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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. 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?
Do any of you have any noteworthy experiences where knowledge of math helped you in an unusual way?
https://en.wikipedia.org/wiki/Monty_Hall_problem
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Get Those Brain Cells Working: The Monty Hall Problem
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squatcho, n.:
The button at the top of a baseball cap.
-- "Sniglets", Rich Hall & Friends
(Score: 2) by engblom on Friday July 29 2016, @10:36AM
This is a classic problem and sadly most people fall into the trap, keeping their first door.
Solution:
First when you pick, you have only 1/3 chance to get the car.
When offered to swap doors, you change your odds from 1/3 to 1/2 if you swap the door as totally there are one car and one goat left and there is 50% chance of car in the other door.
(Score: 0) by Anonymous Coward on Friday July 29 2016, @10:43AM
Are you sure it isn't 2/3 change to get the car? You get the combined chance that it was behind the other two doors if you switch.
(Score: 2) by engblom on Friday July 29 2016, @10:44AM
You are actually right! 2/3 it is. Regardless, it is a win situation to switch door.
(Score: 0) by Anonymous Coward on Friday July 29 2016, @02:33PM
Not always. Therein lies the violation of our intuition. If the contestant could intuitively pick the right door on the first try (something our intuition tells us we can do, regardless of the actual facts that the producers have set things up deliberately to make this effectively impossible), then switching would in fact be a lose situation. Statistically this will occur 1/3 of the time.
So, if you're Clint Eastwood in The Good, the Bad, and the Ugly, you may wish to ignore the mathematical aspect of things and go with your instinct.
As Han Solo said, "Never tell me the odds!"
(Score: 4, Informative) by morpheus on Friday July 29 2016, @10:45AM
When offered to swap doors, you change your odds from 1/3 to 1/2 if you swap the door as totally there are one car and one goat left and there is 50% chance of car in the other door.
I may have misunderstood what you are trying to say here but the odds of winning the car if you always switch are 2/3, not 1/2. Just think of trying to pick the door without the car first. Your chances of doing this are 2/3 and if you are successful (and always switch your choice) you get the only door with the car in the end. If you randomly decide between staying with your original choice and switching, then the probability of picking a car is 1/2.
(Score: 2) by tonyPick on Friday July 29 2016, @11:24AM
Yeah - the trick is that since "Monty" always knows which doors have what (and he's always got an empty/goat door) then he's not changing the selection odds in any way from the initial choice when he does the first part of the reveal, even though the question implies otherwise.
Basically the first choice splits the set into your door ("1/3" chance of having the car) and Monty's two doors ("2/3" chance of having the car), and then you get the chance to swap over to Monty's set: when phrased like that the problem becomes a lot more obvious.
(Score: 2) by Reziac on Saturday July 30 2016, @04:26AM
Quick, James -- move the car!
And there is no Alkibiades to come back and save us from ourselves.
(Score: 2) by tangomargarine on Friday July 29 2016, @02:12PM
I still don't get this.
The host must always open a door that was not picked by the contestant (Mueser and Granberg 1999).
The host must always open a door to reveal a goat and never the car.
The host must always offer the chance to switch between the originally chosen door and the remaining closed door.
If the host has to open a different door, the chance is 1/2 regardless of whether you switch between the remaining two doors.
I also don't get the part on the Wikipedia page that separates the doors into 1 group of 1 and 1 group of 2. Consider all 3 doors separately. The only difference is you go from 1/3 to 1/2 with elimination of the third door; you might as well not pick the initial door at all and randomly select 1 of the 2 remaining for all the good it does you.
In an attempt to clarify her answer she proposed a shell game (Gardner 1982) to illustrate: "You look away, and I put a pea under one of three shells. Then I ask you to put your finger on a shell. The odds that your choice contains a pea are 1/3, agreed? Then I simply lift up an empty shell from the remaining other two. As I can (and will) do this regardless of what you've chosen, we've learned nothing to allow us to revise the odds on the shell under your finger." She also proposed a similar simulation with three playing cards.
Vos Savant commented that though some confusion was caused by some readers not realizing that they were supposed to assume that the host must always reveal a goat, almost all of her numerous correspondents had correctly understood the problem assumptions, and were still initially convinced that vos Savant's answer ("switch") was wrong.
For somebody the article claims supports the conclusion, it sounds to me like Vos Savant is agreeing with *me.*
So I guess I'm saying both viewpoints are wrong. It doesn't matter whether you decide to switch or not; your odds are the same.
"Is that really true?" "I just spent the last hour telling you to think for yourself! Didn't you hear anything I said?"
(Score: 2) by tangomargarine on Friday July 29 2016, @02:14PM
This whole thing smells like one of those word games where they purposely phrase the rules badly and then laugh at you when you're mislead into picking the wrong answer.
How *wouldn't* the initial placement of the car be random?
"Is that really true?" "I just spent the last hour telling you to think for yourself! Didn't you hear anything I said?"
(Score: 2, Informative) by TheRaven on Friday July 29 2016, @02:28PM
If the host has to open a different door, the chance is 1/2 regardless of whether you switch between the remaining two doors.
You can work it through for each case. Let's call the doors A, B, and C, and the car is behind door C.
If you pick A, then the host must close door B (they have no choice), because it is the only door that did not have a car behind it. If you switch, you win, if you don't switch then you lose.
If you pick B, then the host must close door A (no choice again). If you switch, you win, if you don't then you lose.
The final case, you pick C, now the host can choose A or B, but it doesn't matter, because if you switch then you lose and if you stay then you lose.
The only case where you win if you don't switch is if you picked the correct door first. It works because of the asymmetry. You always gain information when the host opens the door.
sudo mod me up
(Score: 2) by tangomargarine on Friday July 29 2016, @02:45PM
But picking a door to begin with is just showmanship; it doesn't affect the odds at all. If you didn't pick a door at all at the beginning and Monty randomly chooses one of the two goat doors, then you choose a door, you'd still have the same odds.
You always gain information when the host opens the door.
The odds get reduced from 1/3 to 1/2. You don't gain any information about the two remaining doors themselves.
The final case, you pick C, now the host can choose A or B, but it doesn't matter, because if you switch then you lose and if you stay then you lose.
If you stay on the car door you still lose? What?
I'm assuming you mean "choose" instead of "close." Rather confusing typo.
"Is that really true?" "I just spent the last hour telling you to think for yourself! Didn't you hear anything I said?"
(Score: 2) by Gaaark on Friday July 29 2016, @04:46PM
I look at it as:
x. ?
y. ?
z. ?
A door is revealed:
You now have 2 doors with no clue/no information.
Are you REALLY any closer to knowing if you should switch?
I see it as going from 33.33333% repeating chance of having chosen correctly to 50% chance of having chosen correctly. Would switching really increase those odds? Isn't it really just 50%?
Has any information been released to increase your odds?
Not from where i'm standing (looking sharp in my banana suit with my stuffed monkey sitting on my shoulder).
--- Please remind me if I haven't been civil to you: I'm channeling MDC. ---Gaaark 2.0 ---
(Score: 2) by VLM on Friday July 29 2016, @05:25PM
Are you REALLY any closer to knowing if you should switch?
Yeah, but your numbers are too small or percentages are too large making it look weird.
Assume it scales. Scale this to a billion doors. You pick a door which almost certainly isn't it. Dude opens a B-2 doors leaving a winner and a loser. The loser is almost certainly the one you picked, the winner is the remaining unopened door.
(Score: 2) by TheRaven on Friday July 29 2016, @05:56PM
sudo mod me up
(Score: 2) by TheRaven on Friday July 29 2016, @05:54PM
But picking a door to begin with is just showmanship; it doesn't affect the odds at all. If you didn't pick a door at all at the beginning and Monty randomly chooses one of the two goat doors, then you choose a door, you'd still have the same odds.
No, because Monty's choice is different. If you didn't pick a door to begin with then you'd have a 1/2 probability of getting the correct answer, not 2/3. Monty would close one of the doors and you'd be in a trivial situation of having to pick between two doors with no knowledge. By picking one of the doors to begin with, you reduce the number of possible moves for Monty. If you (2/3 probability) pick one of the doors that doesn't have a car behind it, then Monty must close the only other car that doesn't have a car behind it. The other door is then the one that has the car. If you picked correctly the first time (1/3 probability), then the Monty can close either door.
This is a bit easier to explain with a little bit of graph theory and game theory (two of the core bits of computer science, so hopefully not too problematic for people here), but difficult without the ability to draw the relevant pictures. The key point is that if you picked incorrectly to start with then Monty will close the door that doesn't have a car behind it and you then have a 100% probability of getting the car if you switch. For the three doors, A, B, and C, with the car behind C, if you picked A or B then he will close the other, if you pick C then he will close either A or B. The set of doors that are not closed and are not the ones that you picked then becomes:
In contrast, the set of doors that are the one that you picked the first time are:
Picking one from the first set gives you a 2/3 probability of choosing the correct one, picking one from the second set gives you a 2/3 probability of picking the correct one. Or, to put it more simply, if you picked correctly the first time then you win by staying but if you picked incorrectly the first time then you win by switching. You have a 1/3 chance of picking correctly the first time and a 2/3 chance of picking incorrectly, therefore it is better to bet that your first pick was incorrect.
If this is still confusing, I present a simple table of all of the possible paths and outcomes:
From this is should be obvious that there is a 2/3 chance of winning if you switch and a 1/3 chance of winning if you stick.
sudo mod me up
(Score: 2, Insightful) by Anonymous Coward on Friday July 29 2016, @02:31PM
Let's try a different explanation, then.
You pick a door.
One third of the time, you have picked the one with the car.
Two thirds of the time, you have picked one with a goat.
Then the rules force Monty to act based on your choice.
In the 1/3 case, Monty can choose either of the other doors (they both have goats), but it does not matter what he chooses. The remaining door will have a goat.
In the 2/3 case, Monty has no choice - he *must* eliminate the only other door with a goat. The remaining door must have the car.
In the 1/3 case, if you switch, you get a goat
In the 2/3 case, if you switch, you get a car
The reason that it is not 50-50 is Monty lack of choice. No matter what he does, he cannot change the situation you find yourself in when the "do you switch"? question is asked. You controlled that with your first choice ... and there were three options at the beginning.
(Score: 0) by Anonymous Coward on Saturday July 30 2016, @09:49AM
"I still don't get this."
No, you sure don't. I knew there would be at least one of you popping up.
It's simple. There are a billion doors and only one car. You pick a door. Is the car behind it? Of course not. There's a goat behind your door. Now the host offers you the chance to have whatever is behind ALL of the remaining doors. The host even generously offers to take the 999,999,998 goats that are behind those 999,999,999 doors off your hands, leaving you with just the car. So do you accept the offer? OF COURSE YOU DO.
What's that? It's only three doors? Fine - do you want your door and its 1/3 chance or do you want the other two doors with their 2/3 chance of revealing a car? YOU DO IF YOU WANT TO IMPROVE YOUR ODDS.
...Fry
What's that? But the host opened one of those other two doors and revealed a goat? SO WHAT? There will always be at least one goat when two doors are involved. THE SHOWING OF A GOAT TELLS YOU NOTHING. The host is offering you the contents of BOTH doors and he's taking the unwanted goat off your hands, leaving you with a 2/3 shot at the car.
There are NEVER any 50-50 odds. Your odds are ALWAYS the odds of the one door you picked or THE COMBINED ODDS OF ALL THE OTHER DOORS.
(Score: 2) by JoeMerchant on Friday July 29 2016, @02:55PM
And a 50% chance of car in the door you picked the first time.
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(Score: 1) by Mike on Friday July 29 2016, @04:05PM
Actually I think the solution is slightly different.
First, what I'll call the "real politic" solution. If the common wisdom is to switch to the other unopened door, then your choice should be based on whether you think they want to give a car away or not (i.e. have they given a car away lately, how frequently do they seem to give a car away etc...). Basically, try to figure how they are gaming the system.
But assuming total random chance, I think the stats are more along the lines of figure what are the odds that the other two doors are both empty: 2/3 * 2/3 = 4/9. The chance that one of the doors has a car is 5/9. Your choice then becomes 4/9 for the first choice and 5/9 for the unopened choice. Close to 50/50, but it's still slightly better odds to go for the other door.
If you've done some research on the game, though, what I called the 'real politic" solution would probably give you better odds than that.
(Score: 1) by toddestan on Friday July 29 2016, @10:50PM
The "real politic" solution is only in play in the host has the option of revealing the goat and allowing you to switch. In that case, your odds of winning when switching can be as low as 0% (the host only offers to switch when he knows you've picked the car) to 100% (the host is out to help you and only offers you the switch if you've picked one of the goats). But in the problem as presented, the host's hands are tied. He must always offer the option to switch and must always open a door that contains a goat. In that case, the odds of winning the car by switching is 2/3. Which is simple if you think about it, as the chances of initially picking the door with the car is 1/3, and that doesn't change when the host reveals a goat behind one of the other doors, which means that if you stay you have the same 1/3 odds, and therefore switching gives you 2/3 odds*.
"Real politic" might come in handy if you've figured out the placement of the car is not random, or there's a pattern to which goat the host reveals when he has a choice. But that information will only help you so you should still have at least 2/3 odds of winning the car.