The Monty Hall probability question
12 September 2013, 09:43 PM
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The Monty Hall probability question
OK this is a famous probability poser. I remember this from university days!
A game show host has 3 doors. Behind one is a car and behind the other two are goats. The host knows which door the car is behind, as the contestent you don't
You are asked to choose a door and then the host will open one of the other two doors to reveal a goat (he knows where the car is remember).
He then asks you if you want to stick with the door you chose or swap to the other door? To increase you chances of winning should you stick or swap to the other door or does it make no difference?
I was remninded of this by reading this article tonight:
Monty Hall problem: The probability puzzle that makes your head melt
A game show host has 3 doors. Behind one is a car and behind the other two are goats. The host knows which door the car is behind, as the contestent you don't
You are asked to choose a door and then the host will open one of the other two doors to reveal a goat (he knows where the car is remember).
He then asks you if you want to stick with the door you chose or swap to the other door? To increase you chances of winning should you stick or swap to the other door or does it make no difference?
I was remninded of this by reading this article tonight:
Monty Hall problem: The probability puzzle that makes your head melt
12 September 2013, 09:48 PM
#2
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The easiest way to visualise the right answer (and often a good trick with any hypothesise) is to test it under extreme conditions
So if you had 10 thousand doors to pick from, and you chose 1, then monty hall opened 9998 other doors to reveal goats, leaving 2 closed doors - would you swap your choice
Yes,
So if you had 10 thousand doors to pick from, and you chose 1, then monty hall opened 9998 other doors to reveal goats, leaving 2 closed doors - would you swap your choice
Yes,
12 September 2013, 09:51 PM
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The easiest way to visualise the right answer (and often a good trick with any hypothesise) is to test it under extreme conditions
So if you had 10 thousand doors to pick from, and you chose 1, then monty hall opened 9998 other doors to reveal goats, leaving 2 closed doors - would you swap your choice
Yes,
So if you had 10 thousand doors to pick from, and you chose 1, then monty hall opened 9998 other doors to reveal goats, leaving 2 closed doors - would you swap your choice
Yes,
12 September 2013, 10:17 PM
#4
No it isnt, you had 3 choices, so that is 33.3% chance of picking correctly,
By removing one of the doors you swap your door and you have a 66.6% chance of been correct. And say hello to your new car!
By removing one of the doors you swap your door and you have a 66.6% chance of been correct. And say hello to your new car!
13 September 2013, 12:06 AM
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13 September 2013, 12:31 AM
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The key component to this problem is that the host always reveals a goat.
At the point of making your initial choice you had a 33% chance of choosing the winner (car) and a 66% chance of losing (goat).
If you choose not to swap those odds don't change.
However, as the host is always forced to show a goat you can see that the only way to lose by swapping is to have initially picked the car (because you swap to a goat), which you had a 33% chance of doing. The other 66% of the time you initially picked a goat meaning that you are always swapping to a car if the other goat has been shown.
At the point of making your initial choice you had a 33% chance of choosing the winner (car) and a 66% chance of losing (goat).
If you choose not to swap those odds don't change.
However, as the host is always forced to show a goat you can see that the only way to lose by swapping is to have initially picked the car (because you swap to a goat), which you had a 33% chance of doing. The other 66% of the time you initially picked a goat meaning that you are always swapping to a car if the other goat has been shown.
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13 September 2013, 01:40 AM
#10
'Clever' people can try to dress this up all they want but the chances of winning the car are always 1/3 in my mind.
3 doors, 1 car and 2 goats. At the first choice, you have a 1 in 3 chance of choosing correctly, by the show host revealing a goat, they are just messing with the contestant's head. By taking one goat out of the equation, they are just visually teasing the contestant and making them doubt their choice.
13 September 2013, 06:50 AM
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Depends how you want to look at it, I suppose.
'Clever' people can try to dress this up all they want but the chances of winning the car are always 1/3 in my mind.
3 doors, 1 car and 2 goats. At the first choice, you have a 1 in 3 chance of choosing correctly, by the show host revealing a goat, they are just messing with the contestant's head. By taking one goat out of the equation, they are just visually teasing the contestant and making them doubt their choice.
'Clever' people can try to dress this up all they want but the chances of winning the car are always 1/3 in my mind.
3 doors, 1 car and 2 goats. At the first choice, you have a 1 in 3 chance of choosing correctly, by the show host revealing a goat, they are just messing with the contestant's head. By taking one goat out of the equation, they are just visually teasing the contestant and making them doubt their choice.
The best explanation I have seen is this;
Let's call the goats David and George. There are then three equally weighted scenarios (i.e. there is a 1 in 3 chance of any of these occurring).
Scenario 1 - the door you initially choose has the car behind it. The host reveals a goat (either David or George, it doesnt matter) and if you swap you get the other goat (either George or David), if you don't you get a car!
Scenario 2 - the door you initially choose has George behind it. The host reveals David and if you swap you get a car, if you don't you get George (a goat).
Scenario 3 - the door you initially choose has David behind it. The host reveals George and if you swap you get a car, if you don't you get David (a goat).
So swapping gets you a car in 2 out of the 3 scenarios whereas not swapping gets you a car in 1 of the 3 scenarios and as the scenarios themselves are equally weighted that means you are twice as likely to win the car by swapping!
13 September 2013, 07:26 AM
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13 September 2013, 07:55 AM
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in really really simple terms by always opening the "other" door Monty ***** the whole 50/50 thing up
you have the grubby hand of human intervention palying with the gods of maths
you have the grubby hand of human intervention palying with the gods of maths
13 September 2013, 08:32 AM
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13 September 2013, 08:56 AM
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13 September 2013, 02:11 PM
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I can't understand how people can still get this wrong after it has been explained carefully and logically? 
Think of it as two games.
Game 1: You take your pick and simply stick with it.
You have a 33% chance of selecting a winner - simple
Game 2: The host must show a Goat and offer a swap
If you stick, nothing has changed since game 1.
However, if you swap then logically only one of the following scenarios can be true:
1. In game 1 you picked the car and swap to Goat 1 or Goat 2 (depending on which Goat the host showed). - you lose.
2. In game 1 you picked Goat 1 and swap to the car because the host shows Goat 2 - you win.
3. In game 1 you picked Goat 2 and swap to the car because the host shows Goat 1 - you win.
Given the above, we know that you'll have picked a car in game 1 33% of the time and either Goat 1 or 2 66% of the time. By applying this logical to Game 2 you can be certain you will win 66% of the time and lose 33% if you swap.
Think of it as two games.
Game 1: You take your pick and simply stick with it.
You have a 33% chance of selecting a winner - simple
Game 2: The host must show a Goat and offer a swap
If you stick, nothing has changed since game 1.
However, if you swap then logically only one of the following scenarios can be true:
1. In game 1 you picked the car and swap to Goat 1 or Goat 2 (depending on which Goat the host showed). - you lose.
2. In game 1 you picked Goat 1 and swap to the car because the host shows Goat 2 - you win.
3. In game 1 you picked Goat 2 and swap to the car because the host shows Goat 1 - you win.
Given the above, we know that you'll have picked a car in game 1 33% of the time and either Goat 1 or 2 66% of the time. By applying this logical to Game 2 you can be certain you will win 66% of the time and lose 33% if you swap.
Last edited by Saxo Boy; 13 September 2013 at 02:13 PM.
13 September 2013, 02:14 PM
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Edited to Say: ...and if you are not I am opening a Casino in your street
13 September 2013, 06:45 PM
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F1, having just read this from the start, At first I got then I read on a bit and then I didn't, now at the end end I feel more knackered than a 5 mile bike ride. Can you try not to do this on a Friday afternoon please 
13 September 2013, 07:30 PM
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13 September 2013, 10:22 PM
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