If you pick 1 out of 100 doors, your odds of picking the prize are 1 in 100. Now, the host eliminates 98 doors. Since you had 1 in 100 chance that the prize was in the first door you chose (extremely low odds), there is a 99 in 100 chance that the last door contains the price. Odds are, you did not pick the right choice in the beginning. This represents why it is not 50/50.

@Mathispower4u

6 ай бұрын

I agree. I address this my longer video on this topic. kzread.info/dash/bejne/iqiTy5KSZsaWh9o.html

@dagmeistr

3 ай бұрын

christ, thank you for a simple the break down, I been trying to wrap my mind around this for 5hrs

@WorldNews-qn9ok

3 ай бұрын

The host doesn't just eliminate 98 doors. He reveals 98 zonks. There's a big difference.

@customarylover3857

3 ай бұрын

​@@WorldNews-qn9ok Couldn't have said it better myself.

@vitorestevam6392

2 ай бұрын

Nice explanation. Now it got clear in my mind 😮❤

How dare you detective Dias, I am YOUR SUPERIOR OFFICER! Bone... BONE!?!?

@buffducks

8 ай бұрын

😂

@RedactedYoutubeBrowser

7 ай бұрын

Just as wunch says when seeing deodorant, I'm not buying it!

@ubberJakerz

5 ай бұрын

Why are we referencing Brooklyn 99??? 🤣

@annabelle3192

2 ай бұрын

I, like others, are here to understand Holt's reference of this in S 4 E8

@codyeveryday5432

2 ай бұрын

​@AverageKZreadBrowser don't count your creepy fish babies before they hatch.

The host will never eliminate your door therefore the odds that’s your door is empty is higher than the door remaining

@jaybefaulky4902

3 ай бұрын

think of it this way: the first round is the real choice. then he showed the fake door which always has a goat. the real choice is now changed to 50/50 because you get to play the second game which is different than the fist round of choices..it's TWO games NOT one..;)

@customarylover3857

3 ай бұрын

​@@jaybefaulky4902 Not true. It's still one game. You'd only get 50/50 odds if he then had everyone close their eyes and randomly rearranged what was behind your door and the other door he didn't open.

@WeebSlayer27

Ай бұрын

The host also can't eliminate the door you initially choose anyway, that door could also be the wrong one because the host doesn't want you to switch. If what you say becomes the consensus, then all hosts have to do is not eliminate any doors and just disqualify the contestant, yall are naive lol.

@WeebSlayer27

Ай бұрын

​@@customarylover3857There was never a 3rd door, there were only two doors, the 3rd door never existed. Capische? If the question is "There are two doors, which one do you pick?" It's a 1/2 chance

@Simonk_6

Ай бұрын

@@WeebSlayer27no?

Best explanation I've found so far!

@Mathispower4u

4 ай бұрын

Wow, thanks!

I just watched three videos on this subject, and this is the first one that made sense, instantly. Nice job!

@Mathispower4u

17 күн бұрын

Thank you for sharing your thoughts.

If you think it’s 50-50, it’s not because you got rid of one of the doors that specifically is the wrong answer. If it was just a random door then yeah, it’s going to be 50-50.

@MrWaltjam

9 ай бұрын

It has tobe as the options changed. There was no certainty that they door choses held the prize or not, until its revealed there is equal uncertainty that the remaining door has the prize. the question that I read asked what should you do, stay on your first choice or select the other door. in the first case you had a 1 in 3 chance of choosing correctly, once the middle door is revealed now there is a 1 in 2 chance...you can be right or wrong in your choice. should you change your mind it is still a 1 in 2 chance of being right or wrong.

@duneblythe

9 ай бұрын

@@MrWaltjam your odds of having chosen the correct door don't magically go up just because you know which unpicked door is wrong. Your choice to switch is not 50/50. Imagine someone was holding a 1000 ticket raffle. You buy one ticket and and someone buys the other 999. If Mr. 999 makes a deal with you that he will let you decide to switch or keep your ticket if he can see the winning number first and throw 998 losing tickets away, you choice isn't magically 50/50. He had a 999/1000 chance of winning, that doesn't change just because he put 998 in the trash.

@marksesl

9 ай бұрын

@@duneblythe If you know that one of the unpicked doors is wrong, then you know you didn't get that zonk, so your odds factually go from 1/3 to 1/2. Your explanation using the lottery tickets introduces a new element if you assume Mr.999 can know before the draw which ticket will won. But you picked a poor example because he wouldn't. If he allowed you to buy some, the odds would just be how many tickets each has.

@duneblythe

8 ай бұрын

@@marksesl fun experiment to prove me wrong. Use python to run a simulation of the monty hall problem. Record how many times switching has you win. Then do it again and record how often staying has you win. Or you could just watch the dozens, if not hundreds, of videos of people doing that exact thing showing that switch gives a 2/3 win rate and staying gives you a 1/3 win rate. Maybe the laws of reality and math are wrong and the intuition the wet bag of electric fat you call a brain is right.

@morbideddie

7 ай бұрын

⁠@@MrWaltjamthis is not correct. You are assuming that all options have an equal chance of winning which is a faulty assumption.

That 1/3 and 2/3 information is only known by the one that knows where the money is. So if you picked door 1 2 or 3 you srill would be faced with the same 1/3 2/3 math but technically you could have picked the right wrong. Making it a 1/1 but the host is making it LOOK like you have a 2/3 chance 1/3 chance but you really have a 50/50 chance because no matter what you pick he's going to make it appear like you didnt pick the right door. You only pick another door if you think you didnt get it right the first time in the 1 out of 3. The math says switch but its really a mind trick

IF YOU'RE HAVING TROUBLE UNDERSTANDING THIS THINK ABOUT IT THIS WAY: The host knows which door is the correct one. He will remove a door that is not the one, BUT he will also not remove the door you chose.

@AttacusPNG

Ай бұрын

Yes, so its always a 50/50 no? Because it was never a 1/3 from the start? Bacause if u chose the right or wrong door theres always 1 door with no money?

@AttacusPNG

Ай бұрын

Nvm i finally understand

Everyone agreeing that it's a 50/50 when it's not lol The hosts knows where the money is and will always open a door with no money behind The chances of picking an empty door are 2/3 The host will then remove the other empty door of the equation. The chances of picking an empty door are still 2/3 and so by switching it means you have 2/3 chance on switching into the door with money

@Mathispower4u

Жыл бұрын

Yes, I agree. I knew this video lead to some interesting comments. :)

@Sarwaan001

Жыл бұрын

You said it better than my comment

@marksesl

9 ай бұрын

The way the video presents it, it IS 50:50. The video just makes it sound as though if it's not behind one door, it's twice as likely to be behind the other. And that is not true. The host has to be consciously avoiding revealing the money to get the 2/3 odds.

@rats7386

8 ай бұрын

If you have 2 options and one of them is right then it's 50/50

@marksesl

8 ай бұрын

@rats7386 Nobody can deny that. If you brought in a new contestant who hadn't any knowledge of what had transpired, that would certainly be true. But that's not the case. The first player chose a door from three doors. The host intentionally revealed a goat from behind one of the other doors. So now the correct choice can be figured out. 🤔 You don't have to just guess.

Best explanation on this...!!! Short and straight...

Knowing this, if i was the game show i would show him the door ONLY if he picked the car bc the player would always change. And if chose a goat, id say 'okay here u go'

Please some one explain why we are treating door 2 as if it still has a chance to have a millions dollars after it has been recealed to have zero. At that point door two is removed from the equation Door 1 and 3 are now 1/2 chances. I have never understood this problem. Why would we group the open door 2 with 3 or 1 as if they both still posses 1/3 odd each when we know for a fact 2 has a 0/3 chance. To prove this point further. Say I choose door 1. You reveal door 2 is a dud. So the claim is my odds improve when I switch to 3? Now, run the same experiment with another person. You keep the prize money behind the same door. They choose door 3. You once again reveal door 2 and so your logic says they increase their odds by switching to 1. This can't be possible for both of us. The situation is the same. Ergo there is no advantage from switching. There is only the illusion of one if treat 2 doors as one and an incorrect door still having a chance to be correct.

@onurrrrr77

14 күн бұрын

I also have the same view as you, this doesn't make sense to me. I see it as: We had 3 doors that might have money behind them, now we have 2, and the denominator should be reconsidered since one of the options is permanently gone.

@twedds9525

13 күн бұрын

hotdog, you are so close to getting it with your ‘proof’. Just add a third person to represent all initial choices. This person starts by selecting door #2, which with your example we know is a dud. Now the host must reveal a door. Since he will reveal the other dud door (host cannot show you the money obviously, and we know we selected dud door #2), switching will result in this person winning. Therefore there are 2/3 scenarios where switching gives you the money, and only 1 when staying.

@arb1ter543

13 күн бұрын

This is a hypothetical scenario. It just so happens to be in this scenario that door 2 is one of the $0 doors. In the game, when you choose a door, the host will intentionally eliminate a door with no prize and reveal its contents (nothing). The host will also never eliminate your door, and the host will never eliminate the door with the prize. However it's still possible that your original choice IS the door with the prize, which is UNLIKELY, and that's the catch. In a 100-door version of this problem, the host will eliminate 98 other $0 doors after making your choice between 100 doors. You're left with your original choice and another door (could be the prize or just another $0 door). Your chances of winning after choosing to switch are 99%, since initially you have a 99% probability of losing (choosing a $0 door).

@Araqius

12 күн бұрын

@@onurrrrr77 Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Switching means the opposite. It's just basic math/logic kids understand.

@Araqius

12 күн бұрын

Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Switching means the opposite. It's just basic math/logic kids understand.

ok so the opposite makes more sense to me. i call it the reverse monty hall problem. there r 2 cars behind 3 doors. 2/3 odds. u chose a door n the host reveals another one with the car. host asks if u wana switch. still 2/3 odds of getting car coz u chose that specific door from the start. 1/3 if u switch to the other door. makes more sense with big numbers. 100 doors n 99 cars. u hav 99% chance to choose the right door. the host opens 98 doors which all of them have cars. u r still left with the door u chose (99% chance car) and the door the host left (which hav 1% chance of car).

Thank you, thank you so much!

So if i picked door 3 on my second chance i should pick door 2 hummm so what if i oicked door 2 now you are going to say pick 3 on the second chance... so thats means the host is going to make it appear like you picked the wrong door no matter what you pick. So pick the one you think it is period! Door 3 has just as much of a chance to be correct as door 2 when you really think about it. its appearing like it doesnt only because he is revealing 1othe doors. But thats lets me know that my door is more likely to be right than wron

So… when I first choose a door, it has 2/3 chance of being wrong. Meaning that the correct door is PROBABLY one of the ones I didn’t pick… The host eliminates one option BUT my door stays the same, 2/3 chance that it’s wrong, because there were three doors when I picked it. Meaning the remaining door has a 2/3 chance of being correct… right? 😐

But, if you say it'S 2/3 cahnce between the eliminated door and my door, and 1/3 to the other door, then I shoudl stay....

Since this is no longer a probability of random facts then it becomes 50% chance. You start a new game once given 2 options

@morbideddie

2 ай бұрын

No, the game is one continuous game, the original placement of the car remains the same. Hence the odds are 2/3 when switching.

@Araqius

2 ай бұрын

Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Switching means the opposite. It's just basic math/logic kids understand.

@WeebSlayer27

Ай бұрын

​@@morbideddieThe door was eliminated, so by logic, a variable was eliminated from the game, and thus it's objectively a different game, just with the same rules. If there are two doors, which one do you pick?

@morbideddie

Ай бұрын

@@WeebSlayer27 ok, you can call it a different game if you want, purely a semantic argument. Either way the location of the car is the same and the information you gained in the initial game can be used in the later game. I’d pick the doors that the host left for me, since that wins in 2/3 games.

@WeebSlayer27

Ай бұрын

@@morbideddie Well yeah, I'd choose any door that the contestant hasn't opened too 😂

No, I dont agree. The collective odds of the two can't just automatically shift like that. The 2/3 probability focuses on the remaining doors only because Monty is avoiding revealing the car. It is his conscious action that is responsible.

@jumex_mango

3 ай бұрын

You don’t agree. That is fine, but that is an incorrect opinion 🙂 Many computational simulations have proven that you have higher odds of winning if you switch, and there are numerous proofs of it. Why do you disagree if your opinion has been proven wrong many many times?

@WorldNews-qn9ok

3 ай бұрын

@jumex_mango You misunderstand. Of course, it's better to switch. It's the explanation I disagree with. The reason the remaining door has 2/3 probability is solely because the host is knowingly avoiding revealing the car (in this case money). Thus, the car has a 2/3 chance of being behind the door the host doesn't open. It's not because the 2/3 collective probability just shifts to the remaining door.

@XavierLikesBLD

15 күн бұрын

@@WorldNews-qn9okright but that’s how this problem works. That is the essence of the problem. The problem is made around this assumption. Your comment leads me to believe that you are assuming the Creator means that in any scenario whether you know or not if the removed door was right or wrong you will be better off switching. That isn’t what he is stating. He clearly states that the second door is wrong.

@crappiesniper

8 сағат бұрын

I agree!!! The host is making it appear to be wrong. The only sure thing is that it's not door 2. So that makes door 3 a stronger possibility not a less. The only way you SHOULD switch is if you are wrong. So since you don't know then you have to ask yourself how confident was I in the first place. Cause if I had picked door 2 you would say switch to the door I actually picked . So no matter what I pick it's suggesting to pick the other door. That's just math possibilities but thats not always right choice to make.

I was confused on how it was 66-33

@wipoute

8 ай бұрын

There are many videos on KZread explaining this (search Monty-Hall and you will find all the answers you need)

what do you mean do you agree 😂 it's a fact

What if it really was in that 3rd door 😂

Maybe it's really behind door 3

@Mathispower4u

7 ай бұрын

It could be, but the ideal is there is only a 1/3 chance of it being door 3.

I still don't get it

But all it is is math. In reality there’s a 50/50 chance.

I was always sceptical of this problem, so once I asked my brother to actually do it to prove me wrong. The first 4 times out of 10 I did not switch my choice and won. And if you find it the first time, that's all that matters.

@RonaldABG

9 ай бұрын

So if you had switched the first time and won, would you automatically have said that switching is the best strategy overall?

@DukeontheThames

9 ай бұрын

@@RonaldABG likely yes, but that did not happen

@RonaldABG

9 ай бұрын

@@DukeontheThames But that way you are not getting what it's more likely to occur. What happened to you is not necessarily which will happen to other persons, so others cannot take your experience as advice. I mean, we agree that if the game consisted in that instead of revealing a door, the host offered you the opportunity to reject your chosen one and instead check inside the other two and take which you prefer from them, it would be obvious that it would be better to switch to the other two. However, you could play once and that time the car happens to be behind your single original door just by chance, not behind any of the other two. But that would not deny that taking the other two would be better strategy, in general, and that's what we are trying to find.

@roccosikora5676

8 ай бұрын

if there were 100 doors your chances to pick the money would be a 1/100 chance if he removes 98 doors you would feel more confident to swap due to the fact that you think you could not have chosen the money the first time around. swapping is always the best answer

@wipoute

8 ай бұрын

​@@DukeontheThames sorry, but here is a way to improve the quality of your experiment : 1. Instead of 10 tries, do 30 or 50 without swapping your choice. 2. Then do the same amount of tries with the swap. The more you increase the amount of experiments, the more likely your results are going to correlate with the probabilities of them happening. 10 tries is not enough and you need to repeat the experiment with both strategies to be able to compare them.

GUYS IT'S NOT 50/50 The probabilities don't shift, they were always there. You have a chance of 1/3 of picking the correct door, so there's a chance of 2/3 that the price is in ANY of the other doors. Because the game host shows you a door that DOESN'T contain the price, you don't have to choose between them, the other door contains the chances of ANY other door, which is 2/3. You can just sum it up with the question: What are the chances that I am incorrect with my first choice?

@Mathispower4u

9 ай бұрын

Yes and here is a longer version in which more than 3 doors are considered. kzread.info/dash/bejne/iqiTy5KSZsaWh9o.html

@DanielMiroballi

9 ай бұрын

Yes but if you look at it the normal way there is one door that has the money and one door that does not. Your probability does not change from picking the first door, this is true, but your choices have gone from a 33.3% to an even 50%. This is a new problem and has nothing to do with picking one out of 3 doors this is a new problem and you are picking from two even doors making it a 50-50.

@DanielMiroballi

9 ай бұрын

Please prove me wrong

@duneblythe

9 ай бұрын

@@DanielMiroballi there are literally thousand of videos running computer simulations millions of times proving you wrong. To be clear, you are not making a 2nd choice at 50/50 odds. There are 3 lotto scratchers that win 1/3. You bought one and your friend bought two. The fact that your friend scratched his 2 first and showed you one was a loser doesn't change anything. He always has at least 1 loser. your ticket was always a 1/3 chance. Your odds never change just because you see a loser you knew he had to have.

@Araqius

8 ай бұрын

​@@DanielMiroballi Let's say the game start with 1 door (1 car, 0 goat). After you make a pick, the host add a goat door and ask whether you want to switch door or not. What is your winning chance if you stay with your first pick? 100% Easy, right? Daniel: Your probability does not change from picking the first door, this is true, but your choices have gone from a 100% to an even 50%. Daniel: This is a new problem and has nothing to do with picking one out of 1 door this is a new problem and you are picking from two even doors making it a 50-50.

What's the difference between predicted and already selected? If you know, that's clearly explained. On the contrary, if you think there is no difference, I can only say that your views and perspectives are very different from mine. The player tells the host his prediction, then removes a door with a sheep, and leaves the rest to the player to choose from, that's it! ! It seems that the host can use many methods and means to manipulate your true choices.

Nope, I don't buy it. Boils down to grouping and inheritance assumptions. Consider. Door 1 starts with a 33% chance, it's true. And door 2 has $0. The first problem is in the grouping. Sure, the door 2-3 combo (at that point in the time) has a 66% combined probability, but so does the 1-2 group! You can't ignore that by insisting 1 has inherited its 33% probability. A 1-2 group is just as likely as a 2-3 group. If both groups are at 66% probability, the groups start out equal. Then you eliminate door 2. The game has changed; you've killed the grouping. Nothing forces door 1 to inherit a 33٪ probability. The 1-2 grouping is just as likely as the 2-3. Since 2 was eliminated, 1 and 3 are both at the same probability. 50/50 or 66/66 depending on your point of view QED

@activision4170

Ай бұрын

Just try it with 1million doors instead of 3. The game doesn't change because the revealed doors are dependent on your first choice. It's a conditional probability. So, first you pick a door. The odds of this door being right is say,

@morbideddie

28 күн бұрын

We don’t need to make any assumptions, it’s a game where we know all variables. Three doors, one prize. Pick a door, host always reveals a goat from the other two then offers the switch. When you pick a goat initially the host reveals the other goat and leaves the car, switching wins. When you pick the car initially the host reveals a goat and leaves a goat, sticking wins. Hence the result of the second decision is determined entirely on the first pick. 1/3 games the contestant picks the car initially and sticking will win. 2/3 games the contestant picks a gay initially and sticking will win. Hence switching wins in 2/3 games. QED

@nimrodszocs2795

21 күн бұрын

Wtf is a 66/66 chance 💀💀

@morbideddie

21 күн бұрын

@@nimrodszocs2795 I missed that 💀 not on only is OP’s comment nonsensical they couldn’t add up to 1. And three people apparently saw it and agreed with it!

@Araqius

12 күн бұрын

Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Switching means the opposite. It's just basic math/logic kids understand. Sadly, it's far too hard for idiots.

If you show that door 2 has nothing wouldn't that then increase the chance of door 1 and 3 to 50% or 1/2 so I would say it doesn't matter either way it's a 50/50 shot which is equal chance which in my mind tells me it's irrelevant if you stick with door 3 or switch to door 1

@user-te8hq6qh4t

Жыл бұрын

Statistically it is always better to switch doors. That's kinda tricky to understand, but probability of door 3 having a prize won't increase after the exclusion of door 2, but door 1 will now have 50/50 chance. Try thinking about it like locking down the probability after your choice.

@wizenedoak5046

Жыл бұрын

@@user-te8hq6qh4t I still dont see it like that if you have 2 options remaining how does only 1 of the 2 increase? There was 3 options all with a random chance of winning one option is eliminated so how does the option you didnt pick have a better chance to win when your left with only 2 options?

@user-te8hq6qh4t

Жыл бұрын

@@wizenedoak5046 No wonder you don't see it because I completely fucked up the explanation. Just imagine there are 100 doors. You pick one and there's probably nothing behind it (1%). But host knows where the money is and he opens 98 empty doors, last one will have the prize. Disregard what I said about 50/50, it's your initial chance vs everything else, since host will open remaining empty doors.

@user-te8hq6qh4t

Жыл бұрын

@@wizenedoak5046 Essentially you choose whether there is something behind 1 specific door or somewhere among all the others

@wizenedoak5046

Жыл бұрын

@@user-te8hq6qh4t uhmm....so I guess by switching your picking 66% instead of 33% if that's right but it is still very odd to me to even try to look at it that way

3 options. 2/3 chance to get it wrong. They remove a choice. Since you were likely to get it wrong the first time, you’re likely to get it right by switching.

@dansir4102

3 ай бұрын

100 options. 99% chance to get it wrong. They remove all other choices leaving 2. Since you had a 99% chance to be wrong on your first pick, switching is likely the right call. It’s not 50/50. It’s a 99% chance that the one left is the right pick.

1 = X + ( Y + Z ) X = 0.33 (probability when you choose 1 out of 3) 0.66 = Y + Z Y = 0 (probability of door Y when host reveal its zonk) 0.66 = Z Host ask, do you want X= .33 or Z= .66 50/50 thinker logic : 1 = X + Y + Z Y = 0 1 = X + Z X = 0.5 (probability when you choose 1 out of 2 which is not what happen) 1 = 0.5 + Z Z = 0.5 Host ask you want X= 0.5 or Z = 0.5

@Araqius

3 ай бұрын

Assume your first pick is D1 (door #1). If the car is in D1, the host can open either D2 or D3. If the car is in D2, the host can only open D3. If the car is in D3, the host can only open D2. Now, let's say you play the game 6 times. The car will be behind D1 twice, the host will open D2 once and open D3 once. The car will be behind D2 twice, the host will open D3 twice. The car will be behind D3 twice, the host will open D2 twice. From these 6 games, the host open D2 = 3 times. One comes from when the car is in D1 and two come from when the car is in D3. This means when the host open D2, it's more likely that D3 is the car. We could also give a rule to the host that "Whenever the player pick D1 (which is the car), the host will always remove/open D2.". If the player pick D1, the host will open D2. Switching means you get D3. If the player pick D2, the host will open D3. Switching means you get D1. If the player pick D3, the host will open D2. Switching means you get D1. Switching win 2 out of 3 games. And if the rule is "Whenever the player pick D1 (which is the car), the host will always remove D3.". If the player pick D1, the host will open D3. Switching means you get D2. If the player pick D2, the host will open D3. Switching means you get D1. If the player pick D3, the host will open D2. Switching means you get D1. Switching also win 2 out of 3 games.

wow finally someone who explains it well 😂 jesus the others are just saying exactly what they heard and have no idea what they are actually talking about lol

yesn't

If I picked the goat at first, would Monty open the car’s door, or would he open the other goat and give me a chance to switch to the car? I believe this only works if I pick the car, so I'm sticking with my choice. Unless Monty doesn’t know and opens it randomly and it was the goat door, now do I have a 2/3 chance or is it still 50/50? My brain froze, and I couldn’t think of any other solution. I feel like I'm the only one who thinks it is still 50/50. Even when it comes to a deck of cards if I pick one card and it has to be the ace of hearts, and then Monty comes to flip 50 cards and none of them is the ace of hearts, and he leaves one face down and I already chose one, so now we have two cards. I still believe it is a 50/50 chance I have the ace of hearts because there is no way the face-down card Monty left has a 96% chance (50/52). Please help me! Because if Monty flips the 50 cards, it won’t be the ace of hearts only 2 times out of 52 so I should switch maybe! This only applies if Monty doesn’t know anything about the cards and where is the car behind which door.

I get the logic but switching might only be a good case if we expand it to more than 3 doors where whatever you do at the beginning, you have a high probability of picking a wrong answer.

@morbideddie

4 ай бұрын

The odds when switching get higher with more initial doors, but it is still in your advantage to switch in the three game scenario.

I disagree

There is also a 2/3 probability for doors 2 and 3. Since door 2 has nothing technically you should stay at 3

@ClipcutYT

Жыл бұрын

He explained it bad, the real way is there is 1 expensive car behind one door and a goat behind the other two doors, you have a 66% chance to get a goat and 33% chance to get a car, you pick door 1 he shows door 2 is a goat, now there is one goat left you were more likely to pick a goat so switch to get higher odds of a car

@ZZubZZero

Жыл бұрын

No, you should always switch since that has higher chances of winning. Double the chance, in fact.

@Araqius

Жыл бұрын

Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Basic math/logic kids understand.

It u don’t understand it it’s because you don’t understand probability u should always swap

No, it does not increase your chances. Dud = no money, winner = money In this scenario, door 2 is and always was going to be a dud. The winner door was decided before you had the chance to choose. By choosing any of the doors, you don’t automatically lump the other two doors into a group. They remain individual doors. The choice does not become 1 door VS 2 doors, it REMAINS 1v1v1. Your error is in lumping doors 1 and 2 together. You already know that at least one of the two doors you DON’T choose has to be a dud. By revealing door 2 to be a dud, you have not actually learned anything. Now, being that door 2 was ALWAYS going to be a dud, the chances of door 2 having the money (33%) don’t get lumps onto door 1, no instead it’s revealed that door 2 never had any chance of having the money. If door 2 never had the chance of having money, then it’s chance of having money originally can’t be added onto door 1 instead. I can prove it doesn’t help you another way. The logic can be applied the opposite direction. Knowing that door 2 was a dud, if you had chosen door 1 instead of door 3, and the host revealed door 2 to be a dud, door 3 now would have a 66% chance of being the dud. But surely that’s absurd, we already established door 1 to have a 66% chance. Either both doors have a 66% chance, or neither do. The answer is both. Since 66% chance = 66% chance, both would default to 50%.

@Araqius

12 күн бұрын

Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Switching means the opposite. It's just basic math/logic kids understand. Sadly, it's far too hard for idiots.

It will be 50-50, as you can see that there will be case in which i picked blue door and monty picked red......At last door blue as 1/3 prob. And door yellow has 2/3 prob. Hence it is opposite to this case to both are equal 50-50!!!!!

@XavierLikesBLD

15 күн бұрын

This is just incorrect. There are 3 options, so 2/3 chance you are wrong right? Correct. Next, he limits the choices down to 2 total choices, one right one wrong. The door you picked at the start had a 66% chance of being wrong, or 33% chance of being right. it’s more likely this door is incorrect. But now if you switch, there is a 66% chance this door is correct. The odds never “changed” they just were skewed by using your perception against you. Simulations have been done on this and it was proofed hundred of times as well.

@Araqius

12 күн бұрын

Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Switching means the opposite. It's just basic math/logic kids understand. Sadly, it's far too hard for idiots.

Its actually very tricky question since when thr first time you see the 3 doors your chance of getting it right is 33.33333% but after there 2 door left there 66.66666% the door thatyou dont peek are correct which mean 43.44444% you correct when you dont switch

@Mathispower4u

Жыл бұрын

You might find this simulation interesting to try. www.mathwarehouse.com/monty-hall-simulation-online/

@morbideddie

Жыл бұрын

33.33333%, not 43.44444%

@Araqius

Жыл бұрын

Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick.

But if you take the yellow door and the red door then you also have 2/3 changes of getting it so how does it work?

@Mathispower4u

Ай бұрын

But you only get to pick 1 door.

@Araqius

Ай бұрын

Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick.

What if I chose A? Should I switch? Or What if I chose B? What now?

@Araqius

4 ай бұрын

"What if I chose A? Should I switch?" If the car is in B or C, switching means you win. "What if I chose B? What now?" If the car is in A or C, switching means you win.

This aint a math equation, it's a psychological one.

@Araqius

2 ай бұрын

It's a math question. Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. It's just basic math/logic kids understand.

Why can't u group your chosen door and the wrong one to give yourself 2/3 isn't that perspective?

@Araqius

4 ай бұрын

Because the host cannot open your door. There is a difference between "cannot be removed" and "can be renmoved but is not removed".

If it’s looked at mathematically from the beginning as if the math can only be based upon “3 door choices minus 1” then the math is obviously correct but (I THINK, Not Know;) the math problem, with it’s required step of eliminating one of the no-money options before the final door choice is made, creates the “paradox” that there were actually never going to be more than 2 doors to choose from to begin with. I think the equation must be evaluated as follows: - The initial (soft) door choice… - then the elimination of one of the remaining (known) no-money doors. At THIS point, once the math reflects these set-up requirements, THEN an equation can be constructed and solved - leaving only 2 doors for the equation to evaluate - a 50/50 chance of winning and no advantage of switching choices. With this “Monte Hall”problem, it is designed that a person was never to have more than 2 door choices so an equation that solves for 2 possible door choices with the odds of 3 if the door choice is switched is a confounding feature of the accepted conclusion. BUT IF, the initial door choice was made AND the elimination of one of the remaining “no-money” doors was done at random, leaving that if the “money door” was opened/chosen for elimination, this would ONLY cause a “push” and the game began again and the game/equation THEN could be solved and then by switching choices to the other door would then give the 2/3rds winning chance even though there were never actually more than 2 real choices. I know I am going against brilliant mathematical minds in my conclusions and I’m probably wrong but this is the only way my little acorn can get wrapped around this problem. I know I’m probably wrong - Somebody Please help me see why the accepted method is correct and I am wrong.

@Mathispower4u

2 ай бұрын

All great thoughts. I appreciate you sharing. I am wondering if have more doors, which I address in my longer video kzread.info/dash/bejne/iqiTy5KSZsaWh9o.html, would change your thoughts at all. To me, there is great value in the discussion, which is sometimes missing in math classes.

@Araqius

2 ай бұрын

Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. It's just basic math/logic kids understand.

Why does this immediately jump to statistics & consider that an independent reality without ever actually explaining the psychological elements of this? Because this completely neglects actual probability.

Not true, it’s a 50/50. Once we know that one door is not the correct option, we can eliminate it entirely from the problem. Leaving us with a choice of only 2 doors. It’s a 50/50

@marksesl

9 ай бұрын

You are taking the obvious answer, which is always wrong in brain teasers.😅

@jonahfiegel8755

9 ай бұрын

@@marksesl except when the brain teaser is mathematically incorrect

@marksesl

9 ай бұрын

@jonahfiegel8755 The so-called "standard assumption" in the puzzle is that Monty would never reveal the car. Thus, the goat wasn't revealed randomly. If the goat had been randomly revealed, you would be correct. This is a non-random game where the host knows where everything is and would consciously be avoiding the car. That makes the other door 2/3 because he would have to avoid it two times out of three.

@jonahfiegel8755

9 ай бұрын

@@marksesl it doesn’t matter if the host knows the location or not. After the first round you are left with two options. It’s either heads or tails. I’m not understanding why everyone includes the first round if the option is taken off the table for us.

@marksesl

9 ай бұрын

@jonahfiegel8755 That's why it's a puzzle. It's not as simple as that. This is not a mathematical problem where randomness is the default position. This is a "non-random" game where the host is purposefully avoiding revealing the car, >> Two times out of three the host will be "forced" to open the door he did because the car will be behind the other door, giving it 2/3 odds of having the car. His opening one door is telling you the car is probably behind the other door, which would factually be true two-thirds of the time. The example of 100 doors makes this abundantly clear. If the host kept revealing goats until only one door was left, where he was consciously avoiding the car each time, that last door would obviously have a 99/100 chance of having the car, with the original choice remaining at 1/100.

i agree

Yes

Very skewed logic. The odds naver change to 2/3 chance, the 2 part of the 2/3’s is simply gone. Once one of the three doors is eliminated, the odds are instantly reduced to 1 in 2. A coin flip. Evens or odds. Binary chance. Like asking if I flip a coin 77 times and it landed on heads all 77 times, what’s the probability it will land on heads again? It’s always going to be 50/50 regardless when you only have two options.

@abdullahmohammed4928

Жыл бұрын

Couldnt have said it better myself

@Mathispower4u

Жыл бұрын

Maybe give this simulation a try. www.mathwarehouse.com/monty-hall-simulation-online/

@morbideddie

Жыл бұрын

Flatly incorrect. Two options do not necessarily need to have an even chance of occurring, and it’s trivially easy to see why. This is a conditional probability problem, not an independent one.

@ZZubZZero

Жыл бұрын

So confident yet so wrong. Your logic is skewed here bro

@sathnidu_athauda

10 ай бұрын

Over the long run, this logic is accurate and 100 percent correct works. If the game is played only 1 time then it's a 50 50 choice.

By this logic, if I'd originally picked Door #1, you'd tell me I should switch to Door #3. This is circular. So my answer is C: It Doesn't Matter

@morbideddie

Жыл бұрын

Yes, the 1/3 games you picked the car initially then switching will lose. Meaning the remaining 2/3 games you didn’t then switching will win.

@Araqius

Жыл бұрын

Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick.

@RonaldABG

Жыл бұрын

The problem is not symmetrical because since the host must always reveal a losing door from the two that you did not pick, then he has two possible options to remove when yours has the prize, but only one when yours hasn't. Take as an example that you pick door 1 and he revealed door 2. Being the car in door 3 would have made him 100% likely to open #2, as he wouldn't have had another option, while being the car in #1 would have only made him 50% likely to choose it, as he could have preferred #3 instead in that case. That's the reason why it is twice as likely that the reason why he opened that option is because the correct is the switching door rather than yours. If you had selected door 3, it would have occurred just the contrary. The car being in #1 would have made him 100% likely to reveal door 2, but being in #3, only 50%.

There is no prefer fot your choice, your choice not effect on probabilities C

You cant better your chance of winning by changing doors. At first its 1 in 3, then a door gets removed so the remaining doors have a 1 in 2 chance. It doesnt matter which door you choose, the best chance you have is 1 in 2. You can't better your chance of winning.

@DarkFingerHD

Жыл бұрын

If you had 10 doors and pick one and the host opens 8 of the other 9 doors, do you switch? This is what this situation represents, you'll have a better chance if you switch.

@Mathispower4u

Жыл бұрын

I have a longer video on the topic that goes into more depth, if interested. kzread.info/dash/bejne/iqiTy5KSZsaWh9o.html

@morbideddie

Жыл бұрын

Incorrect.

@Araqius

Жыл бұрын

Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Basic math/logic kids understand, idiot among idiots doesn't.

@zalavikrantsinh

10 ай бұрын

This is what average person iq is😂😂

Wrong idea here. Odds can't shift. Each door is 1/3. If one door is randomly known to not have the money, the other two are factually 50:50. Monty has to be knowingly avoiding showing the money for the last door to be 1/3.

@MogusaMogusa

9 ай бұрын

But Monty is knowingly avoiding showing the money. That’s part of the game. A wrong door will always be the one opened, never the door with money.

@marksesl

9 ай бұрын

@@MogusaMogusa The explanation doesn't exclude the possibility that a random revealing of a goat could have had the same results. Just saying the host won't reveal the car isn't enough. Everyone knows that and still can't figure it out. The part showing the doors having a collective odds of 2/3, and the showing of a goat shifting its odds to the other door is very misleading and just not what's going on. Odds can't shift. I'm not sure this can even be shown visually. It's more about deductive reasoning.

@ClarelySuperior

4 ай бұрын

@@markseslthis guy said the problem wrong, the point of the problem is monty always knows

@nimrodszocs2795

21 күн бұрын

@@ClarelySuperiorwell what kinda fkin game would it be if he discarded the money door lmao

No, you can’t keep the probability and eliminate a door, the grouping of the second and third door and also applied equally to the grouping of the first and second, it’s still 50-50

@Araqius

12 күн бұрын

Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Switching means the opposite. It's just basic math/logic kids understand. Sadly, it's far too hard for idiots.

@hi-eg8wn

12 күн бұрын

@@Araqius picking goat a first keeps the chances?? This is just a dump problem to make staticians look smart

@Araqius

12 күн бұрын

@@hi-eg8wn lmfao How are you goin to win by staying if your first pick is Goat A? How are you goin to win by staying if your first pick is Goat B?

BUT THERES A 2/3 CHANCE ITS BEHIND THE RED/YELLOW

@neonix2144

9 ай бұрын

Exactly

@EnCroissant

9 ай бұрын

You only picked one though…

@MiddleChildYT

9 ай бұрын

Don’t worry people above me, I later found out how it works

You missed one important part in your explaination: you, as Host of the Game, knew that door 2 was 0$. If you knew the contastant should switch, If you did not know it does not matter.

It's Monty Hall, not 3 card monte This guy doesn't even know the game, yet is going to explain the odds

it does not matter because its 50/50 after you remove door 2

@morbideddie

Жыл бұрын

Incorrect.

@Araqius

Жыл бұрын

Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Basic math/logic kids understand, idiot among idiots doesn't.

@lbright4568

8 ай бұрын

think about it this way: assuming you always switch, every single time you originally pick 0$, you will win the million. therefore, you will win the million 2/3 of the time if you switch.

This isn’t how it works. A mathematician explained it in a video and said that you don’t have a better chance if you switch. You will still have a 50/50 chance not a 66/34 chance

@Game_Ender4

2 ай бұрын

OMG there is no mathematician that thinks it is 50/50 lol. Stop spreading lies. If you want, I can offer multiple explanations to show you why you are wrong, would you like that?

@Araqius

2 ай бұрын

Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Switching means the opposite. It's just basic math/logic kids understand. Sadly, it's far too hard for idiots.

@c0nf1dent1al

2 ай бұрын

@@Araqius I am just a step ahead of everyone else. Everyone who is against it is an idiot.

@Araqius

2 ай бұрын

@@c0nf1dent1al Bark all you want.

@c0nf1dent1al

2 ай бұрын

@@Araqius Mhm

Say that you pick door 3 not because you think the money is there, as in this scenario, but because you know that there is a 2/3 chance that the money is in one of the two doors that is not door 1. The host then shows you door 2. Since there’s a 2/3 chance it’s not in door 1 and you know it’s not in door 2, then you’d be better off staying. But wait, this doesn’t make any sense. Just changing the reason you chose door 3 initially shouldn’t change the probability. Therefore, this is all bogus and it’s 50/50

@RonaldABG

9 ай бұрын

You have not understood the problem. The 2/3 comes because when your selected door has a goat, the host is restricted to reveal one specific door (which has the only other goat), but when your selected one has the car, the host is free to reveal any of the other two, as both would have goats, so we never know which of them he will reveal in that case. (Remember that he must always reveal a goat from the doors that you did not choose). In that way, when you start picking door 1 and he opens door 2, we are sure that in case the car were in #3 he would have opened door 2 with 100% certainty, as he wouldn't have had another option. But in case the car were in door 1 (your choice), it was only 50% likely that he would then open #2, as he could have preferred to reveal door 3 instead. That's why it is easier that once you select #1, he then opens door 2 if the car is in #3 than if it is in #1. If you start selecting door 3, the opposite occurs. If the car is in door 1, he will undoubtely open door 2, but if it is in door 3, we never know if he will prefer to reveal door 2 or door 1 in that case.

@Araqius

8 ай бұрын

Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Switching means the opposite. It's just basic math/logic kids understand.

@wipoute

8 ай бұрын

​@@Araqius sorry but kids don't understand that level of probability problem. You think you look smart by posting this, but you could never have figured that out yourself as a kid.

@Araqius

8 ай бұрын

@@wipoute lmfao. It's just basic math.

@wipoute

8 ай бұрын

@@Araqius yeah, if you say so 🤓

Probability isnt math, its a power of god and only god and let me debunk, you're blind folded with a gun and theres 3 people and 1 of them is your friend, you choose number 1 and the game maker killed number 2 then if number two is gone and got thrown into a dumpster, youre left with 1 and 3 and the game maker gave you another chance, choose 1 or 3, now if you apply this rule, it's funny, because you would sum up a dead guy with a dude thats alive, weird huh. You're 1\3 chance in 1 and 1\3 chance in 2 and 1\3 chance in 3 , but wait 2 got killed, he's LITERALLY GONE, now i have 2 people left, 1 and 3, now before you shoot, how many did you have and how many do you have, and what do you think your chances are, oh right 3 before and 2 now, 1\3 before and 1\2 now because we sum up chances by our TRIES and the TOTAL of things we can choose, not we haven't chosen , if i have 1 shot and 2 people is 1\2 not fkn 1\3 and 2\3 including a dead guy. Sorry this just gets me mad because its a blunt lie that failing mathematicians like, because its no way to math probability, probability is probability just the amount you have againist the amount there is, its ratio if you math it and it's god's power to know the odds we dont know THE ALL KNOWING GOD. So for god's sake stop feeling sorry and try to think for your self. Egyptian 15 year old.{give your best logical explanation after reading my long reply}

@klaus7443

9 ай бұрын

It's no wonder you don't understand the Monty Hall Problem.

@EnCroissant

9 ай бұрын

Statistics is literally a math class you can take

@Araqius

8 ай бұрын

Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Switching means the opposite. It's just basic math/logic kids understand. Sadly, it's far too hard for idiots.

@hassansaied8471

8 ай бұрын

​@@EnCroissantokay fkn explain then.

@hassansaied8471

8 ай бұрын

​@@klaus7443okay genius, tell me how removing a door makes it 2/3 chance when switching

mathematically sure, maybe, however, a game of chance as this one once the middle door is out then the proposition become 50/50 as that just because there once was a 2/3 probability that the money was behind door 1 that there was also a 2/3 probability that it was not there. so the question is there or not, and once a 1 door is revealed as not there remains a there or not proposition . or 50/50 chance you picked correctly. So first guess is usually correct but not always. the first instinct correct is more often than not when watching these games play out.. when players change their mind they lose. depending on the game as well. the LMD show sometimes entices folks away from bad decisions but then switch up to entice them away from correct choices as the hosts do know what is behind doors 1, 2 and 3 and outcomes are based on how easily manipulated the contestant is.

@klaus7443

9 ай бұрын

You're completely lost. When you picked a door the host could have simply asked if you want to exchange it for one that has a different prize.

@MrWaltjam

9 ай бұрын

@@klaus7443true, but what could have happened was not really the question. I am NOT lost. In the first place the probability was a 1 in 3 chance you had the right door. Also 2 in 3 that you had the wrong one. You have to realize that the probability changes once the first empty door is revealed and the question is asked again.. Now it is just as likely that the prize is behind your door than the other. Also just as likely that the prize is not there since there is two doors. They always try to distract you with a "sure thing" sometimes its a "Zonk,"

@klaus7443

9 ай бұрын

@@MrWaltjam "You have to realize that the probability changes once the first empty door is revealed and the question is asked again." Like I said...you're completely lost. When you picked a door the host could have simply asked if you want to exchange it for another with a different prize. There is no difference between.... A) Picking the door with the car, seeing a goat, switching to a goat and... B) Picking the door with the car, switching to a goat And there is no difference between... C) Picking a door with a goat, seeing a goat, switching to the car and... D) Picking a door with a goat, switching to the car Have a nice day!

@rats7386

8 ай бұрын

​@@klaus7443you're the lost one. If you have 2 doors to pick from it's a 50/50 fucking clowns

@Araqius

8 ай бұрын

@@rats7386 Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Switching means the opposite. It's just basic math/logic kids understand. Sadly, it's far too hard for idiots and their stupid parents.

Thats... not how that works

@anthonyjones6870

Жыл бұрын

Yes... It is

@morbideddie

Жыл бұрын

Conditional probability has been understood for hundreds of years. It’s exactly how probability works.

@morrowmorrow4811

Жыл бұрын

@@morbideddie No shot

@morbideddie

Жыл бұрын

@@morrowmorrow4811 you can reject it all you want, but Bayes Theorem is pretty well established. Can you disprove it?

@Araqius

Жыл бұрын

@@morrowmorrow4811 Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Basic math/logic kids understand, idiot among idiots doesn't.