Monty Hall Problem [Video]

by **Nahoj** Sat May 31, 2014 4:05 pm

Hi guys, I found this video and I thought it would be interesting for some of you.
It's about the "Monty Hall Problem", just watch the video you'll see what it is.

She is right, and I am quite curious what you guys think of this

by **Ephemeral** Sat May 31, 2014 4:22 pm

I've seen it, nice trick, but not useful for me (at least in N).

by **Nahoj** Sat May 31, 2014 4:25 pm

SpartaX18 wrote:I've seen it, nice trick, but not useful for me (at least in N).

This is the Off-topic section so it's alright :-)

by **cooperverdon** Sun Jun 01, 2014 5:50 am

This was a shit explanation video for this problem. Great problem though.

@johan1996
I dont know how to reply to you, so hopefully you see this. If you like probability theory, try this question.

Three boxes
Box 1- 2 Blue marbles
Box 2- 2 Red marbles
Box 3- 1 Blue marble, 1 Red marble

After picking a box at random, and picking up a marble from that box at random, that turns out to be blue, what are the chances that the remaining marble in that box is blue?

by **Trollking** Sun Jun 01, 2014 7:42 am

I have seen both these broplems and the answer is

Spoiler:

But the harder variation to the monty hall problem is what if monty hall had a heart attack mid show before revealling a door and the stage crew not knowing what was behind each door decided the show must continue.

So the person chooses door 1 but monty hall is unable to choose the door. The stage crew discuss and decide the 50/50 of opening the other doors. They open door 3 and it isnt the car but a goat (or whatever) should the player switch now or is the 2/3 still in play

by Guest Sun Jun 01, 2014 7:49 am

I think the chance Monty Hall gets a hart attack before opening the door is 1/100, unless you mean he gets a heart attack a week ago and died

by **cooperverdon** Sun Jun 01, 2014 8:02 am

trollking, if you know the answer, dont say it, thats just common decency.

by **Nahoj** Sun Jun 01, 2014 10:29 am

cooperverdon wrote:This was a shit explanation video for this problem. Great problem though.

@johan1996
I dont know how to reply to you, so hopefully you see this. If you like probability theory, try this question.

Three boxes
Box 1- 2 Blue marbles
Box 2- 2 Red marbles
Box 3- 1 Blue marble, 1 Red marble

After picking a box at random, and picking up a marble from that box at random, that turns out to be blue, what are the chances that the remaining marble in that box is blue?

I have read Trollking's answer and yet I have concluded that the answer is 1/2.
How I got the answer:
You choose 1 blue ball so the box with 2 red balls is not an option.
So there are 2 possible boxes that you've got that blue ball from.
Now there is either a red ball or a blue ball left in the box you've picked which makes it a 50/50 chance of it being a blue ball.

Explain me what I did wrong

by **cooperverdon** Sun Jun 01, 2014 10:36 am

johan, i still dont know how to reply.
Eliminate the 2 red ball box.
Now you have 2 boxes. A total of 3 blue, and 1 red.
2/3 of those blue, will yield the box that has the other blue. 1/3 will yield the box that has the other red.

by **EddyMataGallos** Sun Jun 01, 2014 4:01 pm

Let's put some more pro probability stuff here, let's see if you can figure this one out:

We have two boxes, with the following content:

Box 1: 5 blue balls and 5 white ones.
Box 2: 2 blue balls and 8 white ones.

Now, we throw a dice. If we get a 6, we pick a ball at random from the first box. If we get anything else, we pick it at random from the second box.

After picking the first ball, if it was blue, we pick another ball at random from the first box, and if it was white, we pick another ball at random from the second box (in both cases, we don't put the first ball we extracted back in the boxes, of course).

Now, what is the probability of the second ball being blue?

(This is tougher I warn you, but you can solve it if you are able to generalize how this kind of problems work)
-------------------------------------------------------

Trollking wrote:I have seen both these broplems and the answer is
Spoiler:

But the harder variation to the monty hall problem is what if monty hall had a heart attack mid show before revealling a door and the stage crew not knowing what was behind each door decided the show must continue.

So the person chooses door 1 but monty hall is unable to choose the door. The stage crew discuss and decide the 50/50 of opening the other doors. They open door 3 and it isnt the car but a goat (or whatever) should the player switch now or is the 2/3 still in play

So you chose door 1, and then, door 3 is opened randomly. The probability of it containing a goat, if the car is in door 1 is 1, if the car is in door 2, its 1 again, and if its in door 3, its 0.

Since the probabilities must add up to one, it means they actually are 1/2, 1/2 and 0. So in this case, switching and staying is the same. (You have a half on each).

Note that this reasoning also works for the original, 2/3 1/3 version.

by **cooperverdon** Mon Jun 02, 2014 4:14 am

@Eddymatagallos
The Monty Hall problem and Bertrand's box paradox are examples in probability theory whose answers are counter-intuitive to typical human reasoning. Both can be done relatively easily without pen or paper. That is why I like them.
I attempted your problem in my head, but concluded the numbers weren't chosen with much thought, so the fractions one needs to deal with become quite complicated. Not putting the first ball back is where things get complicated...
Ill leave the solving of the problem to Trollking.

@Trollking
If, in the Monty Hall problem, the host (Monty Hall) doesn't know what is behind each door the whole problem gets thrown out the window. It isn't a harder variation at all.

by **Trollking** Mon Jun 02, 2014 4:55 am

Well the answer varies from the 2/3 odds to 1/2 odds as the door was chosen randomly

There was 1/3 chance that it might of been a car but as monty hall problem the chance was 0% thus changing the results to a fifty fifty chance

by **Trollking** Mon Jun 02, 2014 5:00 am

This problem can be described by "is your neighbour a zombie" by jeremy stangroom

Really good book

also wrote "Who owns the golfish"

by **Trollking** Mon Jun 02, 2014 5:20 am

Answer:

by **cooperverdon** Mon Jun 02, 2014 5:33 am

@Trollking
I'm pretty sure your last step should be
5/6 * 8/10 * 2/9 roll 1-5, get white, pick second box again, get blue

by **Trollking** Mon Jun 02, 2014 7:02 am

i forgot to write it but yes
thx coop

by **Nahoj** Mon Jun 02, 2014 8:21 am

The easiest solution to the Monty Hall problem is

Spoiler:

by **EddyMataGallos** Mon Jun 02, 2014 3:15 pm

Well done Trollking, that's correct. Well there's this little miscalculation at the end which Cooperverdon pointed out, which changes the actual final result to 77/270, not 47/270. Other than that, perfect

@Cooperverdon: I agree that while the other marble problems are just a little mental entertainment, you do need some paper here. The reason I like it is that, when you do the easy ones, you're not sure you do understand the principles or if you are just answering by intuition (which tends to be enough for the simple ones). If you solve this, little more complicated ones, you can be sure you understand the principles, which makes it trivial to solve any other problem of its kind. Generalizing is key in maths.

@Johan: Yeah, thats the intuitive solution I came up with when this problem was proposed to me last year. At first, as usual, I found it quite interesting that the answer is not 1/2. But then by doing that reasoning everything becomes absolutely clear.

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