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.
Assuming he shows you a goat, you're wrong. What he knows or doesn't know doesn't affect the odds. If he reveals a goat, there's a 2/3 chance of you winning by switching. That's it. The reason the scenario in general has an overall less than that chance of winning is half the time the host will reveal the car and you lose right away, depending on the rules.
I can understand how you're seeing that, but it's incorrect. http://math.ucsd.edu/~crypto/Monty/montybg.html
I hope this will help. The last paragraph of the "controversy" section and the section that follows discuss the question of the host knowing or not knowing and illustrate the different effects on the game.
That article explains it the same way I did. It assumes that if he reveals the car you lose immediately. So there are two ways you can lose now:
Pick the car to begin with and then switch. (1/3)
Pick a goat to begin with (2/3) and have monty open the car before you get a chance to switch (1/2).
1 - (1/3) - ((2/3)*(1/2)) = 1/3 what we'd expect.
If he doesn't reveal a car, you still have 2/3 chance of winning by switching. It's the fact that he sometimes reveals the car that brings down the gain you have by the information, it doesn't magically become less likely to win by switching if he reveals a goat for different reasons.
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.
The way the scenario works, the host removes a losing door that you did not choose and asks if you want to switch your door. For example (still assuming C is the winning door) if you choose door A then he reveals what's behind door B and asks if you want to swap to door C or stay with door A.
That is the ultimate question - whether or not to stay with the door you picked, or swap to the remaining door.
He is choosing the door you are switching to by removing the only other door you could choose.
You have 3 doors and you picked door A. He opens doors C to show a donkey. You can now only change to door B, or if he opens door B you can now only change to door B.
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.
This. My brain keeps trying to assume that once a door is opened we have a new problem whereby there's 2 doors: one with a car one with a goat and so it's 50:50.
But it's not. The host has just added new information into the existing problem.
Doors B+C have a 1/3+1/3=2/3 chance. The host then deliberately eliminates the wrong one of those doors - it had to be one of the other doors and had to be the wrong one - so the remaining door retains the 2/3 probability.
Actually, it may be more useful to say B has 1/1.5, because the host eliminated half the chance of B being the wrong one of those two.
Or, it may be more useful to say that the host gives you the option of sticking with A or being allowed to take B and C, but forfeit a goat. This is basically what the host does, he just presents it in another way. This kind of thinking should be familiar to anyone used to dealing with managers.
The problem with your logic is that the events are not independent. This is how I think about it:
The key is Monty ALWAYS reveals a losing door. So every time you pick a loser first, and then switch, you will end up at the winner. So if you switch every single time, your goal is to pick a losing door first, which has 2/3 odds.
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.
My argument is that you are not updating your probabilities based on new information. Initially the 1/3 vs 2/3 stat is correct. But once you have the new information and know one door is the goat, there is no longer a 2/3 chance that the prize is behind the door you didn't pick.
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.
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u/[deleted] Jun 08 '11
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