This is a famous problem, but you may not of heard of it. It could be called the Monte Hall problem after the host to an old US TV show "Let's make a Deal".
A game show contestant is allowed to pick one of three prizes which are each hidden behind a door. Behind one of the doors is a new car, behind the other two are joke prizes like maybe a goat.
After the contestant selects a door at random, Monte shows them that behind one of the doors they didn't select is a goat. He then allows them to trade the one they selected with the remaining hidden prize.
Here is the question: Should the contestant trade doors or keep his own? Does it matter?
Assume here that Monte always offers this choice to the contestants and that he is not doing it because he knows the contestant has chosen the correct door.
If you're already familiar with the problem, then you know the answer.
I recently became familiar with the problem and at first thought the answer was that the contestant had just as much chance of winning the car if he traded his pick away as kept it.
Then I had to convince myself that my intuition was wrong and that the math was right and that he has a better chance if he makes the trade.
I find this a fascinating problem because it's sort of like a mental illusion. People's intuition tells them one thing, but the math tells them another. You can then convince your intuition that the math is true by looking at the problem from a different prospective.
So, was I trying to start an argument? Of course not. I was just trying to introduce this problem to people who had not heard of it before so that they could also have the fun of changing their perspective.
But, it seems that probably most everyone in this group has already come across this problem and knows the answer already.
But if you write out the three possible choices the contestent makes you will see that to get the car:
If you don't trade you have a 1/3 chance.
If you do trade, you have a 2/3 chance.
Originally Posted by moeur
This is a famous problem, but you may not of heard of it. It could be called the Monte Hall problem after the host to an old US TV show "Let's make a Deal".
A game show contestant is allowed to pick one of three prizes which are each hidden behind a door. Behind one of the doors is a new car, behind the other two are joke prizes like maybe a goat.
After the contestant selects a door at random, Monte shows them that behind one of the doors they didn't select is a goat. He then allows them to trade the one they selected with the remaining hidden prize.
Here is the question: Should the contestant trade doors or keep his own? Does it matter?
Assume here that Monte always offers this choice to the contestants and that he is not doing it because he knows the contestant has chosen the correct door.
"I like to run on treadmills, because at least I know I'm getting nowhere."
- Me
This is the intuitive answer. There are two doors to choose from, so our chances should be 50/50 for each door; right?
Think about it this way. After you choose a door, suppose Monte says you could trade it for both the other doors. Would you trade? Of course you would because you know that there is a 2/3 chance of winning if you hold two doors as opposed to a 1/3 chance if you keep the one.
Well, this is essentially what's been offered in the original problem except that he has shown you that there is a goat behind one of the other two doors, but you alrady knew that.
Its much easier to draw it out.
Draw out every possibility you can think of, which is three, depending on what object you choose at first.
Then you'll see:
G> G- C trade
G- G> C trade
G G- C> no trade
> is your initial choice
- is the goat the host shows you.
So only 1/3 of the time, a trade will be beneficial.
Originally Posted by moeur
This is the intuitive answer. There are two doors to choose from, so our chances should be 50/50 for each door; right?
Think about it this way. After you choose a door, suppose Monte says you could trade it for both the other doors. Would you trade? Of course you would because you know that there is a 2/3 chance of winning if you hold two doors as opposed to a 1/3 chance if you keep the one.
Well, this is essentially what's been offered in the original problem except that he has shown you that there is a goat behind one of the other two doors, but you alrady knew that.
Does this make sense?
Last edited by capsulecorpjx; Aug 4th, 2005 at 12:04 PM.
"I like to run on treadmills, because at least I know I'm getting nowhere."
- Me
It could be said a bit simpler than that. Ultimately, when you originally chose the door, you had a 1 in 3 chance of picking the correct door. Those odds don't change when somebody tells you what's behind one of the other doors.
Try extending the problem - suppose you had 1,000,000 doors and you pick #1. The host then opens every door apart from #1 and #435,682. Would you swap now?
I don't see why a goat is a "joke prize". There are, literally, millions of people in impoverished parts of the worls to whom a car would be a joke, and a goat would be meat, milk or wool.
Last edited by croxley; Aug 5th, 2005 at 08:14 AM.
I don't see why a goat is a "joke prize". There are, literally, millions of people in impoverished parts of the worls to whom a car would be a joke, and a goat would be meat, milk or wool.
I was expecting that response. I guess when the show plays in those countries, the good prize will be a goat and the dud will be a pile of dirt.
I don't see why a goat is a "joke prize". There are, literally, millions of people in impoverished parts of the worls to whom a car would be a joke, and a goat would be meat, milk or wool.
You are either one, two or all three of the following:
1) A troll
2) A moron
3) Oversensitive
"I like to run on treadmills, because at least I know I'm getting nowhere."
- Me
hi moeur!
wot capsulecorpjx is saying is correct........this is a counter-intuitive problem!!
see this
Gate1 | Gate2 | Gate3
-------|-------|-------
1st case G(o) G P
2nd case G P G(o)
3rd case P G(o) G
Let's assume you choose Gate1 -- you have a 1/3 chance of a good prize.
But (this is key) Monty knows what is behind each door, and shows a bad one.
In cases 1 and 2, he eliminates Gates 2 and 3 respectively (which happen to be the only remaining bad door) so a good Gate is left: SWITCH!
Only in case 3 (you lucked out in your original 1 in 3 chances) does switching hurt you.
So, your probability goes up from 1/3 to 2/3 if you switch after being shown a bad Gate.
but this is not something which can be explained in the first attempt.........u nedd to play the game many times to understand the concept.
check the attached game (its exe.....just telling u if u wish to open).......play the game and remain stick to the door which u selected n switch doors for equal number of times..........say if u play 20 times, then switch doors 10 times and stick to ur selected 10 times.............and the check the result.
Gate1 | Gate2 | Gate3
-------|-------|-------
1st case G(o) G P
2nd case G P G(o)
3rd case P G(o) G
sorry, didn't previewed my post post, but G is Goat, G(o) is the Gate with Goat opened by Hall, n P is Prize gate under Gates 1 2 n 3.
hope u can make it out
i know it is not a good practice to attach an exe in disc forums, but this is wot i got to expalin my point
well thats why i told earlier that i have attached an exe so it is on your risk.... ..........but it is a small game n i can assure u it will pose no threat to ur pc since it didn't posed to mine..........
You can use VB6 or VB.Net.
well i cannot write the entire code for the thing i have posted..............i will take ages to make it plus i cannot devote so much time for that rite now
If seeing is believing, then show us.
the exe i posted will show u the same i wanted to say n its a small game to explain the same
Well, this one really intrigued me because my background is Statistics.
The contestant definately improves their chances to 2/3 by choosing the other door, if they have been shown a third door that is a goat. That's really why it's not 1/2 chance, the conditional probability is based on having been shown a third door that is a goat. It would be different if Monty told you that you won the car every time you picked the correct door from the get-go.
I have a fun little program that proves it...I tried to upload it but I hit the max lines limit. Hopefully, this zip will work...
It will let you run door by door, or in auto mode and in auto mode, let's you try both ways....always choosing the alternate door (66% or always sticking with your original door (33%)
Anybody here a fan of the show Numb3rs? If not, it is a show about a mathematian who uses math to solve crimes for the FBI. Anyways, this same problem was talked about on one of the episodes, and as said by a few of you, choosing the other door does give you a 2/3 chance, contrary to what you would at first think.
The original choice and the fact that there were three original choices is inconsequential. If every time you are given a second choice between two doors, that is what you have to look at.
For your second choice:
Chance that door 1 of 2 has the prize - 50%
Chance that door 2 of 2 has the prize - 50%
Chance that You picked the right door of the two - 50%
Chance that you didn't - 50%
Choosing to switch is just another way of picking between two doors (the one you have already, or the one you don't) so it's still 50%.
Hehehe..it's amazing how many people get tricked by this...
Ok, let's play a modified version:
I'm thinking of a number between 1 and 1,000,000,000
Try and guess what it is....
Done? Ok, now, what if I tell you either you were right, or I chose 441,324.
Now, are you going to stick with your original choice, or change to 441,324??
You obviously change, right?
I mean, if it's either the one you thought of, or 441,324 then you'd pick 441,324.
Hopefully, from that example, it's clearer to see that your chances, if you stick to your original guess, then you've still only got a 1 / 1,000,000,000 chance.
Same thing applies for 1,000,000 doors....or 1,000...or 3!
You'll only have a 1/1,000,000, or 1/1,000, or 1/3 chance
I STILL don't get it. So can someone please knock down my argument by saying what incorrect assumptions I have made etc.
So it ends up with two doors, one has a goat and one has a car. I can pick either one. If i pick door A then it will be a goat or a car. If I pick door B then it could be a goat or a car. The host of the game show has opened the thrid door showing a goat behind it. So for all I care he's opened a million other doors with goats behind it. Fact is I still have 2 doors in front of me to choose from and each door can hold a goat or a car ergo 50% chance!
[finishes ritual] It's alive. ALIIIIIIVE I tell you!
The reason that this is not the case is because of the two doors that remain, you picked one. It is not simply a case of being presented with a million doors, then opening all but two, and then asking you to choose. You have chosen one door to remain shut before any other doors are opened. Hence your choice has had an influence on which doors are subsequently opened, and thus the result cannot simply be pure random chance.
Reply With Quote
Originally Posted by moeur
