Math: The Monty Hall Problem.
This past Tuesday couple of my colleagues at work asked me to solve a math Problem(Everybody is starting to think that I'm smart :)). The question went like this..
"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 #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?"
PAUSE. Try solving the problem before reading further.
Smeagol: No, I wont cos the chances are 50-50.
Colleagues: No! Your wrong.
Right answer: You switch so the probability of winning is 2/3.
They tried to explain why I was wrong while they themselves did not understand the answer.
The answer is quiet intriguing cos when you pick a door in the first place, you think it has the car behind it. When the host opens the second door(knowing whats behind all the doors) which has a goat, you knock it out of your probability of winning and think you only have two doors to choose from. But the fact of the matter is when the host had to open one of the two doors that are left, he knew exactly which one to open so the goat shows up. So the other which he did not open holds the highest probablitiy of having the car behind.
The person who actually solved and proved the answer to be right is Marilyn vos Savant. It took me two days to realize that I was wrong.
The topic was much debated among Ph. D holders from all over the country but it has been experimentally tested in many schools to prove that she is right.
She is in the Guinness book of world records for the person ever recorded to have the highest IQ of 228. She was 10 years old when she scored 228.
They didnt test my IQ when I was 10 :)
Source: Smart Smeagol!
posted by Smeagol @ 11:49 AM
14 comments
14 Comments:
good one dude...jus read thru the entire wikipedia article...n even found a better explanation in this:
http://en.wikipedia.org/wiki/Monty_Hall_problem
but i still have one doubt. if there were 4 doors...n 2 doors were found to hav goats...wats the probability shiji wud come to poconoc trip?:P
I say 100% dude. Shes comin regardless of how many doors there are :P
I know the concept is kinda hard to grasp so I tried my best to explain it :)
Thanks.
there is a huge assumption here at first principles. what if i like goats more than cars and I *want* to win a goat! :D
@n-man
I say...U wud stil try n win the car...sell it...buy a lot more of those goats:)
g1 (goodone).
There's an assumption that's missed: "... the host knows what is behind each door, ALWAYS opens a door revealing a goat, and ALWAYS makes the offer to switch, ..."
Without this assumption, I arrived at the common incorrect answer. Had the assumption been stated, I might have got the same common incorrect answer but with a lot of thinking. :D
link:
http://en.wikipedia.org/wiki/Monty_Hall_problem
Hippy: Thanks.
I think if you really give it a deep thought while tryin to solve, you would have asked for more information/assumptions rather than coming to the wrong conclusion. Cos think abt it with the information given, the whole question is ONE event as far as probability is concerned. So the host opening the door does not eliminate that door out of your probability math. So the denominator for the probability of the winning event(car) should still be 3.
So the common incorrect answer does not sound right anymore.
It is a very tricky question and I think its our understanding about probability that makes us come to a wrong conclusion.
We get the answer by conditional probability. It's two events for switching or not switching.
Without the assumption, odds of wining the car by switching does not always have to be greater. The assumption is necessary.
Events:
1. I choose a door.
2. The host ALWAYS chooses a goat (which is the assumption).
The first event is simple with a P(E1) of 1/3. The second event is tricky, since most of the time, I thought that the second event is "I choose one of the two remainging doors" and the P(E2) would be 1/2 making the P(switching) as 1/3*1/2 and P(NOT switching) = 1/3*1/2. However, with the correct second event "The host ALWAYS chooses a goat", P(E2) can be 1 or 1/2. With all the possible combinations of E1 and E2 broken down, you can calculate the required P. But, I went wrong in the second event, so I said that even with the assumption, I'd have came up with the same incorrect common answer but with a lot of thinking.
@hippy
tats a 'fine english' coated crap u gave. guess watever u wud hav done...u wud hav got it wrong:)
for example...Ur definition of Event1: I choose a door.
The probabilty of such an event occuring is 1 or 100%...unless U want to run away from the game show without choosing a door.
nyways...I dont want to b the jerk who drifted away frm the topic...so lets assume that u in all probabilty hav defined all ur events with a
condition that the contestant(ie. U)...is choosing the correct door(the door with a car bhind it)
if thats the case...the events occuring would be:
E1: U choose the correct door in ur first chance. P(E1) is (1/3)
Now...there is no reason for assuming that the host on his first chance opens the door with the goat(for two reasons):
reason #1: if the host opens the door with car in it...the probabilty of U getting the correct during Event 2 drops to 0 anyways.we dont need to calculate probabilty here!
reason #2: while the problem was putforth by smeagol, it starts with the assumption that the host has opened a door with the goat...Y wud U revert back???
Ok...next comes event 2: which is U choosing the correct door in Ur second chance(either by switching it or by not switching the doors).
E2: if U assume E1 & E2 are totally unrelated(actually they are not), U R right in saying P(E2) is 1/2.
wats up with this multiplication crap? Probabilties of two events...as far as I know are multiplied when U want to calculate the probabilty of
both the events occuring. So during the game show...U want Urselves to choose correct door during E1 and E2? Y? only then will thy giv u the car?
expecting something better nxt time:)
@kripa:
Of course, you are right there. The event would be "Initially choosing the car by selecting one of the doors." DUH. :D
Dear, I would have to disagree with you in this matter. I think the assumptions or constraints are necessary. From the link that you stated read:
1. "The problem, with all constraints explicit."
2. The "History of the problem" in Cecil Adams' column.
As for the P of two events, you can refer the Decision Tree.
--
But here's an interesting part that I left out of yesterday.
We know that:
P(winning a car) = 1/3
P(NOT winning a car) = 2/3
P(winning a car by switching) = 2/3
P(winning a car by NOT switching) = 1/3
In case of 3, see the case where n = 100 and making a similar decision tree, try to find out:
P(NOT winning a car by switching)
P(NOT winning a car by NOT switching)
I think that IF I were able to find these Ps, then I could safetly assume that I have a good understanding of the problem.
Anyway, wish me luck.
But Kripa, I hope you take this as a compliment:
"I love it when you talk DIRTY."
PS: The famous quote by Marino to Dr. Scarpetta from the novels by Patricia Cornwell.
@hippy
2 me Ur comment sounds "Oh shit...Im jus caught blabbering...dont knw wat to reply...oh yah i hav 2 reply to kripa watsoever...oh ok lemme blabber out some think...ok lets make it 100 doors instead...so how about this quote I looked up in my pre-KG textbook yest?...thank god i got something to reply...now i look smart"
n abt the assumtion thing: if U are asked to count the number of apples...ur prob starts frm counting...not frm whr those apples were plucked?
the prob starts frm the point that the host has opened the door wit a goat in it...wats the probabilty if u switch? it doesnt start b4 tat.
tat 1/3 and 2/3 stuff...I already read it n posted the wikipedia link here dude...jus scroll up n read the first comment:)
good luck!
Warning:Above comment hasnt been spell checked. kindly read it @ ur own risk or comment back the spell check version:P
This is simply AUSUMN. With n = 100: After the initial choice of any door is made, the host ALWAYS opens up 98 doors revealing goats so that there's only one door remaining to switch to.
This is what I calculated:
P(winning a car) = 1/100
P(NOT winning a car) = 99/100
P(winning a car by switching) = 99/100
P(winning a car by NOT switching) = 1/100
P(NOT winning a car by switching) = 1/100
P(NOT winning a car by NOT switching) = 99/100
links:
http://www.grand-illusions.com/monty.htm
Its not really a math problem. its a silly trick question............ leave out the sentence "and the host, who knows what's behind the doors" and there is a 50/50 chance. with that sentence in ............. its a silly trick question......why does it need math proffessors and a huge debate to sort out a silly trick question............. for gods sake "lets get the woman with the highest IQ at the age of 10 to solve a trick question" do you know how stupid that sounds
here is my version of 3 solutions
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