Every Weird Math Paradox - Part 2

[@nicolascage5774]

The fact that a video titled "Every Weird Math Paradox" has a Part 2 is a paradox in itself.

[@jacpa2011]

its secretly a part 1

[@currywurst7229]

Isnt that a oxymoron (too)?

[@samu.bionda728]

It isn’t if you take notice of the variable of time; As that video was published at the time t and now we are at the time t+x for a positive x, you can assume the title did not refer to even the future paradoxes. Essentially, by this impossibility, you can deduce that there’s an “up until time t” implicitly written in that title.

[@Epoch11]

It's okay because he's counting in senttimitters.

[@PerriPaprikash]

It's not a paradox, but rather due to the laziness of the authors when making the first video and also due to their greediness in not wanting to loose the view and like counts if they were to upload an extended or corrected version for the first video Also note a lot of the so-called paradoxes are indeed NOT paradoxes at all. Which seems to suggest the authors of the video do not quite understand the meaning of the words paradox and every.

[@adeade3474]

In the example you used for simpson’s paradox, it not possible to have a lower percentage than 70 %. The percentage of combined treated patients that survive will fall between 70 and 80. Check it out

[@aguyontheinternet8436]

yeah that's what I was thinking, the math wasn't mathing out. If you wanted the lowest possible survival rate, you'd look at an all women group. When you mix men in, the survival rate can only climb, not fall

[@kyokajiro1808]

yeah i expect it to be smthn like 50% of women survive and 60% of men survive which is possible unlike the one shown there

[@ThoughtThrill365]

You are correct. My initial example didn't accurately illustrate Simpson's paradox. Here's a clearer explanation: So here is what happens when you combine the data: - Males: - Treated: 70% survive (14/20) - Untreated: 60% survive (3/5) - Females: - Treated: 80% survive (4/5) - Untreated: 90% survive (18/20) Combined Data: - Treated: 72% survive (18/25) - Untreated: 84% survive (21/25) Despite treated males and females having higher survival rates individually, the combined data shows a lower survival rate for treated patients.

[@sensorer]

Thanks for pointing that out! I was sitting staring at my screen for 10 minutes trying to understand how that might happen.

[@drdilyor]

​@@ThoughtThrill365 it's still not quite right yet, untreated females have higher survival rate than treated ones. but i got the point similar things happens in competitions: lets say you are p2 and the results for day1 are: p1: 100 p2: 50 p3: 1 for day2 p1: 1 p2: 50 p3: 100 both days you are in 2nd rank, but in the combined ranking (sum of scores) you are the 3rd.

[@minirop]

2 details that helped me understand some of those: - Monty Hall: since he can't reveal the car, if your original pick was a goat, then Monty is forced to reveal the second goat. So 2/3 of the times, Monty is forced, and 1/3 is free to choose. - the ant: another way without using formulae, what you need to understand is that not only the rope stretches, but also the distance behind the ant, so the ant's progress (in %) is stable if it doesn't move, so with movement added back i, the % can only increase (even if it takes 800 quintillions years to get to 100%).

[@bscutajar]

For the second one, just because a value always increases does not mean it will reach any number, so your intuition is incorrect. Check out Zeno's Paradox.

[@trewajg]

This proves that the percentage is increasing, not that it diverges and, eventually, surpasses 1.

[@katakana1]

Your explanation of the ant implies that if the rope doubles in length each time, it still holds true, but it doesn't! In this case, if we imagine the ant crawls at 1 cm/s and the rope is 2 cm, the ant ends up getting to a distance of about 1.44 cm (exact value seems to be 1/ln(2)) from the start. The distance it has left to travel is now around 2.56 cm. Since the rope's length is multiplied by a constant value each second, it is the same as imagining that time has reset but the ant is now on a 2.56 cm long rope instead of 2 cm. Each second, the ant will get 1.44 cm away from its previous position, and the distance it has to travel will always get longer... Yet when it comes to the percentages, it always increases. If the ant doesn't move, the percentage remains constant. If the ant does move, the percentage increases, but in this case it converges to a constant less than 100%.

[@Rando2101]

​@@bscutajarWhat about Zeno's paradox? (I agree with your statement)

[@Rando2101]

​@@katakana1 it converges to 1/ln(4) I think

[@maniman121]

The percentages you used for combined groups in Simpson paradox can’t occur

[@simonwillover4175]

I know. It's not clarified, but the confounding variable here must be that there is more data than just the original data from the males and females. i.e. a new study was done.

[@Who77777777777]

Cool I’m not the only one who noticed that

[@johannesvanderhorst9778]

It can occur, if you add the data for the non-binary people. 🤪

[@alephnull4044]

Yep, was so confused

[@kommander_]

The percentages are for chance of survival for taking and not taking the treatment; the difference/increase of the survival rate is what the paradox is (the difference between the groups, not the sum). Separately, both groups have a 40% increase in their chance of survival by taking the treatment, versus if you combine the data those chances of survival only increase by 10%, so the treatment seems less effective.

[@quentind1924]

The numbers given for simpson’s paradox can’t be true, and it’s very funny to make that on the paradox for which you said "that’s the reason why we need statistics education in school". But that’s a problem to show an example of that very confusing paradox without actually showing the data for the example that can happen

[@emptyeff]

I see this a lot, same with physics videos explaining, say, the double slit experiment. Purposely confusing or otherwise they do not fully understand and honestly shouldn’t be making a video.

[@simonwillover4175]

7:35 the ant only reaches the end because the ant is moved with the rope as it stretches. i.e. if the left end of the rope is tied to a fixed position, and the ant moves right, when the rope's length is multiplied by s, the ant's coordinate (relative to the left end) is multiplied to. In fact, as the ant approaces the right end of the rope, it's velocity (relative to the left end) will slowly grow to eventually be more than the velocity of the rope's growth, thus the ant will get to the right end.

[@juxx9628]

This example of the ant and the stretching rope takes out all of that considerations. It assumes that the ant has a capped, fixed velocity of 1 cm/s. It just says that if given enough time, the ant can traverse through the entirety of the rope.

[@LevinFroggo-fs7uu]

@@juxx9628but if the ant has a fixed velocity of 1cm/s and the rope gets 1km longer each second the ant can never reach the end because the rope expands faster than the ant moves

[@okbrolmao]

That is the paradox​@@LevinFroggo-fs7uu

[@LevinFroggo-fs7uu]

@@okbrolmao No, the paradox is that if the ant is on the string it can reach the end, contrary to what you would expect

[@isopa2543]

Yea, from my understanding if the ant moves with the rope then it can reach the end because the rope is stationary from its' perspective. But if it doesn't move with the rope then it can never reach the end. As the rope stretches towards infinity the ant will approach being 1/100,000 along the length of the rope.

[@Epic_TanuKitty]

A small thing about the sleeping beauty paradox. So I ran a python program a while back with 1mil iterations using the same coin toss asking 3 different sleeping beauties to see what would happen. This ended with her being awoken just over 1.5Mil times (Which makes sense as ~1/2 the time she's awoken twice). The first was always tails which was a correct guess ~2/3 of the time, the second was always heads which was correct ~1/3 of the time, and the third was decided using python's randint to choose between heads and tails and was correct ~1/2 of the time. The conclusion I came to was that the random guess showed the probability of any given coinflip to be heads or tails, and the other 2 guesses showed the probability of tails or heads given that she's awake (which is always the case), so 2/3 of the time she's awake during tails and 1/3 of the time she's awake during heads. So Technically when she is asked about it being heads the probability is 1/3, as she has to be awake to be asked the question, and given that she's awake, it's 1/3 likely that its heads. Even though the probability of any single coin toss is 1/2. Man conditional probability is weird huh.

[@darrennew8211]

The right answer is that the probability she should answer "heads" is not the probability that the coin came up heads.

[@HectorHernandez-ko9jc]

With the ant case I think it can be stated a little more clearly. If you keep on adding or constructing rope the ant will never reach the end. But by stretching the rope you’re moving the ant alongside with you relative to an absolute point. So make it easier the rope is 2 cm adding 2 cm every second, the ant traverse at 1cm / s. If the rope is being constructed at the end point at 1 second the ant will have traversed 1 cm out of 4. But if the ant is moved when it stretches that means relative to the absolute beginning point it will have moved 2 out of 4 cms

[@benniboiii6893]

Anyone want to point out how he pronounces centimeter like perimeter and not like cent-"e"-"meat"-er. Really caught me off guard.

[@Nelster]

another interesting paradox is the aristotle's wheel paradox - consider 2 points on a wheel where one is on the outer most edge and the 2nd inside of the wheel forming a straight line with the 1st point and the center of the wheel, the linear distance travelled by the wheel and the smaller circle formed by the inner point is the same, even though their circumferences are different.

[@its_lucky252]

your monty hall explanation actually was the only one that made me understand, thank you

[@c.jishnu378]

The thing with the SLEEPING BEAUTY'S experiment is that, the coin is tossed not only once for Wednesday but also for both Monday and Tuesday combined, which increases the odds of waking up on tails but NOT LANDING ON TAILS as it is happening half as much of waking up.

[@warbraid]

The monty hall problem has bugged me for like over 20 years. I always thought the probability would be 50% because now you have half a chance of being right. I still to this day cant wrap my head around why its 66%

[@quentind1924]

Basically, if you start by picking one of the two wrong doors, after the reveal you will be right if you switch and wrong if you stay. If you pick the one right door, you will be wrong if you switch and right ic you don’t. But you’re more likely to start by picking a wrong door because there are two of them

[@kevinderoo3880]

The entire point of the problem is that revealing a goat tricks you into thinking that you are in an entirely new situation (just 2 doors, 50% chance of winning) that has nothing to do with the situation you were just in (3 doors, 33.3% chance of winning). However, you have a 66.7% chance of initially picking a door with a goat. When the host reveals a goat, it must be the last goat 66.7% of the time. Switching doors therefore gives you a 66.7% chance of winning. The mistake people often make is thinking the revealed goat does not influence your chances anymore, but it actually does.

[@insignificantfool8592]

Thats good. The usually accepted solution to the Monty Hall problem would only be correct, if the host is forced to offer a switch like this (every time). A situation like this is practically inexistent in real life. For real life situations similar to Monty Hall, 50/50 is indeed correct.

[@Rando2101]

Imagine a bigger problem: 100 doors with 99 goats and 1 car. You choose the first door. The host opens 98 other doors with goats and ask you to switch. Since the car is way more likely to be in the other 99 doors, switching is the better option (99%) while keeping your choice is only 1%

[@robo3007]

I can explain it in three sentences. If you stick, you will get what's behind the door you originally chosen. If you switch, you will get the opposite to what's behind the door you originally chosen. You are twice as likely to have originally chosen a goat.

[@zorm_]

given any set A, an injection can be built from A to P(A) simply by mapping any element x of A to {x} in P(A)

[@magicmulder]

Yeah the real “paradox” about Cantor was that this also holds for infinite sets, meaning the power set of an infinite set has a “higher infinity” of elements. For finite sets it’s indeed trivial.

[@johannesvanderhorst9778]

But the problem is a surjection from A to P(A). Suppose such an surjection s exists. Let B be the subset of A consisting of exactly all elements b such that b is not an element of s(b). Suppose c is the element of A such that s(c) = B. I can't say that c is an element of B, because otherwise c is an element of s(c) hence by the construction of B c is not an element of B. I can't say that c is not an element of B, because otherwise c is not an element of s(c) hence by construction of B c is an element of B. So there is no element that maps to B, contradiction that s is a surjection.

[@blackeyefly]

Ok cool, so what? That's a good starting point but far from a complete proof

[@suupe4453]

In the ant paradox, the number 1/100000 + 2/200000 + ... is supposed to represent the ants progress, but assuming the ant isnt affected by the movement of the rope, the progress of the ant at X seconds should be X/100000X the length of the rope, which means it would never reach the end.

[@Badspot]

The ant is effected by the movement of the rope, because it is standing on the rope. The entire rope stretches, including the part behind the ant.

[@LaugeHeiberg]

Remember kids: centimeter rhymes with perimeter

[@hughmiller9839]

Hey dawg, I heard you like inifinity so I took your infinite set and made a new set containing infinite subsets of your infinite set to give you more infinity.

[@eliyahzayin5469]

The absent minded driver is an interesting one and something that I've not heard of before. There's a LOT of discussion around it and several solutions (of varying complexity) So far the most convincing arguments I've seen so far is: 1. The concept of an assignable probability of being at x or y is fallacious. (I don't 100% agree but I do think it's ill defined) 2. (From my own consideration) If we treat the probability, A, of being at x as instead the probability of the driver guessing that they're at x, we find that the driver's expected correct guess rate is maximized when they always assume that they're at x. (Which reduces down to the planed strategy) 3. The expectation calculation when at an intersection is wrong. (A commenter on one of the blogs discussing the paradox confirmed this through a simulation. I'm also convinced by the lack of conditional probability appearing in discussions, despite the fact that it clearly plays a roll. For example, you can use it to find the probability, A, as 1-0.5p despite most people quoting it as 1/(1+p)) ETA: Yes, the expectation value calculated at intersections is incorrect. It doesn't account for the law of total expectation or the fact that the probability that the driver is at x or y is not the same for all drivers (100% if they decide to turn at x, otherwise 50%).

[@ariaden]

I agree with argument 1. The same as with the Sleeping Beauty. If you cannot distinguish in which "full game state" you are acting, you have to commit to act with the same probabilities in each indistinguished state. The same as in poker: If you do not see that cards the opponenes have, you cannot decide based on that (you can estimate probabilities, but they do not depend on the actual opponent cards). For absent minded driver, optimal strategy can be computed. Sleeping Beauty has to ask about her final payout (whether guessing right on both days gives her double the winnings).

[@Dodgerific]

For sleeping beauty, i think it depends on the phrase of the question. Imagine instead if heads she is woken up once, but on tails she is woken up 100,000 times. If she wakes up it is much more likely to be one of the 100,000 times, and therefore should guess that tails was the outcome. However the coin itself is fifty fifty.

[@Hambrack]

I don't get how Cantor's Paradox is a paradox. Of course the number of all sets you can make from a given set is bigger than the original. It's just basic logic.

[@fnsdjkovnsdkvn]

It's only unintuitive if the set you start with is infinitely large

[@JH-ty2cs]

The counterintuitive part is infinite sets. At first it seems inuitive that the cardinality of all infinite sets should be the same - e.g, the set of all even numbers is the same size as the set of ALL numbers both odd and even. the intuition is that you can't "run out" of elements of an infinite set, so one infinite set shouldn't be 'larger' than another one. The diagonal theorem proves that actually some infinities are bigger than other ones, and it's in a way most people don't really expect.

[@ptorq]

@@JH-ty2cs Thanks, because I was also wondering how this was a "paradox" and not "something any damn fool ought to intuitively understand." It still seems obvious to me that the cardinality of P(A) cannot POSSIBLY be less than the cardinality of A, because if A has N elements,then there are obviously N subsets each consisting of one of the elements of A, plus the empty set and all of the combinations of two or more elements of A as a bonus. In fact, it seems intuitively obvious that the if the cardinality of A is N, the cardinality of P(A) ought to be 2^N.

[@Leivoso]

The way the paradox is presented is very misleading tbh

[@Qermaq]

I like pallalelograms.

[@whatitmeans]

In Cantor's Paradox you gave an example that violates its Power Law theorem: it is an inside joke? the set A={5,5,5,5,5} has 5 elements but is portrayed with P(A)=1 which violates |A|

[@Leivoso]

But A only has 1 element, 5

[@kommander_]

Everyone confused by the Simpson’s Paradox percentages: the percentage is the chance of survival for that group; so by taking the treatment, if the data is separated into men and women who both took and didn’t take the treatment, you can increase your chance of survival by 40% by taking the treatment for both males and females. However, based on the same data set, if you combine the groups of males and females into all treated and untreated patients, it only seems to increase your chance of survival by 10% if you take the treatment.

[@10.giriashwin9]

Keep such videos as you may not have a large number of views but your content is better than 90% of content creators out there and that matters the most❤

[@SnoFitzroy]

I didn't know there were people who don't know how to pronounce "centimeter"

[@turnerburger]

I am obsessed with his pronunciation of parallelogram and centimeter

[@amogus5902]

Isn't bertrands paradox kind of obvious??? I figured it was 1/3 just by looking at it for a few seconds, or is it the case of right answer but the wrong way to solve it? Because say your line begins with a point directly on B. 2/3rds of the time, it will connect with another point on the circle between B and A, or between B and C. In this case, because the line does not stretch further than going from point B to A, or point B to C, it is not longer than one of the sides. 1/3 of the time, it will connect to a point between A and C. In this case, the line will stretch longer than from point B to A, or B to C. From there, can't we assume it'll work with every point around the circle, not just the starting point B

[@wiener_process]

The paradox talks about the fact that if we just say "random chord" it's not clear what we mean. What you're describing is if we interpret "random chord" to mean choosing uniformly two random points on the circle and connecting them. However Bertrand himself proposes two other different interpretations, which lead to two different results. In fact, it can be shown there's an infinite number of ways to uniformly select a chord per statement of the paradox. In practice, this means you have to state probability problems carefully and precisely, not specifying the distribution properly leads to some really wacky stuff.

[@Epoch11]

Pallolallogram😂, see you got me to comment

[@johannesvanderhorst9778]

Are we sure that the ant reaches the end of the expanding rope? I mean, when the sum finally reaches 1, it is equivalent with when 1 + 1/2 + 1/3 + ... reaches 100.000. Euler's constant teaches that 1 + 1/2 + 1/3 + ... + 1/n - log(n) lies between 0 and 1 (for logarithms, I use e as base.) Now consider n such that log(n) is close to 100.000 It takes more than e^99.999 seconds for the ant to reach the end of the rope, and with our physical knowledge we don't know better than that before that moment we suffer the heat death of the universe, what maybe stops time itself, and in any way stops the ant and the rope from existing.

[@vincehomoki1612]

This is not about if it's possible in the real world, it's about is the time it takes is infinity or not.

[@extazy9944]

yeah he told us bs

[@quentind1924]

It’s a mathematical problem. It’s perfectly fine to have an eternal ant with infinite energy

[@tommyhubbard342]

Isn'1 #1 omitted variable bais? Or does OVB have the potential to lead to simpsons paradox?

[@ralphinoful]

You told the Monty Hall problem slightly incorrect (or just not rigorously enough). You need to make it perfectly clear that Monty purposely chose door B, because he knew B had a goad behind it. The entire paradox hinges behind Monty purposely choosing that door. You can test it yourself, I'll copy/paste the python code below if anyone wants to test it, just put it into an online compiler. If Monty chooses a door at random, and that door "just so happens to have a goat behind it", the odds that you switch into the car is 50%. import random """Incorrect Way of Wording Monty Hall Problem""" cars = 0 valid = 0 trials = 10000 for i in range(trials): doors = [0,0,1] choice_num = random.randint(0,2) choice = doors[choice_num] doors.pop(choice_num) Monty_num = random.randint(0,1) if doors[Monty_num] == 1: continue doors.pop(Monty_num) valid += 1 if doors[0] == 1: cars += 1 print("Incorrect method of wording Monty Hall Problem:") print("Probability counting all trials = {} Probability only when Monty happens to choose a goat = {}".format(cars/trials, cars/valid)) """Correct Way of Wording Monty Hall Problem""" cars = 0 valid = 0 trials = 10000 for i in range(trials): doors = [0,0,1] choice_num = random.randint(0,2) choice = doors[choice_num] doors.pop(choice_num) doors = [sum(doors)] if doors[0] == 1: cars += 1 print(" Correct method of wording Monty Hall Problem:") print("Probability of car = {}".format(cars/trials))

[@sophiacristina]

I think the Monty Hall problem is more about the broken logic behind it than a paradox by itself. I mean, it is technically a paradox, but would be alike the Zeno's Paradox, which is also technically a paradox, but the problem is that in Zeno's Paradox, the logic of cutting smaller time frames is a wrong reasoning. If A have a higher speed than B and the time frame is constant in size, the paradox disappears. In the Monty Hall i see the same, the problem is considering three doors. The correct way to deal with that is to subtract the door and consider only two doors, then it would be 50% chance and any choice would have equal chance, which is what would happen in real life. When you throw three dices, you calculate the chance of the three dices. But if you have one dice that was already roll, to calculate the chance of the roll of the next two dices, you calculate 6^2 and not 6^3.

[@JoeThomas-lu6fy]

In the Monty Hall problem, switching is guaranteed to change which of the prizes you get, since if you picked the car every other door has a goat, and if you picked the goat, the door that doesn't get opened has the car. Since switching changes whether you get the prize, and your chance of picking the door correctly initially is less than 1/2, switching will always improve your chances. You could just make a table or graph of every possible car location, door selected by the player, and choice whether or not to switch, and you would see that there is a 1/3 chance of getting the car if you stay and 2/3 if you switch.

[@Z_E_B_O]

2:40 Nice Hermann grid.

[@cpov1]

but there's two doors?

[@CaritasGothKaraoke]

I would have expected an AI to know how to pronounce “centimetre”

[@Rando2101]

There are 2 ways to perceive the ant one: -Every second, the ant stop, then the line gets stretched, and the ant keeps moving (uses summation) -The line gets stretched continuously while the ant moves (uses integration) Either way, the ant will eventually reach the end of the line

[@Bruh-ui6cs]

Include inspection paradox next!

[@FelixNielsen]

Monty Hall really is not a paradox. It is a matter of presentation, nothing more. You might change it to picking a door after which you are offered the opportunity to pick and additional door, doubling your chances of winning. There is no difference. When playing normally and one of the doors you didn't pick, one with a goat behind it, is opened, no nee information is presented. You already knew that behind at least one of the doors you didn't pick, would be a goat. Granted, you didn't know behind which door, but this is not relevant. Likewise, where you to pick two doors from the start, you would know that behind at least one of them, would be a goat. What matters is if you've picket both goats or not. There is no difference, it really isn't all that difficult to comprehend and if there is a paradox, it is how easily people can be distracted and fooled into thinking otherwise, simply by changing the presentation while providing exactly the same information. You wouldn't say no to pick and additional door, and this is exactly what is offered, only along the way you are reminded that you will win at least one goat.

[@insignificantfool8592]

Of course I would reject two doors instead of one. Because the fact that the two doors were offered is proof that my initial guess was correct. To think the host would offer two doors if he didnt have to is rather naive.

[@bushido-wo4eo]

bruh, from where do you find such amazing content?

[@ToguMrewuku]

why did you need up Simpsons paradox so badly😂

[@Forcommentingpurposes]

Begging pleading hands and KNEES get a pop filter or balance the audio because I love ur videos but the sound gives me a headache

[@fluffysheap]

Sleeping Beauty and Monty Hall are related problems and the solution mostly depends on whether the problem is stated clearly. Sometimes Monty Hall is only stated, confusingly, as "after you pick a door, Monty opens a door and reveals a goat." In this badly stated version of the problem, there's no way to tell whether Monty opened the door because he knew there would be a goat, or if he just opened a door at random, which could have been either the goat or the car. Half of the people then assume it's the second version, and in this case, with "clueless Monty," it doesn't matter whether you switch or not. You only benefit from switching if Monty intentionally revealed the goat. With Sleeping Beauty, similarly, her guess depends on how often the experimenter is allowed to ask her to guess the coin. If he's only allowed to ask her once, no matter how many times she wakes up, it doesn't matter what she guesses. But if he asks her every time she wakes up, she should guess tails. And in both games, if you don't know what the rules are, you should always assume you're playing the one where you have a correct choice!

[@jeffjo8732]

The MHP and SBP are unrelated. The MHP is based on a misunderstanding of conditional probability. And your discussion of the rules is a red herring - everybody who answers 1/2 is assuming the rules you think need to be made explicit. The correct solution is that there are four possibilities that follow after you pick door A: 1) The car is behind door C, and the host opens door B (Probability 1/3) 2) The car is behind door B, and the host opens door C (Probability 1/3) 3) The car is behind door A, and the host opens door B (Probability 1/6) 3) The car is behind door A, and the host opens door C (Probability 1/6) The misunderstanding is that YOU SEE WHICH DOOR HE OPENS. So in a correct solution, you consider only the possibilities where this condition is met. When it is door B, we know we are in either case 1 or case 3. The error is thinking that they are equivalent, but the are not. The probability that the car is behind door A is found by P3/(P1+P3) = (1/6)/(1/3+1/6) = 1/3. For door C, it is P1/(P1+P3) = (1/3)/(1/6+1/3)=2/3. The SBP is based on confusing a random occurrence with the observation of that random occurrence. The confusing part is that you can observe it once or twice, and you don't know how this one observation fits. That is the crux of the 1/3 solution, which is presented incorrectly here. It works out to be the same, but ignores the difference between random occurrence and observation. The trick in the correct solution, as originally posited by Adam Elga in his paper, is to remove the ambiguity between observations. What if we tell you, after you have been awakened, that it is Monday? Then what you have observed is an unaffected coin toss. That is, Pr(Heads|Monday) = Pr(Tails|Monday). Which means, thru the definition of conditional probability, that Pr(Heads&Monday) = Pr(Tails&Monday). And it you are told, instead, that the coin landed on Tails? Then you have no information that would make either day more likely. That is, Pr(Monday|Tails) = Pr(Tuesday|Tails). And that means that Pr(Tails&Monday) = Pr(Tails&Tuesday). But since Pr(Heads&Monday) = Pr(Tails&Monday) = Pr(Tails&Tuesday), and they must add up to 1, each must equal 1/3. This only looks like the solution in the video. There, it was asserted that these three probabilities must be equal. The correct solution proves it. The halfers argument is that these are three different random occurrences, and you can't make the assertion. The (correct) thirder argument is that they represent the combination of observation and occurrence, and that it is proven.

[@whoeveriam0iam14222]

Cent-imeter? Is that how you pronounce that

[@Malik-ul2sk]

"Palallelogram"

[@ogermite888]

nice videos man

[@spencersmustache2844]

Centimitter

[@rubyace7058]

why does bro pronounce centimeter like that

[@johndrico]

Because it's read by an AI. Compare the voice in this video to the previous one.

[@80Mollusc]

Bro also said pallelelogram

[@josephblake5120]

probably a joke about how kilometer is mispronounced - in international English it is spelt centi-metre and the first syllable of "metre" is always stressed in those words. "Meter" refers to something that measures a quantity, e.g. barometer, pedometer, iambic pentameter, etc. and is always unstressed. A "kilometer" would be something that measures thousands, but a "kilometre" is one thousand metres, and pronounced with an unstressed and stressed "me-" syllable respectively, but for some reason a lot of (presumably American) speakers choose to pronounce the distance with an unstressed "me-" syllable, which sounds absurd when you pronounce other SI lengths in that way, like centimetre or millimetre

[@francoisrobidoux7003]

ur monty hall explaination way too over complicated if u chose the right door and switch u lose if u chose a wrong door and switch u win there are 2/3 chance u chose the wrong door so switching is better

[@raresgabriel8991]

Absolutely great content

[@jovetj]

7:32 "sen-TIM-it-ur"? You mean "SEN-tee-mee-tur"?

[@oucyan]

What if I want the Goat?

[@fluffysheap]

As a sheap, I find this solution preferable

[@davidheiser2225]

The Monty Hall Problem isn't a paradox.

[@Rando2101]

Yeah, it's just counter-intuitive, so it feels like a paradox

[@wyattkimball6811]

It may be not but your face is 😂

[@supergalaxycosmo1120]

How do you not have more views wtf first?

[@jaysn1683]

The ant one is sus, just because both the ant‘s travelled distance and the rope‘s length tend towards infinity doesn’t mean they are the same size. The distance between the ant and the rope‘s end tends to infinity after all.

[@Vaaaaadim]

We don't track the ant's travelled distance nor the rope's length. We track the amount of progress that the ant makes as a fraction of the rope's length. This progress value is 0 when the ant has just started, and 1 when it has reached the end of the rope. The amount of progress becomes 1/100,000 then 1/100,000 + 1/200,00 then 1/100,000 + 1/200,00 + 1/300,000 and so on. This corresponds to H(n)/100,000 where H(n) is the n-th harmonic number. H(n) diverges to infinity, so at some point it reaches/goes past 100,00 so at that point the progress value becomes 1 and the ant has reached the end of the rope.

[@LevinFroggo-fs7uu]

@@Vaaaaadimthat is true, but only if the ant on the string is also moved when the string is stretched, which wasn’t explained in the video

[@Rando2101]

​@@LevinFroggo-fs7uudepends on what he meant by "uniformly stretched"

[@fullfungo]

11:10 the shape is called a parallelogram, not palallelogram. From the word “parallel” You misspoke several times.

[@Ru5514n]

Hard to pronounce for some people, man :)

[@bc5852]

one of the worst explanations for the monty hall problem i’ve heard

[@gregc2236]

Please speak slower in your videos, I have to change the speed to 0.75% to follow some of the more complex paradoxes...

[@katakana1]

palallelogram