the card then there has to be definitely a 5 on the other side.
Wow I was right on the first problem. My reasoning is that I should run when my speed is at its lowest
If there were 100 doors, and you picked number 4 lets say, and the Host would open every door except yours and no. 73, it would be better to switch the doors
Yeah, that would be super suspicious haha.
11:20 Except that you don't win a trillion dollars a trillionth of the time. You can't win a trillion dollars in this game. You can win 2^39 (about 0.55 trillion) or 2^40 (about 1.1 trillion dollars). But nothing in between. :P
the first question/puzzle reminds me of one of thy favourite riddles ever
Anne and Frank head off to the airport. Anne jogs the whole way at a constant speed of 8 km/h. Frank walks half of the way at 4 km/h at which point he picks up a bike and cycles the rest of the way at 20 km/h. Who got there faster?
intuitively, for many people, it's frank, but if you think about it, by the time he gets to the bike, Anne will have travelled twice as far as him, so she'd already be finished, so frank could warp to town at the speed of light and still come after Anne
Similar to the travelater question: Assuming rain is fallling constantly and no wind, would you getter wetter if you walked or ran to a shelter?
Does it matter how far the shelter is? The rate of precipitation? The speed you run?
10:24 This is why mathematicians aren’t actually good at gambling
Shut up dog!
Aside from not accounting for acceleration and deceleration, what about rest and recuperation? If you rest on the walkway you are travelling at only half the speed of walking, but you are also replenishing the ability to run again. which quadruples your speed. The optimum achievable time depends on the rate of recovery.
You could even say there is a rate of recovery when walking.
I mean it depends how late you are. I remember needing to catch a bus to the airport [1mile, uphill, from the hotel] at 7.30 AM. Of course it was our last night and we made the most of it, going to bed at about 6. This obviously meant that we awoke at 7.15, hungover [still drunk] and barely able to walk. With 15 mins to get to the bus station we didn't have time to work out the most optimum approach [we didn't even have time to risk calling a taxi], we had no choice but to sprint the entire distance. My maximum sprint capacity is about 150m, i would never have made it using your method. But I did make it, because I ignored what I knew I was capable of, and did what had to be done.
It nearly killed my travelling companion, mind. To this day I wonder of it was fortunate that she manged to resist vomiting until she made it to the bus.
You haven't provided enough information in any of the puzzles to allow us to answer correctly except the Monty Hall question.
Another, simpler?, way to think about the village problem:
P(girls)=1/2
P(boys)=1/2
let x be the number of couples.
Consider the expected number of boys/girls born after each new 'child' in the village. Keep track of the total population
First Child: 0.5x are girls, 0.5x are boys - 0.5x+0.5x = x total population
Second Child: Since only half the original x couples are still having chilldren, 0.125x are girls, 0.125x are boys - x + 0.125x+0.125x - 1.5x total population.
Third Child: Since only a quarter of the original x couples are still having chilldren, 0.0625x are girls, 0.0625x are boys - 1.5x + 0.0625x+0.0625x - 1.75x total population.
This carries on - with each new 'child', the number of new girls born and number of new boys born halves from the previous 'child'.
If we sum this series, we get x +0.5x+0.25x+0.125x..... = Total Population
AND
0.5x+0.25x+0.125x..... = Number of girls
Thus the proportion of girls is (0.5x+0.25x+0.125x.....)/(x +0.5x+0.25x+0.125x.....)
x's cancel to give
0.5+0.25+0.125..../1+0.5+0.25+0.125.....
The sum of 0.5+0.25+0.125..... = 1
Thus 1 / 1+1 = 1/2 = Answer