@kmorgan26 the chance that the door I chose has a car behind it DOES NOT depend on the chance that the door I chose has goat #1 or goat #2 behind it. In this regard, it would be accurate to say that I don't believe the chance that I chose the door with the car behind it is 1/3.
I guess this video is designed for me because I insist on remaining with my original pick. You claim this makes me 'wrong'. I, however, beg to differ. The question you pose is, "Is it to your advantage to switch your choice of doors". I hold that you won't know whether it was to your advantage until the doors are opened. You might say by virtue of this response that I think you an idiot. I don't think your an idiot though and so I refrain from insulting you.
@eeyoreeeyoreuhoh Did you make a Youtube account just to post this comment? I have a feeling you are trying to hide your identity. And if you stay you will only win 1/3 of the time. That's because you have a 1 in 3 chance of picking the correct door on your first pick. It's that simple.
@kmorgan26 Yes, yes I did. Did you go to my channel to try to get an idea about who I am? I have a feeling you did. Well, I won't meddle in your claims that you will only win 1/3 times. I will only say that what the host does regarding the doors does not constitute an intimation of where the car is. Therefore the answer to your question is one won't know whether it was to his advantage until the doors are opened. The reason this is the answer is because the contestant receives no intimation
@eeyoreeeyoreuhoh Have you seen the other video I made regarding this puzzle where I actually "play" the game? I'll send you the video in a message.
Also, I'm thinking of making another video about this, but the video will be of me making a computer program that simulates the puzzle. I have made several of these programs, and it becomes clear that switching is in the interest of the contestant.
@kmorgan26 I think that any door you've chosen, by virtue of the game rules will have either a goat or a car. If you take that to mean that the contestant has a 50/50 chance, so be it.
@eeyoreeeyoreuhoh If your goal is to avoid holding a position on this, then you're succeeding. You commented on this video for a reason. I assumed it was because you disagree with my conclusion...which in this case also happens to be an undeniable fact.
At the point in the game where the decision has to be made to switch or stay, it is in the interest of the contestant to switch. You either agree with that or you don't. Which is it?
@kmorgan26 I do have a position and it has been stated quite clearly. That position is that you won't know whether or not it was to your advantage to switch your choice of doors until the doors have been opened.
Now, you state "At the point in the game where the decision has to be made to switch or stay, it is in the interest of the contestant to switch."
If by 'in the interest of the contestant' you mean 'to his advantage', I would have to disagree with you. Its impossible to tell until after
@eeyoreeeyoreuhoh Run that program 100 times and stay every time and you'll find that you only win 1/3 of the time. If you play it 100 times and switch every time, then you win 2/3 of the time. Why do you suppose that is?
@kmorgan26 I don't suppose why that is. I don't have to. My point argues quite well to conclusion of itself. Moreover, you don't and haven't disagreed with any point I have made about this matter. I can't quite see what your opposition is to what I've said if it is not opposition to any point I have made.
@eeyoreeeyoreuhoh You asked me to quote something I disagreed with and I did that. It's the only time I quoted you so it will be easy to tell.
I'm asking you to clarify your position more clearly and you are avoiding doing that. Let me step through it one step at a time. Do you believe that when you first make a pick, that the chance that you made the correct choice is 1/3?
@kmorgan26 ...the car does not have any "chance" of being behind a certain door. It IS behind a certain door. The door which it is behind is not my concern. I must commit to one. Now that I've committed to a door, since there is nothing the host will do that can be construed as a hint that the door I've committed to is incorrect or that the door I haven't committed to is correct, I have no rational reason to de-commit from the door I have committed to.
@eeyoreeeyoreuhoh Do you not believe in the concept of probability? If there are 3 of something, and only one is right...and you pick one then there is a 1 in 3 chance of you picking the correct one. If you don't believe that, then you're just denying a fact. In the game, you pick door #1, then there is a 1 in 3 chance of you being correct. This isn't my opinion, it's the way it is.
@kmorgan26 "...there is a 1 in 3 chance of you picking the correct one."
Oh but there is also a 1 in 2 chance that you have picked the correct one. The door you picked is either the correct one or its not. You won't know until its revealed by Monty Hall. This too is an undeniable fact
@eeyoreeeyoreuhoh I am either going to be struck by lightning today, or I am not.
One of those will certainly be true, but they are not equally likely. There is a much higher chance that I will not be struck. You are right that there are 1 of 2 possibilities (car or goat) but they are not equally likely. If you can't see that, you're just not very bright I'm afraid. That's not an insult, it's just true.
@kmorgan26 "You are right that there are 1 of 2 possibilities (car or goat) but they are not equally likely."
Well, you are right that they are not equally likely and I see that. However my point is that the more unlikely outcome is equally plausible. If indeed you switch, and you end up with a goat, was it to your advantage to switch? No, because you didn't get the car. If you can't see that, you're just not very bright I'm afraid.
If eventually you are left with only the car and the blue goat (you deduced there was only the blue goat left), you see that the blue goat also had a 1% chance to be behind any specific door. So 1% and 1%, except you got rid of 98 possibilities, so you are left with a 50/50... no?
@MiniGeek31337 no. If you were to randomly open 98 doors, then sure...there would be an even chance that each of the doors held the prize. But the host is opening doors he knows to be goats. He's giving you information you already knew (that there were goats behind 98 of the doors). Your original pick is still 1/100. To switch away from your door is really in exchange for the other 99 doors.
after seeing "21" and had a discussion with my colleagues, i made a little programm that simulates the monty hall problem. while programming i got the idea behind the problem. you don't have to look through the eyes of the guesser but the moderator. then the swap makes perfect sense =D
I wish your simulations weren't deleted. I have to write a C++ program of the Monty Hall Puzzle and I can't understand how to present it with what we learned. Oh, well.
I did some mapping and it looks to me like if you choose door #1, there are seven possibilities as to what the game show host reveals, and what you pick, if you pick randomly. Four of those possibilities are losses, two are winning by swapping, and one is winning by staying. So it is twice the chance there, but only 2 out of 7.
I'm extremely rusty on my probability. Curious, what if you swapped two times out of three instead of swapping every time?
When you choose to swap you're being humble enough to admit that your choice might've been wrong. When all other doors are opened with the goats but one and the one you chose, it's like the gameshow host said: "all right, all of those doors had a goat and I knew it, but I won't tell you whether this one here has one and I respect your decision so much, that I won't tell you that your chosen door has a goat, i.e, your aiming is bad (or not)".
That's interesting to say the least. My friend had a mini program thing with it and you could set it to switch or stay with the door or not. the % gets more clear as the # of times repeated goes up. but yeah, it's a neat puzzle/concept. although I have to say I don't like "probability"... or atleast the class
this is so easy to understand, I just watched this video to see your explanation, I like the way that you explained it with the 100 door problem, I think that people can understant it more easily with that explanation, good job
He never "takes away" a door. He just opens one of the remaining two doors (one that you did not pick). He will never open the door with the car, however.
Even if you switch, no matter what...it is a 50/50. You want to know why? It's because you REMOVE your previous vote, it's not going to win you anything. When you say it's a 2/3 when you switch, you subtract one option and you're left with 1/3 total. With a goat shown, it is a 50/50. You can guess right on the first one as much as you guess wrong, BECAUSE THE HOST WILL ALWAYS SHOW A GOAT. Please explain the flaw in my logic.
Given three options, you will guess wrong twice as often as you guess right (on the first pick), so switching wins 2/3 of the time. This isn't just a hunch of mine. It's true.
Watch the video response to this video too and check the simulation in the description. Lemme know how it went for yah.
why are you so retarded... it's so easy to understand.
theres 2 goat 1 car , you have 2/3 chance of picking goat 1/3 picking car. if you pick goat and switch you win car everytime , as chance of picking goat is 2/3 you should always switch , your first choice is always more likely that you will pick a goat... so it's 2/3 NOT 50/50 !!!!!!!!!!
You're still ignoring the fact that when you switch, you're removing your previous chance. Nothing is gained. Insults doesn't justify your logic, try again.
YungSensei,,, so it's you against the world's top mathematicians...... !!!!! play the game yourself and you will see that you are soooo wrong. you may not understand but just cos you don't , don't say everyone else s wrong.. that is so egotistical and will get you nowhere---listen to the experts and you may learn something in your life
and you have smelly cunt. Childhood name calling over yet? I have a few adult names to call you if you want to get into a debate about the validity of this video's math and thus (gasp) truth.
i know your right, but about the 1 million doors, if i switch ill have 999,999/1000000 chance to win, but if i stay i have 1/1 million? that doesnt seem right to me, it probably is right, but doesnt seem like it... so i have a question. iff i do that with 1000doors, if i do an experiment with a friend. if i choose to switch your saying ill win everytim pretty much.
@RickyAustinR if you do it with a 1000 doors then eliminate 998 doors without revealing the car if you switch every time then yes you will win 99.9% of the time
first of all dude i think you are wrong. 100 doors is a lot different then 3 doors. if you have 3 doors and they take one away. your should look at the problem again and it is still 50% of winning. it is different with 100 doors. but the i did the program and with 3 doors and found i stayed with my same door 100 times stright i was up 54 to 46 so i would have won more cars then you. so whos right
I think you just made the idiot list, the 100 door simulator is just and extreme to explain the problem more and make it more comprehensible, in fact, i tried his simulator and played the 3 door 20 times and got 13 wins and 7 loses, with the 100 door i played 20 times, and got 20 wins, it is just an extreme to show how it all works.
This comment has received too many negative votesshow
I think you're wrong, You initially had 33% chance, but since he eliminated one out of three you have 50% change that your next pick (either the same one or another one) will contain a car.
When he reveals one of the doors, you gain nothing by that information. You already knew that one of the remaining two doors had a goat. What if he gave you the chance to keep your door, or take the other two? This is essentially what he is doing, but just showing you a goat before he offers you the other two doors.
Yeah, I know, but you are still alowed to choose, you can also choose to stick with your current choice, if you on forehand already knew where one of the goats were you would have 66% chance.
But here you have two choices, so either choice is 50% chance.
No. It makes no difference if you know beforehand. You already know beforehand that AT LEAST one of the other two doors has a goat. What is the difference if he shows you? None!
Think about 1 million doors. He shows you 999,998 goats and there are two left? Your original "1 in a million pick" or the other door...still gonna stick with your original?
My logic isn't flawed. In fact, I didn't make this puzzle up. You can debate it forever because you don't understand it.
I could debate forever that the world is flat, but that wouldn't make my argument any stronger. There are plenty of programs online to test this. Why not find one? Google "Monty hall problem simulation" and see if that helps you understand.
This is the golden rule. If you pick a goat, a door is opened to reveal the second goat, swapping will ALWAYS win you the car if you were lucky enough to pick one of the 2 goats.
What are the chances you will pick a goat before he opens the door? 66%
lol when i 1st watched the montyhall problem, i was confused, cuz im like, 12. but u helped! i think about it like:
when u first pick a door, u shud assume that u picked a goat cuz its a higher chance. then the host shows u another goat. so he shows u a goat, and u assume u picked a goat. thats 2 goat. whats left? a car. so make the switch. lol yeah...
Suppose I have 3 cards: an ace and two kings. I deal you a card and I deal myself the other two. Whoever gets the ace wins.
I show you one of my cards. It's a king. No matter what, I have at least one king. Do you want to trade for my other card?
____
Now, same problem but I don't show you any kings. You have 1 card, I have 2. Want to trade my 2 cards for your 1? Of course! The fact that I showed you a king in the previous example is irrelevant!
Really nice explanation. I'm gonna use this on those stubborn folks who still insist on the 50/50 scenario. I've tried the million door thing, and a few other examples, but some people just can't be convinced. This might work. (Of course, I won't get my hopes up).
Yes, the revealed door is irrelevant. The information gained by that door being revealed is not telling. There is always going to be a goat behind the door shown by the gameshow host. Furthermore, of the two doors you do not initially pick, at least one will have a goat. Thus, it's superfluous information.
@jeremyemilio With my parents, I physically put ten things on the table in front of them (we were sitting outside late at night talking) and put something behind one of them when they weren't looking, then after they picked one object, I removed the 8 others that had nothing behind them, and we played the game a few rounds.
The reason why people don't get it is that they can't grasp the whole concept in their mind. They need a physical example to lighten the mental load.
i almost got on the idiot list, but he's right do what he says and play the 100 door game first, then you'll see it's not about 50/50. so i think the movie was wrong about this. thanks dude!
Opening a door is not an honest move. This is based on a con game and the host is a con man. He is opening a door to manipulate you.
If you watched the show, He did not always open a door. Sometimes he brought another box on stage and offered that.
I am not saying the odds are not correct. I am saying that if you act this way the host might only open the door when he knows you picked correct the first time.
You assume that the host will play by the same rules every time. Not the case
Yes. In real life, some people make some of the dumbest fucking comments to try and make themselves look smart. In most cases, this results in total failure.
at least the creator of the video understands the concept
those with a sense of intelligence just give this comment a thumbs up just to prove the amount of people that can back this statement. "this guy is right."
No. It isn't 50/50. Not even close. It's basic statistics. If you pick the car and don't switch, you win 1/3 of the time. If you pick a goat and don't switch, you lose 2/3 of the time. If you pick a goat and switch, you win 2/3 of the time. If you pick a car and switch, you lose 1/3 of the time. Simple.
For those who still don't understand this, think of it as follows:
If you don't change your choice when the host asks you if you want to do so, then your chances remain the same, that is 33%. Now, if there is a 33% chance that the car is behind the door you chose initially, then where can the other 66% be? Well, there is only one possibility, and that is the other door. That's the way i figured it out... hope it helps.
For all of you who fail to understand this, I do not blame you; However, think of it in this perspective:
You are on a game show, hosted by an MC, in which there are 100 doors to choose from. Behind 1 door is a car, behind the rest are goats. The MC says that once you pick your door, he will eliminate all wrong doors but one. Because the MC had 99 doors to begin with, there is a 99% chance that the remaining door is the car. Therefore, there is a 99% chance of winning if you switch to his.
If these were 3 boxes and one of them had $1000 in it while the other two were empty then its the same problem.
If Monty did not open the empty box and asked whether they would like to swap their one box for his two then the result would be the same.
Monty has twice the likelihood of having the prize, since he has two boxes.
One of Monty's boxes will always be empty. So whether he opens the box after, or whether he opens it first and asks you to swap the unopened box, makes no difference.
I've been considering explanations for this for a couple of days, and I've got a couple of simple ones. Here's the first:
1. Two times out of three, the door you choose initially will be a goat. That means that, two times out of three, when Monty opens another door and shows you a goat, the remaining door will be the car. The only time swapping won't win the car is when you've chosen the door with the car initially -- a 1 in 3 chance. So, swapping wins 2 times out of 3.
I think you're misunderstanding the problem. It's 50/50 if you walked into the game show room after a goat door was opened and someone asked you to pick one of the remaining two. But the thing with probability is that you often have to take past events into account. After the host opens a goat door, he leaves you only one choice to switch to, and by taking the switch you're essentially swapping the probability you got it right the first time (1/3) with the probability you got it wrong
Absolutely correct. There are two trials in this game. The second trial is not "fair" because the game show host has prior knowledge of the location of the car. If the game show host had no knowledge, or if you did not see the first pick, then it would be "fair"
the way that seemed easiest for me to understand it is that if u pick door 1 that means that if u put 2 and 3 together its a 2/3 chance that it is behind 2 or 3 so when either 2 or 3 is revealed that 2/3 chance remains with both of them so u would obviously want to pick the one that is not revealed??? i don't know if that makes it easier for anyone
Dude, I just got into an argument about this with my college math teacher last night. I showed him the example on the chalkboard multiple times, raised the number of boxes (what we used) to five and showed him some more. He just couldn't get his logic out the mindset that it was not a 50/50 chance. Needless to say, I left class irritated.
Another simple explanation is, suppose before you even play the game you decide that your strategy is to always switch. Then your first choice of door gives you a 1/3 chance, but your switch effectively gives you 1-1/3, or 2/3.
You should always switch, its simple a math law states that your odds never change, if i had a hat with three colored cards a red green and blue card, and i told you to pick a red card at least once out of 20 tries and you pick 19 green cards in a row, on the last try your odds for picking a red card is still 1/3,in simplified terms if i flip a coin 19 times and in lands heads 19 times the next time i flip the coin my odds are still 1/2 in getting heads again.
afly727.. i agree that sequent coin trous are independent events.
Now, in a dependant chain of events for example i put your cards in a black bag and ask you to remove a red OR green, NOT REPLACING the first you took. the chances of 1st positive result is 2/3, the chance of 2nd positive result depends. If you get a positive first time you are left with 1 in 2. If you didn't get a positive 1sdt time, you'll get one in the 2nd , because you have 2 in 2! If you REPLACE 1st and 2d become independant
Nice job. The variant of using an extreme number of doors, and at the second stage of the game opening -all the others except 1- is a great explanation and it's a shame more people don't use it. It just makes it so clear. The counter-intuition in the problem comes in because in the case of 3 initial doors, "all the others except 1" is just 1 measly door. But in fact, the host reveals almost all wrong doors. And the little app you wrote just drives the point home experimentally, nice.
stay: 1. you will lose; 2. you will lose; 3. you will win. Probability = 1/3.
switch: 1. you will win; 2. you will win; 3. you will lose. The formula for the probability if n is the total number of doors is: for staying, 1 over n. For switching, n-1 over n(n-2). I didn't completely understand kmorgan's point.
I enjoyed your explanation. However the original door that you chose changes from 1 in 3 probability to a 1 in 2 probability of the car being behind your door once the host has changed the number of outcomes for the problem. It is a new problem at that moment. Now you have two people on your idiot list.
I have studied this problem and realize that you are correct. I have accepted the explanation but it still does not "feel" or seem right. I guess that is what makes the problem entertaining. Where are your downloads so that I may play with them? Thanks again!
to everyone that still believes it's 50/50.. you are wrong. it's funny how retards can't comprehend this simple problem even with these videos proving it.
The way I understand it having seen this. When you make your first pick, you have a 2/3 chance of picking goat. When the presenter shows you another goat, your original pick still has 2/3 odds of being goat. And so you pick the other door, which is 1/3 chance of being goat. The trap people fall into is thinking the odds of their 1st pick change after the first door is opened. They forget their first pick always has more chance of being goat, so it makes sense to swap when the other goat is gone.
Well said Kmorgan, it's something I've argued for years.
The way I try to put it is, my original odds of winning are 1 in 3, so the remaining 2 doors combined have odds of 2 in 3. Once I'm offered the chance to switch my choice I then inherit the 2 in 3 odds of winning, the fact that one of those 2 doors has been eliminated bt the host has no relevance. In effect by switching doors I'm swapping one guess for 2 guesses.
Another way of thinking about it is you have a 1/3 chance on your side and 2/3 a chance on the other side. So, if one of the doors on the other side is revealed all of the 2/3 goes to the other side that still has an unrevealed door. So, the door you did'nt originally pick, but is still unopened has a 2/3 chance of being a car.
Thanks for the posting. I watched several other videos here and most of them confused me alot more than help me understand this concept. Keep up the good work!
Does this seem like the right way of thinking on this?: The contestant has a 1 in 3 chance originally, and the game show has a 2 in 3 chance of having the car on their side. So, when they give up one door they gave up 1/3.
The first time I encountered this particular problem was when I watched the movie 21. I have always been exceptionally gifted at math and when that particular part of the movie played, I said 50/50 as well. The entire concept went right over my head but it intrigued me enough to research it and I must say that the Monty Hall Paradox is rather clever in its initial approach but the correct answer is eventually and painfully obvious.
It reminds me of the scene from "A Beautiful Mind" where Nash works out the best strategy for getting laid. That sparked so much interest for me in Prisoner Dilemma's etc.
at 3:20 "because i'm giving you all the other doors"
exactly.
i don't know why some people don't understand this. with 3 doors -- you have 1 and the host has 2. now he tries to trick you by "narrowing down the choices", but the fact of the matter is, you can trade your 1 door for the host's TWO DOORS. this is of course against the rules, so you choose the one that he didn't open.
i shuffle 3 cards (ace of diamonds, ace of hearts, ace of spades) and let you choose one without looking, and i get the other 2. your odds of having the spades ace is 1/3, and my chance of having it is 2/3. let's say i can look at my 2 cards. now it doesn't matter if i reveal a red ace (which i must have), give it to you, eat it, or flush it down the toilet. you still have a 1/3 chance that your hidden card is the spade. 1 - 1/3 = 2/3. there's a 2/3 chance my remaining card is the spade.
God bless you man! Taking it to the idiots and numbnuts out there. It made perfect sense the first time I saw it, and it does whenever I see it again.
I don't think people are as big of idiots as you think they just realize more conditions have to be met for the 2/3 and 1/3 to be correct then are often given in the problem if the problem is stated correctly then your explanation is valid but it many times is left incomplete making the odds of 50/50 or 2/3 not really attainable mathematically in the simulator you created certain aspects are a given you programed them in as constants which is why it works.
oh my bad I didn't watch the video you where commenting on they do give the correct variables that the host knows what is behind each door, has to show you a door first, and has to show you a goat that is all I am saying this problem is often not worded correctly you are correct in this situation people just refuse to see because they are stuck in their real life scenario in there heads where it would remain unknown what was required of the host.
You ALWAYS get a car if you pick a goat first, and you ALWAYS get a goat if you pick a car first. You have to look at conditional probabilities. There is no 50/50.
I can't get your 100 door simulator one to download, but the three door program works great. I actually saw this problem on Numb3rs and thought it was the coolest thing ever. After 100 times of playing the 3-door one and switching every time, it was 63%. That's pretty darn close to the 66.7% Great program!
The third door is only smoke and mirrors. Regardless of you picking right OR wrong, you're going to be revealed a goat. When you apply three variables to two doors, of course your math will be off. Its only a two variable equation.
If you pick right, you see a goat, if you pick wrong, you see a goat. This means that its between the door you picked, and one other door.
This has nothing to do with smoke and/or mirrors. It's a probability problem. It's not 50/50. Download my program (3 door) and try for yourself. Until then, you're on the idiot list.
The work you're doing for the math is right. Your initial math is wrong. You apparently failed word problems in school because you don't know how to set the equation up.
Regardless of how many doors, regardless of your pick, regardless of your programs, you'll still see X goats. So in the case of 100, you'll see 99 goats, NO MATTER WHAT YOU PICKED. If you have 3 doors, you'll be shown one goat, no matter what.
Its 50/50, you fail at comprehension, probability, and setting up a math problem.
Actually, in the 100 door example, you'd be shown 98 goats...not 99. Pretty elementary. So you fail.
This is not my problem...it's a classic one. I didn't "set it up". I just understand that one's odds of picking the car when presented with three doors is one in three...and those odds don't change because doors are opened.
You're right! We'd better get on the phone to every mathemetician that has studied probability for at least the past sixty years (as the three prisoners problem) and let them know. I bow down to your superior mathematical intellect! I'll let all of my fellow mathematics grad student's know of the new development. You fail at life fool.
It may be "smoke and mirrors" to anyone who doesn't truly understand or is willing to accept probability. You do know that the first person who suggested the 2/3 chance of winning for the monty hall problem was a woman with the highest IQ in the world, and after being attacked by countless other mathematicians, she turned out to be correct in the end.
Just for clarification. Marilyn vos Savant may have been "the first person who suggested 2/3 chance winning" for the Monty Hall problem but this problem has several equivalent problems that were solved long before. (see Bertrand's box, three prisoners) So this argument is pointless really. Also, your statement "the highest IQ in the world" is misleading as others may not have been tested etc. Lastly a high IQ is not equivalent to being a good mathematician/statistician.
@kmorgan26 the chance that the door I chose has a car behind it DOES NOT depend on the chance that the door I chose has goat #1 or goat #2 behind it. In this regard, it would be accurate to say that I don't believe the chance that I chose the door with the car behind it is 1/3.
eeyoreeeyoreuhoh 5 months ago
I guess this video is designed for me because I insist on remaining with my original pick. You claim this makes me 'wrong'. I, however, beg to differ. The question you pose is, "Is it to your advantage to switch your choice of doors". I hold that you won't know whether it was to your advantage until the doors are opened. You might say by virtue of this response that I think you an idiot. I don't think your an idiot though and so I refrain from insulting you.
eeyoreeeyoreuhoh 5 months ago
@eeyoreeeyoreuhoh Did you make a Youtube account just to post this comment? I have a feeling you are trying to hide your identity. And if you stay you will only win 1/3 of the time. That's because you have a 1 in 3 chance of picking the correct door on your first pick. It's that simple.
kmorgan26 5 months ago
@kmorgan26 Yes, yes I did. Did you go to my channel to try to get an idea about who I am? I have a feeling you did. Well, I won't meddle in your claims that you will only win 1/3 times. I will only say that what the host does regarding the doors does not constitute an intimation of where the car is. Therefore the answer to your question is one won't know whether it was to his advantage until the doors are opened. The reason this is the answer is because the contestant receives no intimation
eeyoreeeyoreuhoh 5 months ago
@eeyoreeeyoreuhoh Have you seen the other video I made regarding this puzzle where I actually "play" the game? I'll send you the video in a message.
Also, I'm thinking of making another video about this, but the video will be of me making a computer program that simulates the puzzle. I have made several of these programs, and it becomes clear that switching is in the interest of the contestant.
kmorgan26 5 months ago
@kmorgan26 Well, have I said anything that you disagree with? Quote anything in my reasoning that you disagree with and I would happily discuss :)
eeyoreeeyoreuhoh 5 months ago
@eeyoreeeyoreuhoh You said "I hold that you won't know whether it was to your advantage until the doors are opened."
Does this mean that you think it's a 50/50 chance at the point where there are two doors remaining?
kmorgan26 5 months ago
@kmorgan26 I think that any door you've chosen, by virtue of the game rules will have either a goat or a car. If you take that to mean that the contestant has a 50/50 chance, so be it.
eeyoreeeyoreuhoh 5 months ago
@eeyoreeeyoreuhoh If your goal is to avoid holding a position on this, then you're succeeding. You commented on this video for a reason. I assumed it was because you disagree with my conclusion...which in this case also happens to be an undeniable fact.
At the point in the game where the decision has to be made to switch or stay, it is in the interest of the contestant to switch. You either agree with that or you don't. Which is it?
kmorgan26 5 months ago
@kmorgan26 I do have a position and it has been stated quite clearly. That position is that you won't know whether or not it was to your advantage to switch your choice of doors until the doors have been opened.
Now, you state "At the point in the game where the decision has to be made to switch or stay, it is in the interest of the contestant to switch."
If by 'in the interest of the contestant' you mean 'to his advantage', I would have to disagree with you. Its impossible to tell until after
eeyoreeeyoreuhoh 5 months ago
@eeyoreeeyoreuhoh Did you watch the other video or play the game simulator that is linked in that video?
kmorgan26 5 months ago
@kmorgan26 yes.
eeyoreeeyoreuhoh 5 months ago
@eeyoreeeyoreuhoh Run that program 100 times and stay every time and you'll find that you only win 1/3 of the time. If you play it 100 times and switch every time, then you win 2/3 of the time. Why do you suppose that is?
kmorgan26 5 months ago
@kmorgan26 I don't suppose why that is. I don't have to. My point argues quite well to conclusion of itself. Moreover, you don't and haven't disagreed with any point I have made about this matter. I can't quite see what your opposition is to what I've said if it is not opposition to any point I have made.
eeyoreeeyoreuhoh 5 months ago
@eeyoreeeyoreuhoh You asked me to quote something I disagreed with and I did that. It's the only time I quoted you so it will be easy to tell.
I'm asking you to clarify your position more clearly and you are avoiding doing that. Let me step through it one step at a time. Do you believe that when you first make a pick, that the chance that you made the correct choice is 1/3?
kmorgan26 5 months ago
@kmorgan26 ...the car does not have any "chance" of being behind a certain door. It IS behind a certain door. The door which it is behind is not my concern. I must commit to one. Now that I've committed to a door, since there is nothing the host will do that can be construed as a hint that the door I've committed to is incorrect or that the door I haven't committed to is correct, I have no rational reason to de-commit from the door I have committed to.
eeyoreeeyoreuhoh 5 months ago
@eeyoreeeyoreuhoh Do you not believe in the concept of probability? If there are 3 of something, and only one is right...and you pick one then there is a 1 in 3 chance of you picking the correct one. If you don't believe that, then you're just denying a fact. In the game, you pick door #1, then there is a 1 in 3 chance of you being correct. This isn't my opinion, it's the way it is.
kmorgan26 5 months ago
@kmorgan26 "...there is a 1 in 3 chance of you picking the correct one."
Oh but there is also a 1 in 2 chance that you have picked the correct one. The door you picked is either the correct one or its not. You won't know until its revealed by Monty Hall. This too is an undeniable fact
eeyoreeeyoreuhoh 5 months ago
@eeyoreeeyoreuhoh I am either going to be struck by lightning today, or I am not.
One of those will certainly be true, but they are not equally likely. There is a much higher chance that I will not be struck. You are right that there are 1 of 2 possibilities (car or goat) but they are not equally likely. If you can't see that, you're just not very bright I'm afraid. That's not an insult, it's just true.
kmorgan26 5 months ago
This has been flagged as spam show
@kmorgan26 "You are right that there are 1 of 2 possibilities (car or goat) but they are not equally likely."
Well, you are right that they are not equally likely and I see that. However my point is that the more unlikely outcome is equally plausible. If indeed you switch, and you end up with a goat, was it to your advantage to switch? No, because you didn't get the car. If you can't see that, you're just not very bright I'm afraid.
eeyoreeeyoreuhoh 5 months ago
It seems I'll have to download your programs...
luigi7up 7 months ago
@kmorgan26 But of couse!!! Sorry for being too stupid to have understood that from the beginning! Damn, I feel like such an idiot.
MiniGeek31337 7 months ago
Comment removed
MiniGeek31337 7 months ago
If eventually you are left with only the car and the blue goat (you deduced there was only the blue goat left), you see that the blue goat also had a 1% chance to be behind any specific door. So 1% and 1%, except you got rid of 98 possibilities, so you are left with a 50/50... no?
MiniGeek31337 7 months ago
@MiniGeek31337 no. If you were to randomly open 98 doors, then sure...there would be an even chance that each of the doors held the prize. But the host is opening doors he knows to be goats. He's giving you information you already knew (that there were goats behind 98 of the doors). Your original pick is still 1/100. To switch away from your door is really in exchange for the other 99 doors.
kmorgan26 7 months ago
Comment removed
MiniGeek31337 7 months ago
Was it an elektric car? ...otherwise I want the goat....can you win both the goats?
frannsson 9 months ago
@frannsson The goats aren't virgins, so you probably wouldn't be interested.
kmorgan26 9 months ago
@kmorgan26 ...you would know, wouldnt you?
frannsson 9 months ago
haha i laughed so fucking hard that you did put a idiot list xD
yosefmashadani 9 months ago
@yosefmashadani Well, I said I would. I sort of felt obligated. =)
kmorgan26 9 months ago
files seem to be down.
and no im not an idiot.
TheCrazyGameFreak 10 months ago
after seeing "21" and had a discussion with my colleagues, i made a little programm that simulates the monty hall problem. while programming i got the idea behind the problem. you don't have to look through the eyes of the guesser but the moderator. then the swap makes perfect sense =D
darkfiete 10 months ago
@darkfiete Yep! That does help. I've made some programs myself!
kmorgan26 10 months ago
I wish your simulations weren't deleted. I have to write a C++ program of the Monty Hall Puzzle and I can't understand how to present it with what we learned. Oh, well.
paulsimage 11 months ago
I did some mapping and it looks to me like if you choose door #1, there are seven possibilities as to what the game show host reveals, and what you pick, if you pick randomly. Four of those possibilities are losses, two are winning by swapping, and one is winning by staying. So it is twice the chance there, but only 2 out of 7.
I'm extremely rusty on my probability. Curious, what if you swapped two times out of three instead of swapping every time?
theboombody 1 year ago
i want to be on your idiot list, put me on it.
Zralf 1 year ago
HEY DOUCHENOZZLE, RE-UP THE SIMULATIONS.
thisvideosucksss 1 year ago
Common sense to this dude :-p nicely done
Kznarf 1 year ago
Nice work! Also your idiot list is a great idea that I may steal from you in the future :)
FlyingFish112 1 year ago
@FlyingFish112 thanks. Feel free!
kmorgan26 1 year ago
kmorgan26: You're correct. For it to be a 50/50 chance, there had to be two doors from the start.
carchavtube 1 year ago
It's really nice to see you take the time to explain the answer to this problem very clearly for everybody to understand. I love this problem.
lozleigh 1 year ago
@lozleigh Yeah, I love it too. Thanks for the comment!
kmorgan26 1 year ago
When you choose to swap you're being humble enough to admit that your choice might've been wrong. When all other doors are opened with the goats but one and the one you chose, it's like the gameshow host said: "all right, all of those doors had a goat and I knew it, but I won't tell you whether this one here has one and I respect your decision so much, that I won't tell you that your chosen door has a goat, i.e, your aiming is bad (or not)".
valberm 1 year ago
must be studied before saying any nonsense
djpelinn 1 year ago
That's interesting to say the least. My friend had a mini program thing with it and you could set it to switch or stay with the door or not. the % gets more clear as the # of times repeated goes up. but yeah, it's a neat puzzle/concept. although I have to say I don't like "probability"... or atleast the class
MosinSeal 2 years ago
I have written a few programs that simulate this puzzle. Watch the video response I made to this...I use someone's program in that video.
kmorgan26 2 years ago
this is so easy to understand, I just watched this video to see your explanation, I like the way that you explained it with the 100 door problem, I think that people can understant it more easily with that explanation, good job
Mr000Hyde 2 years ago
thanks
kmorgan26 2 years ago
What happens if, after eliminating 1 card (a non-winning card ), somebody comes back 100years later to choose between 2 cards????
...well YOU just don't understand! year from point in time I pick
PeterTheMadCow 2 years ago
Does Monty Hall ever take away the door I picked (winner or not)?
PeterTheMadCow 2 years ago
He never "takes away" a door. He just opens one of the remaining two doors (one that you did not pick). He will never open the door with the car, however.
kmorgan26 2 years ago
Rule is...Monty cannot throw away the winning door if I have already picked it.
PeterTheMadCow 2 years ago
Does Monty Hall ever take away the the winning door if I have picked it???...the full game rules need to be explained!!!
PeterTheMadCow 2 years ago
Does Monty Hall ever take away the winning door???
PeterTheMadCow 2 years ago
Even if you switch, no matter what...it is a 50/50. You want to know why? It's because you REMOVE your previous vote, it's not going to win you anything. When you say it's a 2/3 when you switch, you subtract one option and you're left with 1/3 total. With a goat shown, it is a 50/50. You can guess right on the first one as much as you guess wrong, BECAUSE THE HOST WILL ALWAYS SHOW A GOAT. Please explain the flaw in my logic.
YungSensei 2 years ago
Given three options, you will guess wrong twice as often as you guess right (on the first pick), so switching wins 2/3 of the time. This isn't just a hunch of mine. It's true.
Watch the video response to this video too and check the simulation in the description. Lemme know how it went for yah.
kmorgan26 2 years ago
You are so smart and correct,
eastofakron 2 years ago
Comment removed
christoperhalbower 2 years ago
you should be on the idiot list
why are you so retarded... it's so easy to understand.
theres 2 goat 1 car , you have 2/3 chance of picking goat 1/3 picking car. if you pick goat and switch you win car everytime , as chance of picking goat is 2/3 you should always switch , your first choice is always more likely that you will pick a goat... so it's 2/3 NOT 50/50 !!!!!!!!!!
clemzzz 2 years ago
You're still ignoring the fact that when you switch, you're removing your previous chance. Nothing is gained. Insults doesn't justify your logic, try again.
YungSensei 2 years ago
YungSensei, you do gain by switching...whether or not you understand it is another matter.
I made a video response to this video that has a link to an applet. Go over there and try it out. I'm sure you'll get it then.
kmorgan26 2 years ago
YungSensei,,, so it's you against the world's top mathematicians...... !!!!! play the game yourself and you will see that you are soooo wrong. you may not understand but just cos you don't , don't say everyone else s wrong.. that is so egotistical and will get you nowhere---listen to the experts and you may learn something in your life
clemzzz 2 years ago
Okay, face is a bit red after reading it all over again. I still stand by the smelly cunt statement however.
jediforcecny 2 years ago
and you have smelly cunt. Childhood name calling over yet? I have a few adult names to call you if you want to get into a debate about the validity of this video's math and thus (gasp) truth.
jediforcecny 2 years ago
idiot list
jediforcecny 2 years ago
lol. Nice.
kmorgan26 2 years ago
lol...what did you come comment on this one for?
kmorgan26 2 years ago
i know your right, but about the 1 million doors, if i switch ill have 999,999/1000000 chance to win, but if i stay i have 1/1 million? that doesnt seem right to me, it probably is right, but doesnt seem like it... so i have a question. iff i do that with 1000doors, if i do an experiment with a friend. if i choose to switch your saying ill win everytim pretty much.
RickyAustinR 2 years ago
@RickyAustinR if you do it with a 1000 doors then eliminate 998 doors without revealing the car if you switch every time then yes you will win 99.9% of the time
scoobyj2002 1 year ago
i am not wrong !!
08robertb 2 years ago
dude your waaaay too serious in this video. the days before you met Chad.....
ChadNeverDiedNigger 2 years ago
first of all dude i think you are wrong. 100 doors is a lot different then 3 doors. if you have 3 doors and they take one away. your should look at the problem again and it is still 50% of winning. it is different with 100 doors. but the i did the program and with 3 doors and found i stayed with my same door 100 times stright i was up 54 to 46 so i would have won more cars then you. so whos right
matt02sim 2 years ago
I think you just made the idiot list, the 100 door simulator is just and extreme to explain the problem more and make it more comprehensible, in fact, i tried his simulator and played the 3 door 20 times and got 13 wins and 7 loses, with the 100 door i played 20 times, and got 20 wins, it is just an extreme to show how it all works.
xEscapeFatex 2 years ago
Who's right? Me!
kmorgan26 2 years ago
nope
matt02sim 2 years ago
Walk the earth clueless. See what I care.
kmorgan26 2 years ago
i dont think i am clueless you clown
matt02sim 2 years ago
...and therein lies the problem
kmorgan26 2 years ago
i guess so
matt02sim 2 years ago
This comment has received too many negative votes show
I think you're wrong, You initially had 33% chance, but since he eliminated one out of three you have 50% change that your next pick (either the same one or another one) will contain a car.
RoelorD 2 years ago
When he reveals one of the doors, you gain nothing by that information. You already knew that one of the remaining two doors had a goat. What if he gave you the chance to keep your door, or take the other two? This is essentially what he is doing, but just showing you a goat before he offers you the other two doors.
kmorgan26 2 years ago
kmorgan has stated the solution succintly.
christoperhalbower 2 years ago
Yeah, I know, but you are still alowed to choose, you can also choose to stick with your current choice, if you on forehand already knew where one of the goats were you would have 66% chance.
But here you have two choices, so either choice is 50% chance.
(p.s. don't mind my spelling).
RoelorD 2 years ago
No. It makes no difference if you know beforehand. You already know beforehand that AT LEAST one of the other two doors has a goat. What is the difference if he shows you? None!
Think about 1 million doors. He shows you 999,998 goats and there are two left? Your original "1 in a million pick" or the other door...still gonna stick with your original?
If you do, you're a fool.
kmorgan26 2 years ago
I still don't think your logic is, logical.
see, you had a 1 in a million chance, but if you know which doors not to pick the chance is 1/2.
I could debate this forever, but I still believe that logic is flaud. (unless they expect you to instantly forget when you keep your pick)
RoelorD 2 years ago
My logic isn't flawed. In fact, I didn't make this puzzle up. You can debate it forever because you don't understand it.
I could debate forever that the world is flat, but that wouldn't make my argument any stronger. There are plenty of programs online to test this. Why not find one? Google "Monty hall problem simulation" and see if that helps you understand.
...because you ARE wrong.
kmorgan26 2 years ago
I made a new video about this...should make it more clear.
kmorgan26 2 years ago
This is the golden rule. If you pick a goat, a door is opened to reveal the second goat, swapping will ALWAYS win you the car if you were lucky enough to pick one of the 2 goats.
What are the chances you will pick a goat before he opens the door? 66%
reply if this helps.
NoNameC68 2 years ago
Damn Right dude!!
AH AH :)
You see, I have one of those hard relations with maths... :(
But I love all this *little* things that can surprise me.
The key to this problem is in the host.
He will ALWAYS (100% sure) show you a goat. No matter what.
fiesta181 2 years ago
Hey i got lucky hehe
danielmastia87 2 years ago
If you model it with this
losing doors = O
winning door = X
u have 3 possibilities, in each 1 suppose u choose the one on the right (just to make it easier to show all the 3 possibilities)
so if its
O O X, switch = lose
O X O, switch = win
X O O , swithc = win
those 3 ARE the only 3 possible things that could happen in THIS scenario, u got 2 scenarios for win and 1 for lose,
SousukePanic 2 years ago 2
Writing a program was a good idea, but if you made it a web app, more people would have been able to use it.
Still, thanks for helping to educate the confused!
automin 2 years ago
lol when i 1st watched the montyhall problem, i was confused, cuz im like, 12. but u helped! i think about it like:
when u first pick a door, u shud assume that u picked a goat cuz its a higher chance. then the host shows u another goat. so he shows u a goat, and u assume u picked a goat. thats 2 goat. whats left? a car. so make the switch. lol yeah...
bookwormGrL1 2 years ago
Suppose I have 3 cards: an ace and two kings. I deal you a card and I deal myself the other two. Whoever gets the ace wins.
I show you one of my cards. It's a king. No matter what, I have at least one king. Do you want to trade for my other card?
____
Now, same problem but I don't show you any kings. You have 1 card, I have 2. Want to trade my 2 cards for your 1? Of course! The fact that I showed you a king in the previous example is irrelevant!
christoperhalbower 2 years ago
Really nice explanation. I'm gonna use this on those stubborn folks who still insist on the 50/50 scenario. I've tried the million door thing, and a few other examples, but some people just can't be convinced. This might work. (Of course, I won't get my hopes up).
jeremyemilio 2 years ago 3
Yes, the revealed door is irrelevant. The information gained by that door being revealed is not telling. There is always going to be a goat behind the door shown by the gameshow host. Furthermore, of the two doors you do not initially pick, at least one will have a goat. Thus, it's superfluous information.
christoperhalbower 2 years ago
Comment removed
RoelorD 2 years ago
@jeremyemilio With my parents, I physically put ten things on the table in front of them (we were sitting outside late at night talking) and put something behind one of them when they weren't looking, then after they picked one object, I removed the 8 others that had nothing behind them, and we played the game a few rounds.
The reason why people don't get it is that they can't grasp the whole concept in their mind. They need a physical example to lighten the mental load.
Bobstew68 1 year ago
i almost got on the idiot list, but he's right do what he says and play the 100 door game first, then you'll see it's not about 50/50. so i think the movie was wrong about this. thanks dude!
wwert987 2 years ago
Opening a door is not an honest move. This is based on a con game and the host is a con man. He is opening a door to manipulate you.
If you watched the show, He did not always open a door. Sometimes he brought another box on stage and offered that.
I am not saying the odds are not correct. I am saying that if you act this way the host might only open the door when he knows you picked correct the first time.
You assume that the host will play by the same rules every time. Not the case
earthdog7900 2 years ago
Read the description in the sidebar:
Note: The host will always show you a goat, regardless of your initial pick.
kmorgan26 2 years ago
Knowing how the host is acting is the difference between a math problem and real life.
earthdog7900 2 years ago
Yes. In real life, some people make some of the dumbest fucking comments to try and make themselves look smart. In most cases, this results in total failure.
kmorgan26 2 years ago
LOL. True.
giovanni9107 2 years ago
Comment removed
luvduv1213 2 years ago
i still say stick with the original choice out of intuitiveness.
luvduv1213 2 years ago
at least the creator of the video understands the concept
those with a sense of intelligence just give this comment a thumbs up just to prove the amount of people that can back this statement. "this guy is right."
lpkojiko 2 years ago 10
mate its 50 50 not 33 66
sorry
surfingrox111 2 years ago
No it's not. I'm right. Sorry.
kmorgan26 2 years ago
surfingrox try it with the program
carloshiphop2 2 years ago
LOL @ these people K. It's not a hard concept!
TheBlueDuckMOE 2 years ago
if you dont believe in the program he made try it with cards, after 20 or 30 games the results show clearly that chances are 33 66
giovanni9107 2 years ago
No. It isn't 50/50. Not even close. It's basic statistics. If you pick the car and don't switch, you win 1/3 of the time. If you pick a goat and don't switch, you lose 2/3 of the time. If you pick a goat and switch, you win 2/3 of the time. If you pick a car and switch, you lose 1/3 of the time. Simple.
brianmay9 2 years ago
For those who still don't understand this, think of it as follows:
If you don't change your choice when the host asks you if you want to do so, then your chances remain the same, that is 33%. Now, if there is a 33% chance that the car is behind the door you chose initially, then where can the other 66% be? Well, there is only one possibility, and that is the other door. That's the way i figured it out... hope it helps.
giovanni9107 2 years ago
For all of you who fail to understand this, I do not blame you; However, think of it in this perspective:
You are on a game show, hosted by an MC, in which there are 100 doors to choose from. Behind 1 door is a car, behind the rest are goats. The MC says that once you pick your door, he will eliminate all wrong doors but one. Because the MC had 99 doors to begin with, there is a 99% chance that the remaining door is the car. Therefore, there is a 99% chance of winning if you switch to his.
Goldiney 2 years ago
Nice explanation using the 100 doors simulation!
Btw... what programming language did u use to write the simulations?
yeeen123 2 years ago
Thanks. I made this with MS Visual Basic 6.
kmorgan26 2 years ago
If these were 3 boxes and one of them had $1000 in it while the other two were empty then its the same problem.
If Monty did not open the empty box and asked whether they would like to swap their one box for his two then the result would be the same.
Monty has twice the likelihood of having the prize, since he has two boxes.
One of Monty's boxes will always be empty. So whether he opens the box after, or whether he opens it first and asks you to swap the unopened box, makes no difference.
magicsteviej 2 years ago
Comment removed
mrKirosana 2 years ago
I've been considering explanations for this for a couple of days, and I've got a couple of simple ones. Here's the second:
Query: What happens if the door you chose initially has a goat behind it?
Answer: Monty will open another door with a goat behind it, and the remaining door will have a car behind it.
Query: What are your chances of choosing the door with the goat?
Answer: 2 in 3.
Query: What are your chances of winning if you switch?
Answer: 2 in 3.
gorpusmorpus 2 years ago
You are spot on.
kmorgan26 2 years ago
I've been considering explanations for this for a couple of days, and I've got a couple of simple ones. Here's the first:
1. Two times out of three, the door you choose initially will be a goat. That means that, two times out of three, when Monty opens another door and shows you a goat, the remaining door will be the car. The only time swapping won't win the car is when you've chosen the door with the car initially -- a 1 in 3 chance. So, swapping wins 2 times out of 3.
gorpusmorpus 2 years ago
It's 50/50. Fuck off.
hxnncht 2 years ago
wrong.
kmorgan26 2 years ago
wrong is WRONG.
hxnncht 2 years ago
You're wrong. Download my programs and you'll see.
kmorgan26 2 years ago
No. You're wrong and i'm right and you can't do anything about it.
hxnncht 2 years ago
I have proof in the way of computer programs that I made, and also that I don't suck at math like you. Continue to be ignorant...see if I care.
kmorgan26 2 years ago
Oh.. you're soooo wrong
hxnncht 2 years ago
Well, I'm not the one on the idiot list.
kmorgan26 2 years ago
Yes! Successful troll is successful.
hxnncht 2 years ago
I think you're misunderstanding the problem. It's 50/50 if you walked into the game show room after a goat door was opened and someone asked you to pick one of the remaining two. But the thing with probability is that you often have to take past events into account. After the host opens a goat door, he leaves you only one choice to switch to, and by taking the switch you're essentially swapping the probability you got it right the first time (1/3) with the probability you got it wrong
(2/3).
DMK000 2 years ago
Absolutely correct. There are two trials in this game. The second trial is not "fair" because the game show host has prior knowledge of the location of the car. If the game show host had no knowledge, or if you did not see the first pick, then it would be "fair"
1991707312 2 years ago
the way that seemed easiest for me to understand it is that if u pick door 1 that means that if u put 2 and 3 together its a 2/3 chance that it is behind 2 or 3 so when either 2 or 3 is revealed that 2/3 chance remains with both of them so u would obviously want to pick the one that is not revealed??? i don't know if that makes it easier for anyone
Tylertbon 2 years ago
Dude, I just got into an argument about this with my college math teacher last night. I showed him the example on the chalkboard multiple times, raised the number of boxes (what we used) to five and showed him some more. He just couldn't get his logic out the mindset that it was not a 50/50 chance. Needless to say, I left class irritated.
JeffVoss 2 years ago
Get my simulators I made and bring them in...they are in the sidebar.
...and take comfort in knowing you're right! =)
kmorgan26 2 years ago
Another simple explanation is, suppose before you even play the game you decide that your strategy is to always switch. Then your first choice of door gives you a 1/3 chance, but your switch effectively gives you 1-1/3, or 2/3.
DMK000 2 years ago
It's a simple explanation because you understand the problem! =)
kmorgan26 2 years ago
Thanks and feel free to respond as always.
kmorgan26 3 years ago
You should always switch, its simple a math law states that your odds never change, if i had a hat with three colored cards a red green and blue card, and i told you to pick a red card at least once out of 20 tries and you pick 19 green cards in a row, on the last try your odds for picking a red card is still 1/3,in simplified terms if i flip a coin 19 times and in lands heads 19 times the next time i flip the coin my odds are still 1/2 in getting heads again.
afly727 3 years ago
afly727.. i agree that sequent coin trous are independent events.
Now, in a dependant chain of events for example i put your cards in a black bag and ask you to remove a red OR green, NOT REPLACING the first you took. the chances of 1st positive result is 2/3, the chance of 2nd positive result depends. If you get a positive first time you are left with 1 in 2. If you didn't get a positive 1sdt time, you'll get one in the 2nd , because you have 2 in 2! If you REPLACE 1st and 2d become independant
giovea 3 years ago
Nice job. The variant of using an extreme number of doors, and at the second stage of the game opening -all the others except 1- is a great explanation and it's a shame more people don't use it. It just makes it so clear. The counter-intuition in the problem comes in because in the case of 3 initial doors, "all the others except 1" is just 1 measly door. But in fact, the host reveals almost all wrong doors. And the little app you wrote just drives the point home experimentally, nice.
Phillist 3 years ago
thanks.
kmorgan26 3 years ago
Here's the solution (r = right, w = wrong):
1. 'w' w r
2. w 'w' r
3. w w 'r'
stay: 1. you will lose; 2. you will lose; 3. you will win. Probability = 1/3.
switch: 1. you will win; 2. you will win; 3. you will lose. The formula for the probability if n is the total number of doors is: for staying, 1 over n. For switching, n-1 over n(n-2). I didn't completely understand kmorgan's point.
Dan14159 3 years ago
I enjoyed your explanation. However the original door that you chose changes from 1 in 3 probability to a 1 in 2 probability of the car being behind your door once the host has changed the number of outcomes for the problem. It is a new problem at that moment. Now you have two people on your idiot list.
terrydeannelson 3 years ago
lol. Nope. The probability doesn't change at all. It's not just my opinion, it's just the way it is.
kmorgan26 3 years ago
I have studied this problem and realize that you are correct. I have accepted the explanation but it still does not "feel" or seem right. I guess that is what makes the problem entertaining. Where are your downloads so that I may play with them? Thanks again!
terrydeannelson 3 years ago
OK, I re-uploaded my programs and linked them in the description, so you can download from there. Let me know how it went.
kmorgan26 3 years ago
to everyone that still believes it's 50/50.. you are wrong. it's funny how retards can't comprehend this simple problem even with these videos proving it.
tl2e3le4f 3 years ago
The way I understand it having seen this. When you make your first pick, you have a 2/3 chance of picking goat. When the presenter shows you another goat, your original pick still has 2/3 odds of being goat. And so you pick the other door, which is 1/3 chance of being goat. The trap people fall into is thinking the odds of their 1st pick change after the first door is opened. They forget their first pick always has more chance of being goat, so it makes sense to swap when the other goat is gone.
Inspiration9991 3 years ago
Well said Kmorgan, it's something I've argued for years.
The way I try to put it is, my original odds of winning are 1 in 3, so the remaining 2 doors combined have odds of 2 in 3. Once I'm offered the chance to switch my choice I then inherit the 2 in 3 odds of winning, the fact that one of those 2 doors has been eliminated bt the host has no relevance. In effect by switching doors I'm swapping one guess for 2 guesses.
Kareliasuite 3 years ago
I just had to post this because this really crystalized this concept in my mind. Thanks, again!
johndeerishere 3 years ago
Another way of thinking about it is you have a 1/3 chance on your side and 2/3 a chance on the other side. So, if one of the doors on the other side is revealed all of the 2/3 goes to the other side that still has an unrevealed door. So, the door you did'nt originally pick, but is still unopened has a 2/3 chance of being a car.
johndeerishere 3 years ago
Thanks for the posting. I watched several other videos here and most of them confused me alot more than help me understand this concept. Keep up the good work!
johndeerishere 3 years ago
Thanks!
kmorgan26 3 years ago
Does this seem like the right way of thinking on this?: The contestant has a 1 in 3 chance originally, and the game show has a 2 in 3 chance of having the car on their side. So, when they give up one door they gave up 1/3.
johndeerishere 3 years ago
Yes, it is the right way of thinking on this. The odds don't change by opening a door.
kmorgan26 3 years ago
The first time I encountered this particular problem was when I watched the movie 21. I have always been exceptionally gifted at math and when that particular part of the movie played, I said 50/50 as well. The entire concept went right over my head but it intrigued me enough to research it and I must say that the Monty Hall Paradox is rather clever in its initial approach but the correct answer is eventually and painfully obvious.
nathan3084 3 years ago
It reminds me of the scene from "A Beautiful Mind" where Nash works out the best strategy for getting laid. That sparked so much interest for me in Prisoner Dilemma's etc.
Inspiration9991 3 years ago
at 3:20 "because i'm giving you all the other doors"
exactly.
i don't know why some people don't understand this. with 3 doors -- you have 1 and the host has 2. now he tries to trick you by "narrowing down the choices", but the fact of the matter is, you can trade your 1 door for the host's TWO DOORS. this is of course against the rules, so you choose the one that he didn't open.
jujube4000 3 years ago
i shuffle 3 cards (ace of diamonds, ace of hearts, ace of spades) and let you choose one without looking, and i get the other 2. your odds of having the spades ace is 1/3, and my chance of having it is 2/3. let's say i can look at my 2 cards. now it doesn't matter if i reveal a red ace (which i must have), give it to you, eat it, or flush it down the toilet. you still have a 1/3 chance that your hidden card is the spade. 1 - 1/3 = 2/3. there's a 2/3 chance my remaining card is the spade.
jujube4000 3 years ago
MATH SUCK!! can I be put in the idiot list?? :D
PeopleOfAr 3 years ago
God bless you man! Taking it to the idiots and numbnuts out there. It made perfect sense the first time I saw it, and it does whenever I see it again.
7thRook 3 years ago
I don't think people are as big of idiots as you think they just realize more conditions have to be met for the 2/3 and 1/3 to be correct then are often given in the problem if the problem is stated correctly then your explanation is valid but it many times is left incomplete making the odds of 50/50 or 2/3 not really attainable mathematically in the simulator you created certain aspects are a given you programed them in as constants which is why it works.
timfgates 3 years ago
oh my bad I didn't watch the video you where commenting on they do give the correct variables that the host knows what is behind each door, has to show you a door first, and has to show you a goat that is all I am saying this problem is often not worded correctly you are correct in this situation people just refuse to see because they are stuck in their real life scenario in there heads where it would remain unknown what was required of the host.
timfgates 3 years ago
You ALWAYS get a car if you pick a goat first, and you ALWAYS get a goat if you pick a car first. You have to look at conditional probabilities. There is no 50/50.
P(win on switch)=P(car|goat)x P(goat)=1x2/3=2/3
P(lose on switch)=1-P(win on switch)=1/3
DMK000 3 years ago
I can't get your 100 door simulator one to download, but the three door program works great. I actually saw this problem on Numb3rs and thought it was the coolest thing ever. After 100 times of playing the 3-door one and switching every time, it was 63%. That's pretty darn close to the 66.7% Great program!
Crazy4JaSam 3 years ago
The third door is only smoke and mirrors. Regardless of you picking right OR wrong, you're going to be revealed a goat. When you apply three variables to two doors, of course your math will be off. Its only a two variable equation.
If you pick right, you see a goat, if you pick wrong, you see a goat. This means that its between the door you picked, and one other door.
50/50. Smoke and mirrors.
spammaplease 3 years ago
This has nothing to do with smoke and/or mirrors. It's a probability problem. It's not 50/50. Download my program (3 door) and try for yourself. Until then, you're on the idiot list.
kmorgan26 3 years ago
The work you're doing for the math is right. Your initial math is wrong. You apparently failed word problems in school because you don't know how to set the equation up.
Regardless of how many doors, regardless of your pick, regardless of your programs, you'll still see X goats. So in the case of 100, you'll see 99 goats, NO MATTER WHAT YOU PICKED. If you have 3 doors, you'll be shown one goat, no matter what.
Its 50/50, you fail at comprehension, probability, and setting up a math problem.
spammaplease 3 years ago
Actually, in the 100 door example, you'd be shown 98 goats...not 99. Pretty elementary. So you fail.
This is not my problem...it's a classic one. I didn't "set it up". I just understand that one's odds of picking the car when presented with three doors is one in three...and those odds don't change because doors are opened.
kmorgan26 3 years ago
You're right! We'd better get on the phone to every mathemetician that has studied probability for at least the past sixty years (as the three prisoners problem) and let them know. I bow down to your superior mathematical intellect! I'll let all of my fellow mathematics grad student's know of the new development. You fail at life fool.
nerratt 3 years ago
It may be "smoke and mirrors" to anyone who doesn't truly understand or is willing to accept probability. You do know that the first person who suggested the 2/3 chance of winning for the monty hall problem was a woman with the highest IQ in the world, and after being attacked by countless other mathematicians, she turned out to be correct in the end.
DMK000 3 years ago
Just for clarification. Marilyn vos Savant may have been "the first person who suggested 2/3 chance winning" for the Monty Hall problem but this problem has several equivalent problems that were solved long before. (see Bertrand's box, three prisoners) So this argument is pointless really. Also, your statement "the highest IQ in the world" is misleading as others may not have been tested etc. Lastly a high IQ is not equivalent to being a good mathematician/statistician.
nerratt 3 years ago