A fascinating mathematical problem
Posted by: Minky on 18 April 2004
I've just finished reading "The curious incident of the dog in the night-time" by Mark Haddon. The book has a brief passage on something called "The Monty Hall" problem. Marilyn vos Savant, the smartest person in the world by virtue of her Guinness record IQ, was asked the following question (my words) :
In a game show there are three doors. Behind two of them are goats. Behind the third is a car. You are a contestant trying to win the car. You choose a door and the game show host opens one of the two remaining doors to reveal a goat. Which door is most likely to have the car behind it (should you stick or switch) ?
If you have never encountered this problem before, have a think about it.
When Marilyn vos Savant gave her answer she was quite soundly chastised by some very heavy-hitting brainiacs, and this is at least as fascinating as the problem itself, because (as I have also found) this baby seems to have an almost supernatural ability to turn otherwise mild-mannered people into red-eyed monsters
In a game show there are three doors. Behind two of them are goats. Behind the third is a car. You are a contestant trying to win the car. You choose a door and the game show host opens one of the two remaining doors to reveal a goat. Which door is most likely to have the car behind it (should you stick or switch) ?
If you have never encountered this problem before, have a think about it.
When Marilyn vos Savant gave her answer she was quite soundly chastised by some very heavy-hitting brainiacs, and this is at least as fascinating as the problem itself, because (as I have also found) this baby seems to have an almost supernatural ability to turn otherwise mild-mannered people into red-eyed monsters
Posted on: 19 April 2004 by Two-Sheds
quote:
Now I've finally come around, the important point is that the host knowingly opens a door with a goat behind it.
If you know the host will open a door with a goat in it then by picking door 3 you are actually picking doors 1 and 2.
Why?
I am also assuming that you know the host will open a door with a goat behind it.
The 3 possible combinations to start with are:
goat, goat, car
goat, car, goat
car, goat, goat
I assume you agree that you have a 2/3 chance of picking a goat.
Now then the host will open one of the other doors and reveal a goat. You know the host has picked a goat on purpose. If you are currently on a goat and switch then you will definitely win (since the host has eliminate the other one) and you have a 2/3 chance of being on a goat from the prior to the door being opened so thus by switching you now have a 2/3 chance of being on a car.
Posted on: 19 April 2004 by Haroon
quote:
Originally posted by Two-Sheds:
I am also assuming that you know the host will open a door with a goat behind it.
Yes
quote:
The 3 possible combinations to start with are:
goat, goat, car
goat, car, goat
car, goat, goat
I assume you agree that you have a 2/3 chance of picking a goat.
and 1/3 chance to pick a Car, so 1/3 + 2/3=1
quote:
Now then the host will open one of the other doors and reveal a goat. You know the host has picked a goat on purpose. If you are currently on a goat and switch then you will definitely win (since the host has eliminate the other one) and you have a 2/3 chance of being on a goat from the prior to the door being opened so thus by switching you now have a 2/3 chance of being on a car.
No the 2/3 was prior to opening a door, after you have eliminated one goat as well as one door.
So
goat, goat, car
goat, car, goat
car, goat, goat
no longer holds
its now
goat, car
car, goat
you have 1/2 chance of currently being on a car or a goat
Posted on: 19 April 2004 by Haroon
quote:
Originally posted by Matthew Robinson:
A proof for those who insist on seeing numbers rather than relying on one's intuition:
1. Pick a door (A). Chance of being correct is 1/3
2. Chances of it being one of the other two doors (B, C) is 2/3.
[I assume we all agree to this point]
3. Host eliminates door B which, since it never has the car, has a car probablity of 0.
[This is the bit people get wrong and assume this has a probability of 1/3 -- omitting the fact that the host knows how to pick the one without the car]
4. Probability of it being behind door C is 2/3 - 0 = 2/3. Door A's probability remains unchanged at 1/3.
Matthew
I dont understand this Matthew, your saying
4. Probability of it being behind door C = (B&C) - B
C = 2/3 - 0
C = 2/3.
Door A's probability remains unchanged at 1/3.
How can that be?
You cant subtract B's possibility or chance (zero) from B&C to derive C's as after B is eliminated you know that B&C being 2/3 was wrong, so it was really 1/2 for B&C, not 2/3, and 1/2 for A, as we now B is 0 it leaves us with C 1/2 and A 1/2
OR
After we find B to be nothing or 0 that B&C can not be 2/3 and nor can A be 1/3
B and C were 1/2 and A 1/2
So
B&C = C + B
B&C = 1/2 + 0 and A = 1/2
So B&C = 1/2 and A = 1/2
C = 1/2 - B
Where B is 0
C = 1/2 - 0
C = 1/2
[This message was edited by Haroon on Tue 20 April 2004 at 4:03.]
Posted on: 19 April 2004 by Minky
Haroon,
Pick one of the three doors and stick with it. Forget about opening either of the other doors. Do this 1000 times. Approximately how many cars will you have in your driveway ?
NOW. Pick one of the three doors and stick with it. Open the first of the other doors to reveal a goat, then the second to reveal either a goat or a car. Do this 1000 times. How many cars will you have in your driveway ?
Once you understand that the chance of winning a car doesn't switch to 50:50 after the first door is opened you are home and hosed.
Pick one of the three doors and stick with it. Forget about opening either of the other doors. Do this 1000 times. Approximately how many cars will you have in your driveway ?
NOW. Pick one of the three doors and stick with it. Open the first of the other doors to reveal a goat, then the second to reveal either a goat or a car. Do this 1000 times. How many cars will you have in your driveway ?
Once you understand that the chance of winning a car doesn't switch to 50:50 after the first door is opened you are home and hosed.
Posted on: 19 April 2004 by Two-Sheds
So we are agreed that you have a 2/3 chance of picking a goat first time.
The host opens a door to reveal a goat (and we know that he knew he was opening a door that had a goat behind it) so the two possible current states after one door has been opened are.
state 1 - we have currently selected a goat - 2/3 chance
state 2 - We have currently selected a car - 1/3 chance
The odds have not changed because we know there was no chance the host could open a door to reveal the car the first time.
If we switch then if we are on a goat we will be guaranteed to be on a car (no chance involved). If we switch then if we are on a car we will be guaranteed to be on a goat(no chance involved). So we can take the odds prior to the first door being opened and we can see we are better off switching.
The host opens a door to reveal a goat (and we know that he knew he was opening a door that had a goat behind it) so the two possible current states after one door has been opened are.
state 1 - we have currently selected a goat - 2/3 chance
state 2 - We have currently selected a car - 1/3 chance
The odds have not changed because we know there was no chance the host could open a door to reveal the car the first time.
If we switch then if we are on a goat we will be guaranteed to be on a car (no chance involved). If we switch then if we are on a car we will be guaranteed to be on a goat(no chance involved). So we can take the odds prior to the first door being opened and we can see we are better off switching.
Posted on: 19 April 2004 by Haroon
quote:
Originally posted by Two-Sheds:
So we are agreed that you have a 2/3 chance of picking a goat first time.
Yes but the second time there is one less goat!
quote:
The host opens a door to reveal a goat (and we know that he knew he was opening a door that had a goat behind it) so the two possible current states after one door has been opened are.
state 1 - we have currently selected a goat - 2/3 chance
state 2 - We have currently selected a car - 1/3 chance
The odds have not changed because we know there was no chance the host could open a door to reveal the car the first time.
The odds do change because we know each time one of the two goats will be taken out so you know there is one car and one goat - obviously.
state 1 - we have currently selected a goat - not 2/3 chance as there are only now two options so it becomes 1/2 chance
state 2 - We have currently selected a car - not 1/3 chance as there are only now two options so it becomes 1/2 chance
quote:
If we switch then if we are on a goat we will be guaranteed to be on a car (no chance involved). If we switch then if we are on a car we will be guaranteed to be on a goat(no chance involved).
dont understand that
quote:
So we can take the odds prior to the first door being opened and we can see we are better off switching.
errm we cant take the prior odds as they are based on thinking 1 of 3, take one out its 1 of 2
Posted on: 19 April 2004 by Two-Sheds
one final explanation before bed.
consider all the possibilities (assuming you always switch):
1=goat 2=goat 3=car
you pick 1. host opens 1. you lose.
you pick 1. host opens 2. you switch and win.
you pick 1. host opens 3. you lose.
you pick 2. host opens 1. you switch and win.
you pick 2. host opens 2. you lose.
you pick 2. host opens 3. you lose.
you pick 3. host opens 1. you switch and lose.
you pick 3. host opens 2. you switch and lose.
you pick 3. host opens 3. you win
the lines in italic can never happen since the host will not open your door or reveal the car first. So on the surface it looks like a 50-50 since there a 4 possibilities, 2 we win and 2 we lose.
But we can see that if we pick 1 we win, pick 2 we win, pick 3 we lose. When we pick (before all doors are opened) we know we have a 2/3 chance of picking 1 or 2 and a 1/3 chance of picking 3.
If we stick the possibilities are:
you pick 1. host opens 1. you lose.
you pick 1. host opens 2. you stick and lose.
you pick 1. host opens 3. you lose.
you pick 2. host opens 1. you stick and lose.
you pick 2. host opens 2. you lose.
you pick 2. host opens 3. you lose.
you pick 3. host opens 1. you stick and win.
you pick 3. host opens 2. you stick and win.
you pick 3. host opens 3. you win
now we see that for sticking if we pick 1 or 2 we lose, if we pick 3 we win and we have a 1/3 chance of picking 3.
consider all the possibilities (assuming you always switch):
1=goat 2=goat 3=car
you pick 1. host opens 1. you lose.
you pick 1. host opens 2. you switch and win.
you pick 1. host opens 3. you lose.
you pick 2. host opens 1. you switch and win.
you pick 2. host opens 2. you lose.
you pick 2. host opens 3. you lose.
you pick 3. host opens 1. you switch and lose.
you pick 3. host opens 2. you switch and lose.
you pick 3. host opens 3. you win
the lines in italic can never happen since the host will not open your door or reveal the car first. So on the surface it looks like a 50-50 since there a 4 possibilities, 2 we win and 2 we lose.
But we can see that if we pick 1 we win, pick 2 we win, pick 3 we lose. When we pick (before all doors are opened) we know we have a 2/3 chance of picking 1 or 2 and a 1/3 chance of picking 3.
If we stick the possibilities are:
you pick 1. host opens 1. you lose.
you pick 1. host opens 2. you stick and lose.
you pick 1. host opens 3. you lose.
you pick 2. host opens 1. you stick and lose.
you pick 2. host opens 2. you lose.
you pick 2. host opens 3. you lose.
you pick 3. host opens 1. you stick and win.
you pick 3. host opens 2. you stick and win.
you pick 3. host opens 3. you win
now we see that for sticking if we pick 1 or 2 we lose, if we pick 3 we win and we have a 1/3 chance of picking 3.
Posted on: 19 April 2004 by Haroon
quote:
Originally posted by Minky:
Haroon,
Pick one of the three doors and stick with it. Forget about opening either of the other doors. Do this 1000 times. Approximately how many cars will you have in your driveway ?
333 Cars
quote:
NOW. Pick one of the three doors and stick with it. Open the first of the other doors to reveal a goat, then the second to reveal either a goat or a car. Do this 1000 times. How many cars will you have in your driveway ?
500 Cars
quote:
Once you understand that the chance of winning a car doesn't switch to 50:50 after the first door is opened you are home and hosed.
But it does! As a goat is always eliminated.
The first time you had only one chance at three of car so 333 cars
second time there are two doors, one goat, one car so 1 in 2 chance of it being a car, so 500 cars
[This message was edited by Haroon on Tue 20 April 2004 at 4:34.]
Posted on: 19 April 2004 by Minky
OK.
Posted on: 19 April 2004 by Haroon
quote:
Originally posted by Two-Sheds:
one final explanation before bed.
consider all the possibilities (assuming you always switch):
1=goat 2=goat 3=car
you pick 1. host opens 2. you switch and win.
you pick 2. host opens 1. you switch and win.
you pick 3. host opens 1. you switch and lose.
you pick 3. host opens 2. you switch and lose.
So on the surface it looks like a 50-50 since there a 4 possibilities, 2 we win and 2 we lose.
But we can see that if we pick 1 we win, pick 2 we win, pick 3 we lose. When we pick (before all doors are opened) we know we have a 2/3 chance of picking 1 or 2 and a 1/3 chance of picking 3.
If we stick the possibilities are:
you pick 1. host opens 2. you stick and lose.
you pick 2. host opens 1. you stick and lose.
you pick 3. host opens 1. you stick and win.
you pick 3. host opens 2. you stick and win.
now we see that for sticking if we pick 1 or 2 we lose, if we pick 3 we win and we have a 1/3 chance of picking 3.
If you stick or switch its the same 2 we win and 2 we loose
I think ive started to understand where you are coming from that by sticking one is limiting one self to the 1 in the chance we originally had. Where I think you are wrong is that by switching your odds are 2/3 - but they arent, as shown above you gain nothing by switching, second time you still only get to choose one of two whether you stick to the original or switch to the other your chance is the same.
Posted on: 19 April 2004 by Minky
quote:
Originally posted by Haroon:
I think ive started to understand where you are coming from that by sticking one is limiting one self to the 1 in the chance we originally had.
So if you can understand why sticking at a point where you have a choice between two doors gives you 1:3 chances of winning the car, can't you see that in the same situation choosing the other door MUST give you 2:3 chances ?
Posted on: 19 April 2004 by Haroon
no because that judgement of it being 1/3 was when you did not know what any of them where. When one goat is revealed the 1/3 and 2/3 goes out of the window as the goat revealed tells you it must be one of two, whether you switch or not now does not matter surely.
Posted on: 19 April 2004 by Haroon
Another way
Let say you you know there is 2/3 chance for goats, but you choose say door C - so 1/3 chance car and 2/3 for A and B. But A is revealed as goat now we still go on 1/3 to 2/3 assumptions, but we have two choices: was the 2/3 A&B or A&C and 1/3 C or B - its 50-50!
Let say you you know there is 2/3 chance for goats, but you choose say door C - so 1/3 chance car and 2/3 for A and B. But A is revealed as goat now we still go on 1/3 to 2/3 assumptions, but we have two choices: was the 2/3 A&B or A&C and 1/3 C or B - its 50-50!
Posted on: 20 April 2004 by John Sheridan
quote:
Originally posted by Haroon:
Another way
Let say you you know there is 2/3 chance for goats, but you choose say door C - so 1/3 chance car and 2/3 for A and B. But A is revealed as goat now we still go on 1/3 to 2/3 assumptions, but we have two choices: was the 2/3 A&B or A&C and 1/3 C or B - its 50-50!
so let me get this straight:
you have a 1/3 chance of picking the correct door 1st time and a 1/2 chance if you switch. What happened to the other 1/6? (1/3 +1/2 = 5/6)
Posted on: 20 April 2004 by matthewr
Haroon,
1. Pick door A you have a 1/3 chance
2. I am left with doors B+C and have a 2/3 chance
3. I now don some X-Ray specs and can see where the car is. It's in your door A 1/3 of the time, and my doors B or C 2/3 of the time.
4. In 1/3 of cases I see it is under door A and you will win,
5. In 2/3 of the cases its not under door A and I open the one of B and C with the car and I win.
6. Or, suppose I don't open a door but instead tell you (without lying) which of my two doors B and C (that's the ones with a 2/3 chance) doesn't have a car. Do you now switch?
7. Your argument now requires that me "seeing" where the car is affects your odds of winning. Which is just utterly nonsensical.
Note my program (and Don's spreadsheet) did this 30 Million times and got a 1/3 and 2/3 distribution.
Matthew
1. Pick door A you have a 1/3 chance
2. I am left with doors B+C and have a 2/3 chance
3. I now don some X-Ray specs and can see where the car is. It's in your door A 1/3 of the time, and my doors B or C 2/3 of the time.
4. In 1/3 of cases I see it is under door A and you will win,
5. In 2/3 of the cases its not under door A and I open the one of B and C with the car and I win.
6. Or, suppose I don't open a door but instead tell you (without lying) which of my two doors B and C (that's the ones with a 2/3 chance) doesn't have a car. Do you now switch?
7. Your argument now requires that me "seeing" where the car is affects your odds of winning. Which is just utterly nonsensical.
Note my program (and Don's spreadsheet) did this 30 Million times and got a 1/3 and 2/3 distribution.
Matthew
Posted on: 20 April 2004 by Paul Ranson
1. Mean host, only opens a door and offers the 'switch/stick' if you've already chosen the car. Stick...
2. Nice host, always reveals a goat and offers the 'switch/stick'. Switch...
3. Random host, always opens a door, 1/3 time reveals the car. In this particular case hasn't. Now what to do?
In the latter case I think 'switch/stick' makes no difference. You've already lost 1/3 games, sticking will win 1/2 the remainder, or 1/3 the total, switching similarly.
The 'game' only works if the host knows how to reveal a goat and then always does so.
Paul
2. Nice host, always reveals a goat and offers the 'switch/stick'. Switch...
3. Random host, always opens a door, 1/3 time reveals the car. In this particular case hasn't. Now what to do?
In the latter case I think 'switch/stick' makes no difference. You've already lost 1/3 games, sticking will win 1/2 the remainder, or 1/3 the total, switching similarly.
The 'game' only works if the host knows how to reveal a goat and then always does so.
Paul
Posted on: 20 April 2004 by Rasher
Good morning...It's Groundhog day..!!" 
Posted on: 20 April 2004 by matthewr
PR said "20 heads of tails!"
Sorry?
So we have a £1 bet on 50 coin flips. If there are 23 heads or greater I win £1. If there are 23 heads or less you win £1.
I still don't understand. Surely I must take this bet as the odds are in my favour?
"However, just to prove my point, I flicked 4x20 sets last night (obviously bored) and found the average margin to be 3 points. If you was to flick thousands of times, you would expect the margin percentage to smoothen, but sometimes it can also run away as with pricing derivatives"
Well I don't know about derivatives but in Poker this would make no sense.
The odds of making an flush with four suited cards and one more to come are (roughly) 4 to 1. If the pot is £5 and I have a £1 bet I have to make this bet or I lose money. If I missed my flush 10 times in a row I am still making money on this bet and not making it means I lose 20p per hand (ok there are a number of simplifying assumptions here but essentially that is correct).
One can only make such choices in the long run and I don't see how one can divine short term results from "thinking laterally".
Matthew
Sorry?
So we have a £1 bet on 50 coin flips. If there are 23 heads or greater I win £1. If there are 23 heads or less you win £1.
I still don't understand. Surely I must take this bet as the odds are in my favour?
"However, just to prove my point, I flicked 4x20 sets last night (obviously bored) and found the average margin to be 3 points. If you was to flick thousands of times, you would expect the margin percentage to smoothen, but sometimes it can also run away as with pricing derivatives"
Well I don't know about derivatives but in Poker this would make no sense.
The odds of making an flush with four suited cards and one more to come are (roughly) 4 to 1. If the pot is £5 and I have a £1 bet I have to make this bet or I lose money. If I missed my flush 10 times in a row I am still making money on this bet and not making it means I lose 20p per hand (ok there are a number of simplifying assumptions here but essentially that is correct).
One can only make such choices in the long run and I don't see how one can divine short term results from "thinking laterally".
Matthew
Posted on: 20 April 2004 by matthewr
Excatly my thoughts, Omer.
Although frankly I couldn't really understand exactly what PR was trying to say.
Matthew
Although frankly I couldn't really understand exactly what PR was trying to say.
Matthew
Posted on: 20 April 2004 by Allan Probin
The door problem:
If you always stick you will win 1 in 3 times
If you always switch you will win 2 out of 3 times
If you randomly stick or switch you will win 50% of the time
Therefore everyone is right.
Allan
If you always stick you will win 1 in 3 times
If you always switch you will win 2 out of 3 times
If you randomly stick or switch you will win 50% of the time
Therefore everyone is right.
Allan
Posted on: 20 April 2004 by greeny
It's always amusing this problem.
Trying to get people to accept that changing is better is sometimes quite a challenge.
However when considering the 1,000,000 door example, and the fact that the host will not open the door with the car (as he knows where the car is). Surely noone can think that keeping with their original choice of 1/1000000 is the best thing to do
Trying to get people to accept that changing is better is sometimes quite a challenge.
However when considering the 1,000,000 door example, and the fact that the host will not open the door with the car (as he knows where the car is). Surely noone can think that keeping with their original choice of 1/1000000 is the best thing to do
Posted on: 20 April 2004 by Haroon
lets see
A = 1/3 of the time Car is here AND B&C = 2/3 poss of having goat
OR
B = 1/3 of the time Car is here AND A&C = 2/3 poss of having goat
OR
C = 1/3 of the time Car is here AND A&B = 2/3 poss of having goat
we have three scenarios that are equally valid, our chances of guessing the correct one are 1 in 3 given that we have one choice
You choose one and host eliminates another one with a goat
So lets say that we choose C, we are left with
B = 1/3 of the time Car is here AND A&C = 2/3 poss of having goat
OR
C = 1/3 of the time Car is here AND A&B = 2/3 poss of having goat
Now we have to decide between two scenarios as to which is more likely to have the goat or car our original choice C or switch to B. B has 1/3 chance so does C have a 1/3 chance, you have two choices with the same likelihood of having the car. There is nothing to choose between the two - stick or switch it remains the same
Could also argue as we know that after eliminating A it must be B or C we must revise our odds in light of the revelation
B = 1/2 chance the car being is here AND C = 1/2 poss of having goat
OR
C = 1/2 chance the car being is here AND B = 1/2 poss of having goat
Now if having watched the show several times you found out that B had the car more times than C and even A:
let say:
A = 1/4 of the time Car is here AND B&C = 3/4 poss of having goat
B = 1/2 of the time Car is here AND A&C = 1/2 poss of having goat
C = 1/4 of the time Car is here AND A&B = 3/4 poss of having goat
And we stoopidly choose C and eliminated A, so we have
B = 1/2 of the time Car is here AND A&C = 1/2 poss of having goat
C = 1/4 of the time Car is here AND A&B = 3/4 poss of having goat
Then switching to B would have the better odds as you remember it more likely has the car.
[This message was edited by Haroon on Tue 20 April 2004 at 13:52.]
[This message was edited by Haroon on Tue 20 April 2004 at 14:01.]
A = 1/3 of the time Car is here AND B&C = 2/3 poss of having goat
OR
B = 1/3 of the time Car is here AND A&C = 2/3 poss of having goat
OR
C = 1/3 of the time Car is here AND A&B = 2/3 poss of having goat
we have three scenarios that are equally valid, our chances of guessing the correct one are 1 in 3 given that we have one choice
You choose one and host eliminates another one with a goat
So lets say that we choose C, we are left with
B = 1/3 of the time Car is here AND A&C = 2/3 poss of having goat
OR
C = 1/3 of the time Car is here AND A&B = 2/3 poss of having goat
Now we have to decide between two scenarios as to which is more likely to have the goat or car our original choice C or switch to B. B has 1/3 chance so does C have a 1/3 chance, you have two choices with the same likelihood of having the car. There is nothing to choose between the two - stick or switch it remains the same
Could also argue as we know that after eliminating A it must be B or C we must revise our odds in light of the revelation
B = 1/2 chance the car being is here AND C = 1/2 poss of having goat
OR
C = 1/2 chance the car being is here AND B = 1/2 poss of having goat
Now if having watched the show several times you found out that B had the car more times than C and even A:
let say:
A = 1/4 of the time Car is here AND B&C = 3/4 poss of having goat
B = 1/2 of the time Car is here AND A&C = 1/2 poss of having goat
C = 1/4 of the time Car is here AND A&B = 3/4 poss of having goat
And we stoopidly choose C and eliminated A, so we have
B = 1/2 of the time Car is here AND A&C = 1/2 poss of having goat
C = 1/4 of the time Car is here AND A&B = 3/4 poss of having goat
Then switching to B would have the better odds as you remember it more likely has the car.
[This message was edited by Haroon on Tue 20 April 2004 at 13:52.]
[This message was edited by Haroon on Tue 20 April 2004 at 14:01.]
Posted on: 20 April 2004 by matthewr
Oh I see. Sounds like the sort of Mathematician who ends up working in Finance 
Matthew
Matthew
Posted on: 20 April 2004 by Haroon
quote:
Originally posted by greeny:
It's always amusing this problem.
Trying to get people to accept that changing is better is sometimes quite a challenge.
However when considering the 1,000,000 door example, and the fact that the host will not open the door with the car (as he knows where the car is). Surely noone can think that keeping with their original choice of 1/1000000 is the best thing to do
Tell me about it I really am struggling to believe that changing between two options witht the same odds give you an advantage.
Your above example
your original choice has 1/1000000 but the one you change it to also originally had the chance of 1/1000000, assuming you have a choice between two and the host eliminated 999998 others. the odds between the two are the same 1/1000000, you give up one with a chance of 1/1000000 for another that had the same chance of 1/1000000. But actually take out 999998 you are left with 2 one of which is right so it now becomes 1/2 after the elimination.
Posted on: 20 April 2004 by JeremyD
OK, here's what I hope is an easier to follow explanation for the switcher strategy in the case where the host deliberately chooses a door with a goat behind it:
First, the contestant chooses a door at random.
The probability that the contestant has chosen the door with the car is 1/3.
The frequency with which a contestant will be right is therefore 1/3.
Now the host deliberately chooses a door with a goat.
From the perspective of a newcomer to the problem, with no knowledge of what has happened so far, the probability that the car is behind the contestant's door is indeed now 1/2 (as is the probability that the car is behind the third door).
But the frequency with which the contestant's door has the car behind it remains 1/3. Therefore, the frequency with which the the third door has the car behind it is 2/3.
Hence the probabilities of the car being behind the first door and the third door are 1/3 and 2/3, respectively.
So it makes sense to switch...
[This message was edited by JeremyD on Tue 20 April 2004 at 16:56.]
First, the contestant chooses a door at random.
The probability that the contestant has chosen the door with the car is 1/3.
The frequency with which a contestant will be right is therefore 1/3.
Now the host deliberately chooses a door with a goat.
From the perspective of a newcomer to the problem, with no knowledge of what has happened so far, the probability that the car is behind the contestant's door is indeed now 1/2 (as is the probability that the car is behind the third door).
But the frequency with which the contestant's door has the car behind it remains 1/3. Therefore, the frequency with which the the third door has the car behind it is 2/3.
Hence the probabilities of the car being behind the first door and the third door are 1/3 and 2/3, respectively.
So it makes sense to switch...
[This message was edited by JeremyD on Tue 20 April 2004 at 16:56.]