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# Monty Hall Problem [Video]

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**1**of**1**## Monty Hall Problem [Video]

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

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

**Nahoj**- Admin
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Age : 24

Location : The Netherlands

## Re: Monty Hall Problem [Video]

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

_________________

**Ephemeral**- Admin
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Age : 21

Location : Poland

## Re: Monty Hall Problem [Video]

This is the Off-topic section so it's alrightSpartaX18 wrote:I've seen it, nice trick, but not useful for me (at least in N).

**Nahoj**- Admin
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Age : 24

Location : The Netherlands

## Re: Monty Hall Problem [Video]

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?

@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?

**cooperverdon**- Admin
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Age : 27

Location : Australia

## Re: Monty Hall Problem [Video]

I have seen both these broplems and the answer is

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

- Spoiler:
- 2/3 of the time its blue

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

Last edited by Johan1996 on Sun Jun 01, 2014 9:29 am; edited 2 times in total (Reason for editing : changed 1 word)

**Trollking**- Admin
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## Re: Monty Hall Problem [Video]

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

**Guest**- Guest

## Re: Monty Hall Problem [Video]

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

**cooperverdon**- Admin
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Age : 27

Location : Australia

## Re: Monty Hall Problem [Video]

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

**Nahoj**- Admin
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Age : 24

Location : The Netherlands

## Re: Monty Hall Problem [Video]

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.

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.

**cooperverdon**- Admin
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Age : 27

Location : Australia

## Re: Monty Hall Problem [Video]

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:

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

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

-------------------------------------------------------

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.

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:
2/3 of the time its blue

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.

**EddyMataGallos**- Admin
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## Re: Monty Hall Problem [Video]

@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.

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.

**cooperverdon**- Admin
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## Re: Monty Hall Problem [Video]

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

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

**Trollking**- Admin
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## Re: Monty Hall Problem [Video]

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

Really good book

also wrote "Who owns the golfish"

Really good book

also wrote "Who owns the golfish"

**Trollking**- Admin
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## Re: Monty Hall Problem [Video]

- Answer:
- the answer to the 5/5 and 2/8 problem is 47/270 or 17.4'0'7' chance that it was blue reasoning is

1/6* 5/10* 4/9 rolls a six then blue then blue

+

1/6* 5/10* 2/10 rolls a six red then blue

+

5/6 * 2/10 * 5/10 rolls a 1-5 blue then blue

+

5/6 * 2/10 * 2/9

**Trollking**- Admin
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## Re: Monty Hall Problem [Video]

@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

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

**cooperverdon**- Admin
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## Re: Monty Hall Problem [Video]

i forgot to write it but yes

thx coop

thx coop

**Trollking**- Admin
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## Re: Monty Hall Problem [Video]

The easiest solution to the Monty Hall problem is

- Spoiler:

1/3 of the games you play you will choose the door with the car behind it. So when you change doors, you will lose.

But in 2/3 of the games you play you will choose a goat. And after that guy revealed one of the doors with a goat behind it, the car will be behind the other closed door.

The 3 possible games:

Game 1: car - goat - goat

Game 2: goat - car - goat

Game 3: goat - goat - car

So only when you switch on game 1 you will lose, making it a 2/3 chance of winning if you switch.

**Nahoj**- Admin
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## Re: Monty Hall Problem [Video]

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.

@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.

**EddyMataGallos**- Admin
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