Monty Hall Problem

[calvnandhobbs68]

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If you think like sEgacious, EVERYTHING is 50/50! There is a 50 percent chance of rain today! There is a 50 percent chance of getting into Harvard for crying out loud as long as you apply!! You either get in or you don't! 50%, right?!?

Ah but this is the best part of these kinds of problems. They force people to look at the overarching situation and the events that led to the decision they are making now, rather than just focusing on the decision at hand (which is what most people will naturally do). Helps you to understand that if things were even a little different in the problem, the probability changes.

 

[Diallelus]

Why does this matter? After a goat is revealed, all that tells you is that the revealed door is no longer valid, nothing else.

No, it also tells you that the other door, being a member of the set of doors (the "not chosen" set) with a 2/3 probability of having the car, now has a 2/3 chance of being right all on its own. You always knew there was a dud in that set, and its removal changes nothing.

Imagine a box with 1 gold coin and 2 copper coins. You remove a coin. The box now has one copper coin, and second coin with 2/3 odds of being gold, and 1/3 odds of being copper. The removal of a copper coin from the box does not change the odds for the other coin. The box retains 2/3 odds of having a gold coin, and you should take it.

 

[yankeesmed]

I had a professor that would give this problem to every class he had. After 20ish years of teaching, only one student ever figured it out on his own.

To those that are insisting that after opening the door with a goat, there is still a 50:50 chance you will get a car regardless of if you switch, make three cards, one with a car and two with goats. Grab a friend. Perform experiment. Have friend switch for 100 trials, stay for 100 trials. You'll see.

As people who will be entering the field of medicine, we should understand this. The problem proposed is very simple to construct in your head and see that if you did it 10, 100, 1000 times, that switching would lead to MORE successes. It wouldn't always lead to success, but it would lead to around 66 success vs. 34 failures. So, using an unconventional technique that doesn't quite make sense in your head at first (switching your choice) will lead to success more often.

There will still be failures, but less of them than if you did not have the knowledge you were given by the host.

When it applies to a treatment, riding the probability is sometimes the best you can do. It still will not guarantee success for your patient. You have to use all available knowledge that you have to raise your probabilities.

 

[aSagacious]

If you brought an entirely new contestant in who was not aware of which door was originally chosen, that contestant has a 50/50 shot between the two remaining doors.

This is where I'm getting tripped up. What information is the original contestant privy to that the new contestant is not?

 

[yankeesmed]

This is where I'm getting tripped up. What information is the original contestant privy to that the new contestant is not?

The contestant isn't privy to new information, he is just choosing between 2 doors, not 3.

EDIT: Oh sorry, I kind of misread this at first.

The original contestant knows that the door he didn't choose has a 2/3 chance of being the right one while the new contestant only knows it has a 50% chance.

 

[yankeesmed]

If the second contestant was watching a feed of the game show and is aware of everything that the original contestant is, then he should pick the door the original contestant didn't pick.

 

[rhesuspieces]

This is where I'm getting tripped up. What information is the original contestant privy to that the new contestant is not?

The original contestant knows which door he originally chose. The statistical advantage is in switching from one door to another.

If contestant 2 comes along and has no clue which door contestant 1 originally picked, he has no statistical advantage.

 

[rhesuspieces]

As people who will be entering the field of medicine, we should understand this. The problem proposed is very simple to construct in your head and see that if you did it 10, 100, 1000 times, that switching would lead to MORE successes. It wouldn't always lead to success, but it would lead to around 66 success vs. 34 failures. So, using an unconventional technique that doesn't quite make sense in your head at first (switching your choice) will lead to success more often.

There will still be failures, but less of them than if you did not have the knowledge you were given by the host.

When it applies to a treatment, riding the probability is sometimes the best you can do. It still will not guarantee success for your patient. You have to use all available knowledge that you have to raise your probabilities.

And this is why we do clinical trials! Lesson over, students, see you all next Tuesday and don't forget that your essays on the placebo effect are due Friday at midnight!

 

[aSagacious]

The host is letting you choose both doors, in a sense, because one of the two doors not chosen was bound to be a no-prize door anyway. If he had let you open both doors you would have a 1/3 chance of winning with the first door, and a 1/3 chance of winning with the second door, but one of them had to be a no-prize door, and it doesn't matter whether he shows you which one before or after you choose to switch (as long as he means switching to both doors).

... Just clicked . Thank you everyone for tolerating my stubbornness (except for Yankees, he was pretty quick to invoke Burnett's Law 🙂)

 

[TooMuchResearch]

If you think like sEgacious, EVERYTHING is 50/50! There is a 50 percent chance of rain today! There is a 50 percent chance of getting into Harvard for crying out loud as long as you apply!! You either get in or you don't! 50%, right?!?

There is a 50% chance I'll catch a stray bullet if I stay home all day. Just kidding; it's more than 50%. I live in a terrible neighborhood.

Still, I either get shot or I don't.

 

[Narmerguy]

What's hard about this?

1. You pick a door
2. There is a 66% chance that one of the other two doors is correct
3. He removes a wrong answer from the two other doors.
4. There is still a 66% chance that one of the two other doors is correct, but now you know which of the two doors it can't be so it must be a 66% chance that the only untouched door is the correct one
5. 2-to-1 chance that if you switch you'll get the right door.

 

[TooMuchResearch]

As people who will be entering the field of medicine, we should understand this. The problem proposed is very simple to construct in your head and see that if you did it 10, 100, 1000 times, that switching would lead to MORE successes. It wouldn't always lead to success, but it would lead to around 66 success vs. 34 failures. So, using an unconventional technique that doesn't quite make sense in your head at first (switching your choice) will lead to success more often.

There will still be failures, but less of them than if you did not have the knowledge you were given by the host.

When it applies to a treatment, riding the probability is sometimes the best you can do. It still will not guarantee success for your patient. You have to use all available knowledge that you have to raise your probabilities.

If the paper suggesting the treatment has a higher success rate needs to use a mathematical proof to explain their data, you'd be well advised to check the conflict of interest section of the paper. It isn't that tough to look at a study and decide if it sucks.

 

[yankeesmed]

... Just clicked . Thank you everyone for tolerating my stubbornness (except for Yankees, he was pretty quick to invoke Burnett's Law 🙂)

sEagacious, I love your outlook on lotto. I hope Burnett's law is a keen Yankee's reference and not some scientific term I missed out on since I'm an English major. If it's the Yankees reference, good one!

 

[aSagacious]

sEagacious, I love your outlook on lotto. I hope Burnett's law is a keen Yankee's reference and not some scientific term I missed out on since I'm an English major. If it's the Yankees reference, good one!

Actually, I was pretty happy to hear that AJ Burnett made the starting rotation this year... Chamberlain was painful to watch 🙄

But here's what I was referring to: http://forums.studentdoctor.net/showthread.php?t=528658

 

[boo5000]

What's hard about this?

1. You pick a door
2. There is a 66% chance that one of the other two doors is correct
3. He removes a wrong answer from the two other doors.
4. There is still a 66% chance that one of the two other doors is correct, but now you know which of the two doors it can't be so it must be a 66% chance that the only untouched door is the correct one
5. 2-to-1 chance that if you switch you'll get the right door.

For those of you who feel this way, never teach high school! If I had a dollar for every time I beat my head against a brick wall...

It took a while, but an different perspective made it click. That's how we learn!

 

[Elpenor]

I'm not sure why people are confused by this. If the host always opens a wrong door it is better to switch

if the host opens a random door it doesn't matter

learn up some statistics / logic, bro's

 

[ReptarBar]

With respect, I'm not sure that's right. I had a teacher who explained that the odd probability was due to the fact that the host is actually, in a roundabout way, giving you BOTH of the other doors by allowing you to switch. If he had allowed you to pick both of the other doors, and ignore the one that was wrong, then the odds would clearly be 2/3, and that is actually exactly what he's doing.

If it were possible for the host to reveal a prize the odds should be unchanged (after he fails to do so, of course, but before that the chance that he gives away the game actually helps the player).

Say the player chooses door 1 and the door 3 is the winning door. There's a 50/50 chance the host will give away the prize door, and then the player should obviously switch, and a 50/50 chance the puzzle will proceed as normal, with the player having a 2/3 chance by switching. The intentionality of the host, and the previous chance that he would reveal the prize, are now irrelevant.

wrong. the host's knowledge of where the prize is...is in fact the crux of the entire monty hall problem. map out the possibilities and you will see that if the host could open a correct door, the chances of winning by staying and switching are equal.

here is a crude picture i drew. remember, the key is that the host only removes a wrong door for you.

 

[CaptainSSO]

I'm not sure why people are confused by this. If the host always opens a wrong door it is better to switch

if the host opens a random door it doesn't matter

learn up some statistics / logic, bro's

Yes, exactly. The fact that the host is not randomly opening the doors is why this problem works the way it does. He is essentially condensing two doors down into 1. If a tornado came and opened one of the doors, odds would be 50/50.

 

[Spin]

For the love of God, I am so angry. My math has been violated, and I just want to go hit people out of spite toward this problem. If one of you math saavy people can explain how this doesn't violate the rules of the universe without using a convoluted proof that will make me want to choke you, I will forever be in your debt.

http://en.wikipedia.org/wiki/Monty_Hall_problem

If you don't know what this is, don't look. Save yourself the cognitive misery.

At first, each door has 1/3rd probability of having the car behind it. You pick Door A, which still has 1/3rd chance of having the car behind it. The total probability of the car being behind Door B or Door C is 2/3rds. When the host opens Door C and shows that that particular door doesn't have the car, the 2/3rds probability of the car being behind B or C is "condensed" into just door B.

Recap: Door A had 1/3rd probability. NOT-Door-A had 2/3rds probability. Eliminating a door just made it so that "NOT-Door-A = Door B," rather than "NOT-Door-A = Door B + Door C."

Does that make sense? I'm not sure if there were any other good explanations -- I didn't want to confuse myself by reading the rest of the thread -- but that's how I think of it.

 

[Torr]

At first, each door has 1/3rd probability of having the car behind it. You pick Door A, which still has 1/3rd chance of having the car behind it. The total probability of the car being behind Door B or Door C is 2/3rds. When the host opens Door C and shows that that particular door doesn't have the car, the 2/3rds probability of the car being behind B or C is "condensed" into just door B.

Recap: Door A had 1/3rd probability. NOT-Door-A had 2/3rds probability. Eliminating a door just made it so that "NOT-Door-A = Door B," rather than "NOT-Door-A = Door B + Door C."

Does that make sense? I'm not sure if there were any other good explanations -- I didn't want to confuse myself by reading the rest of the thread -- but that's how I think of it.

Lol. I've avoided this thread while at work cause I knew it would confuse the heck out of me and make me angry. This version makes sense though.

Thanks spin!

 

[Suncrusher]

Someone in my highschool calc class got the answer right

It is super counterintuitive and I love it.

Spin and Namerguy's explanations are really good! I understand where people are coming from with the whole 50% chance thing, but stop trying to look at switching as a brand-new choice between the two remaining doors.

There was a 2/3 chance that you picked wrong originally and should switch to one of the other two doors, but now the extra information eliminated one of the other two doors, so now there is a 2/3 chance that you picked wrong originally and should switch to the other ONE door. Switching = 2/3 chance of winning. Not switching = you still have your original 1/3 chance of winning. The goats/car are still in their original positions even after the new information is revealed; they are not re-randomized. But it is so intuitive to think that you are facing a brand-new choice between the two remaining doors..!

 

[Kgizzle]

Burnett's Law= Awesome!

 

[Diallelus]

wrong. the host's knowledge of where the prize is...is in fact the crux of the entire monty hall problem. map out the possibilities and you will see that if the host could open a correct door, the chances of winning by staying and switching are equal.

You're right. I misinterpreted one of the rules. The interesting thing about what you're saying is that it's true, but only because the "official rules" say that, in a "host has no knowledge" variant, the prize door being revealed results in the whole game resetting (no win, no loss).

If the player were allowed to switch his door like normal when the prize was revealed (and pick the revealed prize), the normal 1/3 to 2/3 odds stand. If the player, as per the accepted rules, must restart the game when the prize is revealed, then the odds indeed revert to 50/50. And, of course, if the game continues as normal, and the player must choose the original door or the unopened one, then the player always loses.

My work, assuming door 1 is the prize door (equivalent scenarios could be drawn for the other two possibilities):

1. Player chooses door 1
a. Door 2 opens (revealing no-prize)
___i. Player Switches: LOSS
___ii. Player Sticks: WIN
b. Door 3 opens (revealing no-prize)
___i. Player Switches: LOSS
___ii. Player Sticks: WIN
2. Player chooses door 2
a. Door 1 opens (revealing prize)
___i. Player Switches: WIN 😎
___ii. Player Sticks: LOSS
b. Door 3 opens (revealing no-prize)
___i. Player Switches: WIN
___ii. Player Sticks: LOSS
3. Player chooses door 3
a. Door 1 opens (revealing prize)
___i. Player Switches: WIN 😎
___ii. Player Sticks: LOSS
b. Door 2 opens (revealing no-prize)
___i. Player Switches: WIN
___ii. Player Sticks: LOSS

Overall WINS: Switch: 4, Stick: 2
If WINS marked with a 😎 are removed: Switch: 2, Stick: 2

As you can see, the normal 1/3 to 2/3 odds apply even when the host can reveal either door, assuming play continues and the player is free to choose the open prize door. If play stops when the prize is revealed, then the odds of winning are even. The rule about stopping play when the host gives away the game essentially robs the player of otherwise winning conditions.

 

[random cell]

How about this-since what is behind the doors is set and doesn't change you can do this:

Door 1: Goat
Door 2: Car
Door 3: Goat

If you pick Door 1, then the host will open door three and switching will cause you to win. Same if you pick Door 3. If you pick Door 2, switching will cause you to lose. So switching gets you 2/3 of the possible winning combinations.

Keep in mind too that the question isn't "What is the probability that the car is behind this door?" but rather "What is the probability that switching is the winning strategy?"

This is a good explanation. What don't you understand about this?

 

[hiyaman]

[YOUTUBE]http://www.youtube.com/watch?v=P9WFKmLK0dc[/YOUTUBE]

Professor Epps!

 

[Spin]

Lol. I've avoided this thread while at work cause I knew it would confuse the heck out of me and make me angry. This version makes sense though.

Thanks spin!

Thanks! I mostly just copied the wikipedia article

 

[TooBright]

People who suck at math make me laugh.



This thread made me laugh.

 

[Laplace]

Too lazy to read the entire thread to see if anyone explained it properly but here's my explanation:

Initially, all doors are 1/3 probability, so the one you pick is also 1/3.

Now, the doors you haven't picked, each having a probability of 1/3 sums to 2/3. Since the one of the doors is opened and it does NOT contain the prize, the entire probability of 2/3 falls upon the un-opened door that you didn't choose.

 

[Spin]

Too lazy to read the entire thread to see if anyone explained it properly but here's my explanation:

Initially, all doors are 1/3 probability, so the one you pick is also 1/3.

Now, the doors you haven't picked, each having a probability of 1/3 sums to 2/3. Since the one of the doors is opened and it does NOT contain the prize, the entire probability of 2/3 falls upon the un-opened door that you didn't choose.

👍 As expected from someone who goes by Laplace.

 

[calvnandhobbs68]

People who suck at math make me laugh.



This thread made me laugh.

People who make a thread asking why you really want to be a doctor for the 100th time make me laugh.

Don't get too mad this time.

 

[cheesier]

Don't feel bad OP. I read a biography of this guy and I remember a section where it talked about how the mathematical explanation for this problem was beyond him.

 

[TOcho118]

Anyone seen the movie "21"? The prof asked this same question in the beginning...stumped me forever. So thanks everyone for all the great explanations! Now imagine this...

So same scenario, you choose door 1, the car is behind 2, and the host opens door 3. He asks you if you would like to switch. Now, at this moment, your friend walks into the room, with no knowledge of the game show. The host then asks your friend "which door would you like to choose to win a car. It's behind one of them." At this point, your friend picking door 2 has a 50% chance, while for you picking door 2 versus door 1 has a 67% chance.

Crazy how two people picking from the same two doors could have different probabilities of winning.

P.S. I understand that you have "extra" knowledge about what's going on (that there was a third door) so you and your friend are not really in equal scenarios. Just interesting to think about...

 

[loveoforganic]

I don't think that's true. Probability remains regardless of perspective.

 

[Diallelus]

I don't think that's true. Probability remains regardless of perspective.

The probability remains the same for you, but for your friend, who has no knowledge of the situation, the decision is arbitrary and therefore 50/50. One door is twice as likely to be right as the other, but if you don't know which door is which that doesn't help you, so your buddy might as well flip a coin.

EDIT: Example: Two contestants are playing. One of them uses x-ray glasses to see through the doors. His odds are 0/1, 0/1, and 1/1 for each door. For the honest contestant, his odds for each door are 1/3, 1/3, and 1/3. Knowledge certainly affects probability.

 

[Spin]

The probability remains the same for you, but for your friend, who has no knowledge of the situation, the decision is arbitrary and therefore 50/50. One door is twice as likely to be right as the other, but if you don't know which door is which that doesn't help you, so your buddy might as well flip a coin.

EDIT: Example: Two contestants are playing. One of them uses x-ray glasses to see through the doors. His odds are 0/1, 0/1, and 1/1 for each door. For the honest contestant, his odds for each door are 1/3, 1/3, and 1/3. Knowledge certainly affects probability.

Just because the situation is arbitrary doesn't mean that the chance of Choice A being right is 50%. I think that's what's hanging a lot of people up. Just because, as far as your friend knows, it's 50%, that doesn't mean that it actually is 50%.

Now, if you asked 100 friends to come and randomly choose between the two remaining doors, 50 of them would choose the door with the car behind it*... so each individual clueless friend has a 50% chance of being right, but that doesn't mean that each door has a 50% chance of having the car behind it.

* Half of them would choose the door you originally chose and have 1/3rd chance of being right, half would choose the other door and have 2/3rds chance of being right, so even in theory, that averages out to about half of them being right/getting the car.

[I'm tired, so I hope this makes sense.]

EDIT: Oh, wait. I just re-read your post. Are you talking about the probability that people will choose certain doors, not the probability that those doors are right?

 

[loveoforganic]

I Read it the way you did spin. Good explanation

 

[Kgizzle]

Simple game theory. I will admit it's hard to explain to other people who dont get it

 

[MustangGtV894]

So then by the same logic on the show "Deal or no Deal" when you are down to the final two cases where the player has a case which he picked and there is one case left to be opened and the only amounts left in play are $1,000,000 and $0.01 should you always go for the switch? Since you randomly picked your case at the beginning out of 30+ cases and you have eliminated around 20+ cases by the end of the game, are the chances of winning by switching much greater than not switching?

Or am I completely off here?

😕😕😕

Edit: Err maybe not I guess, since in this case the host isn't required to open the case with the goat in it.

 

[MonkfishSushi]

When I took a biostatistics class last semester, my professor (http://sph.berkeley.edu/faculty/selvin.php) causally mentioned this problem during one of our lectures. My first thought was that one scene in the movie 21; little did I know that my professor was the one who came up with the problem!

After class that day, I asked him about the origins of how he came up with it. Apparently one weekend while watching televised game shows on his couch, the problem just came to him. He instantly thought of the correct answer in his mind and for some reason, he's puzzled why a large proportion of students struggle with this problem... Haha, it's all good though. He also said that his daughters make fun of him for being too nerdy.

 

[Diallelus]

Just because the situation is arbitrary doesn't mean that the chance of Choice A being right is 50%. I think that's what's hanging a lot of people up. Just because, as far as your friend knows, it's 50%, that doesn't mean that it actually is 50%.

...

EDIT: Oh, wait. I just re-read your post. Are you talking about the probability that people will choose certain doors, not the probability that those doors are right?

The distinction between the "real" probability (that is, the probability based on your knowledge) and the probability from the friend's perspective is the point I was trying to make. The odds are not actually 50/50, but they are from your friend's perspective because he has no knowledge to distinguish them (that is, Door A and Door B have no distinguishing features to him, aside from the arbitrary labels A and B, which could easily be reversed).

In a perfect knowledge situation, all probability (in the macroscopic world) is 0% to 100%, but that doesn't change the fact that perspective-based probabilities are useful and valid. For example, if an audience member peeked behind the doors before the game started, his odds for each door are 0%, 0%, and 100%. This doesn't change the fact that, from the player's perspective, he has a 1/3 chance by sticking and a 2/3 chance by switching. Notice, in this case, that the 1/3 and 2/3 odds are not the real odds, but merely the odds as far as the player knows.

 

[Spin]

The distinction between the "real" probability (that is, the probability based on your knowledge) and the probability from the friend's perspective is the point I was trying to make. The odds are not actually 50/50, but they are from your friend's perspective because he has no knowledge to distinguish them (that is, Door A and Door B have no distinguishing features to him, aside from the arbitrary labels A and B, which could easily be reversed).

In a perfect knowledge situation, all probability (in the macroscopic world) is 0% to 100%, but that doesn't change the fact that perspective-based probabilities are useful and valid. For example, if an audience member peeked behind the doors before the game started, his odds for each door are 0%, 0%, and 100%. This doesn't change the fact that, from the player's perspective, he has a 1/3 chance by sticking and a 2/3 chance by switching. Notice, in this case, that the 1/3 and 2/3 odds are not the real odds, but merely the odds as far as the player knows.

That makes sense. Thanks!

 

[BioBA]

If you switch when given the chance:

2/3 chance of guessing a goat initially, then switching to a car and winning
1/3 chance of guessing a car initially, then switching to a goat and losing

If you don't switch:

1/3 chance of guessing a car and winning
2/3 chance of guessing a goat and losing

Simple as that!

 

[cheesier]

lol at all of the people who are saying this is simple math. True, intuitive responses are pretty simple, but mathematically proving it is a pretty different feat.