The whole point is when things are infinite even if something has almost no probability of happening it almost certainly will if the chance is not zero. It actually makes perfect sense, the whole William Shakespeare thing is meant to be a dummy argument to show that even in ridiculous cases it could still be true.
Doublelift's reaction at @22:45 was my exact reaction when hearing Sneaky argue that changing the door after 98 wrongs ones are revealed is "based on feeling whether you chose the right one at the first try?" LMAO
on another note, I love and respect how calm meteos is and how he helps sneaky out. Sneaky isn't dumb for not understanding a puzzle thats meant to be extremely confusing, just like you're not a loser for being in bronze.
DL and Meteos didnt exactly explain it the best but its like Sneaky is spending all of his brain power trying to think and literally everything that is said to him is just white noise. It's like he is trying to understand so hard that he is missing the relatively simple explanation they are trying to tell him
Not really, monkeys are animals and animals adhere to certain behaviors, so their actions aren't truly random. Like, it'd probably need to be an uncountably infinitely long amount of time (vs countable infinity). Like, Sneaky is right, a monkey smashing a typewriter would do quite a bit wrong, and probably not hit the keyboard much at all unless it's somehow hardwired to do that. Like, it's probably much more understandable if it was "a program spewing random shit of undefined length will eventually output the complete works of Shakespeare"
@@eonnephilim852 Okay, so you're entirely misunderstanding the theorem. First off, this is a monkey that's been trained to type on a typewriter. It doesn't understand what it's typing all it knows is that it must type, be it for rewards or whatever. In the case of infinity, there will be an infinite number of texts written completely by random choice. even if it is infinitesimally small chance it will eventually happen being ran an infinite number of times.
He’s not dumb for not getting the right answer. He’s dumb for not understanding it after it’s explained to him in several different ways for 20 minutes
I feel sneaky is missing the point in the monkey theorem. A computer can simulate random key presses very fast. I'd assume a computer would write Shakespeare before the monkey. The theorem makes a lot of assumptions because obviously the monkey wouldn't live forever or realistically hit the keyboard much. Just like those assumptions I'm certain you are also to assume the randomness of the monkey is truly random. I think the point of the theorem is that you can do a finite amount of things in an infinite amount of times, not that a monkey would write Shakespeare
You're basically reiterating Sneaky's point albeit more eloquently. He already stated the multiple assumptions and the monkey's actions not being truly random.
@@kidathlete I'd have to watch it again but sneaky said it wouldn't happen because a monkey wouldn't truly be random. I'm saying it would it happen because the theorem needs those assumptions to be true. Many math and science problem make a lot of assumptions to make the problem more easily solvable. I believe the theorem was written before computers, so the monkey was just an example of randomness. If the the theorem had been written today I'd bet it would say computer and not monkey. When learning early physics problems lots of times you will ignore friction because it complicates the actual point of the problem.
"I think the point of the theorem is that you can do a finite amount of things in an infinite amount of times, not that a monkey would write Shakespeare" >>> You can do an infinite amount of things in an infinite amount of time.
@@judgedbytime Yes, but the entire works of Shakespeare is still finite. Have you heard of the hyperwebster? It's pretty much the same idea as the infinite monkey theorem except all possible text is in alphabetical order.
Here’s a table for the monty hall problem to prove the logic, depending on which door the goat is behind: Goat Door: Optimal choice for Door 1, Door 2, Door 3 Door 1: Stay, Switch, Switch Door 2: Switch, Stay, Switch Door 3: Switch, Switch, Stay So 3/9 (1/3rd) of the time you should stay, and 6/9 (2/3rds) of the time you should switch So the outcomes pan out. A good way of thinking about it could be the reverse: the door you initially choose has a 2/3rds chance of being a goat. By switching, you lower that to 1/3rd.
My best attempt at simplifyin the Monty Hall problem for people still confused: Divide the problem into two scenarios: "always change your initial pick" versus "never change your initial pick". In the first scenario you win if you pick a goat on your first try, because here we always change the pick to what is necessarily a car. This is a 66% chance to win. Likewise, if you pick the car in this scenario you lose, 33% chance. In the second scenario you win if you pick the car initially, 33% chance. You lose if you pick any of the two goats because we're not changing, 66% chance. Now you can pick what you want to compare: You either compare the chance of winning in the "always change your initial pick" scenario with "never change your initial pick", which is a 66% chance to win vs a 33% chance to win. Or you compare the chance to lose on both scenarios and the lowest one is the best one. Again, 33% chance on scenario 1 vs 66% chance on scenario 2. Meaning "always change your initial pick" is the best situation from a probability standpoint.
If the monkey has an infinite amount of time, then it will eventually write every possible combination of words and sentences. The monkey will have written every thing that’s ever been written before.
No a monkey could not. If it was a computer typing actually randomly then yeah. But no a monkey cannot type words ever. Even with infinite time. That's what sneaky was saying.
@@victorortero8743 why wouldnt it be able too? Assuming the monkey hits a random key every time, in an infinite amount of time the monkey will have hit every possible random combination of letters, including those that make up any of Shakespeare's works. The monkey will have typed every thing possible. even some random shit that you or I wrote out in the 3rd grade.
@@alidasoo6654 The problem is words and sentences aren't random. There are letters that are used more frequently than other letters (that's why when you play hangman if you guess Z to start you're fucking trolling) and words used more than others (and, a, the, etc.). There's no way the monkeys are going to hit the keys in non-random enough ways to actually form an entire story. Think of a different example. If you had an infinite number of dryers drying an infinite amount of clothes for an infinite amount of time, would your clothes ever come out of the dryer already perfectly folded? No, of course not, because a dryer never moves clothes in a way that folds them. Well, a monkey never hits keys in a way that forms sentences so why would you ever expect them to be able to type a story if your dryer will never fold your clothes for you?
If you had a "random word generator" that would chose a random character for infinity, there is a chance that the sequence of characters that are "chosen" during that randomization can equal Shakespeare.
To help them understand door probability they need to imagine how a higher chance just represents the fact that reiterating a certain strategy will lead to a certain amount of successes out of the total attempts. E.g. 66% means changing options each time you are in this situation over ten attempts will lead to approximately 6 successes.
I think the best way to look at the monty hall problem is to think about the "flip" as just inverting your choice. If you chose car initially, you get goat after the switch, if you chose goat first you'll get the car. So really you're just doing your first pick, which has a 2/3 chance of being goat (getting you the car after the switch), and 1/3 of being the car (resulting in a goat). So if you always switch, you have a 2/3 chance of winning the car. If you never switch instead, you have a chance of 2/3 to be on goat, 1/3 to be on car. In total, not switching therefore has 1/3 chance of winning, whereas switching wins in 2/3 of cases.
you have a 2/3 chance of picking a door with a goat. If you pick a goat, he will reveal the other door with a goat, meaning the remaining door has the car. There's only a 1/3 chance that you chose the car initially.
It's like saying Pi has your phone number in it. Because there are infinite numbers in Pi and thus infinite combinations, eventually somewhere it will be there
I've gone over this problem before and also couldn't believe it was real. Made a program to simulate it, ran it 100000 times, switching worked 66.63% of the time. Basically 2/3rds. It doesn't make sense but it's right
It works because of the simple fact that the person revealing the doors *knows* which one is the car. He is intentionally avoiding it. So which is more likely? Your first guess being correct, or the person revealing the doors intentionally dodging the car door?
It makes all the sense if you look at it this way: Divide the problem into two scenarios: "always change your initial pick" versus "never change your initial pick". In the first scenario you win if you pick a goat in your first try, because here we always change the pick to what is necessarily a car. This is a 66% chance to win. Likewise, if you pick the car in this scenario you lose, 33% chance. In the second scenario you win if you pick the car initially, 33% chance. You lose if you pick any of the two goats because we're not changing. Now you can pick what you want to compare: You either compare the chance of winning in the "always change your initial pick" scenario to the"never change your initial pick", which is a 66% chance to win vs a 33% chance to win. Or you compare the chance to lose on both scenarios and the lowest one is the best one. Again, 33% chance on scenario 1 vs 66% chance on scenario 2. Meaning "always change your initial pick" is the best situation from a probability standpoint.
It makes perfect sense, switching always gives the opposite of what you first picked. So [Goat,Goat,Car] essentially becomes [Car,Car,Goat] if you switch. Meaning switching gives you a 2/3 chance of getting a car while not switching gives a 1/3 chance. No need to over-complicate it, just break it down to the most base level of logic.
it makes a lot of sense. if you pick a door and the revealed door is a goat, switching will always make you right 66.6% of the time. Let’s say the 1st door is chosen and the 3rd door is always revealed. [Goat, Car, Goat] [Goat, Goat, Car] [Car, Goat, Goat] Out of the 3 only possibilities, 2/3 of the games would have won if they switch doors from the original choice.
Everyone should look up "21 3 door scene" on TH-cam. It's explains this well and why switching doors is anyways a higher Probability of winning the car.
You have a %66 chance to initially pick a goat. So swap every time because you have a %66 chance that the swap is correct. Right? Idk man I just work here
It suppose to help people make decision not just always swap, I believe the process is you calculate the possibilities then you decided for yourself wether you swap your choice or not. I guess sneaky try to word something like this but he just too mald.
I think sneaky is having a hard time understanding that the first choice affects the second choice when there's only 2 doors left. He still thinks it's a 50-50 because there are 2 doors.
The easiest way to see it is imagine your have 5 billion doors. If the host reveals all but your door and the other then it’s obvious you should swap because your initial choice is basically 0% chance to be the car.
Some facts about the infinite monkey theorem: Not only would it type the complete works of shakespeare, but it would eventually type it all in chronological order by release date without any gaps in between. This would work either with a) one monkey, one typewriter, and an infinite amount of time, b) an infinite amount of monkeys, infinite typewriters, and some of them will succeed on their first try.
This is my favorite content! lol For the Monty Hall Problem: What I don't see mentioned in the comments, that is useful to know. The probability does not become 50/50 after a door is taken away. It would become 50/50 if the car and goat were re-randomized, but they are not. When the initial choice is made, there is a 33% chance of choosing the car. Since the doors are not randomized after opening a goat it is like being able to open 2 doors at once if you switch. Combining that probability of 33% of each door being the car into the one remaining door. Creating the 66% percent chance probability to find the car by switching. Maybe the explanation helps, maybe it doesn't. There are plenty of math classes that dedicate a day to discussing this problem.
Worth noting that it’s would also be 50/50 if Monty could reveal the car or if he could reveal your door. His knowledge is a key part of why the problem works the way it does.
When the Monty Hall problem first became popular in like the 80s or something the woman with the highest IQ in the world gave the answer that you should always switch. People all over the US heard about her answer and told her she was wrong, including dudes with PhD's in Math. So i don't think its unreasonable for Sneaky to not get it instantly. It inherently doesnt make sense to people.
Sneaky's confidence made me revisit the Monty Hall problem and I came to the conclusion that there are three unique, but sensible, interpretations to the problem, as posed, that lead to vastly different results. I will share them here. First, let's establish the exact wording of the Monty Hall problem: "Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?" Here are the three interpretations and the related solutions (i.e. "optimal strategies"): Interpretation 1 (The "classical" interpretation): The game show host knows what's behind the door, and regardless of which door you select, the host is commited a priori to opening a door and opening a door that does not have the car. In this case, all the classical logic holds, the best strategy is to always switch to win 2/3 of the time because the host deliberately removes a wrong answer for you. Interpretation 2 (The "randomized" interpretation): The game show host knows what's behind all the doors but he has no intention of using it. He is going to let you select a door, and then he is going to open one of the two remaining doors at random, even if it contains a car. Looking at the statement of the problem, this scenario is not completely infeasible. In this case, switching or not switching doesn't matter, conditional on seeing a goat (the reason being that in this setting the host is more likely to reveal a goat if you selected a car first, so seeing a goat makes you more likely to believe you already found the car). Both doors have a 50% chance of being the car. Since this is your first time on the game show, you have know way of knowing whether or not the game show host picked the door at random or not, just as Sneaky was arguing. In my opinion, this interpretation seems hard to justify given that the game show host knows what's going on. Interpretation 3 (The "adversarial" interpretation): In this setting, the game show host is doing his best to prevent you from winning a prize. Even more, the game show host is smart, and he has heard of the Monty Hall problem before. Given the fact that it has entered popular culture, he suspects his audience member to know that "switching is always best". With this in mind, the host uses the following strategy to take advantage of these players: If the contestant selects a door with a goat, open it immediately, say "sorry, you didn't win this time" and conclude the game show. Otherwise, if the contest selects a door with the car, open one of the two doors with the goat and offer them a chance to switch (to the door that has the other goat). In this case, switching will guarantee that you lose because you only ever get a chance to switch given that you were about to win. So in this case, the optimal strategy is to *not switch*, which is the opposite of the classical interpretation. This interpretation is completely consistent with the problem as presented in the original quote and unlike interpretation 2 uses the fact that the host has knowledge in a logical way. Hope you guys found this as interesting as I did, feel free to reply if you feel like I said something objectionable. :)
The only big problem is that the host always opens a door with a goat behind it, thats a stated condition. Its entirely a probability problem that looks conditional because it has 2 steps : first choice + choice to keep first choice or change, but in reality boils down to evaluating each scenario ("always change" vs "never change") and realizing that its a 66% to a 33%. It just intuitively seems odd not knowing whats behind your first pick and thats where most ordinary people, most university students and a hand full of mathematicians got stuck on. If the 3 interpretations are just you doing thought experiments then disregard everything I just said. I say this because the motive behind the host's decision to open a door that necessarily has a goat behind it is irrelevant to the problem.
Problem is that those other interpretations are separate problems. The problem assumes the host has to follow the rules and always shows you a goat. You've just kind of made up random scenarios that aren't possible within the confines of the actual question.
I think sneaky understands the infinite monkey theorem, but he just doesn't believe a monkey is true random. The monty hall problem can be easily explained. The chance of initialy being right is 1/3. After the host opens a door, the only time the remaining closed door is wrong, is if your initial pick was correct. Your odds don't get redistrubuted to a 50/50. I somehow hope sneaky reads this and understands, because part of me died inside watching this. I hope I can sleep at night, because the urge to help him understand is eating away at me.
the example with a hundred doors is super simple. The chance that you pick the car initially out of a hundred doors is super small. So it's better to switch, knowing the other one has the same potential as your first choice.
RE: Monty Hall, switching doors or perceiving a difference in odds is the same as picking a lottery number that hasn't come up in the lottery recently. They are isolated events. If you pick a door and then the host removes another door, you now get to choose completely from scratch. Although the question manipulates you with the word "switch", your choice has been reset. Now you get to choose door 1 or 2, and the only difference in odds is whether or not you believe a door that was once correct is either less or more likely to have the correct answer given its success in the past, which of course it isn't. If you "stick" with door one, it has the enhanced odds of knowing despite selecting that door, FOR SURE, it was not the negative result from door 3, just as much as, FOR SURE, door 2 is not the negative result of door 3.
I've never heard of the monkey theorem like that, just call it what it is: Given an infinitely long string of random letters, any possible text will probably appear in it at some point.
The reason the 100 door analogy makes sense is because math is logic. Initially you have 1/x chance of guessing correct given x doors total. Finally you have 1-1/x chance of being correct if you switch because the only way you'd be correct by switching is if you DIDNT originally pick correctly (assuming one of either yours or the one he didnt reveal is always correct). Either you initally guess correct (call this event GC) or you dont initially guess correct (call this event DGC), and since exactly one of these has to be true, we know the sum of their probabilities is 1, or in other words we know 100% that if we guess we will either be right or we will be wrong, that is a logical fact. Converting that logic to math: p(GC)+p(DGC)=1 --> p(DGC)=1-p(GC) --> Where p(X) represents the probability of event X occuring. In the case of the doors, we have p(DGC)=1-1/x Where the probability you initially guessed correct is p(GC)=1/x, because only 1 of the x doors was correct. Thus the probability you did not originally guess correct, which is the same probability that you'd switch to the correct door if you switch, is 1-1/x. And 1-1/x>1/x for all x>2, so your odds are always better if you switch. Examples: 3 doors (x=3): p(GC)=1/3 p(DGC)=1-1/3=2/3 p(DGC)>p(GC) Better off switching 20 doors (x=20): p(GC)=1/20 p(DGC)=1-1/20=19/20 p(DGC)>p(GC) Way better off switching 1000 doors (x=1000): p(GC)=1/1000 p(DGC)=1-1/1000=999/1000 p(DGC)>p(GC) Waaaaaaaay better off switching. The more doors you start with, the more wrong doors you eliminate, and the better your odds become by switching. Going to 100 doors really helps to show that trend, but when you only have 3 doors the difference in probabilities isn't drastic enough for most people to even consider it. Instead most people will tackle the problem using emotion like sneaky, rather than logic and reason. Subconsciously, you know how much more disappointed you'd be if you switched from the winner to the loser so you justify not switching. Sneaky is saying logic and math are different, but they are the same. What he is thinking of as logic is actually human intuition. Humans dont always live their lives in the most logical way possible, but since we continue to live that way, its what makes it feel logical to us
So if you still don't understand the monty hall problem, what you should you is play the same scenario and keep your pick while changing the position of the car, then do the same but switch after the host opens a door with a goat, you'll get better results with the second one than with the first one.
The first time you choose a door, you are essentially choosing a singular door. But once one door is revealed, choosing again is like choosing two doors instead of just 1.
I feel like it’s easier for not math people to just speak what happens, “On my first choice I probably picked a goat because there are more goats than cars. If the other goat is revealed then the last one is probably the car because my first choice was probably a goat.” “Everything happens in an infinite amount of time. If everything happens it doesn’t matter how improbable, it has to eventually happen because everything must happen sometime and infinity is all the times.”
i think the one of the key statement is this is a game showcasing overall averages of games, but the theory does not imply switching will always win. the game is to show that you get a free 33% most of the time if you test this game enough times.
Door: 1 2 3 Scenario1: C G G --> switch from door 1 to door 2 = win goat Scenario2: G C G --> switch from door 1 to door 2 = win car Scenario3: G G C --> switch from door 1 to door 3 = win car Door: 1 2 3 Scenario1: C G G --> stay with door 1 = win car Scenario2: G C G --> stay with door 1 = win goat Scenario3: G G C --> stay with door 1 = win goat After a goat has been revealed behind door 3, there remains a car and a goat behind the other 2 doors. In 2 of the 3 possible scenarios, a goat will be behind door 1, so when 2 doors remain closed in each scenario, switching means switching from a goat to the car in 2 out of 3 cases. Because there is a goat behind one door and the car behind the other door, the brain quickly thinks the odds are 50/50 whether you stay or switch, but this is not the case.
You can definitely turn some of these scenarios into a League context for a game show or even just game theory like Meteos was saying. The Monty Hall Problem could be like: "You are a jungler trying to gank mid, and there were 3 possible routes. One route will be clear of an enemy player, and the other two will have you run into an enemy player; pick a route. Then, your mid laner wards one of your possible routes to reveal an enemy player in that route. Will you keep going on the same route you picked initially or change?" The Infinite Monkey Theorem is basically just me playing ranked, and eventually I'll hit Challenger from Silver if I play an infinite amount of games... Right? :')
This only works if somehow the midlaner knows that the route he warded is NOT the correct route, and of course that he happens to not choose to ward the same route that you picked for some reason
Yes, that should be the go to when initially explaining the problem. For some reason the common initial presentation of three doors trips people up a lot. I would even ditch 100 doors initially and go for something ridiculous like a trillion, so that there would be basically zero chance that someone could get tripped up like sneaky did. Once they understand the concept, they are usually receptive enough to accept the truth when it comes to the three door scenario.
The Monty Hall Problem is based off the statistical assumption that your door does not contain the car, meaning that the other door contains the car and that you should switch.
I understand Sneaky's frustration here and I'm sure having chat malding at him isn't helping lol. When I was presented with the Monty Hall Problem in my Probability class I was also super confused at first, but they also do a really poor job of explaining it until they fully understand it themselves. I think if somebody explained originally that there is no 50/50 ever in the problem then it would've made more sense to him lol.
My favorite explanation for the Monty Hall problem is by altering the game rules while retaining strategic equivalence Original Game: -Player picks door A -Host reveals empty door B -Player can stay with door A, or swap to door C -Player wins prize behind picked door Game 2: -Player picks door A -Host reveals empty door B -Player can stay with door A, or swap to door C -Player wins best prize behind picked doors Game 3: -Player picks door A -Host reveals empty door B -Player can stay with door A, or swap to doors B and C -Player wins best prize behind picked doors Game 4: -Player picks door A -Player can stay with door A, or swap to doors B and C -Host reveals empty door B -Player wins best prize behind picked doors Game 4: -Player picks door A -Player can stay with door A, or swap to doors B and C -Player wins best prize behind picked doors It's clear from Game 4 that, assuming equal probability of each door containing a car, it's better to choose doors B and C, as collectively they have a 2/3 probability of containing the car. It's also quite easy to convinced one's self that each iteration of game is equivalent.
Do the problem with 100 doors pick one, host shows 98 and the logic behind this problem becomes stupidly obvious. If by then you still think its a 50 50 then you're ignoring the fact that when you picked a door you only had a 1% chance of picking the right one to begin and if you don't switch you're insane.
I get the month hall problem, but sneaky does have a point. Intuitively the result shouldn't change no matter if you switch doors or not. I wonder if someone did a lot of simulations of this to get some probabilities
You can write a basic function to test it and it will always come out as switching having 66~% win rate. The logic is sound and intuitively works since switching makes you win if you get a goat, and there's a 2/3 chance of picking a goat.
Klaus 74 Wrong because the POINT is that the car WILL be either behind the door you picked or the door the host left up. He’s purposely giving you better odds. There’s no trick just probability.
I still love to hate on the Monty Hall problem. Probability is whack even after taking it at a college level. You're not alone sneaky, this question got people even halfway through our semester (specifically the half that hadn't already studied it to death cause we all needs lmao.)
It just gets explained like shit. All you need to know is that switching always gives you the opposite of what you picked first time since a goat is eliminated, leaving one goat and the car. So since you have a 2/3 chance of getting a goat on your first try switching gives you a 2/3 chance of getting the car. People confuse themselves by focusing on the probability of the door you switch to when the only actual relevant one is the probability of picking a goat from the initial 3 doors.
@@Sirhaddock yeah, I think a lot of explanations skip out on the key point that Monty always reveals a goat from the remaining doors. It seems small but it makes the difference between 50/50 and most people I have spoken too seem to get it once that has been clarified.
@@victorortero8743 With infinite time, it will eventually hit every possible combination. Its Infinite. They talked about an experiment, it didn't go well, but if they can hit one letter eventually, by pure chance and time, it would happen.
Not necessarily, the odds of your initial choice can change depending on the scenario. Like in deal or no deal, you start with a 1/26 chance but end up with a 1/2 chance when you reach the final two boxes.
This is pretty good, i also already like Meteos but didn't like DoubleLift but on the co streams he seems like a decent human being, this three together works very well. It is interesting to see how the three of them understand this in different ways, you have DL liking what he reads but you dont know if he actually understands the theory behind, you have Meteos thinking out of the box and Seaky in complete denial of everything xD Bring them physics not just mat and probability please ^^
Sneaky was kinda small brain here tbh. If a monkey types random words, there is an incredibly small chance to type a work of Shakespeare. Incredibly small chance with infinite tries will succeed eventually.
If the option is only between 3 doors I agree with Sneaky, it's still a 50/50 and it doesn't matter if you switch or not. If you apply this to 100 doors then you had a 1% of picking the door with the car. So you should switch in this case because it goes up to a 50/50 chance but it's way more likely that the car is behind that door since they revealed 98 incorrect doors. Make sense?
Yeah but in the 3 door scenario, you also had 33% chance of picking the door with the car and 66% chance of picking a goat, so you'll most likely to pick a goat. Therefore, switching is the best option. The chances aren't really high but it is the best option for sure. The only way this is not applicable is that if you picked the car in the first place which is 1 out of 3 chance you'll pick it. Makes sense?
It does matter, you can run a bot that proves it. Switching wins more often since switching always gives you the opposite of what you started with, and you had a 2/3 chance of starting with a goat.
@@lifeofrj9707 I get what you are saying and it does make sense, however mathematically it doesn't change. I understand it's the same logic but in this case with only 3 doors it doesn't change. Yes it technically jumps up to a 50/50 chance but it's basically like the 3rd door was never there in the first place if they show an incorrect 3rd door.
@@IamFallen77 You are contradicting yourself. You said that mathematically it doesn't change but it went up to a 50/50 chance? Wtf? hahahah. The chances went up because the door with a goat (incorrect door) was shown.
So given enough time, Doublelift will eventually get out of groups?
-the infinite Doublelift theorem
No way he wouldn't live long enough
Depends if any of the others can press their F key?
The whole point is when things are infinite even if something has almost no probability of happening it almost certainly will if the chance is not zero.
It actually makes perfect sense, the whole William Shakespeare thing is meant to be a dummy argument to show that even in ridiculous cases it could still be true.
I don't know whether or not this aged well... Because...
I love how amused DL and Meteos are listening to Sneaky's breakdown XD
That's a great name for a podcast with these three: The Infinite Monkey Theorem
The Infinite Monkey Threerem
@sneaky make it happen
I'm so sad you left out the "Triple Monkey Theorem" part, that made me laugh more than anything else lmao
Doublelift's reaction at @22:45 was my exact reaction when hearing Sneaky argue that changing the door after 98 wrongs ones are revealed is "based on feeling whether you chose the right one at the first try?" LMAO
Meteos is actually quite smart and elaborate in how he presents the solution, props to him!
on another note, I love and respect how calm meteos is and how he helps sneaky out. Sneaky isn't dumb for not understanding a puzzle thats meant to be extremely confusing, just like you're not a loser for being in bronze.
bronze mentality
im actually malding at sneakys understanding of this
DL and Meteos didnt exactly explain it the best but its like Sneaky is spending all of his brain power trying to think and literally everything that is said to him is just white noise. It's like he is trying to understand so hard that he is missing the relatively simple explanation they are trying to tell him
Actually so infuriating listening to him talk 😂😂
Literally want to smash everything
Not really, monkeys are animals and animals adhere to certain behaviors, so their actions aren't truly random. Like, it'd probably need to be an uncountably infinitely long amount of time (vs countable infinity). Like, Sneaky is right, a monkey smashing a typewriter would do quite a bit wrong, and probably not hit the keyboard much at all unless it's somehow hardwired to do that.
Like, it's probably much more understandable if it was "a program spewing random shit of undefined length will eventually output the complete works of Shakespeare"
@@eonnephilim852 Okay, so you're entirely misunderstanding the theorem. First off, this is a monkey that's been trained to type on a typewriter. It doesn't understand what it's typing all it knows is that it must type, be it for rewards or whatever. In the case of infinity, there will be an infinite number of texts written completely by random choice. even if it is infinitesimally small chance it will eventually happen being ran an infinite number of times.
He’s not dumb for not getting the right answer. He’s dumb for not understanding it after it’s explained to him in several different ways for 20 minutes
"Thinking of it in terms of just math, it makes zero sense."
-Sneaky on probability.
I don't think the explanations were really helping :p
@@judgedbytime i actually disagree, I think several ways meteos explained should be good enough to understand
Sneaky showed he is the first peak in the Dunning-Kruger graph with his take on the Monty Hall problem.
I love this format. A complication of the Costreams' random discussions is great
As someone who understands the math behind these puzzles it was equal parts painful and hilarious watching this.
I can already tell this will absolutely torture me
I feel sneaky is missing the point in the monkey theorem. A computer can simulate random key presses very fast. I'd assume a computer would write Shakespeare before the monkey. The theorem makes a lot of assumptions because obviously the monkey wouldn't live forever or realistically hit the keyboard much. Just like those assumptions I'm certain you are also to assume the randomness of the monkey is truly random. I think the point of the theorem is that you can do a finite amount of things in an infinite amount of times, not that a monkey would write Shakespeare
I guess the confusing part is saying a monkey can type anything. Obviously it can't.
You're basically reiterating Sneaky's point albeit more eloquently.
He already stated the multiple assumptions and the monkey's actions not being truly random.
@@kidathlete I'd have to watch it again but sneaky said it wouldn't happen because a monkey wouldn't truly be random. I'm saying it would it happen because the theorem needs those assumptions to be true. Many math and science problem make a lot of assumptions to make the problem more easily solvable. I believe the theorem was written before computers, so the monkey was just an example of randomness. If the the theorem had been written today I'd bet it would say computer and not monkey. When learning early physics problems lots of times you will ignore friction because it complicates the actual point of the problem.
"I think the point of the theorem is that you can do a finite amount of things in an infinite amount of times, not that a monkey would write Shakespeare" >>> You can do an infinite amount of things in an infinite amount of time.
@@judgedbytime Yes, but the entire works of Shakespeare is still finite. Have you heard of the hyperwebster? It's pretty much the same idea as the infinite monkey theorem except all possible text is in alphabetical order.
Here’s a table for the monty hall problem to prove the logic, depending on which door the goat is behind:
Goat Door: Optimal choice for Door 1, Door 2, Door 3
Door 1: Stay, Switch, Switch
Door 2: Switch, Stay, Switch
Door 3: Switch, Switch, Stay
So 3/9 (1/3rd) of the time you should stay, and 6/9 (2/3rds) of the time you should switch
So the outcomes pan out.
A good way of thinking about it could be the reverse: the door you initially choose has a 2/3rds chance of being a goat. By switching, you lower that to 1/3rd.
Next week I want the boys take on the trolly problem. Followed by the organ donor problem.
What's the organ donor problem?
No way this is made into a seperate video 😂 I love these co streams so much
All joking aside, Meteos is a great friend here.
i'm hoping for these three to have a discussion like this again this weekend. they're way more entertaining than the LCS games.
Sneaky pronouncing it “shakesphere” is too good
sneaky is monkey king
My best attempt at simplifyin the Monty Hall problem for people still confused:
Divide the problem into two scenarios: "always change your initial pick" versus "never change your initial pick".
In the first scenario you win if you pick a goat on your first try, because here we always change the pick to what is necessarily a car. This is a 66% chance to win. Likewise, if you pick the car in this scenario you lose, 33% chance.
In the second scenario you win if you pick the car initially, 33% chance. You lose if you pick any of the two goats because we're not changing, 66% chance.
Now you can pick what you want to compare: You either compare the chance of winning in the "always change your initial pick" scenario with "never change your initial pick", which is a 66% chance to win vs a 33% chance to win.
Or you compare the chance to lose on both scenarios and the lowest one is the best one. Again, 33% chance on scenario 1 vs 66% chance on scenario 2. Meaning "always change your initial pick" is the best situation from a probability standpoint.
this is what the people needed thank you editor.
If the monkey has an infinite amount of time, then it will eventually write every possible combination of words and sentences. The monkey will have written every thing that’s ever been written before.
yeah thats the thing sneaky don't get it
No a monkey could not. If it was a computer typing actually randomly then yeah. But no a monkey cannot type words ever. Even with infinite time. That's what sneaky was saying.
@@victorortero8743 why wouldnt it be able too? Assuming the monkey hits a random key every time, in an infinite amount of time the monkey will have hit every possible random combination of letters, including those that make up any of Shakespeare's works. The monkey will have typed every thing possible. even some random shit that you or I wrote out in the 3rd grade.
@@alidasoo6654 a lot of ppl really ignore the word infinite or straight up don't get what it means it seems
@@alidasoo6654 The problem is words and sentences aren't random. There are letters that are used more frequently than other letters (that's why when you play hangman if you guess Z to start you're fucking trolling) and words used more than others (and, a, the, etc.). There's no way the monkeys are going to hit the keys in non-random enough ways to actually form an entire story.
Think of a different example. If you had an infinite number of dryers drying an infinite amount of clothes for an infinite amount of time, would your clothes ever come out of the dryer already perfectly folded? No, of course not, because a dryer never moves clothes in a way that folds them. Well, a monkey never hits keys in a way that forms sentences so why would you ever expect them to be able to type a story if your dryer will never fold your clothes for you?
This was unknowingly needed.
Three college dropout who plays games for a living talks about maths. Nice.
And they still have more money in the bank then you.
@@kahrs4465 wait till he finds out that high elo LOL players r actually smart asf lmfao
Infinite monkey theorem has more to do with logic than math fam
@@athena3234 logic is literally a field in mathematics
@@davidwen1900 also philosophy
door 1 is a car is the only bad situation to not change, where as door 2 or 3 car is a win when you change, so its a 66% to win on change
The Triple Monkey Theorem:
Given enough time, Doublelift, Meteos, and Sneaky will produce an LCS costream in which they actually discuss the match.
If you had a "random word generator" that would chose a random character for infinity, there is a chance that the sequence of characters that are "chosen" during that randomization can equal Shakespeare.
Sneaky is the reason that the Monty hall game show made money
To help them understand door probability they need to imagine how a higher chance just represents the fact that reiterating a certain strategy will lead to a certain amount of successes out of the total attempts. E.g. 66% means changing options each time you are in this situation over ten attempts will lead to approximately 6 successes.
I think the best way to look at the monty hall problem is to think about the "flip" as just inverting your choice. If you chose car initially, you get goat after the switch, if you chose goat first you'll get the car. So really you're just doing your first pick, which has a 2/3 chance of being goat (getting you the car after the switch), and 1/3 of being the car (resulting in a goat). So if you always switch, you have a 2/3 chance of winning the car.
If you never switch instead, you have a chance of 2/3 to be on goat, 1/3 to be on car.
In total, not switching therefore has 1/3 chance of winning, whereas switching wins in 2/3 of cases.
you have a 2/3 chance of picking a door with a goat. If you pick a goat, he will reveal the other door with a goat, meaning the remaining door has the car. There's only a 1/3 chance that you chose the car initially.
It's like saying Pi has your phone number in it. Because there are infinite numbers in Pi and thus infinite combinations, eventually somewhere it will be there
There was also a really bad input design which required you to move an 8 digit slider around pi to enter your number
Me: picks car
Host: shows me a goat
Me: goes home with a goat
Also me: F$ck your math homie…
I feel you sneaky.
I've gone over this problem before and also couldn't believe it was real. Made a program to simulate it, ran it 100000 times, switching worked 66.63% of the time. Basically 2/3rds. It doesn't make sense but it's right
It works because of the simple fact that the person revealing the doors *knows* which one is the car. He is intentionally avoiding it. So which is more likely? Your first guess being correct, or the person revealing the doors intentionally dodging the car door?
It makes all the sense if you look at it this way:
Divide the problem into two scenarios: "always change your initial pick" versus "never change your initial pick".
In the first scenario you win if you pick a goat in your first try, because here we always change the pick to what is necessarily a car. This is a 66% chance to win. Likewise, if you pick the car in this scenario you lose, 33% chance.
In the second scenario you win if you pick the car initially, 33% chance. You lose if you pick any of the two goats because we're not changing.
Now you can pick what you want to compare: You either compare the chance of winning in the "always change your initial pick" scenario to the"never change your initial pick", which is a 66% chance to win vs a 33% chance to win.
Or you compare the chance to lose on both scenarios and the lowest one is the best one. Again, 33% chance on scenario 1 vs 66% chance on scenario 2. Meaning "always change your initial pick" is the best situation from a probability standpoint.
It makes perfect sense, switching always gives the opposite of what you first picked. So [Goat,Goat,Car] essentially becomes [Car,Car,Goat] if you switch. Meaning switching gives you a 2/3 chance of getting a car while not switching gives a 1/3 chance. No need to over-complicate it, just break it down to the most base level of logic.
it makes a lot of sense. if you pick a door and the revealed door is a goat, switching will always make you right 66.6% of the time.
Let’s say the 1st door is chosen and the 3rd door is always revealed.
[Goat, Car, Goat]
[Goat, Goat, Car]
[Car, Goat, Goat]
Out of the 3 only possibilities, 2/3 of the games would have won if they switch doors from the original choice.
@@porouscylinder5 "Let’s say the 1st door is chosen and the 3rd door is always revealed."
Then it's 50/50.
Everyone should look up "21 3 door scene" on TH-cam. It's explains this well and why switching doors is anyways a higher Probability of winning the car.
Was about to comment the exact same thing 🤣🤣
Just thank yourself you didn't try cheating off sneaky in school 😂
Edit I'm sorry I actually can't watch this whole vid lol too much
You have a %66 chance to initially pick a goat. So swap every time because you have a %66 chance that the swap is correct. Right? Idk man I just work here
It suppose to help people make decision not just always swap, I believe the process is you calculate the possibilities then you decided for yourself wether you swap your choice or not. I guess sneaky try to word something like this but he just too mald.
I think sneaky is having a hard time understanding that the first choice affects the second choice when there's only 2 doors left. He still thinks it's a 50-50 because there are 2 doors.
@@st3v1sh yes he doesn't understand that when he made the choice he had 3 doors to pick from so no matter what it's still a 1 in 3 chance.
They would freak out if he found out about the fish that beat Pokémon
The easiest way to see it is imagine your have 5 billion doors. If the host reveals all but your door and the other then it’s obvious you should swap because your initial choice is basically 0% chance to be the car.
I haven't been this angry watching a video in about three days. Congratulations Sneaky.
Some facts about the infinite monkey theorem:
Not only would it type the complete works of shakespeare, but it would eventually type it all in chronological order by release date without any gaps in between.
This would work either with a) one monkey, one typewriter, and an infinite amount of time, b) an infinite amount of monkeys, infinite typewriters, and some of them will succeed on their first try.
This is my favorite content! lol
For the Monty Hall Problem:
What I don't see mentioned in the comments, that is useful to know. The probability does not become 50/50 after a door is taken away. It would become 50/50 if the car and goat were re-randomized, but they are not. When the initial choice is made, there is a 33% chance of choosing the car. Since the doors are not randomized after opening a goat it is like being able to open 2 doors at once if you switch. Combining that probability of 33% of each door being the car into the one remaining door. Creating the 66% percent chance probability to find the car by switching.
Maybe the explanation helps, maybe it doesn't. There are plenty of math classes that dedicate a day to discussing this problem.
Worth noting that it’s would also be 50/50 if Monty could reveal the car or if he could reveal your door. His knowledge is a key part of why the problem works the way it does.
Sneaky doesn’t understand gambling or probability it’s hilarious
When the Monty Hall problem first became popular in like the 80s or something the woman with the highest IQ in the world gave the answer that you should always switch. People all over the US heard about her answer and told her she was wrong, including dudes with PhD's in Math. So i don't think its unreasonable for Sneaky to not get it instantly. It inherently doesnt make sense to people.
Sneaky's confidence made me revisit the Monty Hall problem and I came to the conclusion that there are three unique, but sensible, interpretations to the problem, as posed, that lead to vastly different results. I will share them here. First, let's establish the exact wording of the Monty Hall problem:
"Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?"
Here are the three interpretations and the related solutions (i.e. "optimal strategies"):
Interpretation 1 (The "classical" interpretation): The game show host knows what's behind the door, and regardless of which door you select, the host is commited a priori to opening a door and opening a door that does not have the car. In this case, all the classical logic holds, the best strategy is to always switch to win 2/3 of the time because the host deliberately removes a wrong answer for you.
Interpretation 2 (The "randomized" interpretation): The game show host knows what's behind all the doors but he has no intention of using it. He is going to let you select a door, and then he is going to open one of the two remaining doors at random, even if it contains a car. Looking at the statement of the problem, this scenario is not completely infeasible. In this case, switching or not switching doesn't matter, conditional on seeing a goat (the reason being that in this setting the host is more likely to reveal a goat if you selected a car first, so seeing a goat makes you more likely to believe you already found the car). Both doors have a 50% chance of being the car. Since this is your first time on the game show, you have know way of knowing whether or not the game show host picked the door at random or not, just as Sneaky was arguing. In my opinion, this interpretation seems hard to justify given that the game show host knows what's going on.
Interpretation 3 (The "adversarial" interpretation): In this setting, the game show host is doing his best to prevent you from winning a prize. Even more, the game show host is smart, and he has heard of the Monty Hall problem before. Given the fact that it has entered popular culture, he suspects his audience member to know that "switching is always best". With this in mind, the host uses the following strategy to take advantage of these players: If the contestant selects a door with a goat, open it immediately, say "sorry, you didn't win this time" and conclude the game show. Otherwise, if the contest selects a door with the car, open one of the two doors with the goat and offer them a chance to switch (to the door that has the other goat). In this case, switching will guarantee that you lose because you only ever get a chance to switch given that you were about to win. So in this case, the optimal strategy is to *not switch*, which is the opposite of the classical interpretation. This interpretation is completely consistent with the problem as presented in the original quote and unlike interpretation 2 uses the fact that the host has knowledge in a logical way.
Hope you guys found this as interesting as I did, feel free to reply if you feel like I said something objectionable. :)
The only big problem is that the host always opens a door with a goat behind it, thats a stated condition. Its entirely a probability problem that looks conditional because it has 2 steps : first choice + choice to keep first choice or change, but in reality boils down to evaluating each scenario ("always change" vs "never change") and realizing that its a 66% to a 33%. It just intuitively seems odd not knowing whats behind your first pick and thats where most ordinary people, most university students and a hand full of mathematicians got stuck on.
If the 3 interpretations are just you doing thought experiments then disregard everything I just said. I say this because the motive behind the host's decision to open a door that necessarily has a goat behind it is irrelevant to the problem.
Problem is that those other interpretations are separate problems. The problem assumes the host has to follow the rules and always shows you a goat. You've just kind of made up random scenarios that aren't possible within the confines of the actual question.
I'm malding listening to sneaky and I wish I was watching this live, so I can try to explain this to him.
I think sneaky understands the infinite monkey theorem, but he just doesn't believe a monkey is true random.
The monty hall problem can be easily explained. The chance of initialy being right is 1/3. After the host opens a door, the only time the remaining closed door is wrong, is if your initial pick was correct. Your odds don't get redistrubuted to a 50/50. I somehow hope sneaky reads this and understands, because part of me died inside watching this. I hope I can sleep at night, because the urge to help him understand is eating away at me.
''UZI was just a really lucky monkey'' - Doublelift
This is the kind of content I was looking for
Just think of it this way, if you keep you first door, you are wagering you picked the car in your first pick and that’s 1 in however many doors.
the example with a hundred doors is super simple. The chance that you pick the car initially out of a hundred doors is super small. So it's better to switch, knowing the other one has the same potential as your first choice.
RE: Monty Hall, switching doors or perceiving a difference in odds is the same as picking a lottery number that hasn't come up in the lottery recently. They are isolated events. If you pick a door and then the host removes another door, you now get to choose completely from scratch. Although the question manipulates you with the word "switch", your choice has been reset. Now you get to choose door 1 or 2, and the only difference in odds is whether or not you believe a door that was once correct is either less or more likely to have the correct answer given its success in the past, which of course it isn't. If you "stick" with door one, it has the enhanced odds of knowing despite selecting that door, FOR SURE, it was not the negative result from door 3, just as much as, FOR SURE, door 2 is not the negative result of door 3.
It is sad to see sneaky not understand the infinite monkey theorem. Lost respect for him due to lack of intelligence.
I've never heard of the monkey theorem like that, just call it what it is: Given an infinitely long string of random letters, any possible text will probably appear in it at some point.
Statistically impossible and infinite time are fairly mutually exclusive concepts.
The reason the 100 door analogy makes sense is because math is logic. Initially you have 1/x chance of guessing correct given x doors total. Finally you have 1-1/x chance of being correct if you switch because the only way you'd be correct by switching is if you DIDNT originally pick correctly (assuming one of either yours or the one he didnt reveal is always correct). Either you initally guess correct (call this event GC) or you dont initially guess correct (call this event DGC), and since exactly one of these has to be true, we know the sum of their probabilities is 1, or in other words we know 100% that if we guess we will either be right or we will be wrong, that is a logical fact. Converting that logic to math:
p(GC)+p(DGC)=1 -->
p(DGC)=1-p(GC) -->
Where p(X) represents the probability of event X occuring. In the case of the doors, we have
p(DGC)=1-1/x
Where the probability you initially guessed correct is p(GC)=1/x, because only 1 of the x doors was correct. Thus the probability you did not originally guess correct, which is the same probability that you'd switch to the correct door if you switch, is 1-1/x.
And 1-1/x>1/x for all x>2, so your odds are always better if you switch.
Examples:
3 doors (x=3):
p(GC)=1/3
p(DGC)=1-1/3=2/3
p(DGC)>p(GC)
Better off switching
20 doors (x=20):
p(GC)=1/20
p(DGC)=1-1/20=19/20
p(DGC)>p(GC)
Way better off switching
1000 doors (x=1000):
p(GC)=1/1000
p(DGC)=1-1/1000=999/1000
p(DGC)>p(GC)
Waaaaaaaay better off switching.
The more doors you start with, the more wrong doors you eliminate, and the better your odds become by switching. Going to 100 doors really helps to show that trend, but when you only have 3 doors the difference in probabilities isn't drastic enough for most people to even consider it. Instead most people will tackle the problem using emotion like sneaky, rather than logic and reason. Subconsciously, you know how much more disappointed you'd be if you switched from the winner to the loser so you justify not switching. Sneaky is saying logic and math are different, but they are the same. What he is thinking of as logic is actually human intuition. Humans dont always live their lives in the most logical way possible, but since we continue to live that way, its what makes it feel logical to us
So if you still don't understand the monty hall problem, what you should you is play the same scenario and keep your pick while changing the position of the car, then do the same but switch after the host opens a door with a goat, you'll get better results with the second one than with the first one.
The first time you choose a door, you are essentially choosing a singular door. But once one door is revealed, choosing again is like choosing two doors instead of just 1.
The door the host leaves for you to switch with has a 67% chance of having the car behind it.
This is the best idea ever! make it longer like an hour or more!
I feel like it’s easier for not math people to just speak what happens, “On my first choice I probably picked a goat because there are more goats than cars. If the other goat is revealed then the last one is probably the car because my first choice was probably a goat.” “Everything happens in an infinite amount of time. If everything happens it doesn’t matter how improbable, it has to eventually happen because everything must happen sometime and infinity is all the times.”
The host has to know where the car because if a goat is revealed without him knowing where it is then it's 50/50.
@@klaus7443 The host knows where the car is. They will never open a door showing the car. Read the wiki.
@@judgedbytime He said IF the other goat is revealed, he didn't say it would. Read the comment.
@@klaus7443 My bad, the mistake in your sentence gave me the wrong impression.
@@judgedbytime There was no mistake in my sentence.
At 22:45 DL couldn't handle Sneaky lmao
i think the one of the key statement is this is a game showcasing overall averages of games, but the theory does not imply switching will always win. the game is to show that you get a free 33% most of the time if you test this game enough times.
You always have a 2/3 chance of winning by switching regardless as to the amount of times you play.
Door: 1 2 3
Scenario1: C G G --> switch from door 1 to door 2 = win goat
Scenario2: G C G --> switch from door 1 to door 2 = win car
Scenario3: G G C --> switch from door 1 to door 3 = win car
Door: 1 2 3
Scenario1: C G G --> stay with door 1 = win car
Scenario2: G C G --> stay with door 1 = win goat
Scenario3: G G C --> stay with door 1 = win goat
After a goat has been revealed behind door 3, there remains a car and a goat behind the other 2 doors. In 2 of the 3 possible scenarios, a goat will be behind door 1, so when 2 doors remain closed in each scenario, switching means switching from a goat to the car in 2 out of 3 cases. Because there is a goat behind one door and the car behind the other door, the brain quickly thinks the odds are 50/50 whether you stay or switch, but this is not the case.
Best way to explain. Thank you.
You can definitely turn some of these scenarios into a League context for a game show or even just game theory like Meteos was saying.
The Monty Hall Problem could be like: "You are a jungler trying to gank mid, and there were 3 possible routes. One route will be clear of an enemy player, and the other two will have you run into an enemy player; pick a route. Then, your mid laner wards one of your possible routes to reveal an enemy player in that route. Will you keep going on the same route you picked initially or change?"
The Infinite Monkey Theorem is basically just me playing ranked, and eventually I'll hit Challenger from Silver if I play an infinite amount of games... Right? :')
This only works if somehow the midlaner knows that the route he warded is NOT the correct route, and of course that he happens to not choose to ward the same route that you picked for some reason
The thing that makes monty hall click when I explain it to people, imagine there were 100 doors.
Yes, that should be the go to when initially explaining the problem. For some reason the common initial presentation of three doors trips people up a lot. I would even ditch 100 doors initially and go for something ridiculous like a trillion, so that there would be basically zero chance that someone could get tripped up like sneaky did.
Once they understand the concept, they are usually receptive enough to accept the truth when it comes to the three door scenario.
at 18:41 I think meteos unironically does the most effective explanation of the monty hall problem that I've ever heard lmao
Hold on, there was a mythbusters episode about this about the monty hall problem.
SNEAKY WITH GOAT FACTS
The Monty Hall Problem is based off the statistical assumption that your door does not contain the car, meaning that the other door contains the car and that you should switch.
They need to make a fucking podcast
I understand Sneaky's frustration here and I'm sure having chat malding at him isn't helping lol. When I was presented with the Monty Hall Problem in my Probability class I was also super confused at first, but they also do a really poor job of explaining it until they fully understand it themselves. I think if somebody explained originally that there is no 50/50 ever in the problem then it would've made more sense to him lol.
There is an argument to be made that "true" infinity multiplied by 0 is still infinity
This is so great
PLEASE MAKE AN IRL VIDEO OF YOU GUYS TRYING THIS OUT. I WANNA SEE THIS TRICAST TOGETHER
My favorite explanation for the Monty Hall problem is by altering the game rules while retaining strategic equivalence
Original Game:
-Player picks door A
-Host reveals empty door B
-Player can stay with door A, or swap to door C
-Player wins prize behind picked door
Game 2:
-Player picks door A
-Host reveals empty door B
-Player can stay with door A, or swap to door C
-Player wins best prize behind picked doors
Game 3:
-Player picks door A
-Host reveals empty door B
-Player can stay with door A, or swap to doors B and C
-Player wins best prize behind picked doors
Game 4:
-Player picks door A
-Player can stay with door A, or swap to doors B and C
-Host reveals empty door B
-Player wins best prize behind picked doors
Game 4:
-Player picks door A
-Player can stay with door A, or swap to doors B and C
-Player wins best prize behind picked doors
It's clear from Game 4 that, assuming equal probability of each door containing a car, it's better to choose doors B and C, as collectively they have a 2/3 probability of containing the car. It's also quite easy to convinced one's self that each iteration of game is equivalent.
Do the problem with 100 doors pick one, host shows 98 and the logic behind this problem becomes stupidly obvious. If by then you still think its a 50 50 then you're ignoring the fact that when you picked a door you only had a 1% chance of picking the right one to begin and if you don't switch you're insane.
I get the month hall problem, but sneaky does have a point. Intuitively the result shouldn't change no matter if you switch doors or not. I wonder if someone did a lot of simulations of this to get some probabilities
You can write a basic function to test it and it will always come out as switching having 66~% win rate. The logic is sound and intuitively works since switching makes you win if you get a goat, and there's a 2/3 chance of picking a goat.
Yeah they did 6. That's how many it takes to evaluate every possibility here.
I'm surprised that Rick and Morty hasn't made a reference to the Infinite Monkey Theory
dayum i havnt seen meteos for a bit, and he be eating good i see ahhaha
OMG does Sneaky have a friggen body pillow with Meteos' face on it in the background? XDXDXDXD
For many years it's been back there lol true brotherhood
the smart, the dumb and the troller
Sneaky can’t understand infinity
Lol the first examples for probability.
Listening to them talk about this actually melted my brain holy shit.
For the 100 door problem you gotta name your door!
You have a 1% Car door vs a %50 Car door after he takes away the 98 which door you want???
That means there is a 49% chance the car is not behind either door.
Klaus 74 Wrong because the POINT is that the car WILL be either behind the door you picked or the door the host left up. He’s purposely giving you better odds. There’s no trick just probability.
@@VnD77 Do the math Einstein....you have 49% total probability MISSING!!
I still love to hate on the Monty Hall problem. Probability is whack even after taking it at a college level. You're not alone sneaky, this question got people even halfway through our semester (specifically the half that hadn't already studied it to death cause we all needs lmao.)
It just gets explained like shit. All you need to know is that switching always gives you the opposite of what you picked first time since a goat is eliminated, leaving one goat and the car. So since you have a 2/3 chance of getting a goat on your first try switching gives you a 2/3 chance of getting the car. People confuse themselves by focusing on the probability of the door you switch to when the only actual relevant one is the probability of picking a goat from the initial 3 doors.
@@Sirhaddock yeah, I think a lot of explanations skip out on the key point that Monty always reveals a goat from the remaining doors. It seems small but it makes the difference between 50/50 and most people I have spoken too seem to get it once that has been clarified.
Computer randomness isn’t truly random
More podcast edits!
i’ve lost so much respect for sneaky and his intelligence
Dude I'm not even half way in and j can't stop laughing
OMG! Sneaky, dude!! LMAO
Sneaky is right. Monkey isn't random. A random letter generator would write Shakespeare eventually.
Sneaky hurts me @.@ lol He's doesn't understand the concept of infinite time.
Sneaky was saying a monkey will never be able to type. Like the kind of monkey from a zoo. Even with infinite time. A monkey physical can't.
@@victorortero8743 With infinite time, it will eventually hit every possible combination. Its Infinite.
They talked about an experiment, it didn't go well, but if they can hit one letter eventually, by pure chance and time, it would happen.
This was legit painful to listen to
I love how they don't understand the odds locks in as soon as you make the choice, that's why you always want to switch to lock in a better odds
Not necessarily, the odds of your initial choice can change depending on the scenario. Like in deal or no deal, you start with a 1/26 chance but end up with a 1/2 chance when you reach the final two boxes.
This is why we can’t win worlds
This is pretty good, i also already like Meteos but didn't like DoubleLift but on the co streams he seems like a decent human being, this three together works very well.
It is interesting to see how the three of them understand this in different ways, you have DL liking what he reads but you dont know if he actually understands the theory behind, you have Meteos thinking out of the box and Seaky in complete denial of everything xD
Bring them physics not just mat and probability please ^^
Sneaky was kinda small brain here tbh. If a monkey types random words, there is an incredibly small chance to type a work of Shakespeare. Incredibly small chance with infinite tries will succeed eventually.
He makes a good point in the Monty hall problem tho
@@jazeroliversy673 The host can give that whole door to the contestant and it still wouldn't help his staying chances.
Meteos looks fucking weird in this, I dont like it
If the option is only between 3 doors I agree with Sneaky, it's still a 50/50 and it doesn't matter if you switch or not.
If you apply this to 100 doors then you had a 1% of picking the door with the car. So you should switch in this case because it goes up to a 50/50 chance but it's way more likely that the car is behind that door since they revealed 98 incorrect doors. Make sense?
Yeah but in the 3 door scenario, you also had 33% chance of picking the door with the car and 66% chance of picking a goat, so you'll most likely to pick a goat. Therefore, switching is the best option. The chances aren't really high but it is the best option for sure. The only way this is not applicable is that if you picked the car in the first place which is 1 out of 3 chance you'll pick it. Makes sense?
It does matter, you can run a bot that proves it. Switching wins more often since switching always gives you the opposite of what you started with, and you had a 2/3 chance of starting with a goat.
@@lifeofrj9707 I get what you are saying and it does make sense, however mathematically it doesn't change. I understand it's the same logic but in this case with only 3 doors it doesn't change. Yes it technically jumps up to a 50/50 chance but it's basically like the 3rd door was never there in the first place if they show an incorrect 3rd door.
@@IamFallen77 You are contradicting yourself. You said that mathematically it doesn't change but it went up to a 50/50 chance? Wtf? hahahah. The chances went up because the door with a goat (incorrect door) was shown.