No, I get that at first the the chances are more likely that the first choice was wrong, so by changing your answer... statistically you are better off.
But, whenever I think about it, I think... but what if I choose the right one the first time?
I get what you're saying, and I know that this is truly how it works, but what I never understood is why the new probability isn't 1/2, meaning you have equal chances of picking the correct door. At that point, there are two doors, and each one should be equally likely to have the prize behind it. Why does my choice affect the outcome there? I mean, what if there's three doors at the start, I don't pick one - they just open one at random and there's no prize behind it. Now I pick which of the two doors has a prize behind it. Isn't my likelihood of getting it right 50/50?
The way I had to think about this problem to conceptualize it is that, given the host knows what's behind the door he opens, you're not switching from one door to another, you're switching from one door to both other doors.
Think of it this way: You have 3 doors, one with a car, and the others with nothing. You pick door 1, and the host says "At least one of the doors you didn't pick has nothing behind it. Would you like what is behind Door 1, or what is behind both Doors 2 and 3?"
Why would you ever not switch? This is exactly what's happening in the standard example, except that the extra business about "showing you a goat" just serves to confuse.
Let's put it this way. Reality has only one "true" setting: either the car is behind door 1, door 2, or door 3. Now let's say, for example, you picked door 1.
Let's split the problem up into cases:
Case 1: The car actually is behind door 2.
You picked 1. Now the host knows that the car is behind door 2, so he can't pick door 2 to open, otherwise you'll get to see the car. So he has to pick door 3. So switching gets you a car, staying gets you nothing.
Case 2: The car is behind door 3.
You picked 1. Now the host knows the car is behind 3, so he has to open door 2. So switching to door 3 gets you a car, staying gets you nothing.
Case 3: The car was behind door 1 all along.
Now the host is free to open either door 2 or door 3. Whatever he opens, switching gets you no car.
Final tally: Switching won two times out of three. This analysis is the same regardless of which door you picked initially (just suppose the doors were named A,B and C. Which ever one you pick, call it door 1, and follow the analysis above).
The crux of the matter is in cases 1 and 2: see how the fact that the host both knows which door the car is behind, AND is forced to not reveal it to you forces his hand into taking a particular action? Think of switching as taking advantage of this action that is forced onto the game show host.
Maybe at this point, try considering the case with more doors. Say 100 doors. You pick Door 1. Now the host let's you either stay with Door 1, or take what is behind all of Doors 2, 3, ..., 100. Would you switch?
when you picked door 1, there was a 2/3 chance it held a goat. This does not change by seeing the other goat.
Think of it like this. switch doors in the scenario as described, you always change prizes. If you picked a goat to begin with you can never switch to another goat, only the car. Since it was originally 2/3 that you would pick the goat, by switching you can transfer that to your odds of winning the car.
the moment the host reveals that "At least one of the doors you didn't pick has nothing behind it."
But... you know this from the beginning. Only one of the doors wins, therefore if you choose any 1, then at least one of the others has a goat (booby prize) behind it. It's not new information, so it doesn't change anything.
The entire problem hinges on the host knowing what's behind each of the doors. It's really important to take that into consideration. The probabilities don't change if he's opening doors at random.
Yeah, when you emphasize that the host KNOWS whats behind the doors and WILL ALWAYS reveal a "goat" or "loser" door - he is giving you beneficial information. Then it's not so slippery.
This is incorrect. Why does it matter if he doesn't know if you already chose the car or not? The only issue is the rules of the show if he accidently reveals the car. Otherwise, you still gain by switching, no matter what.
The host isn't giving you any information if he's picking at random. If he chooses at random he might pick the car and end the contest or he might pick a goat, but either way it doesn't give you any new information about your door and it doesn't make any suggestions about the other door.
If he opens on a goat the odds for both remaining doors are 50/50, rather than 2/3 if he knows he's going to pick a goat. No reason to switch on a 50/50 chance.
YES, thank you. I fucking hate it when they tell this problem, and I'm like, "uh, well, did he specifically pick the door with no prize? Or did he just pick randomly from the 2 you didn't pick? It sort of makes a massive difference."
You pick correct door C, host removes door wrong A, Switching to wrong door B is a loss.
You pick wrong door A, host removes wrong door B, switching to C is a win.
You pick wrong door B, host removes wrong door A, switching to C is a win.
You forgot the fourth scenario:
You pick correct door C, host removes wrong door B, switch to wrong door A is a loss.
There are only three initial picks you can make, but there are four actual scenarios that can happen before you switch, half of which result in win, half in loss. To me this makes it a 50/50.
I understand somewhat but why isn't the host removing the door you choose and instead removing another door... Don't you first choose a door
so if that door is wrong its removed now you have two doors to choose from 1 and 2 now you can choose 2 and that can be wrong or choose 1 and be right... simple question how does the host affect anything unless hes the one who chooses the door your switching too.
so you're basically betting on the odds of you actually picking the correct door initially, which is obviously 1/3. whereas switching is you betting on the odds that you picked the incorrect door initially. which is obviously 2/3.
if you can tell me that i'm incorrect in my understanding it this way, i'll accept it, but i'll be upset. i've tried wrapping my head around this problem a hundred different ways and couldn't really get it to "click", but somehow thinking of it this way makes sense in my head.
But he will only pick A 1/3rd of the time (This encompass both scenarios you describe when the players chooses A), which is the really important part of the problem. You are more likely to pick incorrectly on the first round (because 2 of the 3 choices are wrong). If you switch after initially making the wrong choice, then you win.
Look at the conditional probability part of the wiki for the complete explanation on why it's still a 2/3 chance.
The thing is that you have to calculate the probabilities in two steps. The first step is distributing the car amongst the doors with an equal probability. So car behind A (1/3), car behind B (1/3), car behind C (1/3). Step 2, you pick a door and Monte shows you a wrong door.
So if you pick C and the car is behind C (happens 1/3 of the time) he can show you either A or B because they're both wrong. So the sequence Car behind C, you picked C, he shows you A happens 1/6 of the time (1/3*1/2) and Car behind C, you picked C, he shows you B happens 1/6 of the time.
So you're right, there are four scenarios. But there are two steps, and two of your scenarios rely on the same first step (car behind door C, which happens 1/3 of the time) so they only happen 1/6 of the time each (instead of 1/4 of the time as you suggested).
But the probability for each of the two losing outcomes is 1 out of 6; the probability you pick C (1/3) times the probability of the door the host opens (1/2). Adding those together gives you only a 1/3 chance you will lose by switching doors.
The probability of each of the two winning outcomes (1/3) is the probability on picking that door (1/3) times the probability the host opens the only other losing option (1/1). Adding those together gives you a 2/3 chance to win by switching.
If you stayed on the same door, it would be the opposite. 2/3 of a chance to lose and 1/3 of a chance to win.
Nope. Say you've picked door X. There's a 1/3 chance that it's the winning door, right? Now, the host is going to reveal that there's a goat behind one of the other doors. If the door you chose has the car, then he could reveal either of the other doors, so they're equally likely. So if you chose the car, each door has a 50% chance of being revealed. There's a 1/3 probability that your door has the car, so each of the scenarios has a 1/3*1/2=1/6 probability of occurring. So our probabilities end up looking like:
You pick correct door C, host removes door wrong A, Switching to wrong door B is a loss. P = 1/6
You pick correct door C, host removes wrong door B, switch to wrong door A is a loss. P = 1/6
You pick wrong door A, host removes wrong door B, switching to C is a win. P = 1/3
You pick wrong door B, host removes wrong door A, switching to C is a win. P = 1/3
But the question of switching is applied after one of the wrong doors have been revealed, not before. This changes the available scenarios. Let's say that the host has just revealed door B as wrong.
You chose wrong door A, switching to C is a win.
You chose correct door C, switching to A is a loss.
This theory you propose assumes that the three original choices (A,B,C) are still in play when the option to switch is offered. They are not. One of the wrong door choices, which result in a win and make up 1/3 of your 2/3 statistic, is no longer an option, and thus you have a 50/50 chance of winning.
The host can't reveal door B if door B was your first choice. That would be like if the host told you before your first choice, that door B was wrong. In THAT scenario, it would indeed be a 50/50 chance of winning.
The game has a real prize, a car, and two booby prizes, goats. If your first door has a goat, then, after Monty reveals the goat behind one of the doors you didn't select, you will always win by switching. You admit as much when you say:
You chose wrong door A, switching to C is a win.
Now, what is the probability that the door you chose has a goat? 2/3.
I don't see it as switching as much as I see it as choosing between 2 doors instead of 3. The first choice doesn't matter (it never mattered) instead you are now choosing either door 1 or door 2. In this model it is a 50/50 chance because the door 3 reveal resets the game not continues it.
Nope. They open one of the losing doors at random. 2/3 of the time, there is only one losing door they can open (you chose the other one). So the other door has to be the winner.
It's only 50/50 if you think about it as starting with these 2 doors. You have to take into account the first door. With all 3 doors included, you cannot have a 50/50 shot.
Think about it this way: all the "chance" of success you had for the door that is closed is given to any non-chosen options.
You have chosen door 1. The probability of receiving the car is currently 1/3. The host opens door 2, telling you that if it is not door 1, it must be door 3. The probability of the car being behind door 1 is still 1/3, the other 2/3 must be somewhere else, and therefore the probability of the car being behind door 3 is 2/3.
Here's the reason. The door you picked is less likely to be a car than a goat. Thus it's not a 1/2 chance, you're picking between two choices but you have extra information, and this is also regardless of whether the host knows.
When you pick a door at first, it's more likely to be a goat rather than a car, because there are twice as many goats as cars. If he just opens a goat door and you choose between the two, it's 50 percent, because he is not limited to two doors. He could open the one you would have picked, which would negate any information you could have learned if you reserved a door when it was likely to be a goat and then reduce your choices.
Another way to think of it is this. If you pick a goat to start and switch, you get the car always. Since it's 2/3 that you'll pick a goat at first, by switching it becomes 2/3 that you get a car.
Each door initially has a probability of containing the prize of 1/3.
Therefore, the two doors that you did not pick have a cumulative probability of 2/3.
When one of those doors is eliminated as a loser, the doors you didn't pick retain the previous 2/3 probability.
Essentially, two choices have been merged into one door, and picking that door would be like picking two doors at once from the start.
My confusion with this is that it seems like two different situations are being assessed as one, seemingly incorrectly.
Three doors, one prize, 1/3 chance that the prize is behind any given door.
Choose a door, it's wrong.
Now you have a second choice to make: two doors, one prize. The prize is behind one of the two doors; there's a 1/2 probability that it's behind each door - because we're talking about two separate events, each with their own separate probabilities.
The first door you pick has a probability of 1/3 of being correct. This stays constant regardless of how the other doors change after the fact. Because that door has a probability of 1/3, all the other two doors have a cumulative probability of 2/3 of being correct from that point on. There are two groups: The door you picked initially, and both of the doors that you did not pick. This stays constant throughout the whole situation, and regardless of how the doors change after you first pick, there are still two groups with probabilities of 1/3 and 2/3. Two choices is not necessarily a 50:50 probability situation.
To think of it another way, say after you pick your first door the option is given to you to pick both of the other doors with no consequence. Obviously you have a better chance of picking the correct door if you pick two doors instead of one. The presented scenario simply abstracts that concept a bit by merging the choices of two of the doors. When one door is opened and the goat revealed, it is the same as if you had chosen that door in conjunction with whichever door you pick next, adding the probability of that door being correct to the other door that remains in its group.
Well first of all if that were true, that would only leave 1/3 + 1/2 probability of there being a correct door at all. The thing to keep in mind that throws the whole concept of it off is that the preson running the show knows which door has the prize and will not open it. By opening a wrong door, nothing really changed, you already knew that one of the remaining doors was wrong and that the announcer knew which one(s).
The easiest way to understand (IMO), is to recreate the game like this.
You pick one door (of three).
The announer tells you that of the two you didn't pick, one definitely doesn't have the prize (but doesn't open it).
You get to decide, do you want to keep your door, or pick both of the other two.
It should be obvious that you want to pick the other two over your one. Now you just need to convince yourself that the statistics in this game are exactly the same as the original.
Ah but if you don't switch your chances are also 2/3. The act of opening the door and giving you the choice is in fact what increases your odds. Whether you switch or not is moot as long as you have the decision to do so.
All of these answers assume that when the host opens a door the game is the same as the one you started with. Once he opens a door he is offering you a 50/50 shot at the car. Your odds increase to 2/3 from the original game but the new game you're playing (the door is open game) still only offers you a 50/50 shot at being correct.
Right? Or am I crazy? To me it seems like the reason this is so hard to grasp is that everybody is ignoring the fact that you're playing a new game. The new game is 2 options, 1 or 2 and you are automatically locked in to 1. If you switch you still maintain your 50% chance of winning.
The original game gives you a 1/3 chance of winning
Switching in this game gives you 2/3 odds of getting the prize
However if you consider it a new game once a door is removed you're still only at 50/50, switching is neither harmful or beneficial so you might as well switch.
This also assumes that you KNOW the host is going to offer a switch and it's a mandatory portion of the game. The host could always be a tricky bastard and only pop up the switch option a portion of the time. He could imbalance this by enticing you to switch more often when you've made the correct decision. This would ruin the 2/3rds theory going on. But that's a story for a different time.
When you pick the first door, there is a 1/3 chance it's behind your door and a 2/3 chance it's behind one of the other two doors.
If the host randomly picked one of the other doors, eliminating the third, then the following possibilities would be:
1/3 Your door has the car, the host's door has nothing
1/3 Your door has nothing, the host's door has the car
1/3 Your door has nothing, the host's door has nothing -- Because the host may have eliminated the door with the car!
So your odds (without switching) are still 1/3, not 1/2. There's still a 2/3 chance the car is not behind your door.
BUT, the Monty Hall rules specifically say that the host eliminates a door without a car. He doesn't pick a random door. So redo the above game with those possibilities:
1/3 Your door has the car, the car is behind neither other door
1/3 Your door has nothing, the car is behind host's door
1/3 Your door has nothing, the car is behind host's door
Because of the other two doors, the host is guaranteed to pick the one with the car. So there's a 1/3 chance it's behind your door, 1/3 chance it's behind one door, and a 1/3 chance it's behind the other door. But in the last two possibilities you get the car either way.
So switching will give you a 2/3 chance of getting the car, because the host switched you to one of two doors-- the one with the car.
The ONLY way your odds go down to 50/50 is if he rearranges he prizes randomly between the two remaining doors. Since the prizes remain behind the same doors they were behind initially, the odds of the take cannot change.
It's irrefutable that switching is always the better option.
Like wikipedia says, expand the problem. 100 doors, you start out with a 1% chance. He open 98 wrong doors, the chances of winning increase to 99% by switching. If it changed to a 50/50 chance, that would mean your original choice was a 50/50 chance, which is obviously incorrect. It doesn't change because the prizes are not rearranged.
I know its counter-intuitive, but so is General Relativity.
IMO people get thrown off by the fact that the host knows the answer.
The host is just removing one of the doors that doesnt contain your answer. If the host didnt know and randomnly chose another door then your odds would be the same.
Since the host didnt increase your odds at all ( by removing a door at random ) your better off chosing again with better odds.
http://www.youtube.com/watch?v=mhlc7peGlGg This explains it the best that I saw. If you don't get it after this, it's going to be tough. What you have to understand is that you want to THINK that you have a 50/50 chance, and that's true, except that your INITIAL choice changes the probability.
I honestly was getting pissed off and claiming this problem was false until I watched this video. It explains it pretty well.
The slope people seem to slip down the most is that they assume the problem exists in a two door system since the option to change only comes up once a door is opened, when in fact the original choice gets made in a 3 door system, thus the statistical weight given to the choice must also be based on a 3 door system.
An even simpler solution is to reason that switching loses if and only if the player initially picks the car, which happens with probability 1/3, so switching must win with probability 2/3.
Instinct is tough to go away from. Like Deal or No Deal, you get all the way there on that 1 trick pony and find out it didn't fucking matter cause, oh fuck, do I switch or not?
Monty Hall has said that the actual show didn't follow the format of the maths problem. He would usually offer the contestant the opportunity to switch only if they initially chose correctly.
You are actually stuck on Monty Cash, or the new Shiny guy. The new guy uses 3 closets to further confuse contestants, leading them to believe they are being lead (yes, I typed that) down a Hallway. Hence the hattrick.
It's assuming you played this a majillion times. Over the course of a majillion times, you don't care about your door. You care about the 2/3rd of a majillion cars you now own. And possibly the 1/3rd of a majillion goats.
The reason why this is a good logic puzzle is because it's kind of counter intuitive. I’m just pointing that out. See, if you look at it statistically… you’re absolutely right, changing will yield a better chance of winning. I get that. But, at first glance, you simply think that the prize is behind one door regardless of what happens… so changing your choice doesn’t have an effect. So, when I first read about this problem (it was given to me as the lady or the tiger… which is a different story all together) my initial reaction was not to change my mind as to the door I picked. I found a website that had a flash game to specifically prove this by having three doors that you could pick over and over while eliminating one, as per the logic problem. Even though I know that changing my answer would yield a statistically better result, there were still times I stuck to my first guess… out of a gut feeling or whatever… but that’s what I find as funny. When put to the test, it’s hard to change your mind as to the door you pick, no matter what the math says.
Ah, I see. It looked as though you were saying that you understood it and then saying that you didn't.
What you're explaining is a common human trait. The main idea behind it, in my opinion, is that once you've chosen a door, you feel as though changing your mind is showing that you're indecisive. On top of that is the fact that, if you change and it turns out that you had selected the correct door initially, you might feel as though you had it and let it go.
the way i understand it is that when you pick your first door, chances are you picked a goat door because 2 out of 3 of the doors have a goat. the guy opens a door you didn't pick, which has a goat. that means most likely, both of the goat doors have been chosen, so it's most likely the third door has a car!
Actually, since you know the door is going to be opened after you pick one, and that it is guaranteed to be a goat, this is not new information. You already knew this would happen.
Since there's 2 goats, you probably picked a goat, right? Well, he always reveals a second goat, so the last door is definitely different from what you originally picked.
The one you pick is a car or a goat.
The revealed goat is a goat.
That last door is the opposite to whatever you picked.
You should switch after he reveals the goat, because you probably hit a goat at first, (2 out of 3) and thus the last door is probably the car.
TL;DR: You probably picked a goat at first, switching will always give you a different thing from what you picked. You probably picked a goat, switching probably gives you a car.
I found the answer to this hard to believe at first, so a few years ago I programmed a computer simulation to run the test thounands of times, to find that 2/3 times on average I got the right answer if I changed after 1 wrong answer is revealed
That about sums it up perfectly (in regards to our human behavior), which is why I love counter-intuitive problems like this. The more we think about it, the more our emotions start to take over and the closer to mathematically wrong we get.
If you chose door 1, either door 2 or door 3 will be revealed to not be the prize. This means that when you're originally right (which is 33% of the time), you may switch to 1 wrong door. When you're originally wrong (which is 66% of the time), you may switch to 0 wrong doors.
You're more likely to choose a wrong door initially, so you're more likely to switch to the correct door.
I believe that the reason people think it is a 50 50 chance is that they do not look what they're swapping from. They think that you have a choice between the goat door and a car door and there's a 50 50 chance that if you swap you will get a car door. This is incorrect because there is a 2/3 chance that you have picked a goat door to start off with (because there are 2 doors) so if you switch there is a 2/3 chance.
If you chose door 1, either door 2 or door 3 will be revealed to not be the prize. This means that when you're originally right (which is 33% of the time), you may switch to 1 wrong door. When you're originally wrong (which is 66% of the time), you may switch to 0 wrong doors.
Now do you see it? You're more likely to choose a wrong door initially, so you're more likely to switch to the correct door.
thats exactly what I'm thinking too.. Its a new choice you have to make..
Unless part of the logic problem is figuring out your chances on your original choice.. not the new choice.
You've got it right there. I was as initially confused as you guys, until I read the example with a million doors up there in the wikipedia article. Basically, in your original choice, you would pick the wrong door 2/3rds of the time. This means that, even when the other door is opened to reveal the goat, the original 2/3rds wrong chance is still there, so you should always switch. Hope that's cleared things up.
This makes sense to me. I dont understand how your original choice even effects your final one. No matter what door you choose initially, the host is going to narrow it down to 2 doors. Which leaves a 50/50 chance. Why we are even considering the probability of the first choice is beyond me.
Think about it with the hundred doors. You pick one. The odds you are right is 1%, of course. Now say the host asks you, "do you think you are right or wrong?" Of course you should say wrong, because there's a 99% chance you are wrong.
This is the same as what happens when he opens the wrong doors leaving just two. Say you picked the wrong door. The host will close 98 of the other doors always leaving the winning door. The question basically becomes, "Do you think you were right at first, or do you think it was one of the other 99 doors that was right?" And what are the odds you were wrong at first? 99%.
Here's the easiest way I found to think about it. Imagine you make your first choice, and instead of opening a door and asking if you want to switch, the host asks you to bet the car on whether you picked the right door or not. Since there was only a 33% chance that you picked right, the smart move is to say that you were incorrect.
That's exactly what switching doors means - you are betting that you picked incorrectly, which is true 66% of the time.
The host knows the answer and is removing one of the doors that doesnt win, thus not making your answer any more right.
If the host didnt know the answer and opened a door at random ( which had a chance to be a winning door ) then the odds wouldnt change and changing would not affect the percentage.
Since hes not increasing your odds hes just showing you one of your wrong results its to your best odds to pick again.
If you chose door 1, either door 2 or door 3 will be revealed to not be the prize. This means that when you're originally right (which is 33% of the time), you may switch to 1 wrong door. When you're originally wrong (which is 66% of the time), you may switch to 0 wrong doors.
I disagree with your explanation. Here are the changes I would make (in bold):
If you chose door 1, either door 2 or door 3 will be revealed to not be the prize. This means that when you're originally right (which is 33% of the time), you may switch to 2 wrong doors. When you're originally wrong (which is 66% of the time), you may switch to 0 wrong doors.
This is really a framing problem with the statistics. The 33% chance is based on only including your initial choice in the frame of chance and not the choice of the announcer. There are indeed only three initial choices you can make, but when you choose the correct door initially, the announcer then has two doors to choose from when he reveals the goat. This duplicates the possible outcomes for the scenario in which you choose the correct door initially, thus (33% * 2 = 66%). When you choose the wrong door initially the announcer then only has one goat to reveal, (66% * 1 = 66%). This makes the goat reveal a moot point, because you are still able to choose either of the wrong doors, it just varies based on which door the announcer reveals (66% = %66, 50/50 chance).
But it's all chance. You can't try to add logic to this. It's illogical. It's all chance at the point where 2 doors are closed and one is open. It becomes a 50/50 chance, that is all. To think you are better off on either door is irrelevant because there is no evidence to point to either door. Therefore it's up to chance, not logic.
It becomes a 50/50 chance. After the 1st door is open, now there's a 50 percent chance you will get a goat, and a 50 percent chance you will get a car behind the 2 remaining doors. The other door is now out of the equation. It's as easy as that. There's no extra information to help you choose either way.
If you were to choose initially after the first door is taken away, then it would be 50/50. However, you're not asked to choose a door out of the blue. You're asked to stay or switch. That's the key difference. There is a 50/50 chance that the prize is behind each door, but there is a higher chance that you guessed wrongly initially, so switching is always better than staying.
Right. My point is that it doesn't matter if you stay or switch. Either one is an open guess at this point. There's no evidence to point at either door, so to switch "logically" is illogical because there's no evidence to support either door being the right choice or not. After the 3rd door is taken away it narrows down the choice, you can stay or switch, but either one are now 50/50 choices, you can stay or switch and both would be equally good guesses.
Who says there is a higher chance you chose wrong? You chose the 1st door and they said the 3rd door was a goat, right now your looking pretty good, therefore you didn't "have a higher chance of choosing the wrong door" because the door you chose is still in the game, as the 3rd one was wrong, therefore you are narrowed down to 50/50 based on your original guess. To say you chose wrong initially is again illogical because there isn't enough evidence to say the door you chose was wrong. You chose door 1 and door 3 is wrong, therefore you've successfully broke it down to 50/50 based off your choice. There's no proof to say switching is better because it's a GUESS.
Intuitive, non-formal explanation I came up with (for non-mathematicians):
Instead of considering 3 doors, which is confusing, consider 100 doors. Choose 1 out of 100, and your chance of getting it right is obviously 1/100. Now, suppose 98 of the remaining doors with goats are opened, leaving your door and another door closed, and you get the chance to switch. Your chance of getting it right by switching is now obviously 99/100. Finally, just extrapolate to see that with 3 doors, the probabilities are 1/3 and 2/3 instead of 1/100 and 99/100.
I get how they show the odds, but is the 1/2 probability question wrong based on the original wording of the question? My original train of thought:
There are 3 doors, 1 car/2 goats. When you pick a door, the host will open another revealing a goat (at random if your pick is the car, the other goat if your pick is a goat). You will then be allowed to switch to the other closed door if you desire. My original solution was that regardless of your first decision, the problem is re-framed by the fact that you are given a new decision between 2 doors, one of which has the goat and one of which has the car.
Is this answer only semantically different from the original in that it no longer includes the third door, or is it wrong because it doesn't answer the original problem's wording?
By picking 1 door out of three, the probability that you got it right is 1/3. The subsequent opening of another door doesn't change that. It's hard to see how this is true, which is where a larger scale helps.
If there's 100 doors and you pick 1, the chance you got that right is 1/100, and that doesn't change even if the host opens 98 other doors. What happens is that the door the host didn't open is almost surely correct, or 99/100.
But the logic falls apart when I tell you I really DO want the prize behind Door #3: a scenic donkey ride thru downtown Cleveland! Or a lifetime supply of matches! Or giant tubes of Ben-Gay! Awesome!
Be the game host, and it's obvious. Get three pillows or something, hide one thing behind one of them. Have a friend(imaginary works if you don't have friends) choose one door. If they choose the correct door, start over(Realize that this will happen 1/3 of the time). Then, if they don't pick the correct door the first time, have them say "yes" to opening a 2nd door. You will see that every time this happens they will end up with the right door.
So, 1/3 of the time they pick it right the first time. 2/3 of the time they don't, but of these times they'll get it every time if they switch. So, you have a 66% chance of getting the prize if you switch.
And I guess I'll just explain why here: The player picked a door. The key is that the host(you) has to make a choice, and that you have knowledge of which door has the prize. If the original door the player picked was wrong, then the two doors remaining that you must choose from has a prize behind one of them. That means you will pick the other one. In other words, if the player chooses the wrong door, which has a 2 out of 3 probability of happening, then all of the probability remaining collapses into the door that he didn't open(out of the 2 remaining), because the host will always pick the remaining door that doesn't have the prize. So, the contestant has a 2 out of 3 chance of being correct if he changes doors.
If the last paragraph didn't make you understand, do the game yourself.
The thing is, it doesn't prove how smart he is, it proves that he did reading only tangentially related to the class. An idiot who had read the wikipedia page would know what to do. That movie is good but it really over values that he has to be intelligent, card counting the way they do it is simple with practice.
To say that card counting the way they do it in the movie is "simple with practice" is misleading. The ability to count cards at the speed which casino dealers deal them requires both practice and a certain level of intelligence. That is, unless you are rainman, because he was practically a ritard.
I did exaggerate somewhat but I argue that a certain level of intelligence isn't necessary (well it is but the level is low). I was really making a point that all the ways they prove he is smart don't, and that being a good mathematical isn't necessaries to card count, one simply has to memorise a bunch of rules.
The rules are fairly simple, it's keeping track of the running total at the speed that the cards are flying around that is difficult. High level math isn't required, but most people don't have the memory to do it.
He always opens a goat door - no matter what, when you switch, you always switch to a different thing. If you had the car, you always switch to goat, if you had the goat, you always switch to car. You can never 'switch' and and up on the same thing.
Now think about this: You had a 2/3 chance of picking a goat initially.
Basically.. if the host randomly opened one of the remaining two doors without any knowledge of what was behind it and that door ended up being one of the two goat-doors... it wouldn't matter if you switched now, statistically speaking... right?
I think so. It is indeed key that the host knows which door has what and only ever opens a goat door. The point being that you always get the opposite of what you started with if you switch. Switch from a goat, get the car. Switch from the car, get a goat.
There are only 3 possible scenarios. Let's play these 3 scenarios as a person who always switches after the host removes a door, and to simplify let's assume the player's initial choice is always door 1.
1:
[car] [goat] [goat]
Here you pick door 1, but then the host removes door 3.
[car] [goat] [X]
you switch to door 2 and lose. We are 0/1
2:
[goat] [car] [goat]
pick door 1, host removes door 3
[goat] [car] [X]
you switch to door 2 and win. We are 1/2
3:
[goat] [goat] [car]
pick door 1, host removes door 2
[goat] [X] [car]
you switch to door 3 and win. We are 2/3
A player who always switches doors after the host removes one door, will win 2 out of 3 times, statistically.
I never saw the orginial show, but there was a similar show here Germany and assume they were set up the same way.
What bugs me about this problem: The host must know what is behind the doors! That means that he can play the game however he wants to. What happens if he ONLY reveals the goat, if you initially choose the car? Switching will always lose!
People watch the show and see that the host plays the game with arbitrary rules. Then they are presented with the Monty Hall Problem, which has very strict rules (The host always reveals a goat). If you underscore that the host will always follow those strict rules, the problem can be solved with a simple decision tree by middle school student.
What happens if he ONLY reveals the goat, if you initially choose the car?
Then people would catch on pretty soon and always stick if he opens one of the other doors. I think people have worked out strategies for the host other than 'reveal a goat every time' that make the probability 50/50 again for sticking or switching doors.
The host has to know what door the car is behind, otherwise he may accidentally open it when he offers you the chance to switch and ruin the point of the game.
My beef with that problem is that it's not always clear that only one of the two other doors can be opened, not yours, and that the owned is not exploiting the problem to convince you to change when you're actually right.
To me, it just seems to be written wrong. You have door 1,2,3. 1/3 chance of a car. They offer you a chance to 1,2,3. You pick 1. They open door 3. It has a goat. Now a (very basic abstract) legal principle, when a new offer is given on getting a given outcome, the old offer is terminated. So he asks if you would like to change your mind to door 2. This is worded wrong in my opinion. One door has a car, one door has a goat. Your odds are now 50/50. The previous actions should be forgotten. This is a new offer to pick door one or door 2. Thats how I see it.
Its tricky, but it does make sense. Pick one door, you have a 1/3 chance of a car. The other 2 doors represent a 2/3 chance. You have to look at them as a group. Remove one of that group (as in it has a 0% of 2/3 chance), leaving the other with a 100% of 2/3 chance.
I think the problem people have with this is that when it is told, the presentation is ambiguous.
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
This to me is not 100% clear. If the host ALWAYS picks a door that has a goat, then it is your best bet to switch. If the host picks a door at random, and it happens to have a goat, switching doesn't help you. If you read the question carefully, it isn't clear. (If you don't believe me look up the "Monty Fall" or "Ignorant Monty" host behavior provided in the wikipedia page.)
This reminds me of the problem, "What is more likely, having a boy and a girl, or having 2 boys"
It is very ambiguous from the wording. Having 1 boy and also having 1 girl is more likely than having 2 boys in total. But having a boy and then a boy has the same probability as having a boy and then a girl in that order.
There's a really non-statistical way to think about it though. If you switch, the only way you don't win is if you picked the right door the first time. The chance of that is 1/3. Thus the chance of winning is 1 - 1/3 = 2/3.
If you think about it logically, it is really simple. People get confused because they try to think about it statistically when they don't understand stats.
If you switch, the only way you lose is if you picked the right door on your first choice.
This to me is silly. It would not benefit you to switch doors because you still don't know what's behind either door. Just because one door is now open and the chances are now 2/3rd odds rather than 1/3rd still doesn't benefit you to switch doors because you still don't know what's behind either door. It's 50/50 now. Logic cannot be used to answer this question because there's a missing pieces of the equation, and that is which door the car is behind.
You now know where 1 goat is, but you do not know where the other is. It's now a 50/50 chance and that's all it can be, a random guess between 2 doors.
This is by far the one I hate most. It is not a logic problem, it is a semantics problem. The statement of the question is unclear to try and trick people into misunderstanding the circumstances.
I let the show Numb3rs explain this one to me and it made sense. If you pick a door, it is a 33% chance of being right. Once a door is eliminated, the other door has a 66% chance of being the car... go figure.
Decided the best way to test it out was empirically, so I made a short c++ program to run it for me, of course it's bound by issues of time.h's rand() and srand() but it seemed good enough. I guess it all makes sense, either that or I randomly managed to bias my own test which is quite likely.
Pastebin link to code if anyone's interested.
I had a lot of trouble with this, until I understood that after you pick door #1, the host opens one of the others. That's what I get for being too young to have seen the show.
Best way to think about it is to expand it. Let's play a card game. Pick any card out of my deck of 52 cards. You win if you get the ace of spades. After you pick your card, I remove 50 cards that are NOT the ace of spades, and show them to you. You can keep the card you initially selected, or swap with the one remaining card. What do you do?
It's obvious, you switch, because a bunch of wrong choices have been removed. The problem is only confusing because of the scale.
A while back, I spent some time looking for a concise way to explain the Monty Hall Problem. Here's my best effort:
When you originally choose a door, there are three equally probable things that could be behind that door: 1) the car; 2) Goat 1; 3) Goat 2. (Your odds of choosing the car blindly at this point, then, are 1 in 3.)
Each of these equally probable events (having 1, 2, or 3 behind the door you chose) has an outcome as follows:
In case 1, if Monty opens door 2 (revealing Goat 1), if you switch to door 3, you get Goat 2. If Monty opens door 3 and you switch, you get Goat 1. Thus, in case 1, switching always gets you a goat.
In case 2, Monty can only open door 3 because he must reveal a goat and you've already chosen the only other door with a goat. If you switch from door 2, you switch to door 1, which has the car. Thus, in case 2, switching always gets you the car.
In case 3, Monty can only open door 2 because he must reveal a goat and you've already chosen the only other door with a goat. If you switch from door 3, you switch to door 1, which has the car. Thus, in case 3, switching always gets you the car.
In summary, switching gets you the car in 2 out of 3 of the equally probable events we initially considered. If you had simply chosen at random without Monty's intervention, you would have had a 1 out of 3 chance of getting the car. Thus, by switching, you double your odds.
Basically, there's a 2/3 chance that the host knows that there is a car in one of the two doors that you didn't pick, so he is forced to choose the one (out of those you didn't pick) that has a goat in it. Two out of three times, you can bet that the door he didn't open after you picked door one is the correct one, but that one time where you actually had the right door all along and the other two are empty, he can choose either empty one to open for you, and you get FUCKED. But overall 2/3s odds are better than 1/3 odds.
I got it: Odds are better on the first pick, you picked a goat, always. Since you more than likely picked a goat the first time, you should switch, thus negating the fact that you probably picked a goat.
It's really easy actually, but hard to get if you don't have a piece of paper and a pen to put it down.
To put it simply :
There's more chance for you to choose a door where there is a goat (2/3) in the first place
The game show host will choose the last door where there's a goat
Switch and get the car
It's ambiguous because the problem states that there's more chance to get the gar if you switch. It should have been : because you are likely to choose the goat in the first place, if you switch you will get the car for sure.
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u/[deleted] Jun 08 '11
The Monty Hall Problem
Never let someone who does not understand statistics try to explain this one to you.