The likelihood of choosing the wrong door is 2/3. Thus, mathematically, there is a 2/3 chance that you DONT have the prize but the game show host does. The opening of one of the doors by the game show host is a useless factor. It doesn't change the original fact that the host have a higher likelihood of holding the prize then you do.
After you choose your first door, the chances of picking the money are 33%, while you picking one of the empty doors 66% (therefor you are probably on an empty door). After he opens an empty door, it is in your interest to switch because him opening the door did nothing. There is still a 66% chance that the current door you are on is empty. So, you should switch because if you do, the chances of you ending up with the money is 66% instead of 33%.
The Monty Hall problem breaks down to this simple question:
"Would you like to have 1 door or 2 doors? If you choose 2 doors, I’ll give you the better prize of those 2 doors. If you choose 1 door, you’ll be stuck with your choice."
This is a "glass half full/half empty" discussion.
Full: You have a low chance of picking the prize door first (1/# of doors). Opening the remaining doors means your chance is higher, since you believe yours was extremely likely to be wrong.
Empty: You start out thinking "no matter what happens, there's going to be 2 doors left, so it's 50/50."
The prize is going to be behind one door, and nothing's going to change that. The only thing that changes is your mindset.
Switch doors. Possible combinations are RWW, WRW, WWR. (R = right, W = wrong). You have a 1/3 probablility of guessing R on your initial pick. Therefore, if you guessed R initially and switched, your odds of losing are 1/3 since the only scenario where that happens is RWW (if we pick leftmost door). Your probability of guessing W initially is 2/3. Therefore, if you guess W initially and switch, your probability of winning is 2/3 since WRW and WWR will both result in a winning scenario.
Chance doesn't really change anything. It's not like the prize will automatically switch to the door you switched to. Chance really has nothing to do with this. I'm serious, it won't change anything to switch doors. The prize still remains behind one of the remaining doors so it doesn't matter whether or not you choose to switch or not. This is what I'm thinking, because this kind of thing really just relies on luck.
@yisoonshin12 Have you ever rolled a six sided die 1000 times? Or even 100 times? You will get each number about 1/6 of the time. That is not luck, it is probability. Yes, you can get lucky and pick the correct door the first time, but you have a 33% chance. That means that on average, out of 100 tries, you will pick the correct door on your first try. However, you have a 66% chance of picking the correct door if you switch.
@dkaufman123 so on my calculator I used the probability simulation to see what happens at 100 times and they weren't that off, but still noticeably off. I agree with you about the dice thing, but that's only if each side is balanced. if one side is even a little heavier then it'll fall on that side more. but that's besides the point. I agree with your statistics, but there's still a portion that relies on luck. Probability won't change what door the prize is behind.
@dkaufman123 But I guess you can use those statistics to your advantage knowing that you will probably pick the prize if you have a feeling you picked an empty door.
@izardadkhan That is false, when you pick door 1, you are making a choice that can be 33% right and 66% wrong. When he reveals that door 3 was empty, then you have 2 left. However, the choice you made in the beginning only had a 33% percent chance of being right, and it maintains the percentage. However, the door 2 now has a 50% chance of being the door with a prize. Of course it is a game of mostly luck, but there is a better chance of winning by picking door 2 because it is 33% v. 50%.
Here's what wikipedia has to say (and I paraphrase):
Imagine the same set up, but with 1000 doors instead of three. After picking a door, Monty then opens up 998 of the remaining doors, revealing them all to be duds. So now there are two doors to choose from. It would seem ludicrous to assume that you picked the right door on your first try (a 1/1000 chance) when there was a 999/1000 chance that it was in the other doors, a probability that is now represented by a single door.
@DementedlyHappy I realise what you are saying here, but it seems like a massive exaggeration. I don't mean to sound like a jerk, please understand that. In the original problem its 1/3 doors. In your explanation (which is obviously helpful) it would actually have been a choice of 333.3333 doors of 1000 in the beginning and then Monty would open another 333 empty doors and you would have the option of keeping the original 333.3333 or the remaining 333.3333 XD
Here's a simplified explanation: you pretty much want to pick the empty door as your first choice. If you select an empty door, monty hall will then reveal the other empty door, thus when you're offered to switch your selection, you will obviously be picking the door with a prize. So then, what are the chances of you originally picking an empty door in the beginning? 2/3.
They reveal one door then give you the choice to pick one of the two remaining, it's 50/50. What would increase the chances of one door having the prize over the other? How does picking a door in the first place make any difference?
@Arthur61987 Your statement is flawed because, when you made your original choice, you had a 33% chance of getting the prize. Once the other door is opened, you should switch and you will have a 66% chance of getting the prize. Here is a way to explain it using three scenarios (you always pick door A)
1. Prize @ door A - Either B/C is revealed, switching doors loses
You choose one door. The guy reveals door 3 that has no prize. There are no benefits if you switch as you now have a 50% chance of winning no matter which door you choose....
Maybe I'm not thinking this problem the way I should. The way I see it is that 1 gets cut, no matter what the probabilities were at the beginning 33% or whatever. At the end, you still end up with 2 doors. And for me it becomes a 50/50 chance of picking the right one since there is only 2 doors left. I see it as a gamble to switch, or to stay
Is in the Players interest to switchhhhhhhhhh! I done the probability calculations already, Hes already giving you two doors if he reveals a goat, Is better to switchhhhhhhhhhhhhhhhh, Come on brilliant people you don't have to brake your headddddddddddddddddddd for this. Grap a Pen, Paper three cards get to work!
HAHAHA this problem is hilarious and so is the answer.
I got a question to this question: case 1: what if there were 3 doors and 2 doors weren't empty but a surprise?1. gives you a girl, 2. gives u a gae dude, 3 empty which one you pick? XD
But then. Case 2: if this wasnt the case and you only get 2 doors u either "try" to calculate it or gamble your luck off. The chance might increase of you winning but also increase chance of losing due to temptation so its a 50/50 if this case was case2. XD
Because one door will always be opened and revealed to be empty, you are actually only choosing between two doors. Your chances are still and always have been 50/50.
Heres how I think of it. On your first pick, you have a 33% chance of choosing the right door. In other words, it is more likely that the prize is behind one of the other two (66% > 33%). Since the choice of "the other two," which had a higher probability to contain a prize gets reduced to one door, I think the probability of winning does increase if you change doors.
After giving this problum some careful speculation, whether one chooses door #2 or remains with their original choice of door #1, there is nothing driven by science suggesting either choice holds the jackpot, therefore either choice is what is in life known as gambling, and choosing the jackpot door is a matter of luck or fate, which falls under the category of random. It's the same as trying to pick a winning scratch ticket from a convenience store showcase. There's just no way of knowing.
Choose door 2. Doors 2 and 3 had a 2/3 probability, and door 1 has a 1/3 probability. You know what's behind 3, so you have the best chances choosing door number 2
So could this extend to Deal or no Deal? In our version of that game there are 22 boxes and you sometimes get the chance to swap the boxes when it gets down to the last two. If there's a very large amount still in play and a lower amount, if you were to gamble, would you be more likely to win the big prize if you swapped?
@birdsofafeather1990 Yes, but they might switch the values inside the cases in between shows. They don't say they do that, but I would think that's a large possibility.
As I understand it: Original percentages door one 33%,Other two 66%. Cross off door three, the percentages remain door 1 33% the other doors (now just door 2) 66%.
I remember to always switch. My Pre Cal teacher mentioned this way back in high school. He also mentioned about some lady with a high IQ tried explaining why it was always better to switch back then but couldn't really get her point across.
... If we pick the door that he didn't reveal we have a ninety percent chance of winning, if not we have a 10 percent chance of winning. I think this makes the problem much more simple and apparent.
IT would only increase the likelyhood if the show host KNOWS behind which doors the actual prize is hiding. You have a 66% chance of winning if you pick the OTHER door in that scenario. IF he does not, the chances are 50% to pick the correct.
If we expand the problem, let's say we have 10 doors. There's a prize behind one of them, the rest are empty. We pick one door, the host reveals 8 of them to be empty, leaving you with the one you picked and the one he didn't reveal....
It matters only if the game show host KNOWS whether or not the door he reveals is empty or not. It he does, you have to account for variable change: the left over door now spouses a 66.6% chance as opposed to the originally picked door's 33.3% chance.
When you first pick a door,and do not swich it with another one, you have a 33 % chance to guess it right. But if you pick a door, another door is revealed as Empty, you are left with 2 doors only, and when switching you have 50 % chance to win.
So you'd better swich it, because the chances increase from 33 % to 50 %.
You should definitely switch. Odds are 66% that the prize is in door 2 (in this example). At first it seems very counter intuitive but the odds do shift in that direction
Either one, Since there are only two doors you have a 50/50 chance. Besides, backstage the prize is being moved to the door you didn't pick, so either one. lol ;)
It also depends what you define as a prize. If I am from deep in the Borneo jungle where there is no gas or roads and I win a car, what good would it do me. It would have no relevance and mean nothing to me. However: If there were a stash of bananas and coconuts behind the door I picked I would have hit the jackpot, whereas to one of us, we would have flunked.
lets simplify a little: X X O X (2/3) O (1/3) at first its 1/3 chance of getting it right and 2/3 chance of getting it wrong. if one door is revealed to be wrong and you are allowed to choose it would be to your benefit to switch because essentially, you are picking 2 out of 3 doors So... its ? ? ? you originally picked: (?) ? ? now you find out: (?) ? X well u can now stay with (?) or change with ( ? , X) (?) this is 1/3 while (? , X ) is 2/3 but i might be wrong :P
I was taiught in college -the little time I spent there- that statistically the first choice you choose in a multiple choice question on a test is usually right, so I would say that the first choice is statistically the best. I don't know. What do you think Pat?
My answer: Yes, this does matter. You have a higher chance of winning if you *keep* the door you picked. this looks a lot like something the dude in numb3rs talked about... love probability. Check out this vid: youtube.com/watch?v=P9WFKmLK0dc
So basically in the beginning, there are 2 no prizes, 1 prize.
you have a higher chance of picking the one without a prize, meaning the other door with the no prize is also revealed, So if you picked it wrong(66% chance) and you swap, you get the prize.
IF YOU DONT SWAP
You still have a 66% chance of getting it wrong if you pick a door. Automatically a door is revealed without a prize, but in the beginning esentially you had a 66% chance of getting it wrong. So if you keep it=1/3
In the beginning you have a 66% chance of getting it wrong.
Therefore, you have a higher chance of picking the first door wrong.
THEREFORE, the revealed door will be empty as well.
Therefore if you swap, you have a 66% chance of getting the prize
If you dont swap, you have a 33% chance of getting the car, since one is revealed and you stick with it, there was still the fact you had a 66% chance of getting it wrong from beginnin
I remember this problem from "21". I think you've got 33.3% of getting the answer right after they switch the doors twice. I really have no clue though.
Also, you should be clear that the host *needs* to open an empty door. Regardless of what door you picked, he cannot open the door with the prize in it. If it were just him opening a random door, I don't think there would be an advantage to switching.
If my strategy is to switch the door everytime, then I am guaranteed to get the prize if I selected an empty door on the first round. The only instance I won't get the prize is if I chose the prize on the first round. Hence, if I switch everytime, 2/3 times I'll pick an empty door, switch, then win the prize. Whereas if I don't switch, 1/3 times I'll pick the prize on the first round, not switch, and win. Hence switching > not switching.
When I was originally offered to choose a door, i had a 33.3 % chance of getting the right door. However, when one of the empty doors were opened, the 33.3 % for this door switched to the benefit of changing my currant choice. Hence, I would get an EXTRA 33.3 % if I change my door.
The only way you lose is if you originally pick the door with the prize behind it which you have a 1/3 chance of doing so. So the only way to lose by switching is if you originally choose the prize. 1-(1/3)=2/3. Therefore, if you switch you have a 2/3 chance of winning the prize because the game show host is effectively telling you where the prize is from the remaining two doors.
The people saying that there is a 50% chance and the odds don't increase didn't actually take the time to solve it.
This is a great conditional probability problem. I think everyone understand that you have a 1/3 chance of winning if you don't switch doors. The reason why its better to switch doors is because the game show host KNOWS which door has the prize. When he reveals another door, he INTENTIONALLY picks the one with nothing behind it. The chance of the game show host picking the door with the prize is 0%. That is why there is a better chance that the prize is behind the other door.
By being able to switch one's answer, and understanding the door she would have to pick, It would present a two thirds chance. It literally switches how everything works. So all losing doors are winning doors, and winning are losing when one incorporates a switch. See, if the door she opened was not "purposefully chosen", then Yes, Your chances would stay the same. But, since it must be controlled, thus to prevent a winning door being opened, the chances change.
Wait, I'm here to change my mind. Like in the movie 21, I can see the benefit of changing doors. The 66% chance gets moved to the door that has not been opened or chosen. But couldn't you say the sum of the percentages of your door and the revealed door to be the 66%? So once door 3 is revealed, door 3 now has a 0% chance and yours has the 66% chance, rather than door 2? If so, I'm going to say that changing doors has absolutely no effect on your chances. It's basically number play.
@TheMadSlick Nope, the percentage definitely shifts to the "offered" door. Out of curiosity, I just made a simulation program to test this, and the results for switching was always around 66% and staying was always around 33%. I think you need to consider it as when there are 3 doors: the 1 door you picked has a 33% chance, the other 2 doors have a 66% chance. By switching your door, you're basically taking the other 2 doors; because one gets revealed, and the other you have now selected
Are you gonna post the solutions too? I was guided by the 1/2 but then, saw all the proofs on wikipedia that it is best to switch cuz you win a 2/3 times
1.Count possible outcomes, like jsobiranski did 17 mins ago. It's valid and his solution is right.
2.You can use conditional probability theory. As it turns out the revelation of the empty door contains information beyond reducing the set of options to 2 doors. Speaking heuristically: The probability of winning from the revealed door is transfered to the other unchosen door, so you should switch, increasing your likelihood of winning from 1/3 to 2/3.
Case 1: door 1 = prize, door 2 = goat, door 3 = goat Case 2: door 1 = goat, door 2 = prize, door 3 = goat Case 3: door 1 = goat, door 2 = goat, door 3 = prize If you stick and case = 1, then you win If you stick and case = 2, then you lose If you stick and case = 3, then you lose If you switch, case = 1, then you lose If you switch, case = 2, then you win If you switch, case = 3, then you win Thus, If you stick, then you win 1/3 time If you switch, then you win 2/3 time
@ivanxdxd2008 If you ever watched the show Let's Make a Deal you would know that Monty Hall would usually reveal a goat behind the door that did not have a big prize like a new car. Another reason I chose "goat" is because it uses fewer characters than "no prize" so that I could stay within the 500 character limit that is imposed upon us for our comments.
Yes, it makes a difference if you stick or switch. You should switch doors because if you stick with the first door you choose then you will only have a 1/3 probability of winning, but if you switch after he selects one of the other doors which do not contain the prize, then you will have a 2/3 probability of winning. We must in each case assume that you first choose door 1. Please see my full explanation in the following comment.
(short continuation) someone says what if there were a million doors and the host removed all but one other. the point is moot, even if there were hundreds of trials beforehand each reducing the choices by half each time, the number that it is potentially out of is irrelevant, all that matters are variables involved. the previous trials with ignored results, don't matter. the 1/2 trial remains 50%
1/3 = 34%; 1/2 = 50%; the likelyhood of getting the right door increases 16%, across trials. those who disagree think of the problem as effectively getting two choices, thus the prob is 66% but it is a mistake. the mistake lies in thinking there were multiple trials, and thus you are getting two choices of the three by changing rather than still having one. the first trial was irrelevant, while you do have a higher chance of guessing 1/2 than 1/3 it is 16% difference not 32~
i've seen this problem discussed in an economics book. It is better to switch and pick door 2. The first time, the likely hood of it being in door 1 was 1/3, but the combined chance of door 2 and 3 was 2/3. Since door 3 is eliminated, there is now a 2/3 chance of it being behind door 2. It's funny b/c the average person thinks that the probability of it being behind a certain door depends only on the number of options a person has, like a coin toss or a dice roll.
@SuperNerd707 I find it hard to believe even now actually...the author went on in the book to claim that humans are not rational beings as conventional economics suggests b/c most of us can't decide what the best thing to do is. And even when we do know, it is still very difficult to believe!:P
It's not 50/50 once the empty door is revealed... you have a higher percentage of guessing correctly if you switch. Think about it this way.. before any doors were revealed you had a 2/3 chance of choosing an EMPTY door. Once one empty door is revealed, it still remains that the door you've chosen would be empty 66.6% percent of the time. By swapping, you actually have a 66.6% chance of getting the prize.
the problem has to do with variable change, not pressure, uncertainty, or psychology what so ever. When asked to pick a door you have a 1/3 chance of guessing correctly, but then once the host eliminates one of your options, and re-offers you the second door, your probability increases to 2/3. therefore you should switch doors.
Statistically, you have a 50/50 chance of getting it right. However, the fact that there was a third option and it has been remove, it may cause you to doubt your door option because of pressure or uncertainty.
I've seen this problem before but it's been awhile. I believe it was if you changed doors your chances went up to 66.6% because you "gain" the 33.3% of the door that was proven to be empty. So theoretically, based on the percentages, I would change doors. In the real situation however, I'm not sure.
You originally start with a 1/3 chance and a 2/3 chance of the one of the other two having the prize. The host will show you one that contains no prize, this then means that the door left is left with a 2/3 in chance of having the prize. There for you switch. Mathematically you win 2/3 of the time when you switch but only 1/3 of the time when you keep your door.
I would switch my door. When you picked the first door you had a 33% chance of winning. The other two doors had a 66% chance of winning. When you revealed the third door, you moved the 66% to the second door. Its not a 50:50 chance. Its a 33:66 chance. I take my brand new car now. haa
yes because the 50:50 chance of winning is only guarenteed to take place if you change your door, otherwise you keep the starting probabilities of 2/3 and 1/3
It won't change anything. You chose door number 1 with a 33% chance of being right, but once a wrong door is eliminated each door now has has a 50% chance of being right.
I believe it's in you benefit to change your choice as before you had only had a 33% chance of winning, but now have a 50% chance of winning. Your probability of losing was at 66% at the beginning. If you don't change you choice, your probability of losing still remains at 66% percent, but changing your choice decreases your chance of losing to only 50%.
@s200960170 Imagine there were a million doors and you picked one. Then the host eliminated all the doors except your pick and one other. Do you still think you picked the prize winning door based upon a million to one chance? I wouldn't.
@seinfan9 If it were to assume there were a million doors, & each time 1 door was eliminated, the probability of you winning keeps increasing. (it increases at a very small range).
However in this case since there are only 3 doors, & when 1 door was picked and eliminated the probability becomes 50%, so it really doesn't matter if you change your choice, the probability is always going to be 50%.
@s200960170 You don't understand the problem. How likely is it that your original pick is the one that has the prize behind it? More doors is supposed to make clearer the line of reasoning for switching. One million doors, only one has a prize. You pick one. All but your pick and one other are eliminated. The doors eliminated had no prize. This scenario was explained in detail in my Discrete Mathematics course last semester. Bottom line, it's better to switch.
@s200960170 Your chance of winning always increases to 2/3. That doesn't mean you'll always win. If you want to go with what is more probable, it's smarter to always switch because you have only a 1/3 chance of winning with your first pick. You can search for a Monty Hall simulation on Google or you can check out the Wiki article. I think the tree map on the probabilities shows it quite well in the case for choosing Door1 first. The probabilities would be the same for choosing Door2 and 3 first.
look at it on a bigger scale, let's say 100 doors, and they reveal that 98 of the doors have nothing behind them after you choose a door. naturally, you had a 99% chance of being wrong with your first pick, so switching on your second pick now gives you a 99% chance of being right.
yes it improves your chances because your have a 1/3 chance of picking your prize on the first shot. So with a 66% chance that you picked a door without the prize. Then when they open a door with out a prize then that means if you switch you will have a better chance at winning
I know that it is 66% per cent blah blah if you change. BUT, I don't believe that. If one of the doors is revealed then you have two left and changing makes NO difference! I'll explain; the supposed 66% can be disproved if for example you made no choice before the door was revealed. This means it is a straight up 50 50 chance of getting the prize. The act of changing the door does not improve chances, as you are still making a CHOICE between TWO, which is definitely 50 50.
@mikemike8000 Originally yes it was a 33% chance, but changing door should not by any means change the percentage, the 66% is psycological due to paranoia and things in the mind.
It would increase chances of winning by switching.
At first you have a 1/3 chance of winning. You have a 66% chance of getting the empty door the first time you pick. When one of the empty doors is revealed. You still have a 66% chance of having your first choice being wrong. So it is best to switch.
Monty Hall need to learn math
REDTEAM22003 1 month ago
Just saw the movie 21 that explained this problem with percentage and variable change .
DarklightALBANIA 5 months ago 2
The likelihood of choosing the wrong door is 2/3. Thus, mathematically, there is a 2/3 chance that you DONT have the prize but the game show host does. The opening of one of the doors by the game show host is a useless factor. It doesn't change the original fact that the host have a higher likelihood of holding the prize then you do.
jonnoj2222 6 months ago
After you choose your first door, the chances of picking the money are 33%, while you picking one of the empty doors 66% (therefor you are probably on an empty door). After he opens an empty door, it is in your interest to switch because him opening the door did nothing. There is still a 66% chance that the current door you are on is empty. So, you should switch because if you do, the chances of you ending up with the money is 66% instead of 33%.
reefreep 8 months ago
If you'd like another way to prove the correct answer to the Monty Hall Problem, see my video "Proving the Monty Hall Problem."
kirkbocek 8 months ago
The Monty Hall problem breaks down to this simple question:
"Would you like to have 1 door or 2 doors? If you choose 2 doors, I’ll give you the better prize of those 2 doors. If you choose 1 door, you’ll be stuck with your choice."
The choice becomes obvious.
133j0hn 8 months ago
it'd be pretty funny if they tricked you by putting nothing behind any door.
yisoonshin12 10 months ago
This is a "glass half full/half empty" discussion.
Full: You have a low chance of picking the prize door first (1/# of doors). Opening the remaining doors means your chance is higher, since you believe yours was extremely likely to be wrong.
Empty: You start out thinking "no matter what happens, there's going to be 2 doors left, so it's 50/50."
The prize is going to be behind one door, and nothing's going to change that. The only thing that changes is your mindset.
RedJoker251 10 months ago
@RedJoker251 That's completely wrong, and I say that in the nicest way possible. Read the two Top Comments, which should explain it nicely.
dkaufman123 10 months ago
Switch doors. Possible combinations are RWW, WRW, WWR. (R = right, W = wrong). You have a 1/3 probablility of guessing R on your initial pick. Therefore, if you guessed R initially and switched, your odds of losing are 1/3 since the only scenario where that happens is RWW (if we pick leftmost door). Your probability of guessing W initially is 2/3. Therefore, if you guess W initially and switch, your probability of winning is 2/3 since WRW and WWR will both result in a winning scenario.
SimplyHerpes 10 months ago
They did this in the movie 21!
briantexts 11 months ago
Chance doesn't really change anything. It's not like the prize will automatically switch to the door you switched to. Chance really has nothing to do with this. I'm serious, it won't change anything to switch doors. The prize still remains behind one of the remaining doors so it doesn't matter whether or not you choose to switch or not. This is what I'm thinking, because this kind of thing really just relies on luck.
yisoonshin12 11 months ago
@yisoonshin12 Unless you have this gift to choose an empty door every single time. Then it does change it a lot to switch doors.
yisoonshin12 11 months ago
@yisoonshin12 Have you ever rolled a six sided die 1000 times? Or even 100 times? You will get each number about 1/6 of the time. That is not luck, it is probability. Yes, you can get lucky and pick the correct door the first time, but you have a 33% chance. That means that on average, out of 100 tries, you will pick the correct door on your first try. However, you have a 66% chance of picking the correct door if you switch.
dkaufman123 10 months ago
@dkaufman123 so on my calculator I used the probability simulation to see what happens at 100 times and they weren't that off, but still noticeably off. I agree with you about the dice thing, but that's only if each side is balanced. if one side is even a little heavier then it'll fall on that side more. but that's besides the point. I agree with your statistics, but there's still a portion that relies on luck. Probability won't change what door the prize is behind.
yisoonshin12 10 months ago
@dkaufman123 But I guess you can use those statistics to your advantage knowing that you will probably pick the prize if you have a feeling you picked an empty door.
yisoonshin12 10 months ago
After door 3 is revealed, there is 50-50 chance that the prize can be behind any door. So it really doesn't change anything by shifting the door.
izardadkhan 1 year ago
@izardadkhan That is false, when you pick door 1, you are making a choice that can be 33% right and 66% wrong. When he reveals that door 3 was empty, then you have 2 left. However, the choice you made in the beginning only had a 33% percent chance of being right, and it maintains the percentage. However, the door 2 now has a 50% chance of being the door with a prize. Of course it is a game of mostly luck, but there is a better chance of winning by picking door 2 because it is 33% v. 50%.
GTSU360 1 year ago
its in the movie "21"
waqasaps 1 year ago
This is actually explained in the movie 21.
Hephestias 1 year ago
if you switch you will have more chances (double) of winningr since the first time you had 66% of loosing
Mazemage 1 year ago
Here's what wikipedia has to say (and I paraphrase):
Imagine the same set up, but with 1000 doors instead of three. After picking a door, Monty then opens up 998 of the remaining doors, revealing them all to be duds. So now there are two doors to choose from. It would seem ludicrous to assume that you picked the right door on your first try (a 1/1000 chance) when there was a 999/1000 chance that it was in the other doors, a probability that is now represented by a single door.
DementedlyHappy 1 year ago 11
@DementedlyHappy I found that explanation animated on stayorswitch (dot) com, it has 100 doors, but it's a good explanation.
mumiscrunk 6 months ago
@mumiscrunk That's good stuff. I'll refer skeptics there now.
DementedlyHappy 5 months ago
@DementedlyHappy Sam Harris explained this in the 'Moral Landscape" video :)
eajoseph217 3 months ago
@DementedlyHappy I realise what you are saying here, but it seems like a massive exaggeration. I don't mean to sound like a jerk, please understand that. In the original problem its 1/3 doors. In your explanation (which is obviously helpful) it would actually have been a choice of 333.3333 doors of 1000 in the beginning and then Monty would open another 333 empty doors and you would have the option of keeping the original 333.3333 or the remaining 333.3333 XD
Natenator77 3 days ago
no you have to switch because if you dont, u have a 33% chance of winning the prize, and if you do swap, you have a 66% chance of winning the prize.
ChelseaFC1250 1 year ago
Here's a simplified explanation: you pretty much want to pick the empty door as your first choice. If you select an empty door, monty hall will then reveal the other empty door, thus when you're offered to switch your selection, you will obviously be picking the door with a prize. So then, what are the chances of you originally picking an empty door in the beginning? 2/3.
itsgot2beME 1 year ago 9
@itsgot2beME This is a great explanation for people who don't really understand probability. Commendations all around.
dkaufman123 10 months ago
They reveal one door then give you the choice to pick one of the two remaining, it's 50/50. What would increase the chances of one door having the prize over the other? How does picking a door in the first place make any difference?
Arthur61987 1 year ago
@Arthur61987 Your statement is flawed because, when you made your original choice, you had a 33% chance of getting the prize. Once the other door is opened, you should switch and you will have a 66% chance of getting the prize. Here is a way to explain it using three scenarios (you always pick door A)
1. Prize @ door A - Either B/C is revealed, switching doors loses
2. Prize @ door B- C is revealed, switching wins
3. Prize @ door C- B is revealed, switching wins
see how you win 2/3 times?
ehud12sc 1 year ago
@ehud12sc
1. Prize @ door A - B is revealed, switch to C loose
2. Prize @ door A - C is revealed, switch to B, loose
3. Prize @ door B - C is revealed, switch to B, win
4. Prize @ door C - B is revealed, switch to C, win
I see 2/4.
Arthur61987 1 year ago
@Arthur61987 Prize at door A is only one scenario, not 2. That's why you are wrong.
ehud12sc 1 year ago
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marianneotv 1 year ago
You choose one door. The guy reveals door 3 that has no prize. There are no benefits if you switch as you now have a 50% chance of winning no matter which door you choose....
YuWallet 1 year ago
switch will give you a better chance of winning because at the beginning, you have 66% of the chance to pick a door with nothing behind it.
skuller1225 1 year ago
this puzzle is answered in one of my favourite movie also
waqasaps 1 year ago
@waqasaps which movie are you taking about? i know it's in the book "The curious incident of the dog in the night time"
skuller1225 1 year ago
@skuller1225 I assume he's talking about 21
ZeroToleranc3 1 year ago
@skuller1225 no he is talking about the movie 21
Indapendent5 1 year ago
Maybe I'm not thinking this problem the way I should. The way I see it is that 1 gets cut, no matter what the probabilities were at the beginning 33% or whatever. At the end, you still end up with 2 doors. And for me it becomes a 50/50 chance of picking the right one since there is only 2 doors left. I see it as a gamble to switch, or to stay
Guitar4Iife 1 year ago
Is in the Players interest to switchhhhhhhhhh! I done the probability calculations already, Hes already giving you two doors if he reveals a goat, Is better to switchhhhhhhhhhhhhhhhh, Come on brilliant people you don't have to brake your headddddddddddddddddddd for this. Grap a Pen, Paper three cards get to work!
RobertoEmilioRomero 1 year ago
HAHAHA this problem is hilarious and so is the answer.
I got a question to this question: case 1: what if there were 3 doors and 2 doors weren't empty but a surprise?1. gives you a girl, 2. gives u a gae dude, 3 empty which one you pick? XD
But then. Case 2: if this wasnt the case and you only get 2 doors u either "try" to calculate it or gamble your luck off. The chance might increase of you winning but also increase chance of losing due to temptation so its a 50/50 if this case was case2. XD
Howin33 1 year ago
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Howin33 1 year ago
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Howin33 1 year ago
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Howin33 1 year ago
Because one door will always be opened and revealed to be empty, you are actually only choosing between two doors. Your chances are still and always have been 50/50.
RedJoker251 1 year ago
Heres how I think of it. On your first pick, you have a 33% chance of choosing the right door. In other words, it is more likely that the prize is behind one of the other two (66% > 33%). Since the choice of "the other two," which had a higher probability to contain a prize gets reduced to one door, I think the probability of winning does increase if you change doors.
fireketchupII 1 year ago
After giving this problum some careful speculation, whether one chooses door #2 or remains with their original choice of door #1, there is nothing driven by science suggesting either choice holds the jackpot, therefore either choice is what is in life known as gambling, and choosing the jackpot door is a matter of luck or fate, which falls under the category of random. It's the same as trying to pick a winning scratch ticket from a convenience store showcase. There's just no way of knowing.
Syruscleat 1 year ago
Choose door 2. Doors 2 and 3 had a 2/3 probability, and door 1 has a 1/3 probability. You know what's behind 3, so you have the best chances choosing door number 2
AContusionInside 1 year ago
wow thanks for posting this we're doing this exact same problem in my logic course at UToronto
Dman40000 1 year ago
No because there is a 50/50 chance the probability will always stay the same so it doesn't really matter
mostafakamaly 1 year ago
So could this extend to Deal or no Deal? In our version of that game there are 22 boxes and you sometimes get the chance to swap the boxes when it gets down to the last two. If there's a very large amount still in play and a lower amount, if you were to gamble, would you be more likely to win the big prize if you swapped?
birdsofafeather1990 1 year ago
@birdsofafeather1990 Yes, but they might switch the values inside the cases in between shows. They don't say they do that, but I would think that's a large possibility.
seinfan9 1 year ago
As I understand it: Original percentages door one 33%,Other two 66%. Cross off door three, the percentages remain door 1 33% the other doors (now just door 2) 66%.
bp56789 1 year ago
well yeah, if ur always unlucky on picking things, then u should switch
if ur pretty lucky in ur whole life, then do not switch lol
shortshortdi 1 year ago
I remember to always switch. My Pre Cal teacher mentioned this way back in high school. He also mentioned about some lady with a high IQ tried explaining why it was always better to switch back then but couldn't really get her point across.
Djlmart 1 year ago
Ciao Patrick! :)
Nicola72av 1 year ago
It's in your favor to change to door #2. Door #2 changes from 33.3% chance of getting the prize to 66.7%.
Henry92RLC 1 year ago
I'd choose door number 2, and in any occasion it's better to change when you can.
hheheks 1 year ago
... If we pick the door that he didn't reveal we have a ninety percent chance of winning, if not we have a 10 percent chance of winning. I think this makes the problem much more simple and apparent.
Merth667 1 year ago
IT would only increase the likelyhood if the show host KNOWS behind which doors the actual prize is hiding. You have a 66% chance of winning if you pick the OTHER door in that scenario. IF he does not, the chances are 50% to pick the correct.
If we expand the problem, let's say we have 10 doors. There's a prize behind one of them, the rest are empty. We pick one door, the host reveals 8 of them to be empty, leaving you with the one you picked and the one he didn't reveal....
Merth667 1 year ago
@Merth667 He has to know where the actual prize is hiding, how else would he be able to pick a door with a goat every time?
ReatuKrentor 1 year ago
yes it is in your interest to switch, thanks for the extra 33.3% chance to win the prize ! :P
NerdfestGames 1 year ago
It matters only if the game show host KNOWS whether or not the door he reveals is empty or not. It he does, you have to account for variable change: the left over door now spouses a 66.6% chance as opposed to the originally picked door's 33.3% chance.
Chokoretokukki 1 year ago
When you first pick a door,and do not swich it with another one, you have a 33 % chance to guess it right. But if you pick a door, another door is revealed as Empty, you are left with 2 doors only, and when switching you have 50 % chance to win.
So you'd better swich it, because the chances increase from 33 % to 50 %.
leppardas 1 year ago
Probably (2/3) you will pick a wrong door, monty hall will reveal the other wrong door, and then you switch to the correct door.
Isn't this the most simple explaination?
borisvankatwijk 1 year ago
if you switch then you have a 2/3 probability to win.
there is an higher probability to choice an empty door, so if you change , you have a higher probability to win.
EmersonAndMe 1 year ago
NO DIFFERENCE
Shigva31 1 year ago
really nice...
sasamuraki 1 year ago
There is no difference what so ever if you switch the doors at any stage! You always have the same chance to win.
adrobotics 1 year ago
if you always switch you will always have a 66% chance to win.
Ervine4 1 year ago
I find this very interesting...
CatacombX32 1 year ago
OMG, smart and logical Youtube comments? The world will definitely end in 2012.
aznlalaland 1 year ago
You should definitely switch. Odds are 66% that the prize is in door 2 (in this example). At first it seems very counter intuitive but the odds do shift in that direction
13tynman 1 year ago
Subtle troll is subtle.
flopski 1 year ago
It increases you likelihood to 66% :) We learned this during the probability unit in precalc, good times
powercrazy11 1 year ago
Either one, Since there are only two doors you have a 50/50 chance. Besides, backstage the prize is being moved to the door you didn't pick, so either one. lol ;)
aetherflow 1 year ago
It also depends what you define as a prize. If I am from deep in the Borneo jungle where there is no gas or roads and I win a car, what good would it do me. It would have no relevance and mean nothing to me. However: If there were a stash of bananas and coconuts behind the door I picked I would have hit the jackpot, whereas to one of us, we would have flunked.
Syruscleat 1 year ago
/watch?v=mhlc7peGlGg
explains it all
Drizzt1343 1 year ago
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Drizzt1343 1 year ago
smexy93kjs 1 year ago
@smexy93kjs lol youtube doesnt add spaces so my explanation looks like a mess. it looked better while i was typing it :(
smexy93kjs 1 year ago
I was taiught in college -the little time I spent there- that statistically the first choice you choose in a multiple choice question on a test is usually right, so I would say that the first choice is statistically the best. I don't know. What do you think Pat?
Syruscleat 1 year ago
My answer: Yes, this does matter. You have a higher chance of winning if you *keep* the door you picked. this looks a lot like something the dude in numb3rs talked about... love probability. Check out this vid: youtube.com/watch?v=P9WFKmLK0dc
Jakathera 1 year ago
@Jakathera yeah, i messed up my answer... should be ***change*** instead of keep... I'ts late, forgive me.
Jakathera 1 year ago
And yes i cheated xD
TheTalmon18 1 year ago
I would switch because the guy in "21" told me to do so... and because I watched the explanation on the related videos =D
HiLoPicoLo 1 year ago
Scenarios:
I chose an empty door (2/3)
-Switching after an empty door is revealed will guarantee that I get the prize
I chose the prize door (1/3)
-Switching after an empty door is revealed will give me nothing.
Switching will give me the prize in 2 out of 3 cases.
MrChessboxing 1 year ago
@MrChessboxing That's a good way of putting it.
DocYoda 1 year ago
@DocYoda
So basically in the beginning, there are 2 no prizes, 1 prize.
you have a higher chance of picking the one without a prize, meaning the other door with the no prize is also revealed, So if you picked it wrong(66% chance) and you swap, you get the prize.
IF YOU DONT SWAP
You still have a 66% chance of getting it wrong if you pick a door. Automatically a door is revealed without a prize, but in the beginning esentially you had a 66% chance of getting it wrong. So if you keep it=1/3
TheTalmon18 1 year ago
Keep the the same door and get an extra 33%
MIT2004 1 year ago
Also, even if you picked the right door, they would still reveal an empty door?
Therefore you already at the beginning have a 50% chance of getin it right.
TheTalmon18 1 year ago
@TheTalmon18 No... in the beginning you picked 1 of 3 doors. 1/3 = 33%.
DocYoda 1 year ago
@DocYoda
One is always revealed empty
But i thought about it
In the beginning you have a 66% chance of getting it wrong.
Therefore, you have a higher chance of picking the first door wrong.
THEREFORE, the revealed door will be empty as well.
Therefore if you swap, you have a 66% chance of getting the prize
If you dont swap, you have a 33% chance of getting the car, since one is revealed and you stick with it, there was still the fact you had a 66% chance of getting it wrong from beginnin
TheTalmon18 1 year ago
In the beginning, you had a 1/3 chance of getting it right.
Then another door is revealed and it says empty.
I wouldnt think of it now as a 50/50 chance.
If you only had one choice, you still didn't pick 3. Therefore its as if it wasnt there.
Because if you picked it, and another revealed empty, it does not change your chances of getting it right because you never picked that door.
TheTalmon18 1 year ago
I remember this problem from "21". I think you've got 33.3% of getting the answer right after they switch the doors twice. I really have no clue though.
aesrp 1 year ago
Also, you should be clear that the host *needs* to open an empty door. Regardless of what door you picked, he cannot open the door with the prize in it. If it were just him opening a random door, I don't think there would be an advantage to switching.
sk8teh14 1 year ago
It's so intuitive.
If my strategy is to switch the door everytime, then I am guaranteed to get the prize if I selected an empty door on the first round. The only instance I won't get the prize is if I chose the prize on the first round. Hence, if I switch everytime, 2/3 times I'll pick an empty door, switch, then win the prize. Whereas if I don't switch, 1/3 times I'll pick the prize on the first round, not switch, and win. Hence switching > not switching.
sk8teh14 1 year ago
When I was originally offered to choose a door, i had a 33.3 % chance of getting the right door. However, when one of the empty doors were opened, the 33.3 % for this door switched to the benefit of changing my currant choice. Hence, I would get an EXTRA 33.3 % if I change my door.
TheFaisal381 1 year ago
Yes, you should switch... Reason being, the host will always choose a NON-PRIZE to win. that is the factor that makes this problem make sense.
If the host had a chance to choose an actual prize then it wouldnt matter whether you switch...
Love the Monty Hall reference --> THANKS STATS PROFESSORS!
ThinKFasT03 1 year ago
The only way you lose is if you originally pick the door with the prize behind it which you have a 1/3 chance of doing so. So the only way to lose by switching is if you originally choose the prize. 1-(1/3)=2/3. Therefore, if you switch you have a 2/3 chance of winning the prize because the game show host is effectively telling you where the prize is from the remaining two doors.
The people saying that there is a 50% chance and the odds don't increase didn't actually take the time to solve it.
shootingrubber1 1 year ago
This is a great conditional probability problem. I think everyone understand that you have a 1/3 chance of winning if you don't switch doors. The reason why its better to switch doors is because the game show host KNOWS which door has the prize. When he reveals another door, he INTENTIONALLY picks the one with nothing behind it. The chance of the game show host picking the door with the prize is 0%. That is why there is a better chance that the prize is behind the other door.
Ru5k1 1 year ago
You better switch!
Kakashinoor 1 year ago
By being able to switch one's answer, and understanding the door she would have to pick, It would present a two thirds chance. It literally switches how everything works. So all losing doors are winning doors, and winning are losing when one incorporates a switch. See, if the door she opened was not "purposefully chosen", then Yes, Your chances would stay the same. But, since it must be controlled, thus to prevent a winning door being opened, the chances change.
ACANOFSODA 1 year ago
Wait, I'm here to change my mind. Like in the movie 21, I can see the benefit of changing doors. The 66% chance gets moved to the door that has not been opened or chosen. But couldn't you say the sum of the percentages of your door and the revealed door to be the 66%? So once door 3 is revealed, door 3 now has a 0% chance and yours has the 66% chance, rather than door 2? If so, I'm going to say that changing doors has absolutely no effect on your chances. It's basically number play.
TheMadSlick 1 year ago
@TheMadSlick Nope, the percentage definitely shifts to the "offered" door. Out of curiosity, I just made a simulation program to test this, and the results for switching was always around 66% and staying was always around 33%. I think you need to consider it as when there are 3 doors: the 1 door you picked has a 33% chance, the other 2 doors have a 66% chance. By switching your door, you're basically taking the other 2 doors; because one gets revealed, and the other you have now selected
DocYoda 1 year ago 22
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marianneotv 1 year ago
@DocYoda You made a program? Just how smart are you?
yisoonshin12 11 months ago
@yisoonshin12 Haha, it's actually pretty easy to make if you have some basic programming skills.
DocYoda 11 months ago
Are you gonna post the solutions too? I was guided by the 1/2 but then, saw all the proofs on wikipedia that it is best to switch cuz you win a 2/3 times
motanium 1 year ago
Who cares? I'm never lucky enough to get a 1 in 3 shot of winning anything to begin with!
RyanEnvy 1 year ago
i would switch because i lack self-respect.
LOL jk i would switch because i know math.
tuestresfat 1 year ago
no difference
3512435124 1 year ago
You can solve this problem 2 different ways:
1.Count possible outcomes, like jsobiranski did 17 mins ago. It's valid and his solution is right.
2.You can use conditional probability theory. As it turns out the revelation of the empty door contains information beyond reducing the set of options to 2 doors. Speaking heuristically: The probability of winning from the revealed door is transfered to the other unchosen door, so you should switch, increasing your likelihood of winning from 1/3 to 2/3.
DenmarkRadar 1 year ago
Wow! Some of these answers are crazy!
crookedtool 1 year ago
@crookedtool yes, i love it!
patrickJMT 1 year ago
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crookedtool 1 year ago
@crookedtool not necessarily. the door they reveal might have the price so at that point you have a 0% chance of winning.
lifemakesmelaf 1 year ago
@lifemakesmelaf prize*
lifemakesmelaf 1 year ago
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crookedtool 1 year ago
@lifemakesmelaf they never reveal the door with the prize (part of the rules)
patrickJMT 1 year ago
@patrickJMT o ok i see. no loophole ):
lifemakesmelaf 1 year ago
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crookedtool 1 year ago
50% it doesn't matter. i don't care who say that if you switch it the odds increase. Make the test n-times like i did a couple years ago.
Dominikenci 1 year ago
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crookedtool 1 year ago
jsobiranski 1 year ago 27
@jsobiranski WTF who said anything about goats??? D:
ivanxdxd2008 1 year ago
@ivanxdxd2008 A running gag on the show was that the "empty" doors would often contain a goat.
Mischlings 1 year ago
@ivanxdxd2008 If you ever watched the show Let's Make a Deal you would know that Monty Hall would usually reveal a goat behind the door that did not have a big prize like a new car. Another reason I chose "goat" is because it uses fewer characters than "no prize" so that I could stay within the 500 character limit that is imposed upon us for our comments.
jsobiranski 1 year ago
Yes, it makes a difference if you stick or switch. You should switch doors because if you stick with the first door you choose then you will only have a 1/3 probability of winning, but if you switch after he selects one of the other doors which do not contain the prize, then you will have a 2/3 probability of winning. We must in each case assume that you first choose door 1. Please see my full explanation in the following comment.
jsobiranski 1 year ago 3
(short continuation) someone says what if there were a million doors and the host removed all but one other. the point is moot, even if there were hundreds of trials beforehand each reducing the choices by half each time, the number that it is potentially out of is irrelevant, all that matters are variables involved. the previous trials with ignored results, don't matter. the 1/2 trial remains 50%
morthim 1 year ago
1/3 = 34%; 1/2 = 50%; the likelyhood of getting the right door increases 16%, across trials. those who disagree think of the problem as effectively getting two choices, thus the prob is 66% but it is a mistake. the mistake lies in thinking there were multiple trials, and thus you are getting two choices of the three by changing rather than still having one. the first trial was irrelevant, while you do have a higher chance of guessing 1/2 than 1/3 it is 16% difference not 32~
morthim 1 year ago
no. it still 50/50.
HappyEarlyBirthday 1 year ago
i've seen this problem discussed in an economics book. It is better to switch and pick door 2. The first time, the likely hood of it being in door 1 was 1/3, but the combined chance of door 2 and 3 was 2/3. Since door 3 is eliminated, there is now a 2/3 chance of it being behind door 2. It's funny b/c the average person thinks that the probability of it being behind a certain door depends only on the number of options a person has, like a coin toss or a dice roll.
SuperNerd707 1 year ago
@SuperNerd707 I find it hard to believe even now actually...the author went on in the book to claim that humans are not rational beings as conventional economics suggests b/c most of us can't decide what the best thing to do is. And even when we do know, it is still very difficult to believe!:P
SuperNerd707 1 year ago
It's not 50/50 once the empty door is revealed... you have a higher percentage of guessing correctly if you switch. Think about it this way.. before any doors were revealed you had a 2/3 chance of choosing an EMPTY door. Once one empty door is revealed, it still remains that the door you've chosen would be empty 66.6% percent of the time. By swapping, you actually have a 66.6% chance of getting the prize.
sjsawyer 1 year ago
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TheMadSlick 1 year ago
the problem has to do with variable change, not pressure, uncertainty, or psychology what so ever. When asked to pick a door you have a 1/3 chance of guessing correctly, but then once the host eliminates one of your options, and re-offers you the second door, your probability increases to 2/3. therefore you should switch doors.
mack1oh 1 year ago
Statistically, you have a 50/50 chance of getting it right. However, the fact that there was a third option and it has been remove, it may cause you to doubt your door option because of pressure or uncertainty.
Valvy 1 year ago
This has been flagged as spam show
I've seen this problem before but it's been awhile. I believe it was if you changed doors your chances went up to 66.6% because you "gain" the 33.3% of the door that was proven to be empty. So theoretically, based on the percentages, I would change doors. In the real situation however, I'm not sure.
needmorereality 1 year ago
You originally start with a 1/3 chance and a 2/3 chance of the one of the other two having the prize. The host will show you one that contains no prize, this then means that the door left is left with a 2/3 in chance of having the prize. There for you switch. Mathematically you win 2/3 of the time when you switch but only 1/3 of the time when you keep your door.
TheMentalistish 1 year ago
I would switch my door. When you picked the first door you had a 33% chance of winning. The other two doors had a 66% chance of winning. When you revealed the third door, you moved the 66% to the second door. Its not a 50:50 chance. Its a 33:66 chance. I take my brand new car now. haa
HowlBlack 1 year ago
50/50. Its basically two doors now and you just pick one..either your first choice or the other
Licronthoup 1 year ago
yes because the 50:50 chance of winning is only guarenteed to take place if you change your door, otherwise you keep the starting probabilities of 2/3 and 1/3
rabidwolf77 1 year ago
It won't change anything. You chose door number 1 with a 33% chance of being right, but once a wrong door is eliminated each door now has has a 50% chance of being right.
Halo3ForumEurope 1 year ago
I'd pry stick with my original choice.
amjPeace 1 year ago
I believe it's in you benefit to change your choice as before you had only had a 33% chance of winning, but now have a 50% chance of winning. Your probability of losing was at 66% at the beginning. If you don't change you choice, your probability of losing still remains at 66% percent, but changing your choice decreases your chance of losing to only 50%.
Charounson 1 year ago
IT DOESN'T MATTER,
The probability is 50%, choosing door 1 or 2 is like flipping a COIN!.
s200960170 1 year ago
@s200960170 Imagine there were a million doors and you picked one. Then the host eliminated all the doors except your pick and one other. Do you still think you picked the prize winning door based upon a million to one chance? I wouldn't.
seinfan9 1 year ago
@seinfan9 haha good analogy.
sjsawyer 1 year ago
@seinfan9 If it were to assume there were a million doors, & each time 1 door was eliminated, the probability of you winning keeps increasing. (it increases at a very small range).
However in this case since there are only 3 doors, & when 1 door was picked and eliminated the probability becomes 50%, so it really doesn't matter if you change your choice, the probability is always going to be 50%.
s200960170 1 year ago
@s200960170 You don't understand the problem. How likely is it that your original pick is the one that has the prize behind it? More doors is supposed to make clearer the line of reasoning for switching. One million doors, only one has a prize. You pick one. All but your pick and one other are eliminated. The doors eliminated had no prize. This scenario was explained in detail in my Discrete Mathematics course last semester. Bottom line, it's better to switch.
seinfan9 1 year ago
@seinfan9 How likely is it that after you switch, the chances of having the price behind the door increases?
[btw im talking about this particular Question]
s200960170 1 year ago
@s200960170 Your chance of winning always increases to 2/3. That doesn't mean you'll always win. If you want to go with what is more probable, it's smarter to always switch because you have only a 1/3 chance of winning with your first pick. You can search for a Monty Hall simulation on Google or you can check out the Wiki article. I think the tree map on the probabilities shows it quite well in the case for choosing Door1 first. The probabilities would be the same for choosing Door2 and 3 first.
seinfan9 1 year ago
it's increasing the chance of winning cuz three doors ..the probability of winning is 1/3 while then now your chance of wining is 1/2...
am i right .. idk
hamburgervisal 1 year ago
No advantage as the probability of the door being correct is more than before for both doors
baggamandem 1 year ago
it increases your chances to switch.
look at it on a bigger scale, let's say 100 doors, and they reveal that 98 of the doors have nothing behind them after you choose a door. naturally, you had a 99% chance of being wrong with your first pick, so switching on your second pick now gives you a 99% chance of being right.
MD3c0 1 year ago
yes it improves your chances because your have a 1/3 chance of picking your prize on the first shot. So with a 66% chance that you picked a door without the prize. Then when they open a door with out a prize then that means if you switch you will have a better chance at winning
LazyCrackerz 1 year ago
this is one of my fav problems, my hs teacher told me about it in math club
aznelf13 1 year ago
I know that it is 66% per cent blah blah if you change. BUT, I don't believe that. If one of the doors is revealed then you have two left and changing makes NO difference! I'll explain; the supposed 66% can be disproved if for example you made no choice before the door was revealed. This means it is a straight up 50 50 chance of getting the prize. The act of changing the door does not improve chances, as you are still making a CHOICE between TWO, which is definitely 50 50.
mikemike8000 1 year ago
@mikemike8000 Originally yes it was a 33% chance, but changing door should not by any means change the percentage, the 66% is psycological due to paranoia and things in the mind.
mikemike8000 1 year ago
i dont think it will increase chances because if you eliminate that door both your original and other box have 50% chance
thebigegg123 1 year ago
it would increase the likely hood of winning.
bilin825 1 year ago
yes i would ,i forgot but u have like a 77% percent to get the right answer if u pick the other one
RAE4eva666 1 year ago
It would increase chances of winning by switching.
At first you have a 1/3 chance of winning. You have a 66% chance of getting the empty door the first time you pick. When one of the empty doors is revealed. You still have a 66% chance of having your first choice being wrong. So it is best to switch.
DearestyouXII 1 year ago
only if i had this video when i was working on my c program
yoyojonathan 1 year ago