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> 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. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you “Do you want to switch to Door Number 2?” Is it to your advantage to change your choice?
The correct answer is that it is in your advantage to switch; it doubles your probability of getting your car from 1/3 to 2/3.
Now can anyone explain to me why? I looked the problem up, at wikipedia and elsewhere, and I just can’t seem to grasp this concept. I think the explanation I am closest to understanding is this one:
![](http://wpcontent.answers.com/wikipedia/commons/thumb/9/9e/Monty_open_door_chances.svg/197pxMonty_open_door_chances.svg.png)
By dividing the choices into two groups, you see that you have a 1/3 chance to have selected the door with the car, and a 2/3 chance of switching to it. It makes sense logically, but I am still having trouble seeing how you can reach this solution.



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I don’t get it, though. Once the door has been opened to show that #3 is a goat, you have a 50/50 chance of getting the car regardless which one you choose.
If you have three slips of paper, two white, and one black, in a hat, and you pull out one white one… you’re left with one white and one black. 50/50. No matter how you turn it.
If there _is_ actually a difference, I would say that it has to do with the fact that he _has_ to show you the goat when the two remaining doors are goat/car. That’s the only difference between this and the slips of paper as far as I can see, though I still don’t see how it’s significant.



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> If there is actually a difference, I would say that it has to do with the fact that he has to show you the goat when the two remaining doors are goat/car. That’s the only difference between this and the slips of paper as far as I can see, though I still don’t see how it’s significant.
Yeah, because there is no way of telling if he picked it randomly and if he picked the one knowing the other is a car. I guess you could try to determine it by the way he chooses them. :P



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I’m looking at it this way. If you switch, you are guaranteed to get the opposite of what you selected. You have a higher chance of starting with a goat (2/3), and thus the same chance (2/3) of switching to something other than a goat (a car). I found a nice illustration using a diamond instead of a car:
![](http://math.ucr.edu/~jdp/Monty_Hall/Monty_Hall_a.GIF)
In other words, the thing you are mostly likely to select is the thing you are least likely to switch to. Since there are two goats but only one car, you are always better off switching.



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If you pick one door randomly out of 3, you have a 1/3 chance of choosing the door with a car behind it.
If you chose the door with the car, and stay with it, you win. If you chose a door with a goat, and stay with it, you lose. Thus, since you had a 1/3 chance of choosing the door with the car, you have a 1/3 chance of winning if you stay.
Conversely, if you choose a door with a goat and switch, you win. If you choose the door with the car and switch, you lose. You had a 2/3 chance of choosing a door with a goat, so switching gives you a 2/3 chance of winning.
It’s a lot more intuitive if you add more doors: Say, 1,000 doors, you pick one, the host opens 998 revealing all goats.



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Hmm. That’s interesting. So it’s the switching that makes it different than the paper slips.



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Yeah. It’s still a bit hard to understand, but I can tell it makes logical sense.



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> *Originally posted by **[SaintAjora](/forums/9/topics/57082?page=1#posts1301261):***
>
> Yeah. It’s still a bit hard to understand, but I can tell it makes logical sense.
I mean, think of it this way. Assume you choose one of the goats (2/3 chance.) Monty Hall would proceed to eliminate the other goat, leaving only the car, and switching would always lead to the car. Otherwise, if you choose the car at first (1/3 chance), switching will always result in a loss.



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I get it now, I think the confusion arises because it is easy to assume that the probability changed just because the number of doors changed.



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Your paper slips don’t apply at all. Envelopes with cards in them would be more appropriate.
It’s easier when we ignore door numbers and give each of the items variables. Goat A, Goat B, and Car C
1/3 Chance that player chooses A, switches to C
1/3 Chance that player chooses B, switches to C
1/3 Chance that player chooses C, switches to A/B
2/3 chance that player switches to C.
1/3 chance that player chooses C.
Another way of explaining it is that the only time the host has a choice is when you have picked the car. Most of the times (2/3), he is forced to reveal the car’s location by removing the only goat he can.



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try to imagine this. instead of 3 doors, we take 1337 doors, 1336 goats and 1 car.
now you pick one.
the monty opens every door except the one you chose and one other door.
now the odds of you having the car are 1/1336. the odds of the car standing behind the other door are a lot bigger.



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> *Originally posted by **[kirdaiht](/forums/9/topics/57082?page=1#posts1302209):***
>
> try to imagine this. instead of 3 doors, we take 1337 doors, 1336 goats and 1 car.
>
> now you pick one.
> the monty opens every door except the one you chose and one other door.
> now the odds of you having the car are 1/1336. the odds of the car standing behind the other door are a lot bigger.
if you have 1337 doors then there is OVER 9000 probability



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You can view this problem in a number of ways.
1. This option is rather simple, and it removes the imaginary “chance” that is suggested here. You only look at the situation _after_ a door with a goat behind it has opened. You start with a clean slate, and your choice of before does not matter, at all.
There is an irrelevant object in your choice, namely a door with a goat behind it. After that, you have a choice between two doors. One door features a goat behind it, the other door features a car behind it. No matter how you look at it, there is a 50% chance you get the car. In this option, I’m not even sure why you would want to look at the case before this. It’s irrelevant. The supposed 2/3 chance is an illusion to make you think you you have a high chance to get the car. The 1/3 chance choice you make in the beginning is simply irrelevant, since you get a second choice after that, which completely negates the first choice.
2. With this option, you _only_ look at you going to switch 100% of the times. In other words, you do not have the choice to switch, you just will. At this point, we’re not ignoring the former situation, but we’re not entirely calculating with it.
You have a 1/3 chance to pick any door. 2/3 chance to pick a goat and 1/3 chance to pick the car. Picking a goat will make sure the host _has_ to show the other goat and thus leaving the car open. This happens 2/3 of the times, and thus when switching you will always get the car. 1/3 of the times you will already have the car and switching will thus give you a goat. So, when always switching, there will be a 2/3 chance to get the car.
3. This is the opposite of number 2. Here you do not switch at all. Basically, in this option, the giving of a second choice to pick is irrelevant, just as the opening of a door with a goat is. You won’t switch any way. You’ll always have a 1/3 chance to get the car.
4. I’ll try to combine the chances now, but not sure if it will come over. I’ll keep it simple and short.
Each door has a 1/3 chance to be picked. After one other door is opened, you have exactly 1/2 chance to get one of the doors. You have a 1/3 chance to pick a car and, when another door has opened, you have a 1/2 chance to get a car. On the other hand, you have a 2/3 chance to pick a goat and, when another door has opened, you still have a 1/2 chance to get a car. _In other words, whichever door you choose, you will always have a 50% chance to get a car after the third door was opened._ This is important. Why? Because I’m arguing what you picked first is irrelevant.
I can make this clearer by the example of having to choose one door out of a thousand. 999 have goats, 1 has a car. When you choose one door, no matter what it is, the host will reveal 998 goats. Whatever was under your door now either is a goat or a car and you now have a 50% chance to get the car.
The chance to get the car is not increased by switching. The chance to get the car is increased due to the host opening up doors that are goats. Yeah, I know I could have just posted that, but I felt like doing the calculations.
EDIT: Basically, this whole issue dwells on the assumption you’re in some state after the goat has been unlocked which makes “switching” look more profitable. But it really is just a fancy and meaningless word. The case before the revelation is simply irrelevant, because an entirely different case has been opened for you.



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Your method is only a way to describe it, and your final statement is true, but you do not have a 50/50 chance at any point, even with 999 goats and 1 car.
999/1000 chances that player chooses a goat and switches to the car.
1/1000 chances that player chooses the car and switches to a goat.
There are only two possible outcomes, but your initial choice does effect what would normally be 50% probability as in your first example. Your initial selection had a pool that favored goats, but your second selection favors that goats have been eliminated, so there is a higher 2/3rds (or 999/1000) chance that the car is in the remaining door. It has nothing to do with 50/50 because 2/3 times the doors are purposefully eliminated (since the only option for the dealer is to remove the other goats) and only when you’ve miraculously selected the car is the selection randomly resized.



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I guess my primary argument is that we shouldn’t look at the act of switching at all, but simply at the fact that 998 choices are removed from the picture. At the very least, it should not matter how many doors with goats we have, because all except one are eventually removed any way. The number of goats is irrelevant and thus the chance of getting the car when switching is as well.
You will always end up with a door having a goat behind it, and a door having a car behind it. More importantly, having chosen that _specific_ door with the goat is exactly of equal chance to having chosen the door with the car. In other words, you have two doors in front of you with equal chances of having chosen those specifically. With the irrelevant goats out of the way, there is only the matter of choosing between either.
As I said, giving you the fancy act of switching doesn’t change any probabilities. Only the removal of doors featuring goats does.



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Read ‘The curious incident of the dog in the nighttime’. It explains the monty hall problem perfectly.



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Right, I’m agreeing with that statement Darkruler, I’m just disagreeing that it’s a 50/50 chance at any one point ever.



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i will find the book now and explain the monty hall problem.



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You are asked to choose a door. door 1, (door with goat behind it) You change: you get a car. You stick: you get a goat. Door 2, (door with goat behind it) You stick:you get a goat. You change:you get a car. Door 3, (with car behind it) You stick: you get a car. You change:you get a goat.
This proves that if you change there is a 2/3 chance that there is a car behind the door if you change. this is illustrated in: [http://images.google.co.uk/imgres?imgurl=http://derivadow.files.wordpress.com/2007/01/montyhallproblem.png&imgrefurl=http://derivadow.com/2007/01/16/themontyhallproblemwinacarnotagoat/&usg=\_\_TzCoVlraPDJCl8tx8b\_F3PrzQ64=&h=294&w=387&sz=34&hl=en&start=2&tbnid=2dV7AERDB0Dg1M:&tbnh=93&tbnw=123&prev=/images%3Fq%3Dthe%2Bmonty%2Bhall%2Bproblem%26gbv%3D2%26hl%3Den%26sa%3DG](http://images.google.co.uk/imgres?imgurl=http://derivadow.files.wordpress.com/2007/01/montyhallproblem.png&imgrefurl=http://derivadow.com/2007/01/16/themontyhallproblemwinacarnotagoat/&usg=__TzCoVlraPDJCl8tx8b_F3PrzQ64=&h=294&w=387&sz=34&hl=en&start=2&tbnid=2dV7AERDB0Dg1M:&tbnh=93&tbnw=123&prev=/images%3Fq%3Dthe%2Bmonty%2Bhall%2Bproblem%26gbv%3D2%26hl%3Den%26sa%3DG)



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> Right, I’m agreeing with that statement Darkruler, I’m just disagreeing that it’s a 50/50 chance at any one point ever.
I’ve just thought it over some more and I’m agreeing now there’s no 50% chance for the car.
I don’t think this is a possible proof, but it worked helping me to understand the problem, so I’ll just post it any way. Taking an infinite amount of doors featuring goats and 1 door featuring a car, your choice of a door will always involve a goat (I know, I can’t mix them like that, because I’m setting the chance to get a goat equal to 100%, while there is still a door with a car). With the goat chosen, the host will remove all other goats and leave your door (with the goat) along with the door featuring the car. Since you had a 100% chance to get a goat, you will _always_ switch to the other door, it will feature the car.
If that’s an impossible proof, you can take any randomly chosen high number. A trillion will be fine. Choosing a door now borders on having an infinitely small chance to choose the car the first time through, so you’re almost at 100% chance to get a goat every time you play it. The host removes the remaining goats, and you will know (trusting the host on his word, of course) that the other door features the car. Switching, in this case, seems to indeed be the preferred option.
However, looking at this in a more simple way, take the same example but now you not choosing a door beforehand. Every time the host opens a door with a goat, your chance to pick the car increases (let’s assume you can shout stop any time and make a pick). When there’s still 100 doors remaining, you have a 1% chance to pick a car. 10% chance at 10 doors remaining. 25% chance at 4 doors remaining. And, whooptedoo, 50% chance at 2 doors remaining. This is more of a sidenote than anything in response to the topic, though. I was just showing where I got my numbers.
EDIT: The above image confuses me somewhat. It seems to imply there’s 6 situations you can end up with, 50% of those getting a car, the other 50% getting a goat. This remains the same when adding another door with a goat. I guess this is somewhat like my above example.



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> No, there is not a 1/2 chance of getting a car. there is a 2/3 chance. is anybody gonna follow what i say, or just ignore me?
Did you even read my post?



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But the above image shows that there is a greater chance of getting a car if you change.



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> But the above image shows that there is a greater chance of getting a car if you change.
What I was saying is that if everything was randomly picked with you not having a choice, you’d have a 50% chance to end up with the car. Thinking logically removes the randomness and gives you a greater chance of winning the car while switching. My point is, though, that this is not because you switch, but because the goats are being removed.



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I love this problem. So easily stated and yet such an unexpected result. :)
In any case, just to add a very short bit of insight, the difference is that, unlike a true random selection, Monty has advanced knowledge of what’s behind the doors. By selecting a goat, you _force_ him to reveal the other goat, thus clarifying which of the other two doors has a car (if a car was available).



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I remember encountering this.
This explanation cleared it up for me:
[http://www.nytimes.com/2008/04/08/science/08monty.html](http://www.nytimes.com/2008/04/08/science/08monty.html)


