Actually 🤓 the problem is actually an extrapolation from the game show. Monty Hall did open a wrong door to build excitement, but offered a known lesser prize - such as $100 cash - rather than a choice to switch doors.
The thing to bear in mind is, the other guy isn't choosing a door to open at random; he's making a point of picking a door behind which lies nothing. You might want to ask yourself why he didn't choose the OTHER available door.
@Esoteric Chaos then why have I simulated this game completely in code and measured win/loss statistics for both stay and switch strategies with 100 million samples each and saw almost exactly 66% of wins came from switching? You can argue the math all you want (though you’d be wrong), but you can’t argue with 100 million sample simulation.
Seen videos on this before, but your explanation using this probabilitty table is the simplest one I've seen yet, while the second example does a great job at making it intuitive. props for that.
One could also see ih from this perspective: First pick a door (Let‘s call it A): P(prize in A) = 1/3 P(prize in B or Price in C) = 2/3 The one of B or C, not containing the prize is opened. Say B was opened. That means that P(prize in B or C) = P(prize in C) = 2/3 since B can’t contain it, otherwise it wouldn‘t have been opened. Loosely speaking: The probability of the price being in the Set {B,C} is collapsed to only {C}.
1. Relevant XKCD: 1282 Assume you NEVER switch. In order to "win", you need to guess the correct door and stick with it. That's a 1/3 chance. Assume you ALWAYS switch. In order to win, you need to guess an incorrect door and switch to the unknown door. That's a 2/3 chance. Switching is better than staying :) Table assuming you always switch, which combos give you the prize _|_1_2__3 1|❌|✅|✅ 2|✅|❌|✅ 3|✅|✅|❌
The interesting thing about this problem is considering the scenario when the host didn't know where the prize is, opened a random door the player didn't choose, and it turned out to be empty. In this case, changing the selection doesn't change anything. It really shocked me back in the day how host's knowledge can actually influence player's odds.
It's not the host's knowledge, per se, though. He could have knowledge, and is under instructions not to *use* the knowledge - to flip a coin after the contestant's first choice. If the coin lands heads, choose the remaining door on the left; if the coin lands tails, choose the remaining door on the right. If the door shows a car, the game-show host says "Well, look, you didn't choose the grand prize - you lose". It's not the possession of knowledge that's key, it's what actions are taken on the basis of that knowledge.
@@hughobyrne2588 That's an abstract point. The host's knowledge is ONLY relevant to the question inasmuch as that knowledge is used IN the question. So a Monty who flips a coin has NO knowledge of the car's position, as far as the scope of the question goes. In that case, the decider is the coin, not Monty. So Monty would be irrelevant, and the contestant could have flipped the coin themselves. So the entity with/without the knowledge is the coin in this hypothetical.
@@jasonl7651 ... I'm not sure what you mean by "That's an abstract point", except in that all communication is abstract. The sentence I was reacting to was "It really shocked me back in the day how host's knowledge can actually influence player's odds". It's not true that "host's knowledge, per se, can actually influence player's odds". He misidentifies what it is that influences. And I gave an example of how, materially, it is not true - it is possible both for a person to have knowledge, and for that person to flip a coin. One does not make the other impossible. It may make the other irrelevant, or outside of the 'scope' of what it takes to correctly apply logic to the givens to reach the correct conclusion, but there's nothing to say that every word of the given question must be relevant or within such 'scope'.
To make the answer intuitively obvious, consider if there were 1000 doors, and after you picked one, 998 empty doors were eliminated, and you had the option of switching to the one remaining door. If you can explain why you'd want to switch in that case, you can probably understand why you'd also want to switch in the 3-door case.
I've seen explanations of the Monty Hall problem enough times that I understand the math behind it quite well, but I don't think I've seen an explanation that used your "picking a single number out of 1000" explanation. I think that explanation really helps better visualize the problem! :D
I remember reading about this interesting scenario. I read that people thought one had a 50% after one of the doors was shown not to have the prize, when it was actually 33.33% (3 recurring).
When you said that revealing the door actually gave us more information, I immediately thought of quantum computers and measuring wave-functions. So I searched "quantum Monty Hall problem" and got a lot more results than I expected... well, down the rabbit hole I guess
Actually the scenario you described in the beginning it really doesn't matter if you switch or not. And yes, I know the logic behind the Monty Hall problem but this isn't the Monty Hall problem because you forgot to include one VERY important detail. In the Monty Hall problem the host knows which door the prize is behind and purposely selects the wrong door. You however, never mentioned this detail. Although it seems like a small nitpick it is crucially important. Your "you should switch because then you will have a 2 in 3 chance of winning instead of a 1 in 3 chance of winning" logic would work IF you purposely selected the wrong door. However, in your scenario you never mentioned that you purposely selected the wrong door, you just selected a door without the prize. In your scenario for all I know you could have selected the door with the car by accident. And that is why it is a 50/50 when you don't purposely select the wrong door. 1/3 of the time you will accidentally reveal the car. Of the 2/3 of the time were you don't, 1/2 the time you should stay and half the time you should switch. This is how the possibility space works for if you choose door 1 and they reveal door 2 Behind door 1: you should stay Behind door 2: you don't even get the option of whether or not you want to stay or switch Behind door 3: you should switch And again, the reason why it is possible in your scenario that they can reveal door 2 even though it's actually behind door 2 is because you never mentioned that they will purposely reveal a wrong door.
another way to look at it is to imagine you got the choice of staying or taking both other doors, in which case obviously you get twice the chance of hitting something (with empty doors just being empty and not forfeiting all wins of course)
You can imagine that there are two sample spaces. One sample space is "what he is chosen" and another sample space is "what he does not choosen". The probability of the prize remains in the second sample space is much bigger than in the 1st sample space. In the process of cancelation the probability of 1st sample space doest not change whereas the probability of 2nd sample space is also remaining same but increasing the probability of the prize in other remaining doors in the second sample space.
TL;DW explanation: If you originally picked the door with the prize, switching gives you nothing. If you originally picked a door with nothing, switching gives you the prize (the other door with nothing was already revealed). You had a 2/3 chance of originally having picked a door with nothing, and therefore you have a 2/3 chance of getting the prize by switching.
Oh, my. Oh, Bri. Let me tell you a little story. It mirrors the first minute of your video, but please bear with me, there's an important question that follows. Let's go to a buffet restaurant. Pick a plate from the dessert table, the plate on the left, the middle, or the right. One of these plates has rhubarb pie, however on the other two plates is apple pie. The plates have these domes on them to keep flies and dirt off, but they have windows in them so you can see what's on the plate - but it just so happens that you're on one side of the dessert table, and I'm on the other, and all the windows are all facing me - you don't know what plate has the rhubarb pie on it but I do. Let's say you select the plate on your left hoping for the rhubarb pie. Before you see what pie it is, I show you what's on the plate in the middle. Turns out, the plate in the middle had apple pie. When you read this story, do you conclude that the way things work at a buffet restaurant is that I will always reveal to you one of the plates you didn't choose that also had apple pie? Isn't 3:50 "I will always reveal to you one of the options you didn't choose that also didn't have the prize" an essential part of the logic you use to reach the conclusion 'probability=2/3', but unfounded by the question asked in the first minute of this video? (Yes yes there are discussions to be had about what Craig Whitaker wrote, there are discussions to be had about what Marilyn vos Savant wrote, but for this comment, I'm just focusing exclusively on this video. On the question asked in the first minute of this video, on the logic used in this video to reach the 'probability=2/3' conclusion, on the constraint added at 3:50 that is not in the first minute of the video - a constraint which makes for a different question. The '2/3' answer is the answer to a question which is different from the question asked in the first minute, is the point I'm trying to make.)
Walking through the possibilities, it's clear that 2/3 of the time switching will have you win and that 1/3 of the time it will have you lose. The intuition is that the host, who knows the correct answer, has to show you a loser. The host's knowledge of the winning door is what makes it valuable to switch.
"the host, who knows the correct answer, has to show you a loser". That is indeed an essential element in the logic of coming to the conclusion that switching is better. The question asked in the first minute of the video, though, says "turns out, door number 2 was a loser". That phrasing is very different from "the host has to...".
This is how I'm making sense of things. So 1/3 chance of picking the right door initially. Let's call that selection Door A. The probability on your initial door selection therefore never changes. There are two doors left, Door B and Door C, with a combined 2/3 chance. The host opens one of them. Let's say they open Door C. It's empty; no prize. The fact that the host didn't open your door (Door A) means the prize could be in there but remember it's only a 1/3 chance. However, even though Door C was open from the Door B and Door C subset with a combined 2/3 chance that 2/3 chance doesn't change just because one of the doors opened. And since 2/3 chance is a greater chance than 1/3, it makes more sense to switch. I think the key to understanding this problem is understand when probabilities change. Apparently they don't always change the way you expect them to!
Actually, the combination of doors B and C is not the reason why Door B now has 2/3 chance. If Door C is opened, you must effectively remove its 1/3 chance. The thing is that in the 1/3 cases in which your chosen Door A is the winner, the host is free to reveal either Door B or Door C, as both have goats, meaning that it is 1/2 likely that he opts for any of them. So we only expect that he reveals Door C in half of those cases that Door A is correct (1/2 * 1/3 = 1/6), not in all. The revelation of Door C eliminated the 1/6 case in which Door A would have been correct but the revealed one would have been Door B. Remember that the host can never reveal your door and neither which has the pirze, so he is limited to one possible door to open if yours is wrong, but two are free for him if yours is the same that has the prize. Therefore, Door A only remains with 1/6 chance while Door B still has its whole 1/3. As the total probabilities must add up 1, you must scale those two values. Applying rule of three, you get that the original 1/6 represents 1/3 at this point, and the original 1/3 represents 2/3 at this point. Compare this with another scenario in which there are 3 pizzas (A, B and C) that will be shared between you and another person (two people). To distribute the total amount equally, each of you two must get 3/2 = 1.5 of pizza, so a pizza and a half. For example, you could take the whole pizza B, the other person could take the whole pizza C, and then each of you two would take half of pizza A. If you notice, from the total amount that you eat, 1/3 of it will come from pizza A while 2/3 will come from pizza B. But that's because you are only eating half of pizza A while you are eating the whole pizza B. The 2/3 does not come from considering that pizza B and C combined represented 2/3 of the total original pizzas.
Picking door 1 means you have a 1/3 chance of winning, since no other information was given at the time. Conversely, that means that doors 2 and 3 together have a 2/3 chance of winning. Opening door 2 and allowing the player to switch is just giving them both doors.
Try to understand it from the perspective of the host. The player provides his prediction to the host. The host opens a door with sheep, leaving only two pairs of doors, one sheep and one car for the player to choose. If the player predicts the door of the car is correct, what will the host do? of couse Want the player to change their choice. The host asks the player to calculate the odds with misleading the player already have once choice before. player swapping his choice, and saying 2/3 chance will win. On the contrary, the player predicted the door with the sheep is correct. What will the host do to make the player still choose the door with the sheep? you can tell. Then..... if you are the player, do you think have any advantage from changing your decision? Why can the probability of getting a sheep or a cart be 2/3 too? ........ Dawyer's door problem, calculate the chance of the host winning. Here is the idea to catculate : The sheep and the car are randomly placed behind the door that you listed. But The player specifies the prediction that there 3 options the position of a doors, so there is still a chance that there is a sheep or a car behind the door.
Outstanding explanation, but how come you didn't include the math? From intuition point of view you explained it beautifully, but the math would have been a really important follow-up.
I disagree because if you go 04:23 min, in the 4th column 2nd line its Win Nothing and no Win prize( if you switcing to 3rd door dont have nothing because in this situation the prize is in 2nd door), then the propability its again 50-50 to stay or no.
Your logic and mathematics are incorrect. The host can simply give you that whole door with a goat instead of opening it and you still would have only a 1/3+0=1/3 chance of having the car.
Это не работает. Так как при таком рассмотрении возможностей, вы ещё не открыли неправильную дверь и вероятность по-прежнему остаётся 33%. Но как только вы откроете неправильную дверь Вариант, в котором эта дверь является выигрышной, уже не учитывается, так что вероятность при таком раскладе 50 на 50
If you're not convinced, think about this example (from the AoPS): What if there was 1000 doors, and you pick one. The host opens 998 other doors and shows they are empty. Do you want to keep your door, or pick the one the host subtly skipped over? Edit: I didn't watch the whole video lmaolmaolmao
Run it backwards There are two doors. Behind one is a car and behind the other is a goat. You have a 50/50 chance of picking the car. You pick Door 1. The host asks you if you want to switch to Door 2. Will this switch increase your chances? No. It still stays 50/50. Now the host shows you a third door, opens it and reveals a goat, and again asks you if you want to switch. Does your chances increase now? No. It still stays 50/50. When you start with 3 doors, your chances are 1/3 to pick the car. Once a door is opened and the goat revealed, that door is out of the play and you divide its 1/3 by two and give 1/6 extra to each remaining door. Door 1: 1/3 + 1/6 = 1/2 Door 2: 1/3 + 1/6 = 1/2
I guess in these situations, whatever you do not choose always have a higher probability of the prize than what you choose. so it makes sense to switch. Probability is sooo whacky💀
Each sheep is also an independent and distinct individual. This game cannot mislead players by treating every sheep as the same thing. There are three options, and two of the sheep also exist independently. If the player can distinguish them, there will be no chance of winning like 2/3. Is the reality of quantum mechanics more logical than what the math? The player never knows which goat is the one revealed.。Don't let the jews fool you.. If life is an endless choice, then if we face it with this way of dealing with things, then what aspect of human nature is eternal? Some people always want to control the world and act as the so-called "God"?
There are a million raffle tickets in a box, you select one, and all but one of the remaining tickets are removed leaving a winning ticket. Do you believe that you have a 50% chance of holding that ticket?
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Actually 🤓 the problem is actually an extrapolation from the game show. Monty Hall did open a wrong door to build excitement, but offered a known lesser prize - such as $100 cash - rather than a choice to switch doors.
The thing to bear in mind is, the other guy isn't choosing a door to open at random; he's making a point of picking a door behind which lies nothing. You might want to ask yourself why he didn't choose the OTHER available door.
@Esoteric Chaos Yep, you can always tell the people who don't understand the problem by how much they imagine they're smarter than everyone else.
@Esoteric Chaos then why have I simulated this game completely in code and measured win/loss statistics for both stay and switch strategies with 100 million samples each and saw almost exactly 66% of wins came from switching? You can argue the math all you want (though you’d be wrong), but you can’t argue with 100 million sample simulation.
@@Adrian-jv2oo dude my brain refuses the fact that it can be 66%
@@Z7youtube that doesn’t make it any less correct my friend
@@Adrian-jv2oo
i know, i didn’t say so, i just mean that the proofs weren’t enough for me
Seen videos on this before, but your explanation using this probabilitty table is the simplest one I've seen yet, while the second example does a great job at making it intuitive. props for that.
Great to hear!
One could also see ih from this perspective:
First pick a door (Let‘s call it A):
P(prize in A) = 1/3
P(prize in B or Price in C) = 2/3
The one of B or C, not containing the prize is opened. Say B was opened. That means that
P(prize in B or C) = P(prize in C) = 2/3
since B can’t contain it, otherwise it wouldn‘t have been opened.
Loosely speaking: The probability of the price being in the Set {B,C} is collapsed to only {C}.
1. Relevant XKCD: 1282
Assume you NEVER switch. In order to "win", you need to guess the correct door and stick with it. That's a 1/3 chance.
Assume you ALWAYS switch. In order to win, you need to guess an incorrect door and switch to the unknown door. That's a 2/3 chance.
Switching is better than staying :)
Table assuming you always switch, which combos give you the prize
_|_1_2__3
1|❌|✅|✅
2|✅|❌|✅
3|✅|✅|❌
As soon as I saw the thumbnail I shouted "GOAT!"
🐐
The interesting thing about this problem is considering the scenario when the host didn't know where the prize is, opened a random door the player didn't choose, and it turned out to be empty. In this case, changing the selection doesn't change anything. It really shocked me back in the day how host's knowledge can actually influence player's odds.
It's not the host's knowledge, per se, though. He could have knowledge, and is under instructions not to *use* the knowledge - to flip a coin after the contestant's first choice. If the coin lands heads, choose the remaining door on the left; if the coin lands tails, choose the remaining door on the right. If the door shows a car, the game-show host says "Well, look, you didn't choose the grand prize - you lose". It's not the possession of knowledge that's key, it's what actions are taken on the basis of that knowledge.
@@hughobyrne2588 That's an abstract point. The host's knowledge is ONLY relevant to the question inasmuch as that knowledge is used IN the question. So a Monty who flips a coin has NO knowledge of the car's position, as far as the scope of the question goes. In that case, the decider is the coin, not Monty. So Monty would be irrelevant, and the contestant could have flipped the coin themselves. So the entity with/without the knowledge is the coin in this hypothetical.
@@jasonl7651 ... I'm not sure what you mean by "That's an abstract point", except in that all communication is abstract. The sentence I was reacting to was "It really shocked me back in the day how host's knowledge can actually influence player's odds". It's not true that "host's knowledge, per se, can actually influence player's odds". He misidentifies what it is that influences. And I gave an example of how, materially, it is not true - it is possible both for a person to have knowledge, and for that person to flip a coin. One does not make the other impossible. It may make the other irrelevant, or outside of the 'scope' of what it takes to correctly apply logic to the givens to reach the correct conclusion, but there's nothing to say that every word of the given question must be relevant or within such 'scope'.
YES
FINALLY
THANK YOU, BRI!
After years trying to understand this, your result table did it for me, why nobody used it before???
I've never seen that 2nd example before, that makes it a lot more intuitive, very interesting!
To make the answer intuitively obvious, consider if there were 1000 doors, and after you picked one, 998 empty doors were eliminated, and you had the option of switching to the one remaining door. If you can explain why you'd want to switch in that case, you can probably understand why you'd also want to switch in the 3-door case.
Correct, just like the example shown in the end of the video.
real
I've seen explanations of the Monty Hall problem enough times that I understand the math behind it quite well, but I don't think I've seen an explanation that used your "picking a single number out of 1000" explanation. I think that explanation really helps better visualize the problem! :D
I remember reading about this interesting scenario. I read that people thought one had a 50% after one of the doors was shown not to have the prize, when it was actually 33.33% (3 recurring).
When you said that revealing the door actually gave us more information, I immediately thought of quantum computers and measuring wave-functions. So I searched "quantum Monty Hall problem" and got a lot more results than I expected... well, down the rabbit hole I guess
Actually the scenario you described in the beginning it really doesn't matter if you switch or not. And yes, I know the logic behind the Monty Hall problem but this isn't the Monty Hall problem because you forgot to include one VERY important detail. In the Monty Hall problem the host knows which door the prize is behind and purposely selects the wrong door. You however, never mentioned this detail. Although it seems like a small nitpick it is crucially important. Your "you should switch because then you will have a 2 in 3 chance of winning instead of a 1 in 3 chance of winning" logic would work IF you purposely selected the wrong door. However, in your scenario you never mentioned that you purposely selected the wrong door, you just selected a door without the prize. In your scenario for all I know you could have selected the door with the car by accident. And that is why it is a 50/50 when you don't purposely select the wrong door. 1/3 of the time you will accidentally reveal the car. Of the 2/3 of the time were you don't, 1/2 the time you should stay and half the time you should switch.
This is how the possibility space works for if you choose door 1 and they reveal door 2
Behind door 1: you should stay
Behind door 2: you don't even get the option of whether or not you want to stay or switch
Behind door 3: you should switch
And again, the reason why it is possible in your scenario that they can reveal door 2 even though it's actually behind door 2 is because you never mentioned that they will purposely reveal a wrong door.
another way to look at it is to imagine you got the choice of staying or taking both other doors, in which case obviously you get twice the chance of hitting something (with empty doors just being empty and not forfeiting all wins of course)
Now this is an intuitive way of looking at it. Props.
Your explanations are soooo good! Thank you.
Scaling it up to the 1000 in the hat FINALLY helped me really understand... thank you! :D
Love your videos
You should do the gamblers fallacy as one of your next projects, that'd be interesting
Awesom,e!!! Thanks
haven't seen the video yet, but i'm pretty sure it's gonna be really nice! love your videos
You can imagine that there are two sample spaces. One sample space is "what he is chosen" and another sample space is "what he does not choosen". The probability of the prize remains in the second sample space is much bigger than in the 1st sample space. In the process of cancelation the probability of 1st sample space doest not change whereas the probability of 2nd sample space is also remaining same but increasing the probability of the prize in other remaining doors in the second sample space.
U r gr8 dude
TL;DW explanation:
If you originally picked the door with the prize, switching gives you nothing. If you originally picked a door with nothing, switching gives you the prize (the other door with nothing was already revealed). You had a 2/3 chance of originally having picked a door with nothing, and therefore you have a 2/3 chance of getting the prize by switching.
Oh, my. Oh, Bri.
Let me tell you a little story. It mirrors the first minute of your video, but please bear with me, there's an important question that follows.
Let's go to a buffet restaurant. Pick a plate from the dessert table, the plate on the left, the middle, or the right. One of these plates has rhubarb pie, however on the other two plates is apple pie. The plates have these domes on them to keep flies and dirt off, but they have windows in them so you can see what's on the plate - but it just so happens that you're on one side of the dessert table, and I'm on the other, and all the windows are all facing me - you don't know what plate has the rhubarb pie on it but I do. Let's say you select the plate on your left hoping for the rhubarb pie. Before you see what pie it is, I show you what's on the plate in the middle. Turns out, the plate in the middle had apple pie.
When you read this story, do you conclude that the way things work at a buffet restaurant is that I will always reveal to you one of the plates you didn't choose that also had apple pie? Isn't 3:50 "I will always reveal to you one of the options you didn't choose that also didn't have the prize" an essential part of the logic you use to reach the conclusion 'probability=2/3', but unfounded by the question asked in the first minute of this video?
(Yes yes there are discussions to be had about what Craig Whitaker wrote, there are discussions to be had about what Marilyn vos Savant wrote, but for this comment, I'm just focusing exclusively on this video. On the question asked in the first minute of this video, on the logic used in this video to reach the 'probability=2/3' conclusion, on the constraint added at 3:50 that is not in the first minute of the video - a constraint which makes for a different question. The '2/3' answer is the answer to a question which is different from the question asked in the first minute, is the point I'm trying to make.)
Is my intuition stupid, because I strongly disagree with you
Walking through the possibilities, it's clear that 2/3 of the time switching will have you win and that 1/3 of the time it will have you lose. The intuition is that the host, who knows the correct answer, has to show you a loser. The host's knowledge of the winning door is what makes it valuable to switch.
"the host, who knows the correct answer, has to show you a loser". That is indeed an essential element in the logic of coming to the conclusion that switching is better.
The question asked in the first minute of the video, though, says "turns out, door number 2 was a loser". That phrasing is very different from "the host has to...".
@@hughobyrne2588 word
Monty Hall Memory (MHM) is here
This is how I'm making sense of things. So 1/3 chance of picking the right door initially. Let's call that selection Door A. The probability on your initial door selection therefore never changes. There are two doors left, Door B and Door C, with a combined 2/3 chance. The host opens one of them. Let's say they open Door C. It's empty; no prize. The fact that the host didn't open your door (Door A) means the prize could be in there but remember it's only a 1/3 chance. However, even though Door C was open from the Door B and Door C subset with a combined 2/3 chance that 2/3 chance doesn't change just because one of the doors opened. And since 2/3 chance is a greater chance than 1/3, it makes more sense to switch. I think the key to understanding this problem is understand when probabilities change. Apparently they don't always change the way you expect them to!
Actually, the combination of doors B and C is not the reason why Door B now has 2/3 chance. If Door C is opened, you must effectively remove its 1/3 chance.
The thing is that in the 1/3 cases in which your chosen Door A is the winner, the host is free to reveal either Door B or Door C, as both have goats, meaning that it is 1/2 likely that he opts for any of them. So we only expect that he reveals Door C in half of those cases that Door A is correct (1/2 * 1/3 = 1/6), not in all. The revelation of Door C eliminated the 1/6 case in which Door A would have been correct but the revealed one would have been Door B.
Remember that the host can never reveal your door and neither which has the pirze, so he is limited to one possible door to open if yours is wrong, but two are free for him if yours is the same that has the prize.
Therefore, Door A only remains with 1/6 chance while Door B still has its whole 1/3. As the total probabilities must add up 1, you must scale those two values. Applying rule of three, you get that the original 1/6 represents 1/3 at this point, and the original 1/3 represents 2/3 at this point.
Compare this with another scenario in which there are 3 pizzas (A, B and C) that will be shared between you and another person (two people). To distribute the total amount equally, each of you two must get 3/2 = 1.5 of pizza, so a pizza and a half. For example, you could take the whole pizza B, the other person could take the whole pizza C, and then each of you two would take half of pizza A.
If you notice, from the total amount that you eat, 1/3 of it will come from pizza A while 2/3 will come from pizza B. But that's because you are only eating half of pizza A while you are eating the whole pizza B. The 2/3 does not come from considering that pizza B and C combined represented 2/3 of the total original pizzas.
Real Legend BriTheMathGuy🥀🥀🥀 love from Bangladesh fan🥀
Picking door 1 means you have a 1/3 chance of winning, since no other information was given at the time. Conversely, that means that doors 2 and 3 together have a 2/3 chance of winning. Opening door 2 and allowing the player to switch is just giving them both doors.
1⁄π you're dividing by the way that you hear differential from your heart and the number of a tree 🌳 in a universe of hexagonal
Nice
Try to understand it from the perspective of the host. The player provides his prediction to the host. The host opens a door with sheep, leaving only two pairs of doors, one sheep and one car for the player to choose. If the player predicts the door of the car is correct, what will the host do? of couse Want the player to change their choice. The host asks the player to calculate the odds with misleading the player already have once choice before. player swapping his choice, and saying 2/3 chance will win.
On the contrary, the player predicted the door with the sheep is correct. What will the host do to make the player still choose the door with the sheep? you can tell.
Then..... if you are the player, do you think have any advantage from changing your decision?
Why can the probability of getting a sheep or a cart be 2/3 too?
........ Dawyer's door problem, calculate the chance of the host winning.
Here is the idea to catculate :
The sheep and the car are randomly placed behind the door that you listed. But The player specifies the prediction that there 3 options the position of a doors, so there is still a chance that there is a sheep or a car behind the door.
How to calculate it mathematically ?
there is no video on the screen
Outstanding explanation, but how come you didn't include the math? From intuition point of view you explained it beautifully, but the math would have been a really important follow-up.
can you tell me what software you use to make such a video?
I can disprove this. We can discuss on it
Ah, the monty hall problem from brooklyn 99
No, the Monty Hall problem from Monty Hall.
That problem is mind boggling. .... Sir your explanation is very precise ...
I disagree because if you go 04:23 min, in the 4th column 2nd line its Win Nothing and no Win prize( if you switcing to 3rd door dont have nothing because in this situation the prize is in 2nd door), then the propability its again 50-50 to stay or no.
Your logic and mathematics are incorrect. The host can simply give you that whole door with a goat instead of opening it and you still would have only a 1/3+0=1/3 chance of having the car.
A, the key here is the host?
Это не работает. Так как при таком рассмотрении возможностей, вы ещё не открыли неправильную дверь и вероятность по-прежнему остаётся 33%. Но как только вы откроете неправильную дверь Вариант, в котором эта дверь является выигрышной, уже не учитывается, так что вероятность при таком раскладе 50 на 50
If you're not convinced, think about this example (from the AoPS):
What if there was 1000 doors, and you pick one. The host opens 998 other doors and shows they are empty. Do you want to keep your door, or pick the one the host subtly skipped over?
Edit: I didn't watch the whole video lmaolmaolmao
Run it backwards
There are two doors. Behind one is a car and behind the other is a goat. You have a 50/50 chance of picking the car. You pick Door 1. The host asks you if you want to switch to Door 2. Will this switch increase your chances? No. It still stays 50/50.
Now the host shows you a third door, opens it and reveals a goat, and again asks you if you want to switch.
Does your chances increase now? No. It still stays 50/50.
When you start with 3 doors, your chances are 1/3 to pick the car. Once a door is opened and the goat revealed, that door is out of the play and you divide its 1/3 by two and give 1/6 extra to each remaining door.
Door 1: 1/3 + 1/6 = 1/2
Door 2: 1/3 + 1/6 = 1/2
* will probably
Hi
I guess in these situations, whatever you do not choose always have a higher probability of the prize than what you choose. so it makes sense to switch. Probability is sooo whacky💀
Each sheep is also an independent and distinct individual. This game cannot mislead players by treating every sheep as the same thing.
There are three options, and two of the sheep also exist independently. If the player can distinguish them, there will be no chance of winning like 2/3.
Is the reality of quantum mechanics more logical than what the math? The player never knows which goat is the one revealed.。Don't let the jews fool you.. If life is an endless choice, then if we face it with this way of dealing with things, then what aspect of human nature is eternal? Some people always want to control the world and act as the so-called "God"?
おはようございます
I learned this game when I was like 10. Now I am studying physics and I still try to use it when answering tests with multiple answers lol
I'm not sure you are BriT-ish
The moment one option gone after revealing the one door. The probability chosing among the two doors become 1/2 not 1/3 rd.
There are a million raffle tickets in a box, you select one, and all but one of the remaining tickets are removed leaving a winning ticket. Do you believe that you have a 50% chance of holding that ticket?
I know you are fraud minded,
Opps it's broadminded 😜
1st